brown's notes 2010

8/15/2019 Brown's Notes 2010

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McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 141 2010 01

CALCULUS 2

Information for Students(Winter Term, 2009 / 2010)

Pages 1 - 20 of these notes may be considered the Course Outline for this course. The page

numbers shown in the table of contents and in the upper right hand corners of pages are not the

same as the numbers of pages in the PDF document. If you wish to print out specific pages, you

should first view the relevant pages at your screen, and determine what are the numbers of the

corresponding PDF pages.

W. G. Brown

April 9, 2010

8/15/2019 Brown's Notes 2010


Information for Students in MATH 141 2010 01

Contents

1 General Information 1

1.1 Force Majeure . . . . . . . . . . 1

1.2 Instructors and Times . . . . . . 2

1.3 Calendar Description . . . . . . 2

1.3.1 Calendar Description . . 2

1.3.2 Late transfer from MATH

151 / MATH 152 . . . . . 3

1.4 Tutorials . . . . . . . . . . . . . 3

1.4.1 Tutorial Times, Locations,

and Personnel (subject to

c h a n g e ) . . . . . . . . . 3

1.4.2 Teaching Assistants (TA’s) 31.4.3 Friday, April 02nd, 2010

and Monday, April 05th,

2010 . . . . . . . . . . 4

1.5 Evaluation of Your Progress . . 5

1.5.1 Your final grade (See Ta-

ble 3, p. 11) In the event

of extraordinary circum-

stances beyond the Uni-

versity’s control, the con-

tent and / or evaluation scheme

in this course is subjectto change. . . . . . . . . 5

1.5.2 WeBWorK . . . . . . . 6

1.5.3 Written Submissions. . . 7

1.5.4 Quizzes at the Tutorials. 7

1.5.5 Final Examination . . . 8

1.5.6 Supplemental Assessments 8

1.5.7 Machine Scoring: “Will

the final examination be

machine scored?” . . . . 9

1.5.8 Plagiarism. . . . . . . . 9

1.5.9 Corrections to grades . . 101.6 Published Materials . . . . . . . 10

1.6.1 Required Text-Book . . 10

1.6.2 Optional Reference Books 10

1.6.3 Recommended Video Ma-

t e r i a l s . . . . . . . . . . 12

1.6.4 Other Calculus Textbooks 13

1.6.5 Website . . . . . . . . . 13

1.7 Syllabus . . . . . . . . . . . . . 141.8 Preparation and Workload . . . 15

1.8.1 Prerequisites. . . . . . . 15

1.8.2 Calculators . . . . . . . 15

1.8.3 Self-Supervision . . . . 16

1.8.4 Escape Routes . . . . . 17

1.8.5 Terminology . . . . . . 18

1.9 Communication with Instructors

and TA’s . . . . . . . . . . . . . 18

1.10 Commercial tutorial and exam

preparation services . . . . . . . 19

1.11 Special Office Hours and Tutorials 20

2 Draft Solutions to Quiz Q1 21

2.1 Instructions to Students . . . . . 21

2.2 Monday Versions . . . . . . . . 21

2.3 Tuesday Versions . . . . . . . . 23

2.4 Wednesday Versions . . . . . . 24

2.5 Thursday Versions . . . . . . . 25

2.6 Friday Versions . . . . . . . . . 26

3 Draft Solutions to Quiz Q2 28

3.1 Instructions to Students . . . . . 28

3.2 Monday Versions . . . . . . . . 293.3 Most Tuesday Versions . . . . . 31

3.4 Most Wednesday Versions . . . 33

3.5 Thursday Versions . . . . . . . 35

3.6 Friday Versions . . . . . . . . . 38

4 References 201

4.1 Stewart Calculus Series . . . . . 201

4.2 Other Calculus Textbooks . . . . 202

4.2.1 R. A. Adams . . . . . . 202

4.2.2 Larson, Hostetler, et al. . 203

4.2.3 Edwards and Penney . . 203

4.2.4 Others, not “Early Tran-scendentals” . . . . . . 204

4.3 Other References . . . . . . . . 204

A Timetable for Lecture Section 001 of

MATH 141 2010 01 1001

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B Timetable for Lecture Section 002 of

MATH 141 2009 01 2001

C Supplementary Notes for Students in

Section 001 of MATH 141 2010 01 3001

C.1 Lecture style in Lecture Section

001 . . . . . . . . . . . . . . . 3001

C.2 Supplementary Notes for the Lec-

ture of January 04th, 2010 . . . 3002

C.2.1 §5.1 Areas and Distances. 3002



C.3.1

§5.1 Areas and Distances

(conclusion). . . . . . . 3005C.3.2 §5.2 The Definite Integral 3007



C.4.1 Summary of the last lec-

tures . . . . . . . . . . . 3013

C.4.2 §5.2 The Definite Inte-

gral (conclusion) . . . . 3014

C.4.3 §5.3 The Fundamental The-

orem of Calculus . . . . 3019


ture of January 11th, 2010 . . . 3021C.5.1 §5.3 The Fundamental The-

orem of Calculus (con-

c lusion) . . . . . . . . . 3021

C.5.2 §5.4 Indefinite Integrals

and the “Net Change” The-

orem . . . . . . . . . . 3023



C.6.1 §5.4 Indefinite Integrals

and the “Net Change” The-

orem (conclusion) . . . 3028C.6.2 §5.5 The Substitution Rule 3033



C.7.1 §5.5 The Substitution Rule

(conclusion) . . . . . . . 3036

C.7.2 5 Review . . . . . . . . 3049


ture of January 18th, 2010 . . . 3052C.8.1 §6.1 Areas between Curves 3052



C.9.1 §6.2 Volumes . . . . . . 3061


ture of January 22nd, 2010 . . . 3069

C.10.1 §6.3 Volumes by Cylin-

drical Shells . . . . . . . 3069

C.10.2 §6.4 Work . . . . . . . . 3075



C.11.1 §6.5 Average value of a

func tion . . . . . . . . . 3077



C.12.1 §7.1 Integration by Parts 3082



C.13.1 §7.1 Integration by Parts

(conclusion) . . . . . . . 3089

C.13.2 §7.2 Trigonometric Inte-

grals . . . . . . . . . . . 3092

C.14 Supplementary Notes for the Lec-ture of February 01st, 2010 . . . 3095

C.14.1 §7.2 Trigonometric Inte-

grals (conclusion) . . . . 3095


ture of February 03rd, 2010 . . . 3101

C.15.1 §7.3 Trigonometric Sub-

stitution . . . . . . . . . 3101


ture of February 05th, 2010 . . . 3106

C.16.1 §7.3 Trigonometric Sub-

stitution (conclusion) . . 3106C.17 Supplementary Notes for the Lec-


C.17.1 §7.4 Integration of Ra-

tional Functions by Par-

tial Fractions . . . . . . 3111

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ture of February 10th, 2010 . . . 3117C.18.1 §7.4 Integration of Ra-

tional Functions by Par-

tial Fractions (conclusion) 3117



C.19.1 §7.5 Strategy for Integra-

tion . . . . . . . . . . . 3124

C.19.2 §7.6 Integration Using Ta-

bles and Computer Al-

gebra Systems (OMIT) . 3129

C.19.3

§7.7 Approximate Inte-

gration (OMIT) . . . . . 3129



C.20.1 §7.8 Improper Integrals . 3130



C.21.1 §8.1 Arc Length . . . . 3140

C.21.2 §8.2 Area of a Surface

of Revolution . . . . . . 3146



C.22.1 §8.2 Area of a Surfaceof Revolution (conclusion) 3147

C.22.2 §8.3 Applications to Physics

and Engineering (OMIT) 3150

C.22.3 §8.4 Applications to Eco-

nomics and Biology (OMIT) 3151

C.22.4 §8.5 Probability (OMIT) 3151


ture of March 01st, 2010 . . . . 3152

C.23.1 §10.1 Curves Defined by

Parametric Equations . . 3152

C.23.2 §10.2 Calculus with Para-metric Curves . . . . . . 3156


ture of March 03rd, 2010 . . . . 3158

C.24.1 §10.2 Calculus with Para-

metric Curves (continued) 3158


ture of March 05th, 2010 . . . . 3164C.25.1 §10.2 Calculus with Para-

metric Curves (conclusion) 3164

C.25.2 §10.3 Polar Coordinates 3165


ture of March 08th, 2010 . . . . 3172

C.26.1 §10.3 Polar Coordinates

(continued) . . . . . . . 3172

C.26.2 §10.4 Areas and Lengths

in Polar Coordinates . . 3179


ture of March 10th, 2010 . . . . 3181


in Polar Coordinates (con-

tinued) . . . . . . . . . 3181


ture of March 12th, 2010 . . . . 3195


in Polar Coordinates (con-

c lusion) . . . . . . . . . 3195

C.28.2 §10.5 Conic Sections . . 3196

C.28.3 §11.1 Sequences . . . . 3197

C.28.4 Sketch of Solutions to Prob-

lems on the Final Exam-ination in MATH 141 2005

01 . . . . . . . . . . . . 3198


ture of March 15th, 2010 . . . . 3207

C.29.1 §11.1 Sequences (conclu-

sion) . . . . . . . . . . 3207

C.29.2 §11.2 Series . . . . . . . 3210


ture of March 17th, 2010 . . . . 3213

C.30.1 §11.2 Series (conclusion) 3213

C.30.2 §11.3 The Integral Testand Estimates of Sums . 3217


ture of March 19th, 2010 . . . . 3219

C.31.1 §11.3 The Integral Test

and Estimates of Sums

(conclusion) . . . . . . . 3219

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ture of March 22nd, 2010 . . . . 3225C.32.1 §11.4 The Comparison

Tests . . . . . . . . . . 3225

C.32.2 Sketch of Solutions to Prob-

lems on the Final Exam-

ination in MATH 141 2006

01 . . . . . . . . . . . . 3230


ture of March 24th, 2010 . . . . 3245

C.33.1 §11.5 Alternating Series 3245

C.33.2 Solutions to Problems on

the Final Examination in

MATH 141 2007 01 . . 3249


ture of March 26th, 2010 . . . . 3265

C.34.1 §11.6 Absolute Conver-

gence and the Ratio and

Root Tests . . . . . . . . 3265


ture of March 29th, 2010 . . . . 3268

C.35.1 §11.6 Absolute Conver-

gence and the Ratio and

Root Tests (conclusion) . 3268

C.36 Supplementary Notes for the Lec-ture of Wednesday, March 31st,

2010 . . . . . . . . . . . . . . . 3272

C.36.1 §11.7 Strategy for Test-

ing Series . . . . . . . . 3272


ture of Wednesday, April 7th, 2010 3278

C.37.1 Final Examination in MATH

141 2008 01 (one version) 3278

C.37.2 Draft Solutions to the Fi-

nal Examination in MATH

141 2009 01 (Version 4) 3293C.38 Supplementary Notes for the Lec-

ture of Friday, April 09th, 2010 . 3297

C.38.1 Final Examination in MATH

141 2009 01 (Version 4,

continued) . . . . . . . 3297

D Problem Assignments from Previous Years 5001

D.1 1998 / 1999 . . . . . . . . . . . . 5001D.1.1 Assignment 1 . . . . . . 5001

D.1.2 Assignment 2 . . . . . . 5001

D.1.3 Assignment 3 . . . . . . 5002

D.1.4 Assignment 4 . . . . . . 5002

D.1.5 Assignment 5 . . . . . . 5002

D.2 1999 / 2000 . . . . . . . . . . . . 5003

D.2.1 Assignment 1 . . . . . . 5003

D.2.2 Assignment 2 . . . . . . 5004

D.2.3 Assignment 3 . . . . . . 5006

D.2.4 Assignment 4 . . . . . . 5007

D.2.5 Assignment 5 . . . . . . 5009

D.2.6 Assignment 6 . . . . . . 5010

D.3 2000 / 2001 . . . . . . . . . . . . 5012

D.4 2001 / 2002 . . . . . . . . . . . . 5012

D.5 MATH 141 2003 01 . . . . . . . 5012

D.6 MATH 141 2004 01 . . . . . . . 5012

D.7 MATH 141 2005 01 . . . . . . . 5013

D.7.1 Written Assignment W 1 5013





D.8 MATH 141 2006 01 . . . . . . . 5021D.8.1 Solution to Written As-

signment W 1 . . . . . . 5021

D.8.2 Solution to Written As-

signment W 2 . . . . . . 5024

D.8.3 Solutions to Written As-

signment W 3 . . . . . . 5025


signment W 4 . . . . . . 5028


signment W 5 . . . . . . 5030

D.9 MATH 141 2007 01 . . . . . . . 5032

E Quizzes from Previous Years 5033

E.1 MATH 141 2007 01 . . . . . . . 5033

E.1.1 Draft Solutions to Quiz Q1 5033



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E.2 MATH 141 2008 01 . . . . . . . 5086E.2.1 Draft Solutions to Quiz Q1 5086




E.3 MATH 141 2009 01 . . . . . . . 5134




F Final Examinations from Previous Years 5152

F.1 Final Examination in Mathemat-

ics 189-121B (1996 / 1997) . . . 5152F.2 Final Examination in Mathemat-

ics 189-141B (1997 / 1998) . . . 5153

F.3 Supplemental / Deferred Examina-

tion in Mathematics 189-141B

(1997 / 1998) . . . . . . . . . . . 5155


ics 189-141B (1998 / 1999) . . . 5156



(1998 / 1999) . . . . . . . . . . . 5158

F.6 Final Examination in Mathemat-ics 189-141B (1999 / 2000) . . . 5160



(1999 / 2000) . . . . . . . . . . . 5161


ics 189-141B (2000 / 2001) . . . 5162



(2000 / 2001) . . . . . . . . . . . 5164


ics 189-141B (2001 /

2002) . . . 5165F.11 Supplemental / Deferred Examina-


(2001 / 2002) . . . . . . . . . . . 5167

F.12 Final Examination in MATH 141

2003 01 . . . . . . . . . . . . . 5169


tion in MATH 141 2003 01 . . . 5171F.14 Final Examination in MATH 141

2004 01 . . . . . . . . . . . . . 5172


tion in MATH 141 2004 01 . . . 5181


2005 01 . . . . . . . . . . . . . 5185


tion in MATH 141 2005 01 . . . 5189


2006 01 (One version) . . . . . 5192


tion in MATH 141 2006 01 . . . 5195


2007 01 (One version) . . . . . 5199


tion in MATH 141 2007 01 (One

version) . . . . . . . . . . . . . 5203


2008 01 (one version) . . . . . . 5207


tion in MATH 141 2008 01 (one

version) . . . . . . . . . . . . . 5217

F.24 Final Examination in MATH 1412009 01 (one version) . . . . . . 5221

G WeBWorK 6001

G.1 Frequently Asked Questions (FAQ) 6001

G.1.1 Where is WeBWorK? . 6001

G.1.2 Do I need a password to

use WeBWorK? . . . . 6001

G.1.3 Do I have to pay an ad-

ditional fee to use WeB-

WorK? . . . . . . . . . 6001

G.1.4 When will assignmentsbe available on WeBWorK? 6002

G.1.5 Do WeBWorK assign-

ments cover the full range

of problems that I should

be able to solve in this

course? . . . . . . . . . 6002

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G.1.6 May I assume that the

distribution of topics onquizzes and final exam-

inations will parallel the

distribution of topics in

the WeBWorK assign-

ments? . . . . . . . . . 6002

G.1.7 WeBWorK provides for

diff erent kinds of “Dis-

play Mode”. Which should

I use? . . . . . . . . . . 6002

G.1.8 WeBWorK provides for

printing assignments in

“Portable Document Format”

(.pdf), “PostScript” (.ps)

and “TEXSource” forms.

Which should I use? . . 6003

G.1.9 What is the relation be-

tween WeBWorK and We-

bCT? . . . . . . . . . . 6003

G.1.10 What do I have to do on

WeBWorK? . . . . . . 6003

G.1.11 How can I learn how to

use WeBWorK? . . . . 6004

G.1.12 Where should I go if Ihave difficulties with WeB-

WorK ? . . . . . . . . . 6004

G.1.13 Can the WeBWorK sys-

tem ever break down or

degrade? . . . . . . . . 6004

G.1.14 How many attempts may

I make to solve a partic-

ular problem on WeB-

WorK? . . . . . . . . . 6005

G.1.15 Will all WeBWorK as-

signments have the samelength? the same value? 6005

G.1.16 Is WeBWorK a good in-

dicator of examination per-

formance? . . . . . . . . 6005

H Contents of the DVD disks for

Larson / Hostetler / Edwards 6101

List of Tables

1 Schedule and Locations of Tu-

torials, as of April 9, 2010. . . . 4

2 Tutors’ Coordinates, as of April

9, 2010 . . . . . . . . . . . . . 5

3 Summary of Course Requirements,

as of April 9, 2010; (all dates

are subject to change) . . . . . . 11

4 Some Antiderivatives . . . . . . 3025

5 VeryShort Tableof Indefinite In-t e g r a l s . . . . . . . . . . . . . . 3026

List of Figures

1 The region(s) bounded by y =

x2 and y = x4 . . . . . . . . . . 3053


sin x, y = sin 2 x between x = 0

and x = π2

. . . . . . . . . . . . 3055


8 − x2, y = x2 between x = ±3 . 3056

4 The region(s) bounded by y =√ x + 2, y = x between x = 0

and x = 4 . . . . . . . . . . . . 3058

5 Regions for Example C.31 . . . 3061

6 The region(s) bounded by x +

y = 3 and x = 4 − ( y − 1)2 . . . 3070

7 The curve x = cos θ +sin2θ , y =

sin θ + cos2θ . . . . . . . . . . 3158

8 The cardioid with equation r =

2(1

−sin θ ) . . . . . . . . . . . 3170

9 The limacon r = 1 − 3cos θ , . . 3172

10 The spiral with equation r = θ ,

(θ ≥ 0) . . . . . . . . . . . . . . 3174

11 The spiral with equation r = θ ,

(θ ≤ 0) . . . . . . . . . . . . . . 3175

12 The full spiral with equation r =

θ , −∞ < θ < +∞ . . . . . . . . 3176

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13 The “4-leafed rose” with equa-

tion r = sin 2θ , . . . . . . . . . . 317714 The “5-leafed rose” with equa-

tion r = sin 5θ , . . . . . . . . . . 3178

15 The lemniscates r 2 = sin 2θ , r 2 =

cos2θ . . . . . . . . . . . . . . 3179

16 Intersecting polar curves r = 1+

sin θ , r 2 = 4 sin θ . . . . . . . . 3182

17 Curves r = sin θ , r = cos θ . . . 3184

18 The strophoid r = 2cos θ − sec θ 3186

19 Curves r = 2 + sin θ , r = 3 sin θ . 3188

20 Intersections of the limacon r =

1

−2cos θ with the circle r = 1 . 3191

21 Intersections of the curve r =

sec θ with the circle r = 1 . . . . 3193

22 The curves with equations r =

1 − cos θ , (θ ≤ 0), and r = 1 +

sin θ , and the point

12

, π2

. . . . 3243

23 The cardioids with equations r =

2 + 2sin θ , r = 6 − 6sin θ . . . . 3262

24 The region bounded by cardioids

r = 2 +2sin θ , r = 6−6sin θ and

containing the point (r , θ ) = (1, 0) 3263

25 The curves with equations r =

4 + 2cos θ , r = 4cos θ + 5 . . . 329126 The limacon r = 1 + 2sin θ . . . 5071

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Information for Students in MATH 141 2010 01 1

1 General Information

Distribution Date: January 04th, 2010

(all information is subject to change)

Pages 1 - 20 of these notes may be considered the Course Outline for this course.

These notes may undergo minor corrections or updates during the term: the defini-

tive version will be the version accessible at

http: // www.math.mcgill.ca / brown / math141b.html

or on myCourses, at

http: // www.mcgill.ca / mycourses /

Students are advised not to make assumptions based on past years’ operations,

as some of the details concerning this course could be diff erent from past years.

Publications other than this document may contain unreliable information about

this course.

All details of the course could be subject to discretionary change in case of force

majeure.

1

1.1 Force Majeure

In the event of extraordinary circumstances beyond the University’s control, all details of thiscourse, including the content and / or evaluation scheme are subject to change.

1Please note that the statements about MATH 141 in an SUS publication called Absolute Zero were not given

to instructors of this course to check, and some of them may not be currently correct.

8/15/2019 Brown's Notes 2010



1.2 Instructors and Times

INSTRUCTOR: Prof. W. G. Brown Dr. S. Shahabi Dr. A. Hundemer(Course Coordinator)

LECTURE SECTION: 1 2 3

CRN: 576 577 578

OFFICE: BURN 1224 BURN 1243 BURN 1128

OFFICE HOURS: W 15:45→16:45 F 09:30→11:30 MW 15:30→16:25

(subject to change) F 10:00→11:00 (tentative)

or by appointment

TELEPHONE: (514)-398-3836 (514)-398-3803 (514)-398-5318

E-MAIL:2 BROWN@ SHAHABI@ HUNDEMER@

MATH.MCGILL.CA MATH.MCGILL.CA MATH.MCGILL.CA

CLASSROOM: ADAMS AUD LEA 219 ADAMS AUDCLASS HOURS: MWF 11:35–12:25 h. MWF 11:35–12:25 h. MW 16:35–17:55 h.

1.3 Calendar Description

1.3.1 Calendar Description

MATH 1414 CALCULUS 2. (4 credits; 3 hours lecture; 2 hours tutorial. Prerequisites:

MATH 139 or MATH 140 or MATH 150. Restriction: Not open to students who have taken

MATH 121 or CEGEP objective 00UP or equivalent; not open to students who have taken or

are taking MATH 122 or MATH 130 or MATH 131, except by permission of the Department of

Mathematics and Statistics. Each Tutorial section is enrolment limited.) The definite integral.

Techniques of integration. Applications. Introduction to sequences and series.

Students Lacking the Prerequisite will, when discovered, be removed from the course.

Students without the prerequisite (or standing in a course recognized by the Admissions Office

as being equivalent to MATH 140) should not assume that, in possibly permitting MINERVA

to accept their registration for MATH 141, the University was tacitly approving their regis-

tration without the prerequisite. In particular, students who obtained a grade of F in MATH

139 / 140 / 150 are expressly excluded from registration in MATH 141, even if they registered in

the course before the failed or missed examination.

2Please do not send e-mail messages to your instructors through the WebCT or WeBWorK3 systems; rather,

use the addresses given in §1.2 on page 2.3E-mail messages generated by the Feedback command in WeBWorK should be used sparingly, and con-

fined to specific inquiries about WeBWorK assignments.4The previous designation for this course was 189-141, and the version given in the winter was labelled

189-141B; an earlier number for a similar course was 189-121.

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# CRN Day Begins Ends Room Tutor

T004 579 Fri 13:35 15:25 BURN 1B39 J. FeysT005 580 Tue 14:05 15:55 ARTS 260 H. Bigdely

T006 581 Tue 16:05 17:55 BURN 1B23 L. Candelori

T007 582 Tue 16:05 17:55 BURN 1B24 F. Castella

T008 583 Thurs 14:05 15:55 BURN 1B39 Y. Canzani

T009 584 Thurs 16:05 17:55 BURN 1B39 X. Zhang

T010 585 Thurs 16:05 17:55 BURN 1B23 A.-P. Grecianu

T011 586 Mon 13:35 15:25 ARTS W-20 J. Macdonald

T012 587 Mon 14:35 16:25 BURN 1B36 M. Prevost

T013 588 Mon 14:35 16:25 BURN 1B24 J. Tousignant-Barnes

T014 589 Wed 13:35 15:25 ARTS W-20 J. Restrepo

T015 590 Wed 14:35 16:25 LEA 14 B. Taji

T016 591 Wed 14:35 16:25 BURN 1B36 A. Tcheng

T017 2071 Mon 13:35 15:25 ENGMD 276 A. Tomberg

T018 2072 Wed 13:35 15:25 BURN 1B24 P. Rempel

T019 8194 Tue 08:05 09:55 BURN 1B23 Y. Rabhi

T020 8840 Wed 15:35 17:25 ENGMD 279 A. Farooqui

T021 8841 Fri 15:35 17:25 ENGMD 256 Y. Zhao

Some of these room assignments could change before or early in the beginning of the term, as we

have a pending request to upgrade some of the rooms. In any case, all assignments are subject to

change.

Table 1: Schedule and Locations of Tutorials, as of April 9, 2010.

• Outside of the normal quiz times in their tutorials, tutors are neither expected nor au-

thorized to administer a special quiz or a quiz that has already been administered to

others.

• Tutors in MATH 141 2010 01 are not permitted to off er paid, private tuition to students

in any tutorial section of this course.

1.4.3 Friday, April 02nd, 2010 and Monday, April 05th, 2010

These two lecture / tutorial days are lost because of the Easter holidays. While the lectures will

resume on Wednesday, April 07th, 2010, and the total number of lecture hours is similar to

past years, there will be some disruption to Monday and Friday tutorials: the Monday tutorials

will meet on Monday, April 12th, 2010, but there is no scheduled Friday available to complete

the Friday tutorials; alternative arrangements for students in Friday tutorials will be announced

later in the term.

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3. The final examination — counting for 70%.

Where a student’s performance on the final examination is superior to her performance on the

tutorial quizzes, the final examination grade will replace the quiz grades in the calculations;

in that case the grade on the final examination will count for 90% of the final grade. It is not

planned to permit the examination grade to replace the grades on WeBWorK assignments.

1.5.2 WeBWorK

1. The WeBWorK system, developed at the University of Rochester — is designed to

expose you to a large number of drill problems, and where plagiarism is discouraged.

WeBWorK is accessible only over the Internet. Details on how to sign on to WeBWorK

are contained in Appendix G to these notes, page 6001.Only answers submitted by the due date and time will count. The WeBWorK assign-

ments which count in your term mark will be labelled A1, . . ., A6.

2. Due dates and times for WeBWorK assignments. Most due dates for WeBWorK

assignments will be on specified Sundays, about 23:30h; last minute changes in the due

dates may be announced either on WeBWorK, on my Courses, or by an e-mail message11

As mentioned in the WeBWorK FAQ (cf. Appendix G), if you leave your WeBWorK

assignment until the hours close to the due time on the due date, you should not be

surprised if the system is slow to respond. This is not a malfunction, but is simply

a reflection of the fact that other students have also been procrastinating! To benefit

from the speed that the system can deliver under normal conditions, do not delay yourWeBWorK until the last possible day! If a systems failure interferes with the due date of

an assignment, arrangements may be made to change that date, and an e-mail message

may be broadcast to all users (to the e-mail addresses on record), or a note posted in the

course announcements on my Courses; but slowness in the system just before the due

time will not normally be considered a systems failure.12

3. Numbers of permitted attempts at WeBWorK questions. While the number of times

you may attempt each problem on WeBWorK An will be limited, there will be a com-

panion “Practice” Assignment Pn (n = 1, 2, . . . , 6) with an unlimited number of attempts

at similar problems, but in which the specific data may be diff erent. Thus you have

group, and observing whether your performance was at an appropriate level. Students who deny themselves this

experience often undergo a rude awakening at the final examination.11Be sure that your e-mail addresses are correctly recorded. See 4, p. 19 of these notes.12Should you find that the system is responding slowly, do not submit your solutions more than once; you may

deplete the number of attempts that have been allowed to you for a problem: this will not be considered a systems

failure.

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some or all problems on quizzes.14

The quizzes may examine on only a sampling of topics. Students should not assume thattopics not examined are in any subsidiary parts of the syllabus.

5. Your tutors will normally bring graded quizzes to the tutorial to be returned to you.

University regulations do not permit us to leave unclaimed materials bearing names and

student numbers in unsupervised locations; you may be able to recover an unclaimed

quiz from the tutor who graded it, during her / his regular office hours. Be sure to attend

the tutorial following a quiz15, as claims of incorrect recording of a quiz or assignment

grade will need to be substantiated by a graded paper.

6. Your quiz grades on assignments and quizzes will be posted on myCourses within about

2 weeks after they become available. Your WeBWorK grades may not be transferred tomyCourses until the end of the term, but will be visible on the WeBWorK site.

1.5.5 Final Examination

A 3-hour-long final examination will be scheduled during the regular examination period for

the winter term (April 15th, 2010 through April 30th, 2010). You are advised not to make any

travel arrangements that would prevent you from being present on campus at any time during

this period.16

1.5.6 Supplemental Assessments

1. Supplemental Examination. There will be a supplemental examination in this course.

(For information about Supplemental Examinations, see

http: // www.mcgill.ca / artscisao / departmental / examination / supplemental / .)

2. There is No Additional Work Option. “Will students with marks of D, F, or J have the

option of doing additional work to upgrade their mark?” No. (“Additional Work” refers

to an option available in certain Arts and Science courses, but not available in MATH

141 2010 01.)

14

In Math 141 the general rule for quizzes is that full solutions are expected to all problems, unless you receiveexplicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in the Student Solutions

Manual [9] to the textbook can serve as a guide to what should be included in a “full” solution.15The return of Quiz Q1 may be delayed to the 2nd week after the quiz was written.16Your instructors learn the date of your examination at the same time as you do — when the Provisional

examination timetable is published.

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You are also referred to the following URL:

http: // www.mcgill.ca / integrity / studentguide /

Other Fraud. It is a serious off ence to alter a graded quiz paper and return it to the tutor

under the pretense that the work was not graded properly.

1.5.9 Corrections to grades

Grades will eventually be posted on my Courses. If you believe a grade has been recorded

incorrectly, you must advise your tutor not later than 4 weeks after the grade has been posted,

and not later than the day before of the final examination whichever of these dates is earlier. It

is hoped that grades will be posted within 2 weeks of the due date. You will have to present thegraded quiz to support your claim, which must be submitted to the tutor that graded the quiz. If

he / she believes there has been an error, the tutor will advise Professor Brown. New corrections

to the my Courses posting will appear the next time grades are uploaded to my Courses.

1.6 Published Materials

1.6.1 Required Text-Book

The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: Early Tran-

scendentals, Sixth Edition, Brooks / Cole (2008), ISBN 0-495-01169-X, [1]. This book is the

first half of J. Stewart, CALCULUS: Early Transcendentals, Sixth Edition, Brooks / Cole(2008), ISBN 0-495-01166-5, [2]; this edition covers the material for Calculus 3 (MATH 222)

as well, but is not the text-book for that course at the present time. The textbook will be sold

in the McGill Bookstore bundled with its Student Solutions Manual (see below). The ISBN

number for the entire bundle is 0-495-42966-X.

1.6.2 Optional Reference Books

Students are urged to make use of the Student Solution Manual:

• D. Anderson, J. A. Cole, D. Drucker, STUDENT SOLUTIONS MANUAL FOR STEW-

ART’S SINGLE VARIABLE CALCULUS: Early

Transcendentals

, Sixth

Edition

, Brooks / Col(2008), ISBN 0-495-01240-8, [3]. This book is also sold “bundled” with the text book;

we expect the Bookstore to stock the bundle numbered ISBN 0-495-42966-X [4].

The publishers of the textbook and Student Solutions Manual also produce

UPDATED TO April 9, 2010

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1.6.3 Recommended Video Materials

Use of the following materials is recommended, but is not mandatory18.

Text-specific DVDs for Stewarts Calculus, early transcendentals, 6th edition [videorecord-

ing]. The publisher of Stewart’s Calculus has produced a series of videodisks, [?]. These will

initially be available for reserve loan at the Schulich Library. There may not be DVD viewing

equipment freely available in the library; the intention is that interested students borrow disks

for viewing on their own equipment at home. Disk 1 covers Chapters 1-6 of the textbook.

Videotapes for Stewart’s Calculus The publisher of Stewart’s Calculus had earlier pro-

duced a series of videotapes, [14] Video Outline for Stewart’s Calculus (Early Transcenden-

tals), Fifth Edition. These will be available for reserve loan at the Schulich Library. There maynot be VCR viewing equipment in the library; the intention is that interested students borrow

a tape for viewing on their own equipment at home.

Larson / Hostetler / Edwards DVD Disks A set of video DVD disks produced for another

calculus book, [28] Calculus Instructional DVD Program, for use with (inter alia) Larson /

Hostetler / Edwards, Calculus of a Single Variable: Early Transcendental Functions, Third

Edition [29] is produced by the Houghton Mifflin Company. A copy has been requested to be

placed on reserve in the Schulich Library. In Appendix H of these notes there are charts that

indicate the contents of these disks that pertain to MATH 141.

Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 6th

Edition, (similar to [15]) 19This CD-ROM is included with certain new copies of the text-

book. It contains, after an enlightening “pep-talk” by the author, a discussion of some of the

worked examples in the text-book, followed by a quiz for each section in the book. Some stu-

dents may find the animations of the examples helpful, although the examples are all worked

in the book. You might wish to try some of the quiz questions using paper and pencil, and

then check your answers with those given on the CD. It is not recommended that you attempt

to enter your answers digitally, as this is a time-consuming process, and uses a diff erent input

method from your WeBWorK assignments, which serve the same purpose.

18

No one will check whether you have used any of these aids; a student can obtain a perfect grade in the coursewithout ever consulting any of them. No audio-visual or calculator aid can replace the systematic use of paper

and pencil as you work your way through problems. But the intelligent use of some of these aids can deepen your

understanding of the subject. However, the most important aid is the Student Solutions Manual to the textbook!19The version of this CD-ROM for the 6th edition is being catalogued by the Library; it may not be available

at the beginning of the term.

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1.6.4 Other Calculus Textbooks

While students may wish to consult other textbooks, instructors and teaching assistants in

Math 141 will normally refer only to the prescribed edition of the prescribed textbook for the

course. Other books can be very useful, but the onus is on you to ensure that your book covers

the syllabus to at least the required depth; where there are diff erences of terminology, you are

expected to be familiar with the terminology of the textbook.20

In your previous calculus course(s) you may have learned methods of solving problems that

appear to diff er from those you find in the current textbook. Your instructors will be pleased

to discuss any such methods with you personally, to ascertain whether they are appropriate to

the present course. In particular, any methods that depend upon the use of a calculator, or the

plotting of multiple points, or the tabulation of function values, or the inference of a trend from

a graph should be regarded with scepticism.

1.6.5 Website

These notes, and other materials distributed to students in this course, will be accessible

through a link on the myCourses page for the course, as well as at the following URL:

http: // www.math.mcgill.ca / brown / math141b.html

The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader, which

many users have on their computers. This free software may be downloaded from the following

URL:

http: // www.adobe.com / prodindex / acrobat / readstep.html 21

The questions on some old examinations will also be available as an appendix to these notes

on the Web.22

Where revisions are made to distributed printed materials — for example these information

sheets — we expect that the last version will be posted on the Web.

The notes and WeBWorK will also be available via a link from the my Courses (WebCT)

URL:

http: // mycourses.mcgill.ca

20There should be multiple copies of the textbook on reserve in the Schulich library.21At the time of this writing the current version appears to be 8.n.22There is no reason to expect the distribution of problems on quizzes or in assignments and examinations from

previous years be related to the frequencies of any types of problems on the examination that you will be writing

at the end of the term.

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When to do the WeBWorK assignment. I recommend that you defer working WeBWorK

problems until you have tried some of the easier odd-numbered problems in the textbook. Forthese you (should) have the Student Solutions Manual to help you check your work. Once

you know that you have the basic concepts mastered, then is a good time to start working

WeBWorK problems. But these should be done first from a printed copy of your assignment

— not worked during real time online.

The real uses of WeBWorK and the quizzes. Students often misunderstand the true signif-

icance of WeBWorK assignments and the quizzes. While both contribute to your grade, they

can help you estimate the quality of your progress in the course. Quizzes are administered un-

der examination conditions, so poor performance or non-performance on quizzes can provide

an indicator of your expectations at the final examination; take proper remedial action if you

are obtaining low grades on quizzes27. Since WeBWorK is not completed under examination

conditions, the grades you obtain may not be a good indicator of your expectations on the ex-

amination; if you require many attempts before being able to solve a problem on WeBWorK,

you should use that information to direct you to areas requiring extra study: the WeBWorK

grades themselves have little predictive use, unless they are unusually low. However, while

both WeBWorK and the quizzes have a role to play in learning the calculus, neither is as im-

portant as reading your textbook, working problems yourself, and attending and listening at

lectures and tutorials.

What to strive for on WeBWorK assignments. Since the practice assignments give you

ample opportunity to experiment, your success rate on the assignments “that count” should beclose to 100%. If you are needing more than 2 attempts to solve a WeBWorK problem, then

you are probably not ready to work the assignment. In order to be able to solve a WeBWorK

problem successfully on the first attempt you will need to check your work, and this is a skill

that you will need on the final examination, and in the advanced studies or the real world where

you may eventually be applying the calculus.

1.8.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experiencing

medical or personal difficulties should not hesitate to consult their advisors or the Student

Aff airs office of their faculty. Don’t allow yourself to be overwhelmed by such problems; theUniversity has resource persons who may be able to help you.

27The worst action is to miss the quizzes, and thereby block out an unwelcome message.

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1.8.5 Terminology

Do not be surprised if your instructors and tutors use diff erent terminology from what you

have heard in your previous calculus course, particularly if that course was at a high school.

Sometimes the diff erences are purely due to diff erent traditions in the professions.

“Negative x” Your instructors and tutors will often read a formula − x as minus x, not as

negative x. To a mathematician the term negative refers to real numbers which are not squares,

i.e. which are less than 0, and − x can be positive if x itself is negative.

However, mathematicians will sometimes refer to the operation of changing a sign as the

replacement of x by “its negative”; this is not entirely consistent with the usual practice, but is

an “abuse of language” that has crept into the professional jargon.

Inverse trigonometric functions A formula like sin−1 x will be read as the inverse sine of

x — never as “sine to the minus 1” or “sine to the negative 1”. However, if we write sinn x,

where n is a positive integer, it will always mean (sin x)n. These conventions apply to any of

the functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolic functions, which we

have met on general functions, so a formula like f 2( x) does not have an obvious meaning, and

we will avoid writing it when f is other than a trigonometric or hyperbolic function.

Logarithms Mathematicians these days rarely use logarithms to base 10. If you were taught

to interpret log x as being the logarithm to base 10, you should now forget that — although it

could be the labelling convention of your calculator. Most often, if your instructor speaks of alogarithm, and writes log x, he will be referring to the base e, i.e. to loge; that is, he is referring

to the function that calculus books call ln. When a logarithm to some other base is intended,

it will either be denoted by an explicit subscript, as log 2, or some comment will be made at

the beginning of the discussion, as “all logarithms in this discussion are to the base 2”. Your

instructors try to think like mathematicians even when lecturing to their classes, and so we use

the language and terminology we use when talking to each other.

1.9 Communication with Instructors and TA’s

1. E-mail messages to your instructor or your TA should be sent to the addresses shown in

Table 1.2 and Table 2. Please show your full name and / or student number, so that wecan clearly identify you.

2. The only messages sent through WeBWorK should be those generated by the Feedback

facility: this means a message that refers to a specific problem on a specific WeBWorK

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1.11 Special Office Hours and Tutorials

The following chart will show any special activities that are scheduled during the term. This

table was last updated on April 09, 2010.

Review Tutorial TA / Instructor location Date Time

Polar Coordinates Dr. Y. Zhao ARTS 145 13 April 15:05–16:55

Sequences + Series J. Feys BURN 1B45 16 April 18:05–19:55


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2 Draft Solutions to Quiz Q1

For each of the days of the week a sample quiz is given below. Our policy is that, to earn part

marks on any separately numbered part of any question a student must have, in the opinion of

the grader, earned at least half of the marks available.

2.1 Instructions to Students

1. Show all your work. Marks may not be given for answers not supported by a full solu-

tion. For future reference, the form of your solutions should be similar to those shown

in the textbook or Student Solutions Manual for similar problems.

2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH

0.) All submissions should carry your name and student number.

3. Time = 20 minutes.

4. No calculators are permitted.

2.2 Monday Versions

1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component

rectangles hang by their upper right-hand corner from the graph) to compute

3 0

(10 x −

2) dx . No other method will be accepted!

Solution: If we divide the interval [0, 3] into n subintervals of equal lengths ∆ x = 3n

, the

right end-points of these intervals are ∆ x, 2∆ x, . . . , n∆ x, so the definite integral is equal

to the following limit of a (right) Riemann sum:

limn→∞

ni=1

(10i∆ x − 2) ∆ x = limn→∞

10(∆ x)2

ni=1

i − 2∆ x

ni=1

1

= limn→∞

10 · 3

2

n2 · n(n+

1)2

− 2 · 3n

· n

= limn→∞

45 · n(n + 1)

n2 − 6 · n

n

= limn→∞

45 · 1 ·

1 +

1

n

− 6

= 39 .

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It would not have been acceptable to evaluate this integral using the Fundamental Theo-

rem, but that theorem could be used by a student to verify her work:3

0

(10 x − 2) dx =5 x2 − 2 x

3

0= 5 · 32 − 2 · 3 = 45 − 6 = 39.

2. [10 MARKS] Compute

(a)

π3

0

(sec x)(7 sec x + 3tan x) dx .

(b)

1 0

√ x ·

5 x2 − 5 x − 4

dx .

Solution: For these problems the use of the Fundamental Theorem was not excluded.

(a)

π3

0

(sec x)(7 sec x + 3 tan x) dx =

π3

0

7sec2 x + 3sec x · tan x

dx

= [7 tan x + 3sec x]π3

0

=

7tan

π

3 + 3sec

π

3

− (7tan0 + 3sec0)

= 7√

3 + 3 · 2 − 7 · 0 − 3 · 1 = 7√

3 + 3 .

(b)

1 0

√ x ·

5 x2 − 5 x − 4

dx =

1

0

5 x

52 − 5 x

32 − 4 x

12

=

5 · 2

7 · x 7

2 − 5 · 2

5 · x 5

2 − 4 · 2

3 · x 3

2

1

0

= 10

7 − 2 − 8

3 = −68

21 .

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=

− e2 x

1 (ln t )2 + 7 dt +

e4 x

1 (ln t )2 + 7 dt

Now we apply the Chain Rule in diff erentiating each of the summand integrals:

d

dx

e4 x

e2 x

(ln t )2 + 7 dt = − d

dx

e2 x

1

(ln t )2 + 7 dt +

d

dx

e4 x

1

(ln t )2 + 7 dt

= − d

du

e2 x

1

(ln t )2 + 7 dt

· du

dx +

d

dv

e4 x

1

(ln t )2 + 7 dt

· dv

dx

= −

(ln e2 x)2 + 7 · 2e2 x +

(ln e4 x)2 + 7 · 4e4 x

=

− (2 x)2 + 7

·2e2 x + (4 x)2 + 7

·4e4 x

= −√ 4 x2 + 7 · 2e2 x + √ 16 x2 + 7 · 4e4 x .

2.5 Thursday Versions

1. [10 MARKS] Use a left Riemann sum (i.e. a Riemann sum where the component rect-

angles hang by their upper left-hand corner from the graph) to compute

2 −1

x2 + 1

dx.

No other method will be accepted!

Solution: If we divide the interval [

−1, 2] into n subintervals of equal lengths ∆ x =

2−(−1)n = 3

n , the left end-points of these intervals are −1, −1 + ∆ x, −1 + 2∆ x, . . . , −1 + (n −1)∆ x, so the definite integral is equal to the following limit of a (left) Riemann sum:

limn→∞

ni=1

(−1 + (i − 1)∆ x)2

+ 1

∆ x

= limn→∞

ni=1

2 − 2(i − 1)∆ x + (i − 1)2(∆ x)2

· ∆ x

= limn→∞

2∆ x

n

i=1

1 − 2(∆ x)2

n

i=1

(i − 1) + (∆ x)3

n

i=1

(i − 1)2

= limn→∞

2 · 3

n· n − 2 ·

3

n

2

· (n − 1)n

2 +

3

n

3

· (n − 1)n(2(n − 1) + 1)

6

= lim

n→∞

6 − 9

1 − 1

n

+

9

2

1 − 1

n

2 − 1

n

= 6 − 9 + 9 = 6 .


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rem, but that theorem could be used by a student to verify her work:

2 −1

x2 + 1

dx = 2

1

3 x3 + x

2

−1

=

8

3 + 2

−−1

3 − 1

= 6 .

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all

critical numbers of the function

9 e x2 +7 x+12

ln t dt .

Solution: We first determine the derivative, using the Fundamental Theorem and the

Chain Rule. Define u = e x2+7 x+12

. Then

du

dx = (2 x + 7)e x2+7 x+12

.

d

dx

9 e x2+7 x+12

ln t dt = d

dx

−e x2+7 x+12

9

ln t dt

= − d

du

e x2 +7 x+12 9

ln t dt

· du

dx

=

− ln e x2+7 x+12 ·(2 x + 7)e x2+7 x+12

= − x2 + 7 x + 12

(2 x + 7)e x2+7 x+12

= −( x + 3)( x + 4)(2 x + 7)e x2+7 x+12 .

This derivative has an exponential factor which cannot equal 0. The function is diff eren-

tiable for all x, and the derivative is 0 when x = −4, −72

, −3, so the latter 3 points are the

critical numbers of the function.

2.6 Friday Versions

1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component rect-

angles hang by their upper right-hand corner from the graph) to compute

2 0

x2 − 3

dx.

No other method will be accepted!

Solution: If we divide the interval [0, 2] into n subintervals of equal lengths ∆ x = 2n

, the

right end-points of these intervals are ∆ x, 2∆ x, . . . , n∆ x, so the definite integral is equal

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to the following limit of a (right) Riemann sum:

limn→∞

ni=1

(i∆ x)3 − 3

∆ x = lim

n→∞

(∆ x)4

ni=1

i3 − 3∆ x

ni=1

1

= lim

n→∞

2

n

4

· n2(n + 1)2

4 − 3 · 2

n· n

= lim

n→∞

4

1 +

1

n

2

− 2

= 4 − 6 = −2 .


rem, but that theorem could be used by a student to verify her work:

2 0

x3 − 3

dx =

1

4 · x4 − 3 x

2

0

= 4 − 6 = −2.

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all

critical numbers of the function f ( x) =

4 x2−4 x−5

sinh t dt .

Solution: Define u = x2

−4 x

−5, so

du

dx

= 2 x

−4. Then

d

dx

4

x2−4 x−5

sinh t dt = d

dx

− x2−4 x−5

4

sinh t dt

= − d

du

x2−4 x−5

4

sinh t dt

· du

dx

= −sinh

x2 − 4 x − 5

· (2 x − 4) .

The sinh function is 0 if and only if its argument is 0, here if and only if ( x − 5)( x + 1) =

x2 − 4 x − 5 = 0, i.e., if and only if x = 5 or x = −1. Thus the derivative — which is

defined for all x — is 0 when x = −1, 2, 5; these are the critical numbers.

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3 Draft Solutions to Quiz Q2

For each of the days of the week a sample quiz is given below. Our policy is that, to earn part

marks on any separately numbered part of any question a student must have, in the opinion of

the grader, earned at least half of the marks available.

3.1 Instructions to Students

1. Show all your work! To be awarded partial marks on a part of a question a student’s

answer for that part must be deemed to be more than 50% correct. For future reference,

the form of your solutions should be similar to those shown in the textbook or Student

Solutions Manual for similar problems.

2. In your folded answer sheet you must enclose this question sheet: it will be returned

with your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH




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3.3 Most Tuesday Versions

1. [10 MARKS] Use appropriate substitutions to compute

(a)

e5t + 5

e5t + 25t dt

(b)

x3

√ x2 − 1 dx (Hint: Start with the substitution u = x2 − 1.)

Solution:

(a) Try the substitution u = e5t + 25t . Then du =5e5t + 25

dt = 5

e5t + 5

dt . Hence

e

5t

+ 5e5t + 25t

dt = 15

du

u= 1

5 ln u + C

= 1

5 ln

e5t + 25t

+ C .

(b) The hint suggests using the substitution u = x2 − 1 ⇒ du = 2 x dx. Then x3

√ x2 − 1 dx =

(u + 1)

√ u · du

2

= 1

5 · u

52 +

1

3u

32 + C

= 15

( x2 − 1)52 + 1

3( x2 − 1)

32 + C .

(There are variations of this substitution that could have been used. For example,

we could try the substitution u =√

x2 − 1, so u2 = x2 − 1, x2 = u2 + 1, u du = x dx.

Then x3

√ x2 − 1 dx =

(u2 + 1)u · u du

=

u4 + u2

du

= u5

5 + u

3

3 + C

= 1

5

x2 − 1

52

+ 1

3

x2 − 1

32

+ C .

2. [10 MARKS] Let R be the region in the first quadrant bounded by the lines y = 0, y = 5 x,

and the curve y = 6

x− 1. Compute the volume of the solid of revolution obtained by

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2. [10 MARKS] Use integration by parts to compute the integral arctan 6

x dx.

Solution: We know that the integration of arctan x can be accomplished by integration

by parts, so we can try the same tactic here, with u = arctan

6 x

, and dv = dx. Then

v = x, and

du = 1

1 +

6 x

2 · −6

x2 dx =

−6

36 + x2 · dx .

Hence arctan

6

x

dx = arctan

6

x

· x + 6

x

36 + x2 dx .

The new integrand can be integrated by using a substitution like w = 36 + x2, where

dw = 2 x dx , so x dx = dw

2 :

arctan

6

x

dx = arctan

6

x

· x + 6

1

2wdw

= arctan

6

x

· x + 3 · ln |w| + C

= arctan

6

x

· x + 3 · ln

36 + x2

+ C .

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Solution: One can start by defining a substitution u = x + 5, but the substitution could

be deferred, or even entirely avoided. For example, just take u = ln( x + 5), dv = x dx, so

du = dx

x + 5 , v =

x2

2 .

Then x ln( x + 5) dx = (ln( x + 5)) · x2

2 −

x2

2 · dx

x + 5 .

The last integral could be integrated by simply using long division, where x2 = ( x +

5)( x − 5) + 25, so

1

2 x2

x + 5 = x

−5

2 dx +

25

2 1

x + 5 d x =

x2

4 − 5 x

2 +

25

2 ln | x + 5| + C . (1)

Hence x ln( x + 5) dx =

x2

2 · ln( x + 5)) − x2

4 +

5 x

2 − 25

2 ln | x + 5| + C 1 .

The long division step is one of the standard operations we will see when we study the

integration of general rational functions. But one could have used, in the second integral

in equation (1), a substitution u = x + 5, to obtain

x2

2( x + 5) d x = (u

−5)2

2u du

=

u

2 − 5 +

25

2u

du

= u2

4 − 5u +

25

2 ln |u| + C =

( x + 5)2

4 − 5( x + 5) +

25

2 ln | x + 5| + C 2 .

The substitution could also have been eff ected at the beginning, and it was this sequence

that was contemplated in the problem:

x ln( x + 5) dx = (u − 5)(ln u) du .

Now we can integrate by parts, taking U = ln u, dV = (u − 5) du, so that dU = du

u,

V = u2

2 − 5u,

x ln( x + 5) dx =

(u − 5)(ln u) du

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= (ln u) u2

2 −5u −

u2

2 −5u ·

du

u

= (ln u)

u2

2 − 5u

−

u

2 − 5

du

= (ln u)

u2

2 − 5u

− u2

4 + 5u + C

= (ln( x + 5))

( x + 5)2

2 − 5( x + 5)

− ( x + 5)2

4 + 5( x + 5) + C 3 .

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3.6 Friday Versions

1. [10 MARKS] Let R be the region bounded by the lines y = 0, y = ln7, x = 0, and

the curve y = ln x. Compute the volume of the solid obtained by revolving R about the

x-axis. [Hint: Sketch the region; then choose the appropriate method.]

Solution:

(a) (Using Cylindrical Shells): We will need to find the height of the cylindrical shell

of radius y, and this will require transforming the equation y = ln x into x = e y.

The distance from x = 0 to x = e y is then e y − 0, so the volume is

ln 7 0

2π y · e y dy ,

which we will integrate by parts, taking u = y, dv = e

y

dy, so that du = dy, v = e

y

:ln 7

0

2π y · e y dy = 2π ye yln 7

0 − 2π

ln 7

0

e y dy

= 2π ye yln 7

0 − 2π [e y]ln 70

= 2π((ln 7) · 7 − 0) − 2π(7 − 1) = 2π(7ln7 − 6) .

(b) (Using Washers): One drawback of this method is that there will be two kinds of

washers: For 0 ≤ x ≤ 1 the washers have no hole in the middle, and the radius is a

constant, ln 7; but for 1 ≤

x ≤

7 there will be a hole whose radius is ln x. The area

is, therefore, a sum 1

0

π

(ln 7)2 − 02

dx +

7

1

π

(ln 7)2 − (ln x)2

dx

=

7

0

π

(ln 7)2 − 02

dx − π

7

1

(ln x)2 dx

= π(ln 7)2

7

0

dx − π

7

1

(ln x)2 dx

= 7(ln 7)2

−π

7

1

(ln x)2 dx .

To integrate (ln x)2 we could apply integration by parts twice. Start by taking u =

(ln x)2, v = 1, so that u = 2ln x

x, v = x, and

7

1

(ln x)2 dx =(ln x)2 · x

7

1− 2

7

1

ln x

x· x dx

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= (ln x)2

· x7

1

−2

7

1

ln x dx

=(ln x)2 · x

7

1− 2[ x · ln x − x]7

1

= [7(ln 7)2 − 0] + 2[(7ln7 − 7) − (0 − 1)]

which leads to the same result as found earlier.

2. [10 MARKS] Use an appropriate substitution to compute the integral

9 0

1

6 +√

xdx .

(Hint: Start with the substitution u = 6 +√

x.)

Solution: One substitution that would simplify this integral is u = 6 + √ x ⇒ √ x =u − 6 ⇒ x = (u − 6)2 ⇒ d x = 2(u − 6) du. Thus

9 0

1

6 +√

xdx =

6+√

9

6

2(u − 6)

udu

= 2

9

6

1 − 6

u

du

= 2[u − 6 ln u]96 = 2(9 − 6 l n 9 − 6 + 6ln6) = 6 + 12ln

2

3 .

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[12] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks / Cole

(2003); bundled with Student Solutions Manual for Stewart’s Single Variable Calculus(Early Transcendentals), Fifth Edition. Thomson * Brooks / Cole (2003). ISBN 0-534-

10307-3.

[13] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcenden-

tals), Fifth Edition. Thomson * Brooks / Cole (2003). ISBN 0-534-39331-4.

[14] Video Outline for Stewart’s Calculus (Early Transcendentals), Fifth Edition. Thomson

* Brooks / Cole (2003). ISBN 0-534-39325-X. 17 VCR tapes.

[15] Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 5th

Edition. Thomson * Brooks / Cole (2003). ISBN 0-534-39326-8.

[16] H. Keynes, J. Stewart, D. Clegg, Tools for Enriching Calculus, CD to accompany [7]

and [8]. Thomson * Brooks / Cole (2003). ISBN 0-534-39731-X.

[17] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition.

Brooks / Cole (1999). ISBN 0-534-35563-3.

[18] J. Stewart, Calculus (Early Transcendentals), Fourth Edition. Brooks / Cole (1999). ISBN

0-534-36298-2.

[19] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Single

Variable Calculus (Early Transcendentals), Fourth Edition. Brooks / Cole (1999). ISBN

0-534-36301-6.

[20] J. Stewart, L. Redlin, S. Watson, Precalculus: Mathematics for Calculus, Enhanced

Review Edition. Thomson * Brooks / Cole. (2006). ISBN: 0-495-39276-6.

[21] J. Stewart, Trigonometry for Calculus. Thomson * Brooks / Cole. ISBN: 0-17-641227-1.

4.2 Other Calculus Textbooks

4.2.1 R. A. Adams

[22] R. A. Adams, Calculus, Single Variable, Fifth Edition. Addison, Wesley, Longman,

Toronto (2003). ISBN 0-201-79805-0.

[23] R. A. Adams, Calculus of Several Variables, Fifth Edition. Addison, Wesley, Longman,

Toronto (2003). ISBN 0-201-79802-6.

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4.2.4 Others, not “Early Transcendentals”

[35] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge University Press

(1967).

[36] H. S. Hall, S. R. Knight, Elementary Trigonometry, Fourth Edition. Macmillan and

Company, London (1905).

[37] S. L. Salas, E. Hille, G. J. Etgen, Calculus, One and Several Variables, 10th Edition.

John Wiley & Sons, Inc. (2007). ISBN 0471-69804-0.

4.3 Other References

[38] G. N. Berman, A Problem Book in Mathematical Analysis. Mir Publishers, Moscow,

(1975) 1977

[39] D. Ebersole, D. Schattschneider, A. Sevilla, K. Somers, A Companion to Calculus.

Brooks / Cole (1995). ISBN 0-534-26592-8.

[40] McGill Undergraduate Programs Calendar 2009 / 2010. Also accessible at

http: // coursecalendar.mcgill.ca / ug200910 / wwhelp / wwhimpl / js / html / wwhelp.htm

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Information for Students in Lecture Section 002 of MATH 141 2010 01 2002

Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAY

MARCH

01 §10.1,§10.2 03 §10.3 05 §10.3, §10.4

WeBWorK Assignment A4 due Mar. 07, 2010

08 §10.4 A4 Q3 10 §11.1, §11.2 Q3 12 §11.2 Q3

15 §11.3 17 §11.4 19 §11.4, §11.5

WeBWorK Assignment A5 due Sunday, Mar. 28, 2010

22 §11.5 A5 Q4 24 §11.6 Q4 26 §11.6 Q4

29 §11.6 31 §11.7, X

APRIL

02 NO LECTURE Q4

This week’s tutorials are the last (except for Monday tutorials).

WeBWorK Assignment A6 due Apr. 04, 2010

05 NO LECTURE 07 X 09 X

12 X 14 X

Notation: An = Regular WeBWorK Assignment An due about 23:30 hours

on the Sunday preceding this Monday

Qn = Quiz Qn will be administered at the tutorials this week.

X = reserved for eXpansion or review

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C Supplementary Notes for Students in Section 001 of MATH

141 2010 01

C.1 Lecture style in Lecture Section 001

Lecture content. The timetable on pages 1001, 1002 will show you approximately what I

plan to discuss at each lecture. I suggest that you look through the material in advance. If

you have time to try some of the exercises, and find some that cause you di fficulty, you are

welcome to bring them to my attention; perhaps I may be able to work some of these examples

into the lecture.

What goes on the chalkboard? — Should I take notes? I believe strongly that students

should not sit in the lecture feverishly copying notes for fear of missing some essential topic;

in this course most of what you need to know is contained in the textbook. You should take

notes, but you should be trying to think at the same time. The chalkboard will be used for

• statement / illustration of specific definitions and theorems

• sketching solutions to problems, or classes of problems

• a scratchpad

Some of this material will be useful to you in learning the material in the course. Even when

the material on the board is equivalent to something in your textbook, the act of writing mayhelp you remember it. But much of the material will be restatements of your textbook, so you

should normally not panic if you miss something.

Graphs My emphasis is on qualitative properties of the graphs of functions, but not on the

production of extremely precise graphs. You can expect to see me draw on the chalkboard

sketches that are extremely crude approximations of functions, sometimes even caricatures

of the true graph. Mathematicians do not base proofs on sketches of graphs — the role of a

sketch is usually only to assist the reader to visualize the verbal or symbolic reasoning which

accompanies it. Sometimes a graph is used help one discover a phenomenon, but the result

would not be acceptable to a mathematician unless it could be proved in a non-graphical way. 29

These supplementary notes Section and paragraph headings follow the order of topics in

the textbook. While some of the comments or explanations may be helpful in understanding the

book, the notes are not required reading for examination purposes. Sometimes it may happen

that the discussion of a topic or an exercise evolves during the lecture into one which requires

29This is why I usually avoid problems in the textbook that appear to be drawing inferences from graphs.

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more detail than is practical to write on the chalkboard. In such cases you may be referred

at the following lecture to supplementary material that will be contained in the notes placedon the Web. Such evolutions are spontaneous and not planned, and cannot be announced in

advance.

Timing and corrections The notes will usually not be posted until after the lecture. While

I do try to check the notes before posting them, there will inevitably be errors: if you see

something that doesn’t look right, please ask. The notes will be progressively corrected as

misprints and other errors are discovered.

C.2 Supplementary Notes for the Lecture of January 04th, 2010

Release Date: January 04th, 2010,subject to revision

Textbook Chapter 5. INTEGRALS.

(There will be examples, etc., in these notes that were not discussed specifically at the lecture,

because of time constraints.)

C.2.1 §5.1 Areas and Distances.

When, in [1, §2.1], the textbook was motivating the diff erential calculus, it presented two appli-

cations: “The Tangent Problem”, which was geometric; and “The Velocity Problem”, which

was physical. Now, in motivating the integral calculus, the author presents two analogous

problems: “The Area Problem”, which is geometric; and “The Distance Problem”, which is

physical.

The Area Problem. The textbook discusses approximation of the area under the graph of

y = f ( x) between x = a and x = b — more precisely, the area between the graph, the x-axis,

and the vertical lines x = a and x = b, as a limit of a sum of areas of narrow vertical rectangles.

The approximation is first motivated with simple functions where the area can be bounded

above and below by easily computable sums, which together converge to the same value as

their number approaches ∞ and their width approaches 0. You should become comfortable

using the “sigma notation”, where, for integers n1 and n2 with n1 ≤ n2, we writen2

i=n1

f (i)

to mean the sum f (n1) + f (n1 + 1) + f (n1 + 2) + . . . + f (n2) . The textbook then proposes the

following definition of the area between the graph of a function, two vertical lines, and the

x-axis:

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cases of the following:k + i=k

(rai + sbi) = r

k + i=k

ai + s

k + i=k

bi .

I may have more to say about the sigma notation after I discuss [1, §5.5], where we shall

encounter properties of the integral that have analogues for sums. For the present let it be

noted that the symbol i ink + i=k

ai is not a “free” variable, in that you cannot assign any values to

it: it performs a function in the symbol, but that function would be performed equally well if

we replaced i by any other symbol that is not already in use, e.g.,k + u=k

au,k + λ=k

aλ,k + ♥=k

a♥ .

The Midpoint Rule. The “Midpoint Rule” is an approximation formula for definite inte-grals. Use of an approximation formula entails a willingness to accept an error in the cal-

culation. Mathematicians normally expect to see an estimate of how good or how bad an

approximation can be before recommending their use. A partial justification of the Midpoint

Rule is contained in [1, §7.7], a section that is to be omitted from the syllabus. For that reason

you are asked to omit this subsection: you will not be expected to know anything about the

Midpoint Rule.

Properties of the Definite Integral – Linearity Properties. The textbook lists many prop-

erties of the Definite Integral, proving some of them.

1.

b a

c dx = c(b − a)

2.

b a

[ f ( x) + g( x)] dx =

b a

f ( x) dx +

b a

g( x) dx

3.

b a

c f ( x) dx = c

b a

f ( x) dx

4.

b a

[ f ( x) − g( x)] dx =

b a

f ( x) dx −b

a

g( x) dx

for any real numbers a, b, c, and any continuous functions f , g.

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Release Date: Friday, January 08th, 2010, subject to correction

C.4.1 Summary of the last lectures

1. The objective was to define what we mean by the (signed) area under the graph of a

function; or, equivalently, the (signed) distance travelled by a moving particle, giving its

instantaneous velocity.

2. For a function f continuous on an interval [a, b] I defined the definite integral

b a

f ( x) dx

as a kind of limit limmax ∆ xi → 0

n → ∞

ni=1

f ( x∗i )∆ xi , called a Riemann sum, and mentioned

that this definition and introduction was simplified for your first encounter, and some-

what lacking in rigour. The treatment of the textbook, which is restricted to continuous

functions, does not require the possibility that the width ∆ xi of the ith interval be pos-

sibly diff erent from the that of other subintervals; thus we will be considering only one

width, and denote it simply by ∆ x.

3. I mentioned that it can be shown that the sum must have the same limit for all sub-

divisions of [a, b] into subintervals [a = x0

, x1

], [ x1

, x2

], . . ., [ xi−1

, xi], [ x

i, x

i+

1], . . .,

[ xn−1, xn = b], and all choices of x∗i in the ith subinterval (i = 1, . . . , n).

4. I illustrated computations with Riemann sums using the function f ( x) = x2 over the

interval [a, b]; the variation I considered had the rectangles “hanging” from the graph

of f by their upper right corners, but I suggest that students should rework the example

with the rectangles hanging by their upper left corners. In my notes, but not discussed in

class, was the example of a similar computation for the function f ( x) = x2 + 2 x − 5 over

the interval [1, 4]; for part of the interval f ( x) < 0, and the value of the integral includes

a cancellation of “negative” and “positive” contributions to area under the graph of f .

5. I reminded you of the formulæ for summing the 0th, 1st, 2nd, and 3rd powers of the first

n natural numbers, and indicated their use in the preceding computation.

6. I indicated that computation of the integral in this way will normally not be necessary,

as we will be meeting a theorem today which will enable much practical computation

for many functions. However, students are still expected to be able to carry out the

calculation of the value of a definite integral using Riemann sums.

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7. I stated basic properties of the definite integral, most of which can be derived from (17),

(26), (23): b

a

(r · f ( x) + s · g( x)) dx = r

b

a

f ( x) dx + s

b

a

g( x) dx c

a

f ( x) dx =

b

a

f ( x) dx +

c

b

f ( x) dx

C.4.2 §5.2 The Definite Integral (conclusion)

Properties of the Definite Integral – Comparison Properties. The textbook lists 3 proper-

ties, which are interrelated — each of them can be used to prove the other 2.

f ( x) ≥ 0 for a ≤ x ≤ b ⇒ b

a

f ( x) dx ≥ 0 (25)

f ( x) ≥ g( x) for a ≤ x ≤ b ⇒ b

a

f ( x) dx ≥ b

a

g( x) dx (26)

m ≤ f ( x) ≤ M for a ≤ x ≤ b ⇒ m(b − a) ≤ b

a

f ( x) dx ≤ M (b − a) . (27)

While the integral

b

a

f ( x) dx has, thus far, been defined only for a function f which is con-

tinuous on [a, b], we will eventually permit generalizations to that definition. Under those

generalizations, properties (25), (26), (27) will continue to hold wherever they “make sense” 33.

Some worked examples based on problems in an earlier edition of your textbook.

Example C.2 ([7, Exercise 18, p. 391]) “Express the limit as a definite integral on the given

interval: limn→∞

ni=1

e xi

1 + xi

∆ x, on the interval [1, 5].”

Solution: This problem is not stated in perfect mathematical language, but we know what the

textbook means. We are to consider the interval 1 ≤ x ≤ 5 to be subdivided into n subintervals

of equal length ∆ x, so ∆ x = 5−1n

. Then we are to interpret e xi

1+ xias f ( x∗

i), the value of a function

at a point x∗i chosen in the ith subinterval, so

1 + (i − 1)∆ x ≤ x∗i ≤ 1 + i∆ x .

33We will even have generalizations permitting infinite values for the integral, and the properties will hold

there, provided we don’t have to work with “values” like ∞ − ∞ or 0 × ∞.

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Solution: The integrand, sin x, is an increasing function when x is in the 1st quadrant; this

implies that, for

π

6 ≤ x ≤ π

2 ,

1

2 = sin

π

6 ≤ sin x ≤ sin

π

2 = 1 .

The length of the interval over which the integral is being evaluated is π

2 − π

6 =

π

3 . The value

of the integral is then bounded by the areas of two rectangles on the base

π

6, π

2

: the lower

bound is given by the rectangle whose height is the minimum value of the function, the value at

x = π6

; the upper bound is given by the rectangle whose height is the maximum value, attained

at x = π2

:

1

2 · π

3 ≤

π2

π6

sin x dx ≤ 1 · π

3 .

The exact value of the integral will eventually be seen to be√

32

.

Example C.5 ([7, Exercise 58, p. 392]) “Use (27) to estimate the value of the integral 2

0

( x3 − 3 x + 3) dx .”

Solution: On the real line the given function has critical numbers at ±1; of these only x = +1

is in the interval of the integral. By the 2nd Derivative Test x = 1 is a local minimum: thefunction value there is 1. At the end-points of the interval of integration, 0 and 2, the function

has values 3 and 5. We conclude that, on the given interval, the function values are bounded

as follows:

1 ≤ x3 − 3 x + 3 ≤ 5 .

The length of the interval is 2 − 0 = 2. The value of the integral is, therefore, bounded between

2 and 10. (Eventually we shall be able to evaluate this integral exactly, and shall be able to

show that its value is 24

4 − 3

2 · 22 + 3 · 2 = 4 − 6 + 6 = 4 .

5.2 Exercises

[1, Exercise 22, p. 771] “Use the form of the definition of the integral...to evaluate the integral 4

1 ( x2 + 2 x − 5) dx.”

Solution: Here we are finding the area under the graph of the polynomial x2 + 2 x − 5, a

continuous function, on the interval 1 ≤ x ≤ 4, i.e., the area of the region bounded by the

graph on top, the lines x = 1 and x = 4 on the two vertical sides, and the x-axis on the

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bottom. We divide the interval into n parts of length x = 4−1n

, and the end-points of the

ith subinterval are xi−1 = 1 +(i−1)∆ x = 1 +(i−1)·3

n on the left and xi = 1 +i · x = 1 +i·3

non the right.

If we choose the sample points to be the right end-points of the subintervals, i.e., to find

the limit of the right Riemann sum, we find that 4

1

( x2 + 2 x − 5) dx

= limn→∞

ni=1

( x2i + 2 xi − 5)∆ x

= limn→∞

ni=1

1 +

3i

n2

+ 2

1 +

3i

n − 5

3

n

= limn→∞

ni=1

1 + 2 · 3i

n+

9i2

n2

+ 2

1 +

3i

n

− 5

3

n

= limn→∞

ni=1

1 + 6

n

ni=1

i + 9

n2

ni=1

i2 + 2

ni=1

1 + 6

n

ni=1

i − 5

ni=1

1

3

n

= limn→∞

ni=1

1 + 6

n

ni=1

i + 9

n2

ni=1

i2 + 2

ni=1

1 + 6

n

ni=1

i − 5

ni=1

1

3

n

= limn→∞(1 + 2 − 5)

ni=1

1 +

6 + 6

n

ni=1

i +

9

n2

ni=1

i2 3

n

= limn→∞

−2n +

12

n· n(n + 1)

2 +

9

n2 · n(n + 1)(2n + 1)

6

3

n

= limn→∞

−2n + 6(n + 1) +

3n +

9

2 +

3

2n

3

n

= 21

If we choose the sample points to be the left end-points, i.e., to find the limit of the left

Riemann sum, we obtain a very similar sum. We can write the sum, like the preceding,as a sum over the index i ranging from i = 1 to i = n, but where the function is evaluated

at the left end-points: 4

1

( x2 + 2 x − 5) dx

= limn→∞

ni=1

( x2i−1 + 2 xi−1 − 5)∆ x

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= limn→∞

n

i=1

1 + 3(i − 1)

n

2

+ 2 1 + 3(i − 1)

n −

53

n

= limn→∞

ni=1

1 + 2 · 3(i − 1)

n +

9(i − 1)2

n2

+ 2

1 +

3(i − 1)

n

− 5

3

n

= limn→∞

ni=1

1 + 6

n

ni=1

(i − 1) + 9

n2

ni=1

(i − 1)2 + 2

ni=1

1 + 6

n

ni=1

(i − 1) − 5

ni=1

1

3

n

At this point we can proceed in a variety of ways. One is to observe that the first term

in the sums of powers is 0, so that we are summing only n − 1 non-zero powers. A more

formal way to to that is to define

j = i − 1 (28)

and to rewrite each of the sums as a sum over j, replacing the value i = 1 by j = 0 andthe value i = n by j = n − 1. We then obtain 4

1

( x2 + 2 x − 5) dx

= limn→∞

n−1 j=0

1 + 6

n

n−1 j=0

j + 9

n2

n−1 j=0

j2 + 2

n−1 j=0

1 + 6

n

n−1 j=0

j − 5

n−1 j=0

1

3

n

= limn→∞

(1 + 2 − 5)

n−1

j=0

1 + 6 + 6

n

n−1

j=0

j + 9

n2

n−1

j=0

j2

3

n

= limn→∞

−2n +

12

n· (n − 1)(n)

2 +

9

n2 · (n − 1)n(2n − 1)

6

3

n

= limn→∞

−2n + 6(n − 1) +

3n − 9

2 +

3

2n

3

n

= 21

again. The definition of j in (28) is analogous to the changes of variables that we will be

making in definite integrals in [1, §5.5]: there, as here, we have to change the limits of

the integral to correspond to the new values of the variable of integration. What we have

here is an example of the “finite diff erence calculus”, where there are results similar to

those that we will be developing in the “infinitesimal calculus”.

[1, Exercise 40, p. 378] “Evaluate the integral by interpreting it in terms of areas:10 0

| x−5| dx.”

Solution: (While we could evaluate this integral using Riemann sums, the intention is

that the student interpret the area under the curve in terms of familiar geometric objects,

and use known formulæ to determine the value.)

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The portion of the integral from 0 to 5 is the area of a right-angled triangle whose hy-

potenuse is on the line y = −( x − 5), with base of length 5, and height 5, so its area is52

2 =

25

2 . The portion of the integral from 5 to 10 is the mirror image of the triangle

described above, this time with hypotenuse along the line y = x − 5; its area is the same

as the previous one, so the value of the integral is 25.

C.4.3 §5.3 The Fundamental Theorem of Calculus

In a number of areas of mathematics there are theorems that have acquired the name “The

Fundamental Theorem of...”. The present section is devoted to such a theorem, also known as

“The Fundamental Theorem of (the) Integral Calculus”, one part of which relates the value of

a definite integral to antiderivatives of its integrand, and provides a method for evaluating suchintegrals without the need for computing limits of complicated sums. (The formulation of the

theorem as being divided into two parts is not completely standard.)

Diff erentiation and Integration as Inverse Processes.

Theorem C.6 (The Fundamental Theorem of Calculus) If f is continuous on [a, b], then

1. d

dx

x

a

f (t ) dt

= f ( x).

2. F = f ⇒ b

a f ( x) dx = F (b) − F (a) .

Definition C.2 We may represent a diff erence F (b) − F (a) by

[F ( x)]ba

or even more briefly by

F ( x)]ba

if the latter expression is unambiguous.

I sometimes use a notation which is not standard, but is unambiguous, and write the pre-

ceding diff erence as

F ( x)] x=b x=a

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Example C.7 Following my discussion of [1, §5.2] in these notes there is35 a solution of

[1, Exercise 22, p. 377], in which the integral

4 1

x2 + 2 x − 5

dx is evaluated “from first

principles”, proving that its value is 21. Let’s now verify that result using the Fundamental

Theorem. One antiderivative of x2 + 2 x − 5 is x3

3 + x2 − 5 x. The value of the integral is,

therefore, x3

3 + x2 − 5 x

4

1

=

64

3 + 16 − 20

−

1

3 + 1 − 5

= 21 .

Example C.8 ([7, Exercise 12, p. 402]) “Use Part 1 of the Fundamental Theorem of Calculus

to find the derivative of the function F ( x) = 10

xtan θ d θ .”

Solution: First observe that F ( x) = − x

10 tan θ d θ . Having written the integral in the form to

which the Fundamental Theorem applies, i.e., with the variable in the upper limit, we may

apply that theorem: the derivative is minus the value of the integrand, tan θ , evaluated where

θ = x, i.e., − tan x.

(Eventually we will see that F ( x) = ln cos x − ln cos(10); again students may then verify

that the Fundamental Theorem is giving us the correct derivative.)

35on pages 3016–3018

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Release Date: Monday, January 11th, 2010

updated on 12 January; subject to further updates and corrections

C.5.1 §5.3 The Fundamental Theorem of Calculus (conclusion)


to find the derivative of the function g(u) =

u

3

1

x + x2 d x.”

Solution: The derivative is the value of the integrand, 1

x + x2 evaluated at the upper limit of the

integral, i.e., where x = u: g(u) = 1

u + u2.

(Eventually we will see that g(u) = ln u + 1

u− ln

4

3; students may then verify that the

Fundamental Theorem is giving us the correct derivative.)

5.3 Exercises

[1, Exercise 54, p. 389] “Find the derivative of the function y =

5 x

cos x

cos(u2) du.”

Solution: Two observations are necessary:

• The Fundamental Theorem is concerned with an integral whose upper limit is vari-

able; if we wish to apply that theorem here, we shall need to transform the problem

to one where only the upper limit of the integral(s) is variable.

• The Fundamental Theorem is concerned with an integral whose upper limit is the

independent variable under consideration; should we wish to permit the upper limit

to vary in a more complicated way, we will need to apply the Chain Rule.

1. We shall transform the given integral into a sum of two where the lower limit of

each is constant. We do this by splitting the interval of integration, [cos x, 5 x] into

two parts at a “convenient” point. It is not even necessary that the point we choose

be inside the interval, since the property we are applying, (18) on page 3011 of

these notes, does not require that fact. I will choose the constant 0 to be the pointwhere the splitting occurs: 5 x

cos x

cos(u2) du =

0

cos x

cos(u2) du +

5 x

0

cos(u2) du

= − cos x

0

cos(u2) du +

5 x

0

cos(u2) du

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Hence

d dx

5 x

cos x

cos(u2) du = − d dx

cos x

0

cos(u2) du + d dx

5 x

0

cos(u2) du (29)

2. Each of these derivatives will be computed using the Chain Rule, with the interme-

diate variable being the function appearing as the variable upper limit. In the first

case, if we take the intermediate variable to be, say z = cos x, we have

d

dx

cos x

0

cos(u2) du = d

dz

z

0

cos(u2) du · dz

dx

= cos( z2) · dz

dx

= cos(cos2 x)·

(−

sin x)

For the second integral we take the intermediate variable to be w = 5 x. Here

d

dx

5 x

0

cos(u2) du = − d

dw

w

0

cos(u2) du · dw

dx

= cos(w2) · dw

dx

= cos(25 x2) · 5

Combining the two results gives

d

dx 5 x

cos x cos(u

2

) du =

− d

dx cos x

0 cos(u

2

) du+

d

dx 5 x

0 cos(u

2

) du

= − cos(cos2 x) · (− sin x) + cos(25 x2) · 5

= cos(cos2 x) · sin x + 5 cos(25 x2)

[1, Exercise 58, p. 389] “Find the interval on which the curve y =

x

0

1

1 + t + t 2 dt is concave

upward.”

Solution: By the Fundamental Theorem, y( x) = 1

1 + x + x2; hence y( x) =

−1 − 2 x

(1 + x + x2)2.

The denominator of the second derivative is a non-zero square, so it is always positive;

the function will be positive whenever −1 − 2 x > 0, i.e., whenever x < −1

2 : this is wherethe graph is concave upward.

(Eventually we will be able to evaluate the integral explicitly, showing that the curve is

y = 2√

3

arctan

2 x + 1√ 3

− π

6

.

Students may diff erentiate to check that I have found the correct second derivative.)

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[1, Exercise 74, p. 390] Suppose that

b

0

e

x

dx = 3

a

0

e

x

dx . Express b in terms of a.

Solution: Substituting the values of the given integrals, from Part 2 of the Fundamental

Theorem, yields

e x]b0 = 3

e x]a

3

,

which implies that

eb − e0 = 3ea − e0

⇒ eb = 3ea − 2

⇒ b = ln (3ea

−2) .

Problems Plus

[1, Exercise 12, p. 413] Find

d 2

dx2

x

0

sin t

1

√ 1 + u4 du

dt .

Solution:

d 2

dx2 x

0 sin t

1

√ 1 + u4 du dt =

d

dx d

dx x

0 sin t

1

√ 1 + u4 du dt

= d

dx

sin x

1

√ 1 + u4 du

=

1 + sin4 x · cos x .

C.5.2 §5.4 Indefinite Integrals and the “Net Change” Theorem

Indefinite Integrals The traditional symbol for a “general” antiderivative F ( x) of a function

f ( x) (i.e., some function with the property that F ( x) = f ( x)) is

f ( x) dx, which is called the

indefinite integral of f ( x). Since two antiderivatives diff er by a constant (by a corollary to theMean Value Theorem), we usually write statements in the form

f ( x) dx = F ( x) + C

where F ( x) is one specific antiderivative, and C is a constant of integration, intended to range

over all real numbers. Once a particular antiderivative F has been chosen, the particular real

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number C that applies in a particular situation has to be determined from additional information

that is usually available in the problem at hand. Much of this course will be concerned withmethods for finding indefinite integrals. While the finding of the indefinite integral may be a

difficult problem, the verification that a function F that is claimed to be an antiderivative of f

is not, since all that needs to be done is to di ff erentiate F and to check whether the derivative

is f . (In principle, if a function F has the property that F = f , then F is an antiderivative of

f : thus it is possible to find an antiderivative by guessing, or by simply copying the answer

from the back of the textbook or from your neighbour’s work. The intention in the course is

that you should normally be expected to be able to show a systematic way of determining an

antiderivative; there will be a very few special situations where you will be presented with an

antiderivative without a convincing way of finding it.)

Note the same symbol, a stylized letter S — called the integral sign — is used for

both the indefinite integral and the definite integral; while they are related by the Fundamental

Theorem, they are diff erent operations.

Even though the two operations are diff erent, they share some similar properties. For

example, parallel to property (17) of the definite integral on page 3011 of these notes, we can

also prove that (r f ( x) + sg( x)) dx = r

f ( x) dx + s

g( x) dx , (30)

where r , s are any constants. Equations like the preceding, in which an indefinite integral ap-

pears on both sides of the equal sign, are normally written without any constant of integration.

The “Net Change” Theorem This is simply the author’s name for the second part of the

Fundamental Theorem. It is not a term in standard usage, and I am not likely to use it. As

a beginning in the development of general techniques for determining indefinite integrals, we

can reformulate results that we developed for derivatives. For example, since we know that

d

dxsin x = cos x ,

we can reformulate this result as

cos x dx = sin x + C .

Some reformulations require minor changes, e.g., division by an appropriate constant. From

the result thatd

dx xn = n · xn−1 when n 0

we can conclude that xn−1 dx =

1

n xn + C when n 0

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Function One antiderivative

f ( x) F ( x)g( x) G( x)

f ( x) + g( x) F ( x) + G( x)

xn (n −1) xn+1

n + 11

xln | x|

e x e x

cos x sin x

sin x − cos x

sec2 x tan x

sec x tan x sec xcsc x cot x − csc x

1

1 + x2 arctan x

1

1 + x2 −arccot x

1√ 1 − x2

arcsin x

1√ 1 − x2

− arccos x

Table 4: Some Antiderivatives

or, after the substitution of m + 1 for n, xm dx =

1

m + 1 xm+1 + C when m −1 .

Thus Table C.5.2 of antiderivatives on page 3025 of these notes, which I included last semester

in my notes in MATH 140 2009 09, can now be recast in the form of Table C.5.2 on page 3026

below.

Applications While we have been using the 2nd part of the Fundamental Theorem to express

the value of a definite integral in terms of the “net change” in the antiderivative, we can also

apply the result in the opposite way: that the net change can be found by evaluating the integral.

This is the spirit of the motivation that the author called “The Distance Problem” (see above,

page 3005, or the solution below, on page 3029 of [1, Exercise 44, p. 397]]). We will see

another example below in the solution of [1, Exercise 62, p. 398].

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[r f ( x) + sg( x)] dx = r f ( x) dx + s g( x) dx k dx = kx + C xn dx =

xn+1

n + 1 + C (n −1)

1

xdx = ln | x| + C

e x dx = e x + C

a x dx =

a x

ln a+ C (a > 0)

sin x dx = − cos x + C

cos x dx = sin x + C

sec2 x dx = tan x + C

csc2 x dx = − cot x + C

sec x tan x dx = sec x + C csc x cot x =

−csc x + C

1

1 + x2 = arctan x + C

1

1 + x2 = −arccot x + C

1√ 1 − x2

= arcsin x + C

1√

1 − x2dx = − arccos x + C

Table 5: Very Short Table of Indefinite Integrals

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Example C.10 ([7, Exercise 10, p. 411]) “Find the general indefinite integral of x2 + 1 +

1

x2 + 1

dx .”

Solution: Break the integrand into two parts: the summand at the end is recognizable as the

derivative of arctan x; the two terms at the beginning are multiples of powers of x, and we have

observed earlier how to integrate them. Thus x2 + 1 +

1

x2 + 1

dx =

x2 + 1

dx +

1

x2 + 1 d x

= 1

3

x3 + 1

1

x + arctan x + C

where the letter C represents the constant of integration. Even though there are two indefinite

integrals on the right side of the equation, only one constant is needed: if we were to include

two constants, as +C 1 + C 2, we would not gain any more freedom.

Example C.11 ([7, Exercise 11, p. 411]) Find the general indefinite integral of 2 − √

x2

dx .

Solution: One might be tempted, at first, to consider the possibility that the antiderivative of

the given 2nd power is a multiple of the 3rd power of 2 − √ x; unfortunately that temptationwill have to be resisted, as the resulting functions will not have the correct derivative. The

simplest approach is to expand the square: since (2 − √ x)2 = 4 − 4 x

12 + x,

2 − √ x2

dx =

4 − 4

√ x + x

dx

= 4 x − 4 · 2

3 x

32 +

1

2 x2 + C

8/15/2019 Brown's Notes 2010




Release Date: Wednesday, January 13th, 2010, subject to correction

C.6.1 §5.4 Indefinite Integrals and the “Net Change” Theorem (conclusion)

5.4 Exercises

[1, Exercise 34, p. 397] “Evaluate the integral 9

1

3 x − 2√ x

dx.”

Solution: Simplify the integrand by dividing the denominator into the two summands of

the numerator: 9

1

3 x − 2√ x

dx =

9

1

3

√ x − 2√

x

dx

=

3 · 2

3 · x 3

2 − 2 · 2 · x 12

9

1

=2 x

32 − 4

√ x9

1

= (2 · 27 − 4 · 3) − (2 · 1 − 4 · 1) = (54 − 12) − (2 − 4) = 44 .

[1, Exercise 38, p. 397] “Evaluate the integral π3

0

sin θ + sin θ · tan2 θ

sec2 θ d θ .”

Solution: When the integrand involves trigonometric functions, one may have to apply a

familiar identity to simplify the integration. There is often more than one way to do this.

In the present example


sec2 θ = sin θ · cos2 θ + sin3 θ

= sin θ · (cos2 θ + sin2 θ ) = sin θ

so π3

0


sec2 θ d θ =

π3

0

sin θ d θ

= [− cos θ ]π3

0 = −cos

π

3 + cos0 = −1

2 + 1 =

1

2.

Eventually you will know how to evaluate the parts of this integral separately, but the

present solution is faster than evaluating and adding the parts.

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[1, Exercise 44, p. 397] Evaluate the integral

3π2

0

| sin x| dx.

Solution: The integrand is continuous, so we know the integral exists. However, it is

not convenient to work with an antiderivative of | sin x |. So we split the interval of

integration into the parts [0, π] andπ, 3π

2

where the integrand is respectively positive

and negative, and thereby avoid working with the absolute value. For the integrands,

respectively sin x and − sin x, we know antiderivatives − cos x and + cos x, so we may

apply the Fundamental Theorem to each part separately:

3π

2

0 |sin x

|dx =

π

0

sin x dx + 3π

2

π

(

−sin x) dx

= [− cos x]π0 + [cos x]

3π2

π

= (−(−1) + 1) + (0 − (−1)) = 3.

[1, Exercise 58, p. 398] The velocity function is v(t ) = t 2 − 2t − 8 for a particle moving along

a line. Find

(a) the displacement; and

(b) the distance travelled by the particle

during the time interval 1 ≤

t ≤

6.

Solution: Denote the position of the particle at time t by x(t ); then v(t ) = d dt

x(t ), so x is

an antiderivative of v.

(a) The displacement is simply the diff erence between initial and final positions of the

moving particle; it is equal to the area under the graph of the velocity function

between the appropriate times.

displacement = x(6) − x(1) = [ x(t )]61 =

6

1

v(t ) dt

= 6

1 (t

2

− 2t − 8) dt

=

1

3t 3 − t 2 − 8t

6

1

= (72 − 36 − 48) −

1

3 − 1 − 8

= −10

3

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(b) The distance travelled is equal to the area under the graph of the speed function

between the appropriate times; the speed is the magnitude of the velocity.

distance travelled =

6

1

t 2 − 2t − 8 dt =

6

1

|(t + 2)(t − 4)| dt .

The function (t + 2)(t − 4) changes sign at t = −2, which is outside the interval of

integration, and again at t = 4, which is inside the interval of integration. We can

break the integral up into two parts at x = 4, and then we can express each of the

parts without using absolute value symbols: 6

1

|(t + 2)(t − 4)| dt =

4

1

|(t + 2)(t − 4)| dt +

6

4

|(t + 2)(t − 4)| dt

= − 4

1

(t + 2)(t − 4) dt + 6

4

(t + 2)(t − 4) dt

= − 4

1

(t 2 − 2t − 8) dt +

6

4

(t 2 − 2t − 8) dt

= −

1

3t 3 − t 2 − 8t

4

1

+

1

3t 3 − t 2 − 8t

6

4

= −[−18] +

44

3

=

98

3

Note that the factorization of the quadratic was needed in order to determine where the

split the interval of integration, but it did not help in the actually integration operation,and had to be reversed at that stage.

[1, Exercise 62, p. 398] “Water flows from the bottom of a storage tank at a rate of r (t ) =

200 − 4t litres / min, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank

during the first 10 minutes.”

Solution: Let’s denote by W (t ) the amount of water (measured in litres) that has left the

tank by time t . We are told that

d

dt W (t ) = r (t ) = 200 − 4t (0 ≤ t ≤ 50).

Then

W (10) − W (0) =

10

0

dW

dt dt

=

10

0

(200 − 4t ) dt =200t − 2t 2

10

0

= (200 · 10 − 2 · 102) − 0 = 1800 ,

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so the total amount of water leaving the tank in the first 10 minutes is 1800 litres.


to evaluate the integral, or explain why it does not exist:

3 −2

x−5dx.”

Solution: The integrand is not defined at the point 0 in the interval [−2, 3]. This means that

we cannot even speak of the integral at this time. (Later we will generalize our definition of

integral to permit us to consider certain “improper” integrals where there is an infinite discon-

tinuity. But even that generalization will not apply to this problem, although it is premature

to consider it here. Look at this problem again when we study [1, §7.8].) The theorem is not

applicable because the integrand is not defined at one point in the domain, and the disconti-

nuity is neither removable nor a jump discontinuity: a removable discontinuity would have noeff ect at all, and a jump discontinuity could be accommodated by the method of Example C.14

above, i.e., by splitting the integral into two parts at the jump discontinuity.

Example C.13 ([7, Exercise 42, p. 403]) Let f ( x) =

x if −π ≤ x ≤ 0

sin x if 0 < x ≤ π . Use Part 2 of

the Fundamental Theorem of Calculus to evaluate the integral

π −π

f ( x) dx, or explain why it

does not exist.

Solution: The function f is defined piecewise, by gluing together one function defined on

[−π, 0] and another defined on (0, +π]; however, it is continuous, as lim x→0− x = 0 = lim x→0+ sin x,and f is defined at x = 0 and equal to the common value of the two one-sided limits. We know

that one antiderivative of x is 1

2 x2; and that one antiderivative of sin x is − cos x. Consider the

function F ( x) =

x2

2 if −π ≤ x ≤ 0

1 − cos x if 0 < x ≤ π . This function is evidently diff erentiable at all

points except possibly 0; and F is an antiderivative of f on the interval −π < x < +π, with the

possible exception of the point x = 0. At x = 0 the two one-sided limits of diff erence quotients

are

lim x→0−

F ( x)

−F (0)

x=

lim x→0−

x2

3 − 0

x

= lim x→0+

x

3 =

1

3 · 0 = 0

lim x→0+

F ( x) − F (0)

x = lim

x→0+

(1 − cos x) − 0

x

= lim x→0+

1 − (1 − 2 sin2 x2

) − 0

x

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= lim x→

0+ sin x

2 x

2 ·sin

x

2

= lim x→0+

sin x2

x2

· lim x→0+

sin x

2 = 1 · 0 = 0 ,

so the function is also diff erentiable at x = 0, where the derivative is equal to 0, i.e., to f (0);

thus F is an antiderivative of f on the interval (−π, +π). Thus we can apply the Fundamental

Theorem:π

−π

f ( x) dx = F (π) − F (−π) = (1 − cos π) −

(−π)2

2

= 2 − π2

2 .

But finding the antiderivative was a complicated computation, and rendered the problem more

difficult than necessary. Instead, one should proceed as follows, applying (23) in these notes,page 3011:

π −π

f ( x) dx =

0 −π

f ( x) dx +

π 0

f ( x) dx

=

0 −π

x dx +

π 0

sin x dx

= x2

20

−π+ [− cos x]

π

0

=

0 − (−π)2

2

+ (− cos π + cos 0)

= −π2

2 + (−(−1) + 1) = 2 − π2

2 .

The lesson to be learned from this example is that there are often advantages to splitting up an

integral, even if it is theoretically possible to evaluate it without doing so.

Integration of “piecewise continuous” functions. The definition [1, Definition 2, p. 366]

of a definite integral given by the textbook, is more restrictive than necessary. This definitionrequires that the integrand be continuous throughout the interval of integration. In fact, the

definition can be weakened to apply to a broader class of functions. While we don’t require

full generality in this course, we do wish to be able to apply the theory to functions that have

isolated “jump” discontinuities; (we can handle removable discontinuities just be “removing”

them, i.e., by extending the function to the appropriate value at the points missing in the origi-

nal definition). If a function f is continuous on an interval [a, c], except for a point b in (a, c)

8/15/2019 Brown's Notes 2010



where lim x→b−

f ( x) and lim x→b+

f ( x) both exist, but are not equal, we will define

c

a

f ( x) dx =

b

a

f ( x) dx +

c

b

f ( x) dx (31)

that is, we will define the integral to be such as to satisfy the additivity property (23) we saw

earlier in these notes, on page 3011. Note, however, that we cannot apply to the whole interval

the Fundamental Theorem to evaluate integralb

a

f ( x) dx if f has a discontinuity at a point c

such that a < c < b.

Example C.14 Suppose that

f ( x) = −1 if x ≤ 0

1 if x > 0

Evaluate

2

−1

f ( x) dx.

Solution: We cannot apply the Fundamental Theorem to the entire interval [−1, 2], as the

integrand is discontinuous at point 0. So we split the integral at the point x = 0, which

is the point of discontinuity. The two integrals we obtain now satisfy the conditions of the

Fundamental Theorem:

2

−1

f ( x) dx = 0

−1

f ( x) dx + 2

0

f ( x) dx

=

0

−1

(−1) dx +

2

0

1 dx

= [− x]0−1 + [ x]2

0 = (−0 + (−1)) + (2 − 0) = 1.

C.6.2 §5.5 The Substitution Rule

The “Substitution Rule” is a reformulation, in terms of the integral, of the Chain Rule.

Theorem C.15 Let u = g( x) and f ( x) be respectively di ff erentiable and continuous on a given

interval; then ( f g)( x) · g( x) dx =

f (g( x)) · g( x) dx =

f (u) du .

In applying this theorem we usually begin with a “complicated” integral, whose form we try

to interpret like the left side of the above equation, and try to find an appropriate “substitution”

of the form u = g( x) which will transform the integral into one whose integrand is one that we

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are able to integrate. In practice one works with the diff erentials d x and du in a “mechanical”

way that can be justified by the theorem.Proof: Let F and g be diff erentiable. Then

F (g( x)) · g( x) = d

dxF (g( x)) .

Integrating with respect to x, we have F (g( x)) · g( x) dx =

d

dxF (g( x)) dx = F (g( x)) + C .

If we define u = g( x), f = F , then

f (g( x)) · g( x) dx =

d

dx F (g( x)) dx = F (u) + C =

f (u) du .

In practice this “substitution” is often applied by “mechanically”, substituting functions

and diff erentials, and such operations can be shown to be fully justifiable. The general idea in

looking for substitutions is to try to reduce the complication of the original indefinite integral.

This is a subjective term, and diff erent users may find a variety of distinct substitutions which

they will find helpful in evaluating an indefinite integral.

Finding the appropriate substitution is one step in solving a problem. In the first prob-

lems in the list of exercises the author suggests a substitution which will be helpful; eventually

students are expected to find an appropriate substitution on their own — often there are severalpossible choices. You should be experimenting with diff erent substitutions and getting to know

the types of problems each of them is useful in solving.

Example C.16 Earlier, in Example C.11 on page 3027 of these notes, I considered [7, Exercise

11, p. 411], which was concerned with the indefinite integral of

2 − √

x2

dx. We evaluated

this integral by expanding the square and then integrating the powers of x separately. Could

we use the same methods for the indefinite integral

2 − √

x10000

dx?

Solution: While we could expand the integrand in this case too, the result would have 10,001

terms, each of which would have to be integrated. The result we would obtain would not

be very useful. Consider the following alternative approach, that could also have been used

when the exponent was 2. We have here an integral of the form

( f g)( x) · g( x) dx, where

f ( x) = x10000, and g( x) = 2 − √ x. Define u = g( x) = 2 − √

x, so that du = − 1

2√

xdx, so

dx = −2√

x du = 2(u − 2)du .

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2 −

√ x

10000dx = 2 u10000(u

−2) du

= 2

u10001 − 2u10000

du

= 2

10002u10002 − 4

10001u10001 + C

= 2

10002(2 − √

x)10002 − 4

10001(2 − √

x)10001 + C

= (2 − √ x)10001

− 2

√ x

10002 − 4

(10001)(10002)

+ C

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Release Date: Friday, January 15th, 2010 (subject to revision)

C.7.1 §5.5 The Substitution Rule (conclusion)

Review The Substitution Rule (cf., p. 3033) states that

Let u = g( x) and f ( x) be respectively diff erentiable and continuous on a given

interval; then ( f g)( x) · g( x) dx =

f (g( x)) · g( x) dx =

f (u) du

In Example C.16, discussed at the end of the last lecture, I showed how we could evaluate 2 − √

x10000

dx using the substitution u = g( x) = 2 − √ x, obtaining

dx = −2√

x du = 2(u − 2)du

⇒

2 − √ x10000

dx = 2

u10000(u − 2) du

= 2

u10001 − 2u10000

du

=

2

10002 u

10002

− 4

10001 u

10001

+ C

= 2

10002(2 − √

x)10002 − 4

10001(2 − √

x)10001 + C

= (2 − √ x)10001

− 2

√ x

10002 − 4

(10001)(10002)

+ C

Example C.17 [1, Example 6, p. 403] “Calculate

tan x dx.”

Solution: This problem does not, at first, appear to be a candidate for a substitution; but, by

expressing the tangent as sin x

cos x, the textbook shows that it can be simplified by treating cos x

as the new variable. That is, by defining u = cos x, which implies that du =

−sin x dx, we can

evaluate the integral as follows: tan x dx =

sin x dx

cos x

= −

d (cos x)

cos x

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which has the form

− du

u, which we should recognize as

− d (ln

|u

|). Thus one sub-

stitution that is indicated is u = cos x: the integral becomes −

d (ln |u|) = − ln |u| + C =

− ln | cos x| + C . The integral could also be expressed as ln | sec x| + C or as ln |k sec x|, where

C or k are constants of integration. Three other indefinite integrals can be evaluated in similar

ways; perhaps you should remember36 them together:37 tan x dx = ln | sec x | + C = − ln | cos x | + C (32) cot x dx = − ln | csc x | + C = ln | sin x | + C (33)

tanh x dx = ln cosh x + C (34) coth x dx = ln | sinh x | + C (35)

Example C.18 ([7, Exercise 2, p. 420]) “Evaluate the integral

x(4 + x2)10 dx by making the

substitution u = 4 + x2.”

Solution: First we should observe that we could solve this problem without any substitution

at all. We could expand the 10th power of the binomial into a polynomial of degree 20,

multiply each of the terms by an additional x, and then integrate the resulting sum of 11 terms

one by one. The resulting polynomial would be the correct solution. But what would you

do it the exponent were not 10, but 10000? Applying the method of substitution, we set

u = g( x) = 4 + x2, f (u) = u10, so du

dx= g( x) = 2 x. Then

x4 + x2

10dx =

4 + x2

10 · 1

2

d

dx

4 + x2

dx

= 1

2

d

dx

4 + x2

11

dx

= 1

22

4 + x2

11+ C .

In practice we often operate mechanically with diff erentials; from u = 4 + x2 we have du =

2 x dx, so x

4 + x2

10dx =

xu10dx =

u10 · x dx

=

u10 · 1

2 du =

1

22u11 + C .

36This doesn’t mean to memorize them — just to remember the “trick” needed to evaluate them.37Why are absolute signs missing in one of these cases?

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But it would be bad form to stop here, since our answer has been expressed in terms of u, not

the original variable x; so we continue

= 1

22

4 + x2

11+ C .

Example C.19 ([7, Exercise 30, p. 421]) Evaluate the indefinite integral ax + b√

ax2 + 2bx + cdx .

Solution: Later in the term we will consider the integration of any function of the form

kx +

√ ax2 + 2bx + c ,

but the present problem concerns a special case which is easy to handle. (Note that the textbook

should have insisted that not all of a, b, c can be 0, as then the ratio is not defined.)

Again, we begin by attempting to simplify the integrand. It would appear that the quadratic

function whose square root appears in the denominator is the feature we should attempt to

simplify. One way to do this is to define u = ax2 + 2bx + c. This implies that

du = 2(ax + b) dx,

so we obtain ax + b√

ax2 + 2bx + cdx =

1

2

du√

u

= u12 + C

= (ax2 + 2bx + c)12 + C .

The problem could also be solved by the substitution v = (ax2 + 2bx + c)12 .

Example C.20 ([7, Exercise 38, p. 421], slightly changed) Evaluate the indefinite integral

7√

x3 + 1

· x5 dx .

Solution: The method I propose to use here is based on the fact that

d ( xn) = n xn−1dx

for any integer n. When an integrand is expressible in terms of a power of the variable, short by

just 1, then this type of substitution is often a good way to begin simplifying it, although further

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steps might be needed. So, in this case, I begin with u = x3, which implies that du = 3 x2 dx,

i.e., that x2

dx =

1

3 du. I then obtain 7√

x3 + 1 · x5 dx = 1

3

7√

u + 1 · u du

after which I consider further simplification. At this point I would like to simplify the 7th root.

I can do this by either setting v = u + 1, or w = 7√

u + 1. In the first case I would obtain dv = du,

while, in the second, u + 1 = w7, so du = 7w6 dw.

1

3

7√

u + 1 · u du = 1

3

7√

v · (v − 1) dv

= 1

3 v87 − v

17 dv

= 1

3

7

15v

157 − 7

8v

87

+ C

= 1

3

7

15(u + 1)

157 − 7

8(u + 1)

87

+ C

= 1

3

7

15( x3 + 1)

157 − 7

8( x3 + 1)

87

+ C

or 1

3

7

(u + 1) · u du = 1

3

w ·

w7 − 1

· 7w6 dw

= 7

3 w14

−w7 dw

= 7

3

1

15w15 − 1

8w8

+ C

= 7

45(u + 1)

157 − 7

24(u + 1)

87 (u + 1)

87 + C

= 7

45( x3 + 1)

157 − 7

24( x3 + 1)

87 ( x3 + 1)

87 + C .

Definite Integrals Thus far I have been applying the Substitution Rule to indefinite integrals.

Substitution may be applied to a definite integral

b

a

f (g( x)) g( x) dx

in two ways:

• Apply the Substitution Rule to the corresponding indefinite integral, f (g( x))g( x) dx ;

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then apply the second part of the Fundamental Theorem to the resulting antiderivative;

or

• A variant of the Substitution Rule can be formulated specifically for definite integrals.

Using the same functions as described earlier, it is b

a

f (g( x))g( x) dx =

g(b)

g(a)

f (u) du .

That is, change the limits of the new integral to the values that u has when x = a and

x = b.

These two methods are equivalent — use whichever you prefer.

Symmetry The textbook reviews the definitions of even and odd functions, and considers

their definite integrals over a finite interval centred at the origin. The author shows that the

definite integral of an even function f over the interval −a ≤ x ≤ +a will be twice the integral

over the interval 0 ≤ x ≤ a; and the integral of an odd function over the same interval will be

0, because the integral to the left of 0 will cancel the integral to the right of 0. The proofs are

simple applications of the Substitution Rule.

Example C.21 Evaluate the definite integral

π6

− π6

tan2 θ d θ .

Solution: Normally we will evaluate an integral of this type by replacing the integrand, tan2

θ ,by sec2 θ − 1. In this case we can also use the symmetry of the integrand and the interval of

integration to further simplify the calculations:

π6

− π6

tan2 θ d θ =

π6

− π6

sec2 θ − 1

d θ

= 2

π6

0

sec2 θ − 1

d θ

since the integrand is an even function (Prove this!)

= 2 [tan θ

−θ ]

π6

0

= 2

1√ 3

− π

6

.

5.5 Exercises

[1, Exercise 28, p. 407] “Evaluate the indefinite integral

tan−1 x

1 + x2 dx.”

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Solution: In looking for a substitution, our general intention is to try to simplify the

integrand. In the present integrand, we might be expected to consider the arctangentfactor the most complicated, so I will try to simplify by the substitution u = tan−1 x.

This implies that du = 1

1 + x2dx, which is also present in the integral. The integral

becomes tan−1 x

1 + x2 dx =

u du =

1

2u2 + C =

(tan−1 x)2

2 + C .

We can verify our work by diff erentiating the function we claim to be an antiderivative:

d

dx

1

2 tan−1 x

2

= 1

2 · 2 tan−1 x · 1

1 + x2 =

tan−1 x

1 + x2 .

[1, Exercise 24, p. 407] Evaluate the indefinite integral

(1 + tan θ )5 sec2 θ d θ .

Solution: In this case one should observe that there is a factor sec 2 θ , which we recognize

as the derivative of tan θ . A first simplification would be obtained by the substitution

u = tan θ

since then du = sec2 θ d θ , or d θ = cos2 θ du. The integral transforms to

(1 + tan θ )5 sec2 θ d θ = (1 + u)5 sec2 θ cos2 θ du

=

(1 + u)5 du .

One can observe that the integral here will be a constant multiple of (1 + u)6, and then

replace u by tan θ . If you don’t make the observation, a second substitution would be in

order. I would observe that, in the last mentioned integral, the “most complicated” factor

is 1 + u; so a substitution v = 1 + u could be attempted. This yields

dv = 0 + du

⇒ (1 + u)5 du = v5 dv

= 1

6 · v6 + C

= 1

6 · (1 + u)6 + C

= 1

6 · (1 + tan θ )6 + C .

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[1, Exercise 36, p. 407] “Evaluate the indefinite integral sin x

1 + cos2 xdx.”

Solution: First we see the cosine in the denominator; then we see in the numerator

sin x dx, which is −d (cos x). This suggests a substitution u = cos x. The integral be-

comes

−du

1 + u2 du = −arctan u + C = −arctan(cos x) + C .

[1, Exercise 42, p. 407] “Evaluate the indefinite integral

x

1 + x4 d x.”

Solution: (I am going to begin the solution by making a poor choice for a substitution,

a choice that I will then “fix” by following it with a second substitution. Then I will

observe what would have been a better first choice.) Examining the integral we see that

the powers of x present are x

1

in the numerator, and x

4

in the denominator. A first ideamight be to try u = x4. That would yield x

1 + x4 dx =

1

4

1√

u(1 + u)du

which looks more complicated than before. However, we could then undertake a second

substitution, v =√

u, which would correspond to an original substitution of v = x2; this

would produce x

1 + x4 dx =

1

2

1

1 + v2 dv

= 12

arctan v + C = 12

arctan x2

+ C

An experienced student would have been able to see the substitution v = x2 imme-

diately: a substitution u = x2 is indicated when the entire integrand can be expressed in

simple terms of x2, with the exception of one extra single power of x that is “left over”.

[1, Exercise 57, p. 407] Evaluate the definite integral

π6

− π6

tan3 θ d θ .

Solution: Eventually you will be able to integrate the cube of the tangent; but, at this

point, you may not be able to do that. However, you can observe that the integrand is an

odd function of θ . (Why?) Hence the integral from −π

6 to 0 will be equal in magnitudebut opposite in sign to the integral from 0 to π

6; so the given integral is 0.

[1, Exercise 57, p. 407] “Evaluate the definite integral

π6

− π6

tan3 θ d θ .”

Solution: Eventually you will be able to integrate the cube of the tangent; but, at this

point, you may not be able to do that. However, you can observe that the integrand is an

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odd function of θ . (Why?) Hence the integral from −π6

to 0 will be equal in magnitude

but opposite in sign to the integral from 0 to π

6 ; so the given integral is 0.

[1, Exercise 65, p. 407] “Evaluate the definite integral

2

1

x√

x − 1 dx.”

Solution: This integral can be simplified by defining u = x − 1, so du = d x.38 Then x

√ x − 1 dx =

(u + 1)

√ u du

=

u

32 + u

12

dx

=

2

5 u

52

+

2

3u

32

+ C

= 2

5( x − 1)

52 +

2

3( x − 1)

32 + C

Taking the diff erences of the values of this antiderivative (i.e., with any specific value

for C , e.g., with C = 0) at the limits yields2

5( x − 1)

52 +

2

3( x − 1)

32

2

1

= 2

5(2 − 1)

52 +

2

3(2 − 1)

32 −

2

5(1 − 1)

52 +

2

3(1 − 1)

32

=

2

5 +

2

3

− (0 + 0) =

16

15

Alternatively, the Definite Integral version of the Substitution Rule could have been used,

to obtain 2

5u

52 +

2

3u

32

2−1

1−1

=

2

51

52 +

2

31

32

−

2

50

52 +

2

30

32

etc.

The Logarithm Defined as an Integral (material in §5.6 of the 5th Edition) 39While it is

theoretical, the discussion below is an essential part of the course; its purpose is to substantiatesome of the discussions in [1, §§1.5, 1.6] by replacing the weakest parts of the definitions given

at that point in the textbook. This is one of several possible ways of treating exponentials and

38An even better substitution would be v =√

x − 1, which would imply that v2 = x − 1, so d x = 2v dv, and the

integral would be equal to1

0

(v2 + 1)v · 2v dv etc.

39The textbook material has been moved in the 6th edition to [1, Appendix G, pp. A48-A55]

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logarithms “properly”; the treatment in [1, §§1.5, 1.6] was necessarily inadequate, as there

was a perceived urgency to make logarithms and exponentials available to students who didn’thave the background for a more substantial treatment.

The Natural Logarithm Previously we had “defined” the logarithm to be the inverse of

the exponential function, whose definition was, at best, intuitive. Now we sketch very briefly a

more rigorous definition of both functions, beginning with the logarithm. This theory has been

delayed until now because parts of the proofs require the concept of substitution in a definite

integral.

Henceforth we take the following as our primary definition:

Definition C.3 Let x be any positive real number. Define

ln x =

x

1

1

t dt

i.e., it is the area under the right branch of the hyperbola y = 1

t , between t = 1 and t = x.

Which Definition is “Correct”? The definitions in [1, §§1.5, 1.6] were all we had avail-

able at that time to follow the objective of the textbook to introduce the exponential and log-

arithm functions “early”. We had to rely largely on intuition in deriving properties of these

functions. Now that we have the integral available, we can return and replace those inadequate

definitions by some that are more rigorous. Even the present definitions have some issues,since we haven’t rigorously proved all the properties that we are using for the definite integral.

While most mathematicians today would probably follow the present development — of intro-

ducing the logarithm first, as a definite integral, and the exponential as its inverse — there are

other possible ways of defining the functions. Yet another way will be available to you if you

follow the theory of infinite series beyond what we will be doing in [1, Chapter 11].

Properties of the Logarithm From the first part of the Fundamental Theorem we imme-

diately obtain the fact that

Theorem C.22

d dx

ln x = 1 x

. (36)

We can also prove the following, using basic properties of the integral:

Theorem C.23 Let x, y be positive real variables. Then

1. ln( xy) = ln x + ln y

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2. ln 1 = 0

3. ln 1 x

= − ln x

4. d

dxln | x| = 1

x

5. lim x→∞

ln x = ∞

6. lim x→0+

ln x = −∞

While I do not suggest you memorize the proofs, except possibly that of the first part below,

they provide simple exercises on various aspects of the integral.

Proof:

1. This is a simple application of properties of the definite integral, and in the Substitution

Rule.

ln xy =

xy

1

1

t dt

=

x

1

1

t dt +

xy

x

1

t dt

by decomposing the interval [1, xy] into [1, x] and [ x, xy]

= x

1

1

t dt +

y

1

1

xu x du

using the substitution t = xu

=

x

1

1

t dt +

y

1

1

udu = ln x + ln y

2. ln 1 =

1 1

1

t dt = 0 since the integration is over an interval of length 0.

3. Using a substitution u = 1

t , where dt = − 1

u2du, we obtain

ln 1

x=

1 x

1

1

t dt

=

x

1

u

− 1

u2

du

= − x

1

1

udu = − ln x

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4. Recall thatd

dx | x| = +1 if x > 0

−1 if x < 0 .

It is convenient to write these values in the form

d

dx| x| =

| x| x

if x 0

undefined if x = 0.

Hence, by the Chain Rule and the Fundamental Theorem,

d

dx(ln | x|) =

1

| x| · | x|

x=

1

x

when x 0.

5. We will reconsider this proof when we study “harmonic series” in [1, §11.2, Example 7,

pp. 691-692]. Consider the Riemann sum obtained for the definite integral

2n 1

1

t dt . We

will hang the rectangles by their upper right-hand corners, so that the Riemann sum is

clearly less than the integral. Then the area under the curve is greater than the sum

1

2 +

1

3 +

1

4 + . . . +

1

2n .

Let’s collect these terms into groups ending with the reciprocal of each power of 2. We

obtain

1

2 =

1

21

3 +

1

4 >

1

4 +

1

4 =

1

21

5 +

1

6 +

1

7 +

1

8 >

1

8 +

1

8 +

1

8 +

1

8 =

1

2. . . . . .

1

2n−1 +

1

2n−1 + 1 + . . . +

1

2n > 2n−1 · 1

2n =

1

2

We see that the area of the rectangles under the curve is n2 . As we allow n → ∞ thislower bound for the area approaches ∞, implying that the full area surely → ∞.

6. Let’s make the substitution y = 1

x, so that, as x → 0+, y → +∞. Then

lim x→0+

ln x = lim y→+∞

ln 1

y= − lim

y→+∞ln y = −∞ .

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7. I presented the preceding proof because it is classical. The definition of the logarithm

as a definition integral permits a much simpler proof of the same result. Since

1

t ≤ 1 for

t ≥ 1,

x 1

1

t dt ≥

x 1

1 dt = x − 1 → ∞ as x → ∞.

The construction began with a formal definition for the (natural) logarithm. At this point we

can define the (natural) exponential function as the inverse of the logarithm. We begin by

calling the function exp( x); only after we prove that it behaves the way we expect a power to

behave, do we revert to the familiar notation.

The Natural Exponential Function We could now justify the various equations that we

used intuitively in the “early transcendentals” treatment in [1, §§

1.5, 1.6]. As mentioned, we

begin by showing that ln, as defined above, is invertible. Then we temporarily call the inverse

function, i.e., ln−1 x, exp( x). Thus we have,

Theorem C.24

exp x = y ⇔ ln y = x (37)

exp(ln x) = x (38)

ln(exp x) = x (39)

exp( x + y) = (exp x) · (exp y) (40)

exp( x − y) = exp x

exp y(41)

(exp x) y = exp( xy) (42)

following which, knowing that the exponent rules are satisfied, we define

Definition C.4

e = exp1 (43)

and change our notation by recognizing that

exp x = e x . (44)

We also show that

Theorem C.25d

dx (exp x) = exp x . (45)

Using standard properties of the definite integral we can also show that

Theorem C.26

lim x→−∞

exp x = 0 (46)

lim x→∞

exp x = ∞ (47)

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General Exponential Functions The preceding constructions have been for the natural

logarithm and the natural exponential. We may now extend these definitions to exponentialsto any positive base a 1.

Definition C.5 For any positive real number a and any real number x we define

a x = e x ln a . (48)

This leads to the exponent rules for general (positive) bases:

a x+ y = a x · a y (49)

a x− y = a x

a y (50)

(a x) y = a xy (51)

(ab) x = a x · b x (52)

and to the derivative propertyd

dx(a x) = a x · (ln a) (53)

General Logarithmic Functions Finally the inverse function of a x is named loga, and

its familiar properties are developed, e.g.,

Theorem C.27

loga x = y ⇔ a y = x (54)

d

dx

loga x

=

1

x ln a(55)

The Number e Expressed as a Limit

Theorem C.28 Let a > 0. Then lim x→0

(1 + ax)

1 x

= ea

Proof:

lim x→0

(1 + ax)

1 x

= lim x→0

eln(1+ax)

1

x

since exponential and logarithm are inverses

= lim x→0

e

1 x·ln(1+ax)

by the exponent rules

= lim x→0

e

ln(1+ax) x

= lim

x→0

e

ln(1+ax)−ln(1+0· x) x

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= e

lim x→

0

ln(1 + ax) − ln(1 + 0 · x)

x

by the continuity of the exponential function

= e

d

dxln(1 + ax)

x=0

by definition of the derivative at x = 0

= e

a

1 + ax

x=0 by the Chain Rule

= ea

C.7.2 5 ReviewTrue-False Quiz Students tend to avoid the True-False questions because these are unlikely

to appear on quizzes or examinations. It’s true that I would not wish to have a simple True-

False question in any test, since there would be a 50% chance of a correct guess. However,

the True-False questions in your textbook are much more difficult than that, as they ask “De-

termine whether the statement is true of false. If it is true, explain why. If it is false, explain

why or give an example that disproves the statement.” This requirement of a proof or a coun-

terexample makes these problems very challenging. The odd-numbered problems are solved

in your Student Solutions Manual [3]. I consider some of the even-numbered problems below.

[1, True-False Exercise 2, p. 409] If f and g are continuous on [a, b], then b

a

[ f ( x) · g( x)] dx =

b

a

f ( x) dx · b

a

g( x) dx

Solution: This statement is FALSE. That is, it is not true for all constants a, b and for all

functions f , g continuous on [a, b]. There will, of course, be functions and intervals for

which the statement is true, but the generalized statement, quantified for all functions

and intervals is not true. While you would have know to suspect this after we study

[1, §7.1], but you should be suspecting it already, and be able to construct a simple

counterexample. Here is one:

Define f ( x) = g( x) = x. Then b

a

f ( x) dx =

b

a

g( x) dx = x2

2

b

a

= b2 − a2

2 .

b

a

[ f ( x) · g( x)] dx =

b

a

x2 dx = x3

3

b

a

= b3 − a3

3

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While the polynomials b2 − a2

2 ·

b2 − a2

2

and b3 − a3

3

look diff erent, that does not consti-

tute a counterexample, as it could happen that we have two algebraic functions that are

equal by virtue of some identity that we don’t happen to recognize at the moment. So,

in order to complete the counterexample, we need to find specific values of a and b that

will entail that b

a

f ( x) dx · b

a

g( x) dx

b

a

[ f ( x) · g( x)] dx ,

i.e., thatb2 − a2

2 · b2 − a2

2

b3 − a3

3 .

One such example is a = 0, and b having any value except 0, 4

3:

b2

2 · b2

2 =

b4

4

b3

3 ;

for example, take b = 1.

But the counterexample given is far from the simplest! For example, take f ( x) = g( x) =

1 for all x. Then the left side of the claimed equation is equal to b − a, while the right

side is (b − a)2, which will be diff erent from b − a except where b − a = 1 or b = a.

[1, True-False Exercise 4, p. 409] This is very similar to [1, True-False Exercise 2, p. 409];

you should have no trouble constructing a counterexample.

[1, True-False Exercise 8, p. 409] If f and g are diff erentiable, and f ( x) ≥ g( x) for a < x < b,

then f ( x) ≥ g( x) for a < x < b.

Solution: This statement is false. Think geometrically. The condition that f ( x) ≥ g( x)

tells us that the graph of f is above the graph of g; the condition that f ( x) ≥ g( x) tells us

that the graph of f is steeper than the graph of g. So we can construct a counterexample

by finding, for a convenient function g, a function that is larger, but whose graph is not

so steep. For example, take g( x) = 0 for all x; now all we need is a positive function f

whose graph is sloping downward; for example, f ( x) = e− x will work, with any interval

a

≤ x

≤ b. A simpler counterexample is f ( x) = a + b

− x, g( x) = 0.

[1, True-False Exercise 14, p. 409] All continuous functions have antiderivatives.

Solution: This problem is interesting, as it follows [1, True-False Exercise 13, p. 409],

which states that “All continuous functions have derivatives.” That preceding statement

is false, and you should be able to give counterexamples (e.g., | x| is continuous, but lacks

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Release Date: Monday, January 18th, 2010 (subject to revision)

Textbook Chapter 6. APPLICATIONS OF INTEGRATION.

C.8.1 §6.1 Areas between Curves

If, for a ≤ x ≤ b, continuous functions f and g have the property that f ( x) ≥ g( x) — i.e., if

the graph of f is not lower than the graph of g for that interval, then the area bounded by the

graphs and the vertical lines x = a and x = b is given by the integral

b

a

( f ( x) − g( x)) dx .

Before working some examples, I make several observations:

• When g is the 0 function — i.e., when the lower boundary of the region is the x-axis —

this formula reduces to the definite integral b

a f ( x) dx.

• We can find the areas of certain regions by decomposing them into parts that can be

described as above. Sometimes there is more than one “natural” way to decompose the

region, but all decompositions should yield the same area.

• An analogous formula can be applied if we consider the “region” bounded by graphs of

the form x = f ( y), x = g( y), for a ≤ y ≤ b. In that case we say that we are integrating

with respect to y or integrating along the y-axis.

• Often we wish to find the area between two graphs determined by points where they

intersect. In these cases the vertical sides of the region have zero length.

• In solving problems it is useful to make a sketch showing an “element of area” — a

thin rectangle whose width is shown as d x or ∆ x in the case of integration with respect

to x, and analogously for the case of integration with respect to y. It is difficult for me

to include such sketches in these notes, but they will be shown on the chalkboard. The

sketch is not a formal part of the solution, but you are likely to find it helps you formulate

your solution.

• Where two graphs cross in several places, so that the area between them is in severalpieces, be sure you understand what you intend if you write an integral that extends

over several pieces: if the curves interchange positions, with the upper one becoming

the lower, then the signs of the areas will change, and some of the areas will cancel. If

this is not what you intend, you must either write the integrand with absolute signs, as

| f ( x)−g( x)|, or, equivalently, find the area of each of the pieces, and add their magnitudes

so as to prevent cancellation.

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Example C.29 ([7, Exercise 8, p. 442]; see Figure 1 on page 3053) To find the area of the

1.0

x

0.5

0.50.0

0.25

−0.5

1.0

0.75

0.0

−1.0

Figure 1: The region(s) bounded by y = x2 and y = x4

region bounded by the curves y = x2 and y = x4.

Solution: Note that the text-book used the word “region” in a general way. Not all authors use

the word in this way. For this problem there are two connected regions bounded by the curves,and the intention of the textbook was that you find the areas of both of them together.

To determine where the curves intersect, we solve their equations, obtaining ( x, y) = (0, 0)

and ( x, y) = (±1, 1). Note that it is not enough to guess the coordinates of these points of

intersection from the graph: you must determine the coordinates by rigorous solution of the

equations!

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First solution: Integration with respect to x.

2

1

0

x2 − x4

dx = 2

x3

3 − x5

5

1

0

= 2

3 − 2

5 =

4

15.

Second solution: Integration with respect to y. I find the area of the region in the first quad-

rant. The equations of the curves are, respectively, x = √

y and x = 4√

y. The area is

1

0

y

14 − y

12

dy =

4

5 · y 5

4 − 2

3 · y 3

2

1

0

= 4

5 − 2

3 =

2

15.

Now double this area. Note that the order of the curves in this integral is the reverse

of that used when we integrated with respect to x, because we now order them by theirrelative distance from the y-axis.

While the limits of the two integrals look as though they are the same, the limits in the first

case refer to the extreme values of x, while the second refer to the extreme values of y.

Example C.30 ([7, Exercise 22, p. 442]) (see Figure 2 on page 3055) To find the area of the

region bounded by the curves y = sin x and y = sin2 x between the vertical lines x = 0 and

x = π2

.

Solution: To determine the points of intersection of the curves, we solve y = sin x with y =

sin2 x, and solve sin x

− sin 2 x = 0. Since sin 2 x = 2(sin x)

· (cos x), the intersections must

satisfy sin x · (1 − 2cos x) = 0: either

2cos x = 1

or we must be unable to divide by sin x, because

sin x = 0 .

In the interval 0 ≤ x ≤ π2

, we must therefore have either x = π3

(to make cos x = 12

) or x = 0

(to make sin x = 0). This yields the points of intersection (0, 0), and

π3

,√

32

. The

regions to be considered therefore consists of a lune-shaped region with one end at the origin

and the other at

π3

, √ 32

; and a second region which begins at

π3

, √ 32

and ends at the vertical

line x = π2

. We will find the areas of each of the regions and add them, since that appears to

be what the textbook expects. (The wording of the question is not totally unambiguous; some

readers might be justified in assuming that the author intended signed areas to be used, rather

than the absolute values, as I am taking.)

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1.51.250.75 1.00.50.0 0.25

0.25

0.5

0.0

0.75

1.0

Figure 2: The region(s) bounded by y = sin x, y = sin 2 x between x = 0 and x = π2

For the left-most region the curve y = sin 2 x is above y = sin x, so the area will be

π

3

0

(sin 2 x − sin x) dx =

−1

2 cos 2 x + cos x

π3

0

=

−1

2 ·−1

2

+

1

2

−−1

2 + 1

=

1

4

For the right region the orders of the curves are reversed, and the area is

π2

π3

(sin x − sin2 x) dx =

− cos x +

1

2 cos 2 x

π2

π3

8/15/2019 Brown's Notes 2010



= −0 + 1

2 ·(

−1) − −

1

2

+ 1

2−

1

2 =

1

4Thus the total area is 1

4 + 1

4 = 1

2.

6.1 Exercises

[1, Exercise 18, p. 420] (see Figure 3 on page 3056) To find the area of the region bounded

−2.5

31 20

10.0

−1

2.5

−3

0.0

5.0

−2

7.5

Figure 3: The region(s) bounded by y = 8 − x2, y = x2 between x = ±3

by the curves y = 8 − x2 and y = x2 between the vertical lines x = −3 and x = 3.

Solution: Let’s first determine where the curves intersect. Solving the equations shows

that the intersections occur when x = ±2, y = 4, i.e., in the points (±2, 4). But we are

asked to find the area from x = −3 to x = 3. Thus this region has three parts:

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8/15/2019 Brown's Notes 2010



Solution: (see Figure 2 on page 3058) We can interpret this as the area of the region

2 40−1 1

4

1

0

2

3−2

3

Figure 4: The region(s) bounded by y =√

x + 2, y = x between x = 0 and x = 4

bounded by the curve y =√

x + 2, the line y = x, and the vertical lines x = 0 and x = 4.

If we square the former equation, we obtain y2 = x+2, which can be seen to be a parabola

that is symmetric about the x-axis, with its vertex on the line x = −2 and opening to theright. Squaring the equation caused the new equation to include the lower branch of this

parabola; the original equation represents only the upper branch. The upper branch of

the parabola and the line y = x meet both in the origin and in the point (2, 2). (The other

point of intersection of the parabola and the line does not lie on the upper branch of the

parabola, and so is extraneous to this discussion.)

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The simplest way of evaluating this integral is to break it up into two parts: 4

0

√ x + 2 − x

dx =

2

0

√ x + 2 − x

dx +

4

2

x −

√ x + 2

dx

=

( x + 2)32

32

− x2

2

2

0

+

−( x + 2)32

32

+ x2

2

4

2

=

2

3 · 4

32 − 2

−

2

3 · 2

32 − 0

+

−2

3 · 6

32 + 8

−−2

3 · 4

32 + 2

= 16

3 − 2 − 2

3 · 2

32 − 2

3 · 6

32 + 18 +

16

3 − 2

=

44

3 − 4

3

√ 2 − 4

√ 6

[1, Exercise 50, p. 421] 1. “Find the number a such that the line x = a bisects the area

under the curve y = 1

x2, 1 ≤ x ≤ 4

2. “Find the number b such that the line y = b bisects the area in part (a).”

Solution:

1. We solve for a the equation

a

1

1

x2 d x = 4

a

1

x2 dx

to obtain a = 85

. It follows that the area of half of the region is the common value

of the two integrals,

1 x

a

1= 3

8.

2. The right side of the region has an irregular boundary: the upper part has equation

y = 1 x2 , i.e., x = 1√

y; the lower part is on the line x = 4. The two parts of the

boundary meet in the point

4, 116

. Thus the area of the rectangle bounded by the

lines x = 1, x = 4, y = 0, y = 116

is 316

. As this is less than half the total, we know

that b > 116

. This fact is important, as it tells us that we can represent the upper half

of the area — the portion above the line y = b, by the integral 1

b

1√

y− 1

dy ,

where the −1 in the integrand represents the lower boundary of the region — now

viewed as being “under” x = 1√ y

and “over” x = 1. Setting this integral equal to 38

8/15/2019 Brown's Notes 2010



and solving, we obtain 2√

y − y1

b= 3

8, which evaluates to

1 − 2√

b + b = 3

8

⇔ (√

b − 1)2 = 3

8

⇔√

b = 1 ±

3

8

⇔ b = 11

8 ±

3

2

One of the values we obtain is greater than 1, which contradicts our assumption

that 0 ≤ b ≤ 1. The other gives the bisecting line

y = 11

8 −

3

2 ≈ 0.150255129.

The value we rejected has a geometric significance: it is the height of a horizontal

line at which the area bounded by the curve y = 1 x2 , the line x = 1 and that line is

equal to 38

.

And what would have happened if we failed to observe that b ≥ 116

? We would

have obtained an equation for b that could not be solved; it would be equivalent to

requiring that 5

8 − 3b =

5

16 when 0 ≤ b ≤ 1

16 , which are contradictory statements.The area which I found to equal 1 − 2

√ b + b by integrating with respect to y could

also be evaluated by integrating with respect to x: here the line y = b meets y = 1 x2

in the point

1√ b

, b

, so the area is

1√ b

1

1

x2 − b

dx = b − 2

√ b + 1 = (

√ b − 1)2 ,

which we must again equate to 38

and solve.

8/15/2019 Brown's Notes 2010




Release Date: Wednesday, January 20th, 2010 (subject to revision)

C.9.1 §6.2 Volumes

Just as we defined the area of a region as the limit of a sum of narrow rectangles, we can

define volumes as limits of sums of thin elements; these elements can be assembled in various

ways. In this course you will see volumes expressed either as sums of thin slices with planar

sides (called laminæ), and — for solids with rotational symmetry (called solids of revolution)

— as sums of thin shells with cylindrical sides. Some problems will lend themselves only

to one type of dissection; where a problem can be approached in more than one way, it is

instructive to try it in several ways, in order to verify your answer and also to gain experience

in choosing the method that is more efficient for diff erent types of problems. In this section we

will be considering dissections into thin slices; where the slice has the shape of a disk with a

concentric disk cut from its centre, the author uses the term washer 40.

Example C.31 1. (see Figure 1 on page 3053) ([7, Exercise 2, p. 452]) “Find the volume

1.

2.5

1.5

2

1

0

10.80.40.60.2

0.5

0

4.

7

5

1

6

4

0

210 .5 1 .50

2

3

Figure 5: Regions for Example C.31

of the solid obtained by rotating the region bounded by the curves y = e x, y = 0, x = 0,

x = 1 about the x-axis.” We will first solve the problem as stated, and then consider some

related problems obtained by changing some of the data.

40Students whose native language is not English sometimes cannot understand why the author will use the

name of a household appliance here; this is another meaning of the English word washer , which refers to a thin

disk with a hole in the middle

8/15/2019 Brown's Notes 2010



Solution: (Remember to come back to this problem when we study [1, §6.3], to solve it

using the methods of that section.)We will decompose the solid into laminæ that are thin disks. The “washers” will be

obtained by rotating about the x-axis the element that we would have used for the area

if we had found the area by integrating with respect to x. For the disk whose faces are

centred at points ( x, 0) and ( x + ∆ x, 0), the volume obtained by hanging the element from

its upper left corner on the curve is π y2∆ x = πe2 x∆ x. By the same reasoning that led us

to express areas as definite integrals, we have

Volume =

1

0

πe2 xdx

= π

2 · e2 x1

0=

π

2 · (e2 − 1)

2. Now let us change the problem, asking that the solid be rotated about the y-axis. Here

the description of the cross sections will depend on the height of the cross section. We

will be expressing everything in terms of y, not x, and integrating “with respect to y”.

For y ≤ 1 (the height where the curve cuts the y-axis), the cross sections are disks of

radius 1, and the volume of that part of the solid is 1

0

π12dy = π y]10 = π.

For y ≥ 1 the cross sections are “washers”, with outer radius 1 and inner radius x; to

express this in terms of y we need to rewrite the equation of the curve y = e x in an

equivalent way that expresses x in terms of y, as x = ln y. The washers at height y have

volume

π(12 − x2)∆ y = π(1 − (ln y)2)∆ y

so this part of the solid of revolution has volume

π

e

1

(1 − (ln y)2) dy .

where the upper limit of integration is the ordinate of the point where the line x = 1meets the curve y = e x. You are not quite ready to complete this integration. You can

make a substitution like x = ln y, under which the integral transforms to π 1

0 (1− x2)e x dx,

and you know how to integrate part of this integral, as

π

1

0

1e x dx = πe x]10 = π(e − 1) ,

8/15/2019 Brown's Notes 2010



but you are not ready to integrate π 1

0 x2e x dx. We will, in [1, §7.1], develop a method

to determine that

x2

e x

dx = ( x2

− 2 x + 2)e x

+ C (which is obvious by diff erentiation).With that we know that

Volume = π[e x − ( x2 − 2 x + 2)e x]10 = π[(− x2 + 2 x − 1)e x]1

0 = π .

3. As another variant on this problem, consider the following: “Find the volume of the

solid obtained by rotating the region bounded by the curves y = e x, y = 0, x = 0, x = 1

about the line y = −3.” The analysis is similar to what we did originally, except that the

laminæ now have a hole of radius 3 in the middle, and the outer radius is 3 units larger.

This leads to an integral

Volume = 1

0π(e x + 3)2 − 32

dx =

1

0πe2 x + 6e x

dx

= π

1

2e2 x + 6e x

1

0

= π

1

2e2 + 6e

−

1

2 + 6

=

π

2

e2 + 12e − 13

4. Finally, suppose that the right boundary of the region generating the solid by revolution

changed from x = 1 to y = e2( x − 1). This new right boundary also passes through (1 , 0),

but meets the curve in the point ( x, y) = 2, e2

. The volume will be

1

0

πe2 x dx +

2

1

π

(e x)2 −

e2( x − 1)

2

dx

= π

2

e2 − 1

+ π

2

1

e2 x dx − πe4

2

1

( x − 1)2 dx

= π

2

e2 − 1

+ π

e2 x

2

2

1

− πe4

1

3( x − 1)3

2

1

= π

e4 − 1

2

− π

e4

3 = π

e4

6 − 1

2

Example C.32 ([7, Exercise 8, p. 452]) Find the volume of the solid obtained by rotating the

region bounded by the curves y = sec x, y = 1, x = −1, x = 1 about the x-axis.

Solution: The lines x = ±1 intersect y = sec x in the respective points (1, ± sec 1). The area of

the washer centred on the x-axis between cross sections x = X and x = X + X is approximately

π(sec2 X − 12) · X . The volume of revolution will be 1

−1

π

sec2 x − 1· dx = π[tan x − x]1

−1 = 2π(tan 1 − 1).

8/15/2019 Brown's Notes 2010



(Had the integral involved arctan 1, you would have been expected to simplify it further; but

you cannot evaluate tan 1 without calculators or techniques that you will not meet until Calcu-lus 3.)

Example C.33 ([7, Exercise 56, p. 454]) Here is a problem where the solid is not generated

by revolving a plane region about an axis. “Find the volume of the solid S: the base of S is the

parabolic region ( x, y)| x2 ≤ y ≤ 1 ; cross-sections perpendicular to the y-axis are equilateral

triangles.”

Solution: The cross-section of the base at level y has ends with coordinates (± √ y, y), so the

length of the base is 2√

y, and the area of the triangular cross-section is 12·2 √

y·√

3√

y =√

3· y.

Integrating along the y axis we find that

Area = 1

0

√ 3 y dy = √ 3

2 y2

1

0

=

√ 3

2 (12 − 02) =

√ 3

2

Example C.34 ([7, Exercise 15, p. 458]) (Where this problem appeared in the cited textbook,

students were asked to solve it using the method of “shells”. Let’s see if we can solve it using

the method of “washers”.) To find the volume of the solid of revolution generated by revolving

the region bounded by y = x2, y = 0, x = 1, x = 2 about the axis x = 1.

Solution: The washers generated by elements of area parallel to the x-axis will have two kinds

of descriptions, depending on whether their cross sections are above or below the point where

x = 1 meets y = x2, i.e. (1, 1): below y = 1 the cross sections are rectangles of width 2 −1 = 1;above y = 1 the cross sections are rectangles obtained from such a rectangle as below y = 1

by cutting away a rectangle of length x2 starting at the left end. But, in evaluating the limit

of the sum of the volumes of these washers, we shall be integrating with respect to y, so we

need to express the dimensions and position of the rectangular element in terms of y. The

equation of the right branch of the parabola y = x2 is x = √

y. The hole in the washer has

radius x − 1 = √

y − 1; the washer will have area

π(12 − (√

y − 1)2) = π(2√

y − y) .

The line x = 2 meets the parabola in the point (2, 4), so we shall integrate for y ranging from

0 to 4: from 0 to 1 using the constant integrand π12, and from 1 to 4 using the integrandπ(2

√ y − y). The volume of revolution is

1

0

π12 dy +

4

1

π(2√

y − y) dy = π y1

0 + π

4

3 y

32 − y2

2

4

1

= π + π

4

3 · 8 − 8

− π

4

3 · 1 − 1

2

=

17π

6

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which is the same result given in the textbook for the solution that could be obtained in the

next section using the method of cylindrical shells.

6.2 Exercises

[1, Exercise 12, p. 430] Find the volume of the solid obtained by rotating the region bounded

by the curves y = e− x, y = 1, x = 2 about the line y = 2.

Solution: Through the point ( x, 2) on the line y = 2 the cross section will be an annulus

(ring) with outer dimension 2−e− x and inner dimension 2−1 = 1; the annulus is bounded

by concentric circles centred at the point ( x, 2). The volume will be

2

0

π

2 − e−22

− π12

dx

= π

2

0

3 − 4e− x + e−2 x

dx .

The first summand of the integrand has antiderivative 3 x; the second summand can be

integrated by using a substitution u = − x, which leads to an antiderivative +4e− x; the

last summand can be integrated by using a substitution u = −2 x, which leads to an

antiderivative −12 · e−2 x. Putting these three components together we find the value of the

integral to be

π

3 x + 4e− x − 1

2 e−2 x2

0= π

6 +

4

e2 − 1

2e4 − π

0 + 4 −

1

2

= π

5

2 +

4

e2 − 1

2e4

.

[1, Exercise 39, p. 431] The textbook asks you to use a Computer Algebra System to find the

volume of the solid generated by revolving about the line y = −1 the region bounded by

y = sin2 x, y = 0, for 0 ≤ x ≤ π. No Computer Algebra System is needed , although we

haven’t yet seen how to integrate this. Here is the full solution. The cross sections are

washers with centre on the line y = −1, outer radius ending on the curve y = sin2 x, and

inner radius ending on the line y = 0 (the x-axis). The volume is thus

π

π

0

(sin2 x − (−1))2 − 12

dx = π

π

0

1 − cos2 x

2 + 1

2

− 1

dx

= π

4

π

0

(3 − cos2 x)2 − 4

d x

= π

4

π

0

5 − 6c os2 x + cos2 2 x

dx

8/15/2019 Brown's Notes 2010



= π

4 π

05 − 6c os2 x +

1 + cos 4 x

2 dx

= π

8

π

0

(11 − 12cos2 x + cos4 x) d x

= π

8

11 x − 6sin2 x +

sin 4 x

4

π

0

= π

8 · 11π =

11π2

8

[1, Exercise 44, p. 431] Describe the solid whose volume is represented by the integral

π

π2

0

[(1 + cos x)2 − 1] dx .

Solution: The d x tells us that the integration is along the x-axis. That is, the planes of the washers are perpendicular to the x-axis. The integrand is the diff erence of 2 squares,

multiplied by π. We may interpret this as the area of a washer whose outer radius is

1 + cos x, and whose inner radius is 1. If we interpret 1 + cos x as cos x − (−1), we

can interpret this as the radius of a disk whose centre is at the point ( x, y) = ( x, −1),

generated by a radius extending from that centre to the point ( x, cos x) on the graph of

the cosine function. The subtracted term −π12 can be interpreted as the area of a disk

whose centre is at the same point, but its radius extends from that point ( x, −1) to the

point ( x, 0) above it on the x-axis. Thus the integral represents the solid of revolution

about the line y = −1 of the region bounded by the graph of the cosine function, and the

x-axis, between x = 0 and x = π

2

. (This integral is not difficult to evaluate. We will see

in [1, Chapter 7] that, if we replace cos 2 x by 12

(1 + cos2 x), the value of the integral is

π

2sin x +

x

2 +

sin 2 x

4

π2

0

= 2π + π2

4 .

This is not the only way of interpreting the integral. We could also reason that it rep-

resents the volume of a region rotated about the x-axis, bounded by the y-axis, the line

y = 1, and the graph of y = cos x + 1.)

[1, Exercise 49, p. 431] Find the volume of a right circular cone with height h and base radius

r .

Solution: This is a standard problem that every student should be able to work.

A cone is a surface generated by joining to all points curve in a plane a fixed point (called

the apex outside of the plane. A cone is circular if the base curve is a circle. It is right

circular if the apex is located on the normal to the plane of the curve through the centre

of that base circle.

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It is convenient to set up coordinate axes so that the cone is generated by a right angled

triangle with height h and base r , i.e., with hypotenuse along the line

x

r +

y

h = 1, whichtriangle is to be rotated about the y-axis. The horizontal rectangular elements of area,

with height dy and length x are rotated about the y-axis to generate disk-shaped laminæ.

Expressing x as a function of y for the hypotenuse, we have x = r

1 − y

h

, so the cross-

sectional area at height y is πr 2

1 − y

h

2

. Hence the volume must be

h

0

πr 2

1 − y

h

2

dy .

We know how to evaluate an integral of this type by expanding the square. But it is

easier to make a change of variable: u = 1 − y

h, so du = −1

h dy, dy = −h du and

Volume = − 0

1

πr 2u2h du = πr 2h

u3

3

0

1

= 1

3(πr 2h) .

(cf. [9, p. 47]). If you remember this as 13

π × the area of the base, you will know this

as a special case of a general theorem. In fact, it is not hard to show that the area is

not aff ected by the fact that the cone is a right cone: even if the apex were moved to a

location not over the centre of the circular base, the area would not change. Moreover,

it can be shown that even the fact that the base is circular is not relevant! The volume of any “cone” is as shown:

1

3 × π × Area of base × height .

For example, the case of a square base is discussed in [1, Example 8, p. 429].

[1, Exercise 55, p. 432] Find the volume of the solid S which is a tetrahedron with three mu-

tually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4

cm, and 5 cm.

Solution: I find it convenient to locate the three perpendicular sides along the coordi-

nate axes, with one vertex at the origin. So I locate the vertices of the tetrahedron at

(0, 0, 0), (3, 0, 0), (0, 4, 0), (0, 0, 5). This tetrahedron is just a pyramid or cone on a trian-

gular base; thus we know by the theory of [1, Example 8, p. 429] that the volume will

be 13 ×

12 × 3 × 4

× 5 = 10. But I will pretend we don’t know that.

Consider cross sections perpendicular to the z-axis. These are triangles whose x-dimension

at height z will be 35

(5 − z) (by similar triangles in the xz-plane); and whose y=dimension

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at height z will be 45

(5 − z), (again by similar triangles). The area of the triangle will be1

2 (5 − z)2

, so the volume will be 5

0

1

2 · 12

25 · (5 − z)2 dz =

6

25

−1

3(5 − z)3

5

0

= − 2

25

0 − 53

= 10 .

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C.10 Supplementary Notes for the Lecture of January 22nd, 2010

Release Date: Friday, January 23nd, 2010, subject to revision

C.10.1 §6.3 Volumes by Cylindrical Shells

The integrand when using cylindrical shells The volume of a right circular cylinder (i.e.,

with a disk as base, axis perpendicular to the base) is — as you can easily prove using washers

or otherwise — the product of the area its base and its height. If we consider a hollowed out

cylinder of radius r 2, in which an inner cylinder of radius r 1 is removed, then the volume will

be, if the height is h,

πr 22h − πr 21h = π

r 22 − r 21

h .

Let’s assume that r 2 = r 1 + r , and expand this product:

πr 22 − r 21

h = π

(r 1 + r )2 − r 21

h

= πr 21 + 2r · r 1 + (r )2 − r 21

h

= 2πhr · r 1 + πh(r )2

Both of these products approach zero as we allow r → 0. If we wish to determine the volume

of a solid of revolution by decomposing it into cylindrical shells about the axis of revolution,

it would appear that we should add elements of volume

2πhr · r 1 + πh(r )

2

and then permit the number of shells to approach infinity, and the width ∆r → 0. It can be

shown that, in any such limiting process, the sum of the terms of type πh(r )2 approaches

0; that is, not only do the individual terms πh(∆r )2 approach 0, as we permit r → 0, but

even the sum of these terms, increasing arbitrarily in number as r → 0, also approaches 0.

Thus the volume sought can be viewed as a Riemann sum, leading to a definite integral of the

form

2πr 1h dr . Thus, to find the volume of a solid that we can decompose into elements

which are cylindrical shells, we need only consider an integral related to terms of the first type.

The integrand can be interpreted as the product of 2πr 1 and h, i.e., as the area of a rectangle

obtained by “cutting open” the inner surface of the cylindrical shell and “unrolling” it; then the

r can be interpreted as the thickness of a thin rectangular lamina based on that rectangle. Of course, when you attempt to do that, you find that there will be a small error in that product,

but, as I have mentioned, it can be shown that the totality of these errors approaches 0 as we

replace the sum of volumes of elements by the definite integral. All that remains to be done is

to express h in terms of the radius, and to determine the appropriate limits for integration.

If you choose to approach these problems by substituting in formulæ, you are urged to

remember how to generalize to situations where the axis of circular symmetry is parallel to

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but diff erent from one of the coordinate axes. I do not recommend memorizing these formulæ.

Among the following examples are some that were not discussed in the lecture. One of these,which you should certainly read shows how to find the volume of a sphere; instead of this

example, I worked a more difficult — but much more instructive — example which finds the

volume of a torus.

6.3 Exercises

[1, Exercise 14, p. 436] (see Figure 6 on page 3070) “Use the method of cylindrical shells to

4

2

3

1

-1

0

431 20-2 -1

Figure 6: The region(s) bounded by x + y = 3 and x = 4 − ( y − 1)2

find the volume of the solid obtained by rotating the region bounded by the given curves

about the x-axis. Sketch the region and a typical shell: x + y = 3, x = 4 − ( y − 1)2.”

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Solution: The parabola meets the line in the points (0, 3) and (3, 0), on the coordinate

axes.

1. First we follow the instructions, using the method of cylindrical shells. The ele-

ments of area that generate the shells will be narrow horizontal rectangles; at height

y the rectangle has width (4 − ( y − 1)2) − (3 − y), obtained by expressing the equa-

tions in the form x =function of y. The circumference of the base of the generated

cylinder is then 2π y, and the volume is

2π

3

0

(4 − ( y − 1)2) − (3 − y)

y dy

= 2π 3

03 y2

− y3 dy

= 2π

y3 − y4

4

3

0

= 27π

2 .

2. Now let us compute the volume using washers. The equation of the line may be

rewritten as y = 3 − x; but the parabola now splits into 2 curves — the upper

branch has equation y = 1 +√

4 − x, and the lower has equation y = 1 −√

4 − x.

The description of the element of area that generates the washer will change at

x = 3. To the left of x = 3 the element of area at horizontal position x has height

(1 +√

4 − x) − (3 − x). We need to compute the area of the annulus the element

generates. For that purpose the length of the element is not enough, as we need to

know the distance from the axis about which it is revolving. The outer radius of the washer is 1 +

√ 4 − x, and the inner radius is 3 − x, so the area of the annulus is

the diff erence between the areas of two disks:

π(1 +√

4 − x)2 − π(3 − x)2

and the volume of the solid generated up to x = 3 is

π

3

0

(1 +

√ 4 − x)2 − π(3 − x)2

dx .

The elements of area to the right of x = 3 have inner radius 1 −√

4 − x, and outer

radius 1 + √ 4 − x, so the area of the annular cross-section is

π1 −

√ 4 − x

2 − π1 +

√ 4 − x

2= 4π

√ 4 − x ,

and the volume is 4π 4

3

√ 4 − x dx. The total volume by the method of washers is

π

3

0

(1 +

√ 4 − x)2 − π(3 − x)2

dx + 4π

4

3

√ 4 − x dx

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= π 3

0−( x

−1)( x

−4) + 2

√ 4

− x dx + 4π

4

3

√ 4

− x dx

= π

1

4

(u − 3u2 + 2√

u) (−1) du + π

0

1

4√

u(−1) du

using the substitution u = 4 − x, du = − x, x = 4 − u

= 27π

2 as before.

[1, Exercise 20, p. 437] “Use the method of cylindrical shells to find the volume generated

by rotating the region bounded by the given curves about the specified axis. Sketch the

region and a typical shell: y = x2, x = y2, about y = −1.”

Solution:

1. At height y the cylinder is generated by an element of area whose horizontal di-

mension is √

y − y2, and whose vertical dimension is ∆ y; the circumference of the

circle generate by a point at one end of this element under revolution is 2π( y−(−1))

(since the radius is the distance between a point ( x, y) and the point ( x, −1) below

it), so the volume is 1

0

√ y − y2

2π( y + 1) dy

= 2π 1

0 y

32

− y3 +

√ y

− y2 dy

= 2π

2

5 y

52 − 1

4 y4 +

2

3 y

32 − 1

3 y3

1

0

= 29

30π .

2. It wasn’t asked for in the problem, but let’s find the same volume by the method of

washers. The external radius of the vertical washer at x is√

x − (−1) =√

x + 1; the

internal radius of that washer is x2 − (−1) = x2 + 1. The area of the cross section

will be π times the diff erence of the squares of the radii, i.e.,

π( √ x + 1)

2

− ( x2

+ 1)2

.

The integral from x = 0 to x = 1 is again equal to 2930

π.

[1, Exercise 29, p. 437] The following integral represents the volume of a solid; describe the

solid: 3

0

2π x5 dx .

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Solution: Since the problem appears in this section, the textbook expects you to interpret

the integral as the result of application of the method of cylindrical shells. If we interpret2π x5 · ∆ x as (2π x) · x4 · ∆ x, we see that it is the volume of the solid generated by rotating

about the y-axis the region bounded by the x-axis and the curve y = x4 from the point

(0, 0) to the line x = 3.

But the author should not have used the definite article the, since the integral can be

interpreted in other ways. For example, we can interpret it as resulting from application

of the method of washers also. This time we interpret 2π x5 · ∆ x as π√

2 x52 · ∆ x: it

is the area of the solid obtained by rotating about the x-axis the region bounded by that

axis and the curve y =√

2 x5 between x = 0 and x = 3.

[1, Exercise 32, p. 437] The integral

π4

0

2π(π − x)(cos x − sin x) dx represents the volume of a

solid. Describe the solid.

Solution: The same integral could easily represent more than one solid. One interpre-

tation is the following: The solid is a solid of revolution around the vertical line x = π,

generated by revolving the region bounded by the y-axis and the graphs y = cos x and

y = sin x up to the point where they intersect, ( x, y) =

π4

, π4

. But the region could be de-

formed vertically without changing the volume. For example, the region could be taken

to be bounded by the lines y = 0, x = 0,

y = cos x − sin x = sin π

2 − x + x2

· cos

π2 − x − x

2 = 1√

2cos

π4 − x

from x = 0 to x = π

4.

[1, Exercise 43, p. 437] Use cylindrical shells to find the volume of a sphere of radius r .

Solution: Here is a case where the solid is prescribed, but not the way of generating it. A

sphere can be generated by the rotation of a half-disk around its diameter. I chose to take

the diameter as the y-axis, and the equation of its boundary as x2+ y2 = r 2, more precisely,

as x = +

r 2 − y2 where −r ≤ y ≤ r . The length of a vertical rectangular element of

area a distance of x from the axis is the distance between the points ( x,√

r 2

− x2) and

( x, −√ r 2 − x2), i.e., 2 √ r 2 − x2. This is the height of the cylindrical generated when the

element is rotated about the axis. The base of the shell is at a distance of x from the

axis, so it generates a circular band (annulus) of radius 2 x, hence of circumference 2π x.

Here I have not been precise about whether this is the inner or outer circumference of

the annulus, since, in the limit, these distinctions have no e ff ect on the calculations. The

thickness of the annulus is represented by the diff erential, dx, in the integral, and we

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obtain a volume of r

0

2π x(2 √ r 2 − x2) dx = −4π3

(r 2 − x2)

32

r

0= 4πr 3

3 . (56)

Of course, this volume could also be evaluated by the method of washers, see [1, Exam-

ple 1, pp. 423-424]. If you found the number of variables in equation (56), you could

amplify the notation, and write x=r

x=0

2π x(2√

r 2 − x2) dx = −4π

3

(r 2 − x2)

32

x=r

x=0=

4πr 3

3 for all r . (57)

[1, Exercise 44, p. 437] Use cylindrical shells to find the volume of the solid torus generated

by rotating a disk of radius r around a line located a distance R from its centre.

Solution: We can take the disk to be bounded by ( x − R)2 + y2 = r 2, and the axis of

rotational symmetry of the torus (doughnut) to be the y-axis. The boundary of the disk

is the graphs of 2 functions, y = ±

r 2 − ( x − R)2, so the height of the element of area is

2

r 2 − ( x − R)2. The volume is R+r

R−r

2π x · 2

r 2 − ( x − R)2 dx = 2π

r

−r

(u + R)2√

r 2 − u2 du

under the substitution u = x − R

= 4π

r

−r

u√

r 2 − u2 du + 4π R

r

−r

√ r 2 − u2 du

• Here the first integral can be seen to be the area under the graph of an odd functionfrom −r to the symmetrically located value +r , so the volume is 0; it may also be

integrated as

4π

3 · (r 2 − u2)

32

r

−r

= 0.

This integral could also be evaluated “naıvely” (as I did in the lectures), by using

one of several possible substitutions, or by observation, since the integrand is ev-

idently the derivative of 3

2 ·−1

2

· (r 2 − u2)

32 , which, when evaluated between −r

and +r gives a diff erence of 0.

• The second integral is 2π R times the area under the curve y =√

r 2 − u2 from u = −r

to u = r , which can be seen to be the area of a half disk of radius r , which we knowto be

πr 2

2 . Thus the volume is 2π2 Rr 2.

[1, Exercise 45, p. 437] I began the lecture by finding the volume of a right circular cone

(already considered in [1, Exercise 49, p. 431], solved in the notes of the previous lecture.

A solution to this variation of the problem can be found in the Student Solutions Manual

[3]).

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C.10.2 §6.4 Work

This section has been omitted from the syllabus because it involves physical concepts that

some students from outside of the Faculties of Science and Engineering might not be prepared

for. If you are a Science or Engineering student, you are urged to peruse the section and try

the problems. Your instructors and TA’s will be happy to help you with any di fficulties.

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Release Date: Monday, January 25th, 2010, subject to further revision

Review of preceding two lectures We have studied how to use a definite integral to deter-

mine volumes of solids, by expressing the solid as the limit of a union of thin layers: in [1,

§6.2] as the union of thin laminæ with planar sides and, in [1, §6.3], for solids of revolution, as

the union of thin circular cylindrical shells. Where the solid is obtained by revolving a plane

region about a line in that plane the solid can be called a solid of revolution about that line, and

the methods of both sections are, in principle, applicable. I do not recommend attacking these

problems by substitution in formulæ, as there are many variants to be considered, and the blind

use of formulæ often leads to disaster when the wrong formula is applied, or the right formula

is applied incorrectly. I list some formulæ below just to summarize the nature of the resultswe have found, with the intention that, in each case, you decompose the solid in one of the

methods that is applicable and set up the integral by carefully examining the decomposition.

1. When a solid is decomposed into thin laminæ between parallel planes which are per-

pendicular to the x-axis: the volume is expressible in the form

b a

A( x) dx, where the

decomposition ranges between values x = a and x = b, and A(u) is an expression for the

cross-sectional area at x = u. (If the decomposition is along the y-axis, then the limits of

the integral will be the limiting values of y, and we will usually express the integral in

terms of y; of course, the x or y that appears in the integral is irrelevant, since these arebound or “dummy” variables, that are part of our notation, and don’t actually a ff ect the

numerical value of the integral.)

2. When a decomposition as in item 1 above has circular symmetry as a solid of revolution

about a line x = c of a region in the xy-plane, the laminæ can be interpreted as being

generated by revolving a thin rectangle, and will consist of disks, possibly with a hole

in the middle — the textbook calls them “washers”. The integrand will be an expression

of the form π (r 2( x))2 − π (r 1( x))2, where r 2 and r 1 are the outer and inner radius of the

disk, determined by the distances of the two ends of the thin rectangle from the point of

intersection of the extended rectangle with the axis of rotational symmetry of the solid.

Here we need to be prepared to work with either x or y, and possibly to consider rotationabout a line parallel to but distinct from the coordinate axes.

3. In the case of decomposition into cylindrical shells of a solid with rotational symmetry

about an axis, the cylindrical shells can also be interpreted as being generated by rotating

a thin rectangle about the axis — but here the long dimension of the rectangle is parallel

to the axis. The integrand may also be interpreted as the area of a “flattened” cylindrical

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shell, obtained by slitting the shell open and unrolling it; one dimension will be the

circumference of the circle generated by revolving any point of the cylinder around theaxis of symmetry — this factor will be of the form 2π x when we are at distance x from

the axis of rotational symmetry; the other dimension will be the height of the cylindrical

shell. Again, we need to adjust this formula according to the orientation of the axis of

rotational symmetry, and according as the axis is or is not a coordinate axis.

Before proceeding with the topics scheduled for today, I reminded students of the ease in which

we can determine the volume of a sphere of radius r . A solution using cylindrical shells ([1,

Exercise 43, p. 437]) can be found beginning on page 3073 of these notes. We can find the

volume using washers by taking the equation of circle to be x2 + y2 = r 2, and revolving the

upper semi-disk around the x-axis, obtaining −

r r 2π(r 2 − x2) dx = 2π

r 2 x − x3

3

r

−r

= 2π

r 3 − r 3

3

− 2π

−r 3 +

r 3

3

=

4

3 · πr 3 .

C.11.1 §6.5 Average value of a function

In this section the textbook defines what is meant by the term average of a continuous function

or a function “pieced” together from continuous functions over an interval a ≤ x ≤ b. The

definition is a generalization of the definition familiar to you of the average of a finite number

of numbers y1, y2, . . . , yn, i.e.,

average =

n

i=1

yi

n .

If, in the finite case just mentioned, we treat each of yi as the height of a rectangle of unit width,

situated so that its base is placed on the x-axis with its corners at (i − 1, 0) and (i, 0), and with

height yi, then the sum

ni=1

yi is the area under the graph of the function f defined by

f ( x) =

y1 if 0 ≤ x ≤ 1

y2 if 1 < x ≤ 2

. . .

yn if n − 1 < x ≤ n

For any function that is piecewise continuous on the interval a ≤ x ≤ b, we define

average of f over [a, b] = 1

b − a

b

a

f ( x) dx ,

=

b

a f ( x) dx b

a 1 dx

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and so our definition is consistent with the earlier definition when the function is defined at a

finite number of points x1, . . . , xn: we are simply extending the definition of that function byextending the value at any integer point to the entire interval of unit length to the left of the

point.

The Mean Value Theorem for Integrals If f is continuous on [a, b], the Mean Value The-

orem may be applied to the function g( x) = x

a f (t ) dt , which we know from the Funda-

mental Theorem to be diff erentiable. It asserts the existence of a point c in (a, b) such thatg(b) − g(a)

b − a= g(c), i.e., such that

b

a

f (t ) dt = g(b)−

g(a) = f (c)·

(b−

a) ,

or f (c) =

b

a f ( x) dx b

a 1 dx

=

b

a f ( x) dx

b − a.

[1, Exercise 23, p. 445].

As mentioned in connection with the Mean Value Theorem in Math 140, the spirit of this

theorem is in the existence of the point c, not in the specific values that c takes. The theorem

is not constructive: it proves the existence without telling you how to find the point. (A

constructive proof of a theorem proves existence by providing a way of finding the point; the

proof we have given for the present theorem is not constructive.)

Example C.35 [1, Exercise 13, p. 445] The textbook asks you to prove that, if f is any contin-

uous function for which

3

1

f ( x) dx = 8, then f takes on the value 4 somewhere in the interval

1 ≤ x ≤ 3. The MVT for Integrals tells you that there is a point c such that 1 ≤ x ≤ 3 and

1

3 − 1

3

1

f ( x) dx = f (c)

and the left side of this equation is exactly 82

= 4.

What if we were to ask the same question with 4 replaced by another real number, e.g.,

4.001? The answer would now be negative: as a counterexample41 take the constant function f ( x) = 4. This function is continuous and has the desired area over the interval 1 ≤ x ≤ 3; but

it never assumes the value 4.001.

41an example to disprove a general statement: since the statement refers to all functions with a given property,

we can disprove it by exhibiting just one example.

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Average velocity Suppose that the position at time t of a particle moving along the x-axis is

f (t ). In [1, §3.7, p. 221] the textbook has defined the Average Velocity of the particle over atime interval a ≤ t ≤ b to be

∆ x

∆t =

f (b) − f (a)

b − a.

You may recall being told by your instructor that the use of the word average there was also

a generalization of the traditional meaning recalled above for the average of a finite set of

numbers. You can now see that, since

f (b) − f (a)

b − a=

b

a

d dt

f (t ) dt

b − a,

that use of the word is consistent with the use we have defined here. In other words, the Average

Velocity is, in fact, the average of the velocities. So the earlier use of the word “average” was,

though premature, consistent with the generalization that was planned. Previously “Average

Velocity” was a two-word name for a concept, and you would not be justified in treating the

first word as a modifier of the second; now you may interpret that as a conventional use of

language, where average is an adjective. When mathematicians name concepts we try to make

the nomenclature intuitive, and consistent with earlier usage.

6.5 Exercises

[1, Exercise 4, p. 445] Find the average value of the function g( x) = x2√

1 + x3 on the interval

[0, 2].Solution:

average = 1

2 − 0

2

0

x2√

1 + x3 dx

We apply the substitution u = x3, so du = 3 x2 dx and

average = 1

2 − 0

23

0

√ 1 + u · 1

3 du

= 1

6 · 2

3 (1+

u)

32 8

0

= 1

99

32

− 1

32

= 26

9

(We could also use the substitution u =√

1 + x3.) Had the problem asked us to “sketch

a rectangle whose area is the same as the area under the graph of g,” we could take a

rectangle based on the interval 0 ≤ x ≤ 2, with height 269

.)

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Example C.36 (taken from a problem book for students in Russian technical universities) [38,

Problem 1647, p. 126] Suppose that a trough has a parabolic cross-section with equation of the form y = K x2, and measures 1 meter across the top and is 1.5 meters deep. Determine the

average depth.

Solution: From the data we know the cross section passes through the point

12

, 32

, so 3

2 =

K

32

2, and K = 6. The depth of the trough at position x is 3

2 − 6 x2, so

average depth =

12

− 12

32 − 6 x2

dx

12 −−1

2

= 2

12

03

2 −6 x2 dx

by symmetry, since the integrand is an even function

= 2

3 x

2 − 2 x3

12

0

= 1

so the average depth is 1 meter, i.e., two-thirds of the way to the bottom of the trough.

6 Review

[1, Exercise 32, p. 447] “Let R1 be the region bounded by y = x2, y = 0, and x = b, where

b > 0. Let R2 be the region bounded by y = x2, x = 0, and y = b2, to the right of the line

x = 0.

1. “Is there a value of b such that R1 and R2 have the same area?

2. “Is there a value of b such that R1 sweeps out the same volume when rotated about

the x-axis and the y-axis?

3. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated

about the x-axis?

4. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated

about the y-axis?”

Solution: Note that the statement of the problem in the textbook is ambiguous, as the

description of R2 applies to one region to the right of the y-axis and another to the left;for that reason I have added the italicized words. Now the two regions combine to form

the rectangle ( x, y)| 0 ≤ x ≤ b, 0 ≤ y ≤ b2.

1. We are asked to investigate solutions of the equation b

0

x2 dx =

b

0

(b2 − x2) dx ⇔ b3

3 =

2b3

3 ,

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which has no positive solution. (We could have evaluated either or both of these

areas by integration along the y-axis, writing the equation of the right branch of theparabola as x =

√ y. With both integrations along that axis the equation would be

b2

0

(b − √ y) dy =

b2

0

√ y dy .

2. I will find the volume about the x-axis using washers, and the volume about the

y-axis using the same elements of area, using cylinders. Equating the two areas I

obtain the equation

b

0

π x22dx =

b

0

2π x

· x2 dx

⇔ π x

5

5

b

0

= 2π x4

4

b

0

which is equivalent to b4(2b − 5) = 0 and has one positive solution, b = 52

. (This

problem could be solved by evaluating either of the integrals in the “other” way.)

3. We have determined the volume swept out by rotating R1 about the x-axis to beπb5

5 . If we compute the volume obtained by rotating R2 about the x-axis, we find it

to be b

0

π

b22 −

x22

dx = π

b4 x − x5

5

b

0

= 4π

5 b5 using washers, and

b2

0

2π y · √ y dy = 2π · 25

y52

b2

0= 4π

5 b5 using cylindrical shells.

Equating the volume obtained by rotating R1 about the same axis to this, we obtainπb5

5 = 4πb5

5 , which has no positive solution.

4. The volume obtained by rotating R1 about the y-axis has been determined above to

be π2

b4. The volume obtained by rotating R2 about the y-axis is

b2

0

π(√

y)2 dy = π

b2

0

y dy = π

y2

2

b2

0

= π

2b4

using washers, or b

0

2π x

b2 − x2

dx = 2π

b2 x2

2 − x4

4

b

0

= π

2b4

using shells. Here we see that the two volumes of revolution are equal for all values

of the parameter b.

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Release Date: Wednesday, January 27th, 2010,

subject to further revision

Textbook Chapter 7. TECHNIQUES OF INTEGRATION.

C.12.1 §7.1 Integration by Parts

Earlier we developed the Substitution Rule for evaluating integrals, from the Chain Rule for

diff erentiation. Now we will develop another procedure for evaluation of integrals, called

Integration by Parts, based on the Product Rule for diff erentiation. As with the Substitution

Rule, this rule will be applicable to both definite and indefinite integrals. If does not aff ect the

(“independent”) variable and so there will be no change to limits in definite integrals.

Starting from the Product Rule,

d

dx[ f ( x) · g( x)] =

d

dx f ( x) · g( x) + f ( x) · d

dxg( x) ,

we integrate all 3 members with respect to x: d

dx[ f ( x) · g( x)] dx =

d

dx f ( x) · g( x)

dx +

f ( x) · d

dxg( x)

dx

and observe that the integral on the left is simply f ( x)

· g( x) + C . Moving the terms around

gives the Rule of Integration by Parts:

f ( x) · g( x) =

d

dx f ( x) · g( x)

dx +

f ( x) · d

dxg( x)

dx

⇔ d

dx f ( x) · g( x)

dx = f ( x) · g( x) −

f ( x) · d

dxg( x)

dx ,

f ( x) · d

dxg( x)

dx = f ( x) · g( x) −

d

dx f ( x) · g( x)

dx ,

or, compactly, f dg = f g

− g d f .

Traditionally we often name the functions u and v, or variations of these symbols42 — since

the solution of a specific problem may require multiple applications of integration by parts.

These equations are always true for diff erentiable functions, but we shall be applying them

when they tend to replace a difficult integral by one that is “easier” to evaluate. Usually the

42like u1, v1, U , V , u, v, . . .

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applications will be such that the integrand admits a “natural” factorization into two factors,

where either one of them becomes “simpler” under diff erentiation, or one becomes “simpler”under integration. I begin with an example where the use of Integration by Parts is obviously

indicated.

Example C.37 To integrate

xe x dx.

Solution: If we factorize the integrand into u = x, v = e x, then u is “simplified” by diff er-

entiation, since u = 1, while v is not “complicated” by diff erentiation, where v = e x. We

obtain xe x dx = xe x −

1e x dx

= xe x

− e x

+ C .

Example C.38 Sometimes the factorization of the integral is less than obvious. Consider the

problem of integrating ln x (cf. [1, Example 2, p. 454]). The “factorization” we choose is

u = ln x, dv = d x . (58)

By (58),

du = dx

x, and

v = x

where we have chose one convenient antiderivative. ln x dx = (ln x) x −

x · dx

x

= (ln x) x − x + C .

This is a derivation you should remember, as we often need an antiderivative of a logarithm. 43

Example C.39 Similar to the preceding example is the integration of arctan x (cf. [1, Example

5, p. 456]). Here again the function does not admit a well defined factorization, but we can try

(58) and obtain arctan x dx = x · arctan x −

x

1 + x2 dx .

43Perhaps one should remember it in the “more general” form ln | x| dx = x ln | x| − x + C .

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Now apply a substitution to the remaining integral, either

v = x2 ⇒ dv = 2 x dx or

w = 1 + x2 ⇒ dw = 2 x dx

Hence arctan x dx = x · arctan x −

1

wdw

= x · arctan x − ln |w| + C

= x · arctan x − ln |1 + x2| + C .

Of course, the absolute signs are not needed here because the argument of the logarithm func-tion is evidently non-negative.

There are several “standard” situations where we need to use integration by parts. Suppose

we need to integrate a function of one of the following forms, where P( x) is some polynomial:

P( x) · sin x, P( x) · cos x, P( x) · e x, P( x) · cosh x, P( x) · sinh x.

In each of these cases the derivative of the polynomial is “simpler” (here meaning “of lower

degree”), while the integral of the other factor is “not more complicated”. Repeated applica-

tions cause the polynomial to disappear, leaving only an integral involving the second factor.

Here again one should remember the derivation, but not memorize the formulæ, since they can

be easily reconstructed, and there are too many variations to memorize.

The rule of integration by parts need not be used in isolation: it may be necessary to

precede or follow its use by substitutions, and several applications of integration by parts could

be needed to complete the solution to a problem.

Two applications of Integration by Parts? We saw in connection with substitutions that

we might need to use the procedure more than once. Can that occur with Integration by Parts?

It can always occur, but, if not done carefully, the second iteration will reverse the action of the

first, and return us to the integral that we began with. Where the purpose of using Integration

by Parts is to truly simplify the integrand, then it is unlikely you will make this tactical error.

But consider the first example in the next paragraph.

Solving an equation to evaluate an indefinite integral Sometimes the application of inte-

gration by parts does not appear to make any progress, but a second or more applications may

eventually produce a constraint on the integral, which enables one to evaluate it. This idea will

be extended later.

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Example C.40 To evaluate e x cos x dx.

Solution: Set u = e x, dv = cos x, so that du = e x dx, v = sin x. Then e x cos x dx = e x sin x −

e x sin x dx .

The new integral is of similar difficulty to the old one. We apply integration by parts again:

U = e x, dV = sin x, dU = e x dx, V = −cos x: e x cos x dx = e x sin x −

e x sin x dx

= e x sin x− −e x cos x + e x cos dx

= e x sin x + e x cos x −

e x cos dx

Could this be an instance of the pitfall that was described in the preceding paragraph? Fortu-

nately not. The same integral appears on both sides of the equation, but with diff erent coeffi-

cients. If we move the integral from the right side to the left, we obtain

2

e x cos x dx = e x sin x + e x cos x dx + C (59)

which implies that

e x cos x dx = 1

2(e x sin x + e x cos x) + C (60)

Several comments are appropriate:

1. If we had taken the second application of Integration by Parts as U = sin x, dV = e x dx,

dU = cos x dx, V = e x, then we would have obtained e x cos x dx = e x sin x −

e x sin x dx

= e x sin x

− e x sin x

− e x cos dx= e x sin x − e x sin x +

e x cos dx =

e x cos x dx ,

which is a tautology. The statement obtained would not be incorrect, but we would have

wasted our time to obtain an end result that could have been stated immediately, and

which does not get us any closer to solving the problem at hand.

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2. Why did the constant of integration appear in equations (59), (60) but not in the preced-

ing equation? The preceding equation was of the form f ( x) dx = g( x) +

h( x) dx .

When both sides of an equation contain an indefinite integral, one understands that each

side is a set of functions which diff er by a constant; no more generality is achieved if we

add the notational comment that one may add a constant to one side. But, when one side

would consist of a single function — here it is e x sin x + e x cos x — the inclusion of a

constant changes the meaning from one specific function to all functions obtained from

it by adding any real number.

3. When we integrate the dv term, we appear to be selecting a specific antiderivative. Isthis restrictive? Should we be including a constant of integration? Try to convince

yourself that he selection of a specific antiderivative is not at all restrictive; that is, if you

do include a constant of integration, the changes to the formula will cancel each other

out. For that reason you should always choose the simplest antiderivative of v that is

convenient.

Reduction Formulæ Since integration by parts is helpful when the integrand is a product of

two functions, one of which does not become “more complicated” under diff erentiation, and

the other of which does not become “more complicated” under integration, this method should

be able to assist in the integration of the product of a polynomial and a sine, cosine, exponen-tial, or hyperbolic function. But what happens if we have to integrate other products of these

functions? In some cases the diff erentiations and integrations do not produce major changes

of simplicity, but they can lead to information that enables us to determine the integral, in the

way in which the preceding example was solved. This method is particularly important when

we wish to obtain an algorithm for evaluating certain general classes of integrals. Sometime

we will need to use the methods of the preceding paragraph in this connection, and sometimes

not.

Example C.41 Find a general formula for evaluating I (n) =

xne− x dx where n is a positive

integer.

Solution: Let u = xn, dv = e− x dx, so du = n xn−1 dx, v = −e− x. Then xne− x dx = − xne− x + n

xn−1e− x dx . (61)

or I (n) = − xne− x + n · I (n−1) . For any specific value of n, this formula may be used to evaluate

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the integral recursively. For example, if we need to know x5e− x dx, we have

x5e− x dx

= I (5) = − x5e− x + 5 · I (4)

= − x5e− x + 5(− x4e− x + 4 · I (3))

= − x5e− x + 5(− x4e− x + 4(− x3e− x + 3 · I (2)))

= − x5e− x + 5(− x4e− x + 4(− x3e− x + 3(− x2e− x + 2 · I (1))))

= − x5e− x + 5(− x4e− x + 4(− x3e− x + 3(− x2e− x + 2(− x1e− x + 1 · I (0)))))

= − x5e− x − 5 x4e− x − 5 · 4 x3e− x − 5 · 4 · 3 x2e− x − 5 · 4 · 3 · 2 xe− x − 5!e− x + C

= − x5 + 5 x4 + 5 · 4 x3 + 5 · 4 · 3 x2 + 5 · 4 · 3 · 2 x1 + 5 · 4 · 3 · 2 · 1 x0

e− x + C

Example C.42 [1, Exercise 50, p. 458] Let n be an integer greater than 1. Find a procedure

— i.e., a reduction formula — that can be used to evaluate sec n x.

Solution: secn x dx =

secn−2 x ·

sec2 x dx

Applying integration by parts with u = secn−2 x, dv = sec2 x dx, we set

du = (n

−2) secn−3 x

·sec x tan x dx = (n

−2) secn−2 x

·tan x dx, v = tan x .

We obtain secn x dx = secn−2 x · tan x − (n − 2)

tan2 x · secn−2 x dx

= secn−2 x · tan x − (n − 2)

(sec2 x − 1) · secn−2 x dx

= secn−2 x · tan x − (n − 2)

secn x − secn−2 x

dx

which may be solved for the desired indefinite integral. First we move all copies of the same

integral to one side of the equation:

(n − 1)

secn x dx = secn−2 x · tan x dx + (n − 2)

secn−2 x dx .

Dividing by n − 1 yields the reduction formula secn x dx =

1

n − 1 secn−2 x · tan x +

n − 2

n − 1

secn−2 x dx . (62)

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Since we know the integrals of the 0th and 2nd powers of the secant we can now find the

integral of any even positive power. For the odd positive powers we may reduce the problemto the integration of sec x, which has not been achieved yet.

Example C.43 ([7, Exercise 36, p. 480]) Follow a substitution by integration by parts to inte-

grate

x5e x2

dx.

Solution: The obvious substitution is u = x2, so du = 2 x dx. Then x5e x2

dx = 1

2

u2eu du

to which we apply integration by parts with

U = u2

, dV = eu

du, dU = 2u du, V = eu

= 1

2

u2eu − 2

u du

= 1

2u2eu −

ueu du

to which we apply integration by parts with

u = u, dv = eu du, du = du, v = eu

= 1

2u2eu −

ueu −

eu du

= 1

2 u

2

e

u

− (ue

u

− e

u

)+

C

=

1

2u2 − u + 1

eu + C

=

1

2 x4 − x2 + 1

e x2

+ C

Your solution is incomplete unless you express the integral in terms of the original variable.

(This section will be discussed further at the beginning of the next lecture.)

8/15/2019 Brown's Notes 2010




Release Date: Friday, January 27th, 2010, revised 01 February, 2010,

subject to further revision

C.13.1 §7.1 Integration by Parts (conclusion)

Recapitulation In the last lecture I introduced “Integration by Parts” — an integration tech-

nique related to the Product Rule of diff erentiation. I discussed the routine types of applications

we will see, as well as application to the integration of ln x and arctan x. I ended with an appli-

cation which required two successive applications of the technique, followed by the solution

of an equation.

7.1 Exercises

[1, Exercise 10, p. 457] “Evaluate the integral

arcsin x dx.”

Solution: Since you probably don’t know an antiderivative of the inverse sine, but do

know its derivative, we can try integration by parts with u = arcsin x, dv = dx, so

du = 1√

1 − x2dx and v = x. (I say try because not every attempt to apply one of the

integration rules will be successful: the rules are valid, and do convert the given integral

into another; but if you are unable to evaluate the new integral, you haven’t fully solved

the problem, and sometimes you may have made it even more difficult to solve.) arcsin x dx = x arcsin x −

x√

1 − x2dx

to which we apply the substitution w = 1 − x2, so dw = −2 x dx

= x arcsin x −

1√ w

−1

2 dw

= x arcsin x + 1

2 · 2

√ w + C

= x arcsin x +√

1 − x2 + C

Other substitutions that could have been used are w = x2, w = √ 1 − x2.

[1, Exercise 30, p. 458] Several methods suggest themselves for evaluating

1

0

r 3√ 4 + r 2

dr .

One solution using integration by parts can be based on

u = r 2, dv = r √

4 + r 2dr ⇒ du = 2r dr , v =

√ 4 + r 2 .

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Hence 1

0

r 3√ 4 + r 2

dr =r s

√ 4 + r 2

1

0− 2

1

0

r √

4 + r 2 dr

=r s

√ 4 + r 2

1

0− 2

1

2 · 2

3(4 + r 2)

32

1

0

=

r 2 − 8

3 ·

√ r 2 + 4

1

0

= 16 − 7

√ 5

3

which is approximately 0.115841, where the second integral was found by observationagain.

Some students may have difficulty observing the integral of dv, but a substitution could

make that phase easier.

For a solution that does not use integration by parts, try the substitution u =√

4 + r 2.

Then u · du = r dr , and r 2 = u2 − 4.

1

0

r 3√ 4 + r 2

dr =

√ 5

2

(u2 − 4) du

=u3

3 − 4u

√ 5

2

= 16 − 7

√ 5

3

as before.

Alternatively, one can use the substitution v = r 2 + 4, so dv = 2r dr . Then 1

0

r 3√ 4 + r 2

dr =

5

4

v − 4√ v

dv

2

=1

3 · v3

2 − 4 √ v5

4

=

5√

5

3 − 4

√ 5

−

8

3 − 8

,

which is the same value as obtained earlier twice.

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[1, Exercise 44, p. 458] 1. Prove the reduction formula cosn x dx = 1

ncosn−1 x · sin x + n − 1

n

cosn−2 x dx . (63)

2. Use Part 1 to evaluate

cos2 x dx .

3. Use Parts 1, 2 to evaluate

cos4 x ds.

Solution: First observe that the textbook has overlooked the restriction that n ≥ 1; the

equation is not meaningful when n = 0. We do know that, when n = 1, cos1 x dx =

1

1 sin x + 0. (64)

1. For n

≥ 2, we may take u = cosn−1 x, dv = cos x; then

du = (n − 1) cosn−2 x · (− sin x) dx

by the Chain Rule, and v = sin x. Let’s denote

cosn x dx by I n. Then a first

application of Integration by Parts yields cosn x dx = cosn−1 x · sin x + (n − 1)

cosn−2 x · sin2 x dx

= cosn−1 x · sin x + (n − 1)

cosn−2 x · (1 − cos2 x) dx

= cosn−1 x · sin x + (n − 1) ( I n−2 − I n) .

Collecting all terms in I n to the left side of the equation, and dividing by n − 1, weobtain

I n = 1

n· cosn−1 x · sin x +

n − 1

n I n−2

for n ≥ 2, as desired.

2. When n = 2, the reduction formula reduces to

I 2 = 1

2 cos x · sin x +

1

2

dx

= 1

2 cos x · sin x +

x

2 + C .

3. When n = 4, a second application gives

I 4 = 1

4 cos3 x · sin x +

3

4 I 2

= 1

4 cos3 x · sin x +

3

4

1

2 cos x · sin x +

x

2

+ C

= 1

4 · cos3 x sin x +

3

8 · cos x sin x +

3

8 · x + C

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C.13.2 §7.2 Trigonometric Integrals

This section is concerned with integrating functions that can be expressed simply in terms of

trigonometric functions. The techniques rely on heavy use of familiar trigonometric identities.

In particular, integration of the following types of functions is considered:

• products of non-negative powers of sin x and cos x

• products of non-negative powers of tan x and sec x

The integration of other functions that can be reduced to functions of these two types is also

considered. Strategies are developed for products of these types. However, students may well

be able to use techniques already seen to integrate certain integrals of these types in ways other

than those suggested here. To reiterate: YOU MAY BE ABLE TO INTEGRATE CERTAIN

FUNCTIONS OF THE TYPES LISTED BY USING OTHER METHODS. The objective inall of these procedures is to “simplify” the integration; where the function is a product of

trigonometric functions, this “simplification” is usually measured by a reduction in the total

degree of the product, i.e., in the total number of trigonometric factors. Since the total is finite,

repeated applications of such procedures will eventually result in successful integration.

Two ways of integrating

sin2 x d x and

cos2 x d x: One way of evaluating these integrals

is to use one of the double angle formulæ from trigonometry:

sin2θ = 2sin θ · cos θ (65)

cos2θ = cos

2

θ − sin

2

θ = 2 cos

2

θ − 1 = 1 − 2sin

2

θ (66)which follow from the formulæ for the sine and cosine of a sum. Two formulæ involving cos 2θ

may be rewritten as

sin2 θ = 1

2(1 − cos2θ ) (67)

cos2 θ = 1

2(1 + cos2θ ) (68)

From these we obtain

sin2 x dx = 1

2 (1

−cos2 x) dx

= 1

2

x − 1

2 sin 2 x

+ C

cos2 x dx = 1

2

(1 + cos2 x) dx

= 1

2

x +

1

2 sin 2 x

+ C

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As we saw earlier in connection with cos2 x dx, another way to evaluate these integrals is

through integration by parts. See [1, Exercises 43(a), 44(a)(b) p. 458] which describe howto use the reduction formula in [1, Example 6, p. 457] and an analogue for cosines for these

purposes.

Strategy for evaluating

sin m x · cos n x d x

0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing an

integral that may be evaluated immediately, the last 2 steps will lead to a simplification

in the integrand, after which the procedure is begun again.

1. If n is odd, use the identity cos2 x = 1

−sin2 x to convert all but one of the cosine factors

into a function of sines. Then apply the substitution u = sin x with du = cos x dx.

2. If m is odd, proceed analogously to the preceding: use the same identity to convert all but

one of the sine factors into a function of cosines. Then apply the substitution u = cos x,

with du = −sin x dx.

3. If both m and n are even, use the identities cos2 x = 12

(1+cos2 x) and sin2 x = 12

(1−cos2 x)

to express the integrand as a sum of products of sines and cosines of 2 x. The degrees

of these terms will be less than the degree of the integrand we started with, so that we

have simplified the problem, and can repeat the procedure until we have completed the

integration.

4. When both m and n are odd, other identities may also be used to simplify the integrand;

for example, we can use sin 2 x = 2sin x · cos x combined with the two double angle

formulæ mentioned immediately above.

Example C.44 To determine

sin4 x · cos2 x dx.

Solution: sin4 x · cos2 x dx =

1 − cos2 x

2

2

· 1 + cos2 x

2 dx

= 1

−cos2 x

−cos2 2 x + cos3 2 x

8

dx

= 1

8 · x − 1

16 sin 2 x −

1 + cos4 x

16 dx

+ 1

16

1 − sin2 2 x

· d

dxsin(2 x) dx

etc. (Use the substitution u = sin 2 x to evaluate the last integral.)

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(to be continued)

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C.14 Supplementary Notes for the Lecture of February 01st, 2010

Release Date: Monday, February 01st, 2010,

subject to revision

C.14.1 §7.2 Trigonometric Integrals (conclusion)


sin m x · cos n x d x (Repetition from last lecture).

0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing an

integral that may be evaluated immediately, the last 2 steps will lead to a simplification

in the integrand, after which the procedure is begun again.

1. If n is odd, use the identity cos2

x = 1 − sin2

x to convert all but one of the cosine factorsinto a function of sines. Then apply the substitution u = sin x with du = cos x dx.

2. If m is odd, proceed analogously to the preceding: use the same identity to convert all but

one of the sine factors into a function of cosines. Then apply the substitution u = cos x,

with du = −sin x dx.

3. If both m and n are even, use the identities cos2 x = 12

(1+cos2 x) and sin2 x = 12

(1−cos2 x)

to express the integrand as a sum of products of sines and cosines of 2 x. The degrees

of these terms will be less than the degree of the integrand we started with, so that we

have simplified the problem, and can repeat the procedure until we have completed the

integration.

4. When both m and n are odd, other identities may also be used to simplify the integrand;

for example, we can use sin 2 x = 2sin x · cos x combined with the two double angle

formulæ mentioned immediately above.

Example C.45 To determine

sin4 x · cos2 x dx.

Solution: sin4 x · cos2 x dx =

1 − cos2 x

2

2

· 1 + cos2 x

2 dx

= 1

−cos2 x

−cos2 2 x + cos3 2 x

8

dx

= 1

8 · x − 1

16 sin 2 x −

1 + cos4 x

16 dx

+ 1

16

1 − sin2 2 x

· d

dxsin(2 x) dx

etc. (Use the substitution u = sin 2 x to evaluate the last integral.)

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tan m x

·sec n x d x. Earlier I discussed an algorithm for integrating

products of non-negative integer powers of sines and cosines. I now consider another important

class of products of trigonometric functions.

1. The basis of the solution I propose for most of these cases is to use the facts that

d (tan x) = sec2 x dx

d (sec x) = sec x tan x dx

sec2 x = tan2 x + 1

2. If n is even and n ≥ 2, then we can replace the factor secn x by

sec2 x

n−22 d

dxtan x =

tan2 x + 1

n−22 · d

dxtan x

which, when multiplied by tanm x — where m has any value — yields an integral that is

simplified by the substitution u = tan x.

3. If n = 0 we have

tanm x dx:

(a) When m = 0, the solution is

dx = x + C .

(b) When m = 1, the function integrates as tan x = ln

|sec x

|+ C =

−ln

|cos x

|+ C .

(c) When m ≥ 2, one may detach 2 powers of the tangent from the others, replacing

them by sec2 x − 1, thereby reducing the problem to the integration of a lower

power of the tangent, and an integral of the form

tanm−2 x · sec2 x dx, which can

be integrated following a substitution u = tan x.

Henceforth we may assume that n is odd.

4. If m is odd, we may detach one power of tan x from tanm x and one power of sec x from

secn x, and write the integrand as

tan2 x

m

−1

2 · secn−1 x · d dx

sec x

which is equal to sec2 x − 1

m−12 · secn−1 x · d

dxsec x

and can be integrated after a substitution u = sec x.

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5. In the only remaining cases m is even and n is odd. One method would be to transform

the entire integrand into powers of sec x and then to use the reduction formula [1, Exer-cise 50, p. 458] to reduce everything to the problem of integrating sec x. Other reductions

are possible: for example, into a function expressible in terms of cos x, with one factor

cos x left over — this would permit a substitution u = cos x that converts the problem to

the type we shall meet in [1, §7.3]; alternatively, one may express the integrand as a sum

of powers of tan x, and develop a reduction formula for them (cf. [1, endpapers, item 75

p. 9.]).

The integral

tan x dx. As observed above,

tan x dx = − ln | cos x| + C = ln | sec x| + C .

The integral

sec x dx. The textbook observes that

sec x dx = ln | sec x + tan x| + C .

Do not forget the absolute value signs when you quote these results, although you may expect

that in many of the problems you consider, where sec x + tan x is positive, omission of the signs

may not produce any visible error.While it is possible to prove the validity of the preceding equation simply by diff erentiation

of the alleged antiderivative, a more direct proof requires some ingenuity.44


cot m x · csc n x d x. Analogues of the preceding techniques can

simplify integrals of these types.

44One way to derive this result is to observe that

sec x = 1

cos x=

cos x

1

−sin2 x

= cos x(1 − sin x)(1 + sin x)

= 1

2

cos x

1 − sin x +

cos x

1 + sin x

using ideas of partial fractions that we shall be meeting in [1, §7.4]. Each of these summands can be integrated

as a logarithm (See [33, pp. 505-506 ].)

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“Other” trigonometric identities. By adding and subtracting the expansions of sin( A ± B)

and cos( A ± B), one may obtain the following identities

sin A · cos B = 1

2(sin( A − B) + sin( A + B)) (69)

sin A · sin B = 1

2(cos( A − B) − cos( A + B)) (70)

cos A · cos B = 1

2(cos( A − B) + cos( A + B)) (71)

These identities permit the derivation of another class of useful identities. If we replace A − B

and A + B respectively by U and V — equivalently, if we replace A by V + U

2 and B by

V − U

2 ,

we obtain

sin U + sin V = sin V + U 2

· cos V − U 2

(72)

sin U − sin V = sin −V + U

2 · cos

V + U

2 (73)

cos U + cos V = cos V + U

2 · cos

V − U

2 (74)

cos U − cos V = sin V + U

2 · sin

V − U

2 (75)

Example C.46 Integrate

(sin 50 x · cos 12 x) dx.

Solution: (One possible solution) (sin 50 x · cos 12 x) dx =

1

2

(sin 38 x + sin 62 x) dx

= −1

2 · 1

38 · cos 38 x − 1

2 · 1

62 · cos 62 x + C

Example C.47 ([7, Exercise 62, p. 488]) Find the volume obtained by rotating the region

bounded by the curves y = cos x, x = 0, y = 0, x = π2

about the axis y = 1.

Solution: I give a solution, by washers. (The problem can be solved by cylindrical shells also,

but that method is much more difficult.)

Volume = π

2

012

−(1

−cos x)2 dx

=

π2

0

(2 cos x − cos2 x) dx

=

2sin x − x

2 − sin 2 x

4

π2

0

= π

2 − π

4

.

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7.2 Exercises

[1, Exercise 30, p. 466] To evaluate

π3 0

tan5 x sec6 x dx.

Solution: This integral can be evaluated in at least two diff erent ways.

1. Since the exponent of the secant is even, we can use a substitution u = tan x, so

du = sec2 x dx.

π3

0

tan5 x sec6 x dx = tan π

3

tan0

u5

u2 + 1

2du

=

√ 3

0

u5

u2 + 12

du

=

√ 3

0

u9 + 2u7 + u5

2du

=

u10

10 +

u8

4 +

u6

6

√ 3

0

= 243

10 +

81

4 +

27

6 =

981

20

2. Alternatively, since the exponent of the tangent is odd, we can use a substitution

v = sec x, so dv = sec x · tan x dx:

π3

0

tan5 x sec6 x dx =

2

1

(v2 − 1)2v5 dv

=

v10

10 − v8

4 +

v6

6

2

1

= 981

20

[1, Exercise 48, p. 466] To evaluate

dx

cos x − 1 .

Solution: This integral is not of any of the forms shown in the chapter, so some ingenuity

is needed. I give more than one solution.

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1. First solution, using double angle formula. When you see cos x − 1, that should

suggest the identity cos 2θ = 1 − 2sin

2

θ . Applying that identity here, with θ =

x

2 ,yields

dx

cos x − 1 =

dx

−2 sin2 x2

= −1

2

csc2 x

2 d x = cot

x

2 + C

2. Second solution, using the identity sin2 x + cos2 x = 1. The idea here resembles

the rationalization of fractions involving square roots, seen earlier.

dx

cos x − 1 =

1

cos x − 1 · cos x + 1

cos x + 1

dx

=

cos x + 1

− sin2 xdx

= −

cos x

sin2 xdx −

csc2 x dx

The first integral may be evaluated in several ways, for example by using the sub-

stitution u = sin x, so du = −cos x dx. That integral becomes

cos x

sin2 xdx = du

u2

= −1

u+ C 1 = −csc x + C 1.

The second integral can be seen immediately to be −cot x + C 2. The two together

give us dx

cos x − 1 = csc x + cot x + C .

But, are the two answers equal? This can be seen through trigonometric identities. For

example

csc x + cot x = 1 + cos x

sin x =

2cos2 x

22 sin x

2 · cos x

2= cot

x

2

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C.15 Supplementary Notes for the Lecture of February 03rd, 2010

Release Date: Wednesday, February 03rd, 2010,

subject to revision

C.15.1 §7.3 Trigonometric Substitution

In this section we describe a type of substitutions which simplify certain commonly met inte-

grals. We approach these substitutions in the reverse direction from that used earlier. Whereas

earlier we investigated substitutions of the form u = g( x), this time we will usually formulate

our substitutions first in the form x = h(u); that is, we will look for a substitution that will

simplify the integrand, and then try to implement it. We usually proceed “mechanically” in

these problems; but, in principle, we are postulating the existence of an inverse function, and

should be checking that there really is an inverse whenever we use the method. I will try to

go through some of those steps in the first examples we consider, but, in practice, we often

become careless and don’t check everything unless something indicates a problem. This is

unwise; the reason that it “works” is that we usually confine the substitutions to certain well

understood pairs of functions / inverses, where all of the snags have already been worked out.

Example C.48 [1, Exercise 4, p. 472] To evaluate

2√

3

0

x3

√ 16 − x2

dx.

Solution: A first look at the integrand suggests that the complication comes from the ex-

pression√

16 − x2 in the denominator. We could try to simplify by giving this a new name,

u =

√ 16 − x

2

. That leads todu = − x√

16 − x2dx ⇒ x dx = −u du

and so the integral becomes 2

4

(u2 − 16) du =

u3

3 − 16u

2

4

= 40

3 .

But we would like to illustrate the notion of trigonometric substitution here (in a prob-

lem where it isn’t really needed!) The component√

16 − x2 suggests that we might wish

to interpret the problem geometrically, with this component arising from an application of

Pythagoras’s Theorem to a right-angled triangle; equivalently, from the identity that

sin2 θ + cos2 θ = 1 .

To do this, we can first divide out the factor 16, which, when it leaves the square root, will

reappear outside as 4:√

16 − x2 = 4

1 − x2

16 = 4

1 −

x

4

2

.

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This suggests a substitution

x

4 = sin v or, equivalently, x = 4 sin v

that will make x

4 into a sine or a cosine — either one will work. When we express the substi-

tution that way we are really working with the inverse — to conform with our earlier theory

about substitutions we should be expressing the new variable in terms of the old; so we should

be starting with

v = arcsin x

4 .

The range of values of x that interest us is 0

≤ x

≤ 2

√ 3, equivalently 0

≤ x

4 ≤

√ 3

2

, and

we know that the inverse sine function is defined over this domain. So, beginning with the

substitution above, we obtain

dv = 1

1 −

x4

2 · 1

4 d x =

dx√ 16 − x2

,

and the integral transforms as follows:

2

√ 3

0

x3

√ 16

− x2

dx =

arcsin 2

√ 3

4

arcsin 04

64 sin3 v dv

= 64

π

3

0

sin3 v dv

which we proceed to evaluate using the methods of the preceding section:

64

π3

0

sin3 v dv = 64

π3

0

sin v

1 − cos2 v

dv

= 64

− cos v +

1

3 cos3 v

π3

0

= 40

3

In practice this method works smoothly, but one must occasionally be careful about theevaluation of the inverse function, remembering our original definitions of the restricted inter-

val where the inverse was taken. The method is indicated whenever we see expressions like√ 1 − x2 or, more generally,

√ a2 − x2, since we can transform the latter into the former by di-

vision by a positive real number. The inverse cosine could be used instead of the inverse sine,

and the results would be no more difficult.

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Example C.49 ([7, Exercise 8, p. 494]) Evaluate √

x2 − 4

x4

dx.

Solution: We need a substitution that will convert x2 to 4 times the square of a secant. 45 One

way to achieve this is to make u have the property that

x = 2 sec u ;

so the actual substitution will be

u = arcsec x

2 ,

which implies that

dx = 2 sec u tan u du .

Under the substitution what happens to

√ x

2

− 4? It becomes

√ 4tan

2

u, i.e., 2| tan u|. Do weneed the absolute signs? Recall that the inverse secant takes its values in the two intervals

0 ≤ u < π2

and π ≤ u ≤ 3π2

. In these two intervals the tangent is always positive, so the absolute

signs may be dropped.

√ x2 − 4

x4 dx =

2 tan u

16 sec4 u· 2sec u · tan u du

= 1

4

sin2 u · cos u du

= 1

12 sin3 u + C

We can’t leave the answer in this form, as it must be expressed in terms of the original variable

x. Since

sin u = tan u · cos u

= tan u

sec u

=

12

√ x2 − 4

x2

,

112

sin3 u = 112

· ( x2 − 4)3

2

x3

Hence √ x2 − 4

x4 dx =

( x2 − 4)32

12 x3 + C .

45Alternatively, we could make x2 four times the square of a hyperbolic cosine.

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Example C.50 I have avoided computing the area of a disk until now. It is trivial with a

trigonometric substitution. If the disk is the set of points ( x, y) such that x2

+ y2

≤ R2

, then it isthe region bounded by the 2 graphs y = ±√

R2 − x2. Its area is R

− R

√ R2 − x2 −

−

√ R2 − x2

dx =

R

− R

2√

R2 − x2 dx

=

R

0

4√

R2 − x2 dx ,

since the integrand is even, and the interval of integration is symmetric around 0. The sub-

stitution u = cos−1 x R

for 0 ≤ x ≤ R implies that cos u = x R

or x = R cos u, where the

interval of integration is now from u = cos−1 0 = π2

to u = cos−1 1 = 0; dx = − R sin u du;

√ R2 − x2 = R| sin u| = R sin u, since the sine is positive in this interval. The integral transformsto

4

R

0

√ R2 − x2 dx = 4

0

π2

| R sin u|(− R sin u) du

= −4 R2

0

π2

sin2 u du

since the sine is positive for 0 ≤ u ≤ π2

= −4 R2

u

2 − sin

2u

4 0

π2

= −4 R20 − π

4

= π R2

If we had not used the evenness of the integrand to reduce the original problem to integrating

over the interval 0 ≤ x ≤ R, there would have been a serious difficulty. That is because theinverse cosine function takes its values between 0 and π.

If we had naively carried through the substitution over the entire interval − R ≤ x ≤ R,

we would have obtained

4

R

− R R2 − x2 dx = 4

− π2

π2


= 4

0

π2

| R sin u|(− R sin u) du + 4

− π2

0


= 4

0

π2

( R sin u)(− R sin u) du + 4

− π2

0

(− R sin u)(− R sin u) du

= −4 R2

0

π2

sin2 u du + 4 R2

− π2

0

sin2 u du

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=

−4 R2

0

−π

2

sin2 u du + 4 R2 π

2

0

sin2 u du .

But note that the integrand is an even function, so the two integrals would cancel, and

the answer would be 0. This is clearly incorrect, but what went wrong? The error was

in attempting to replace x by R cos u: u is not uniquely defined for − π2 ≤ x ≤ π

2! We

could, though, have used a substitution u = sin−1 x R

over the full interval, and the correct

answer would have been obtained. We would have defined u = arcsin x R

, so sin u = x R

,

cos u du = 1 R · dx.

4

R

− R

R2 − x2 dx = 4

π2

− π2

( R cos u) R cos u du

= 4 R2 π

2

− π2

1 + cos 2u

2 du

= 2 R2

u +

sin wu

2

π2

− π2

= 2 R2

π

2 + 0

−−π

2 + 0

= π R2 .

8/15/2019 Brown's Notes 2010



C.16 Supplementary Notes for the Lecture of February 05th, 2010

Release Date: Friday, February 05th, 2010

C.16.1 §7.3 Trigonometric Substitution (conclusion)

Table of trigonometric substitutions I can expand the table of substitutions of this type

given in the textbook:

Expression Inverse Substitution Substitution Identity√

a2 − x2 θ = arcsin xa

x = a sin θ 1 − sin2 θ = cos2 θ

(−a ≤ x ≤ +a) −π2 ≤ θ ≤ π

2√ a2 − x2 θ = arccos xa

x = a cos θ 1 − cos2 θ = sin2 θ

(−a ≤ x ≤ +a) (0 ≤ θ ≤ π)

√ a2 + x2 θ = arctan x

a x = a tan θ 1 + tan2 θ = sec2 θ

(−∞ < x < +∞)−π

2 ≤ θ ≤ π

2

√

x2 − a2 θ = arcsec xa

x = a sec θ sec2 θ − 1 = tan2 θ

(−∞ < x ≤ −a orπ ≤ θ < 3π

2 or

a ≤ x < ∞) 0 ≤ θ < π2

I have shown both sine and cosine versions of the first substitution, and could similarly haveproduced a cotangent version of the tangent substitution, and a cosecant version of the secant

substitution; in practice the latter two variants are not used frequently, and usually o ff er no

advantages over the substitutions shown.

Another trigonometric substitution will be discussed in the next section. It is used not to

simplify a square root, but to simplify denominator terms of the form x2 + a2.

7.3 Exercises

[1, Exercise 24, p. 472] Evaluate 1√ t 2

−6t + 13

dt .

Solution: I shall complete the square of the quadratic polynomial in the denominator

in order that, after a first substitution, this integral will be of a type that we recognize.

Since

t 2 − 6t + 13 = (t − 3)2 + 4 = 4

t − 3

2

2

+ 1

,

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a first substitution u = t − 3

2 , which implies that dt = 2 du, could be applied:

dt √ t 2 − 6t + 13

=

du√

u2 + 1.

Now I take u = tan θ , i.e., θ = arctan u. Thus −π2

< θ < + π2

.

du√

u2 + 1=

sec2 θ

| sec θ | d θ

=

| sec θ | d θ

=

sec θ d θ since − π

2 < θ < π

2

= ln | sec θ + tan θ | + C

= ln

± 1 + tan2 θ 1

2+ tan θ

+ C

where the + sign is taken since |θ | < π2

and the cosine

and secant are positive in Quadrants ##1,4,

= ln

1 + u2 1

2+ u

+ C

= ln 1 + t − 3

2

2

12

+ t − 3

2 + C

= ln

t 2 − 6t + 13 1

2+ (t − 3)

+ (C − ln2)

And I could rename the constant with a single letter, e.g., K = C − ln2.

Hyperbolic substitutions It is possible to achieve the same sorts of simplifications by using

inverse hyperbolic functions. Since we have spent little time in becoming comfortable with

the hyperbolic functions, I will not discuss these substitutions in general, but may apply them

in specific cases.

Example C.51 This is [1, Exercise 31, p. 472].

1. Use trigonometric substitution to show that dx√

x2 + a2= ln

x +

√ x2 + a2

+ C

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Example C.52 ([7, Exercise 26, p. 494]) Evaluate x2

√ 4 x − x2

dx.

Solution: Completion of the square yields

4 x − x2 = −( x2 − 4 x) = 4 − ( x2 − 4 x + 4) = 4 − ( x − 2)2 = 22

1 − x − 2

2

2 .

Hence x2

√ 4 x − x2

dx =

x2

2

1 −

x−2

2

2dx

= (2u + 2)2

2 √ 1 − u2

2 du

under substitution u = x − 2

2

= 4

sin2 v + 2sin v + 1

dv

under substitution v = arcsin u = arcsin x − 2

2

= 4

1 − cos2v

2 dv − 8cos v + 4v

= 6v − sin 2v − 8cos v + C

= 6v − 2(4 + sin v)cos v + C

Now the arcsine function takes values between −π2

and π2

, in which interval the cosine is posi-

tive; hence

cos v = +

1 − sin2 v =

√ 1 − u2 =

1

2

√ 4 x − x2 .

We conclude that x2

√ 4 x − x2

dx = 6 arcsin x − 2

2 − x + 6

2 ·

√ 4 x − x2 + C .

Example C.53 ([7, Exercise 28, p. 494]) Evaluate 1

5 − 4 x − x2 5

2 dx.

Solution: As in the preceding example, I shall complete the square of the quadratic polynomial

in the denominator in order that, after a first substitution, this integral will be of a type that we

recognize. Since

5 − 4 x − x2 = −( x2 + 4 x − 5) = 9 − ( x + 2)2 = 9

1 − x + 2

3

2 ,

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a first substitution u = x + 2

3 , which implies that d x = 3 du, could be applied:

1

5 − 4 x − x2 5

2

dx = 1

81

du

(1 − u2)52

.

Now we can apply a second substitution u = sin φ — actually it is φ = arcsin u – where

d φ = du1 − u2

12

. We obtain

1

81

du

(1 − u2)52

= 1

81

d φ

cos4 φ

= 181

sec4 φ d φ

= 1

81

tan2 φ + 1

sec2 φ d φ

= 1

81

1

3 tan3 φ + tan φ

+ C .

The function φ was defined to be an arcsine, so its values are in the interval −π2 ≤ φ ≤ π

2, in

which the cosine is positive. Hence

tan φ = sin φ

cos φ=

sin φ

+

1 − sin2 φ

= u

√ 1 − u2

=

x+23

1 −

x+23

2=

x + 2√ 9 − 4 x − x2

and 1

81

1

3 tan3 φ + tan φ

=

( x + 2)3

2435 − 4 x − x2

32

+ x + 2

815 − 4 x − x2

12

8/15/2019 Brown's Notes 2010




Release Date: Monday, February 08th, 2010

C.17.1 §7.4 Integration of Rational Functions by Partial Fractions

In this section we shall see that an entire class of functions can be integrated by a systematic

algebraic decomposition procedure, followed by specific methods for the various components

into which we decompose the functions.

An example to illustrate the general procedure Before describing the general procedure

let us consider some examples that will make it easier to comprehend.

[1, Exercise 14, p. 482] Evaluate the integral

1

( x + a)( x + b) d x.

Solution: There will be two quite diff erent solutions, depending on whether a = b.

Case a = b: This can be integrated directly by observation, or by using the substitution

u = x + a. 1

( x + a)2 d x = − 1

x + a+ C .

Case a b: We try to decompose the integrand into the sum of fractions whose denom-

inators are, respectively, x + a, and x + b. The degrees of the numerators must be

less than these polynomials of degree 1, so they have to have degree 0, i.e., theyhave to be constants. So suppose that

1

( x + a)( x + b) =

α

x + a+

β

x + b.

It can be shown algebraically that such a decomposition is always possible; all that

is missing is to know the values of the constants α and β. The last equation, after

multiplication of both sides by ( x + a)( x + b), yields

1 = α · ( x + b) + β · ( x + a) .

Here are 2 ways to find α and β:

1. Express both sides of the equation as polynomials in x, and equate the coeffi-

cients of corresponding powers of x. The left side is 1 x0 + 0 x1. Thus

degree 0: 1 = α · b + β · a

degree 1: 0 = α + β

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Solving these equations gives

α = 1

b − a

β = 1

a − b

2. This equation must be true for all values of x. Give x “convenient” values to

obtain equations in α and β, and solve those equations. Two “convenient” val-

ues are x = −a and x = −b. They give, respectively, the following equations:

1 = α · (−a + b) + β · 0 ⇒ α = 1

b − a

1 = β · (0) + β · (−b + a) ⇒ β =

1

a − b

We may now complete the integration: 1

( x + a)( x + b) d x =

1

a − b

− 1

x + a+

1

x + b

dx

= 1

a − b(− ln( x + a) + ln( x + b)) + C

= 1

a − bln

| x + b|| x + a| + C

=

1

a − b ln x + b

x + a

+ C

This family of functions could be written in other ways. For example, since C · (a−b) = ln eC ·(a−b), we could call eC ·(a−b) K , and write the family as

1

a − b

ln

x + b

x + a

+ ln K

=

1

a − bln

K · x + b

x + a

where K serves as the constant of integration.

Polynomials Recall that a polynomial [1, p. 28] is a function of the form

P( x) = an xn + an−1 xn−1 + · · · + a1 x1 + a0 x0

where a0, . . . , an are real numbers, called the coe fficients of the polynomial. Except for the

zero polynomial, which is the constant function 0, all polynomials will have a largest integer

m such that am 0; m is called the degree of the (non-zero) polynomial, and am is called the

leading coefficient; a0 is called the constant term. We usually write x0 simply as 1, and x1

simply as x.

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Rational Functions A rational function is a ratio of polynomials of the form

A( x) B( x)

where A and B are polynomial functions. The function will have as discontinuities the roots of

B, if there are any.

The goal of partial fraction decompositions In a partial fraction decomposition we express

a ratio of polynomials as a sum of “partial” fractions — fractions that have special properties.

In these “partial” fractions the denominator polynomials will always be powers of one poly-

nomial that is irreducible, i.e., it cannot be factored further (unless we move beyond the real

number system to, for example, the complex number system — with which you are not ex-

pected to be familiar). It is a theorem of algebra that

Theorem C.54 If a non-zero real polynomial is irreducible then it must be of one of the fol-

lowing forms:

1. a non-zero constant

2. a polynomial of degree 1, of the form ax + b, where a 0.

3. a polynomial of degree 2 without real roots, i.e., of the form a x2 + bx + c where a 0

and b2 < 4ac.

The only type of ratio to which we apply this decomposition is one where the degree of the

numerator is strictly less than the degree of the denominator; if the rational function that we

start with does not have this property, then we will have to preprocess it to obtain a ratio of this

type; after the procedure of partial fraction decomposition is applied, the resulting “partial”

fractions will have the same property — that the degree of their numerator will always be

less than that of the denominator. We will then show that we are able to integrate all partial

fractions of these types, and so we will be able to integrate all rational polynomials A( x)

B( x).

The procedure we shall develop depends on having such a factorization of the denominator

of the given ratio into irreducible polynomials. In this course we shall not be concerned with

finding the factorization itself, beyond knowing

• that P(a) = 0 ⇒ x − a divides P( x) — the so-called “Factor Theorem”;

• how to factorize quadratic polynomials, both by using the quadratic formula, and by

“completing the square”;

• that, for any positive integer n,

an − bn = (a − b)(an−1 + an−2b + . . . + abn−2 + bn−1)

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8/15/2019 Brown's Notes 2010



Suppose that, instead of the given integral, we wished to integrate x3 + (a + b) x2 + abx + 1

( x + a)( x + b) dx .

This function cannot be expanded into partial fractions until it is arranged that the degree

of the numerator — presently 3 — be less than the degree of the denominator — 2. If

we divide the denominator into the numerator, we find that

x3 + (a + b) x2 + abx + 1 = x · ( x + a)( x + b) + 1 .

Hence the integral may be expressed as

x +

1

( x + a)( x + b)

dx, and its value will be

x2

2 +

1

a − bln

x + b

x + a

+ C

What would have happened if we had attempted to expand this function into partial frac-

tions? No such decomposition can exist. Using the first method (equating coe fficients of

corresponding powers) we would have obtained 4 equations which would overdetermine

the constants α and β, and which would be inconsistent — there would be no solution.

But, if we used the second method, and didn’t take enough equations, we might not no-

tice that there was an inconsistency, and the alleged partial fraction would be incorrect.

The general procedure. The general procedure has many facets, and some will not be ex-

plicitly discussed in the lecture; you are expected to work many problems in this section of the

textbook, not only problems discussed in the lecture or appearing on WeBWorK or the quizzes.

Remember the steps we need to follow:

1. The first step in any of these problems is to ensure that the degree of the

numerator must be less than that of the denominator. If it is not, you must

divide the denominator into the numerator, obtaining a quotient and a re-

mainder. The quotient integrates as a polynomial, and we are left with the

ratio of the remainder to the denominator, for which the methods we are

discussing will enable a complete solution.

2. Factorize the denominator into a product of polynomials of degrees 1 and 2which are irreducible, i.e., which do not factorize further into lower degree

polynomials. The quadratic factors will be those having no real roots. Group

factors which are exactly the same together, so that your denominator is a

product of powers of distinct, irreducible polynomials.

3. Then the fraction must be decomposed into partial fractions using methods

we have illustrated before, and continue in this lecture.

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Example C.55 [1, Exercise 28, p. 482] Evaluate the integral x2 − 2 x − 1

( x − 1)2

( x2

+ 1)

d x.

Solution: The denominator has two distinct irreducible factors: x − 1, of degree 1 and multi-

plicity 2, and x2 + 1, an irreducible quadratic factor of degree 2 and multiplicity 1. The degree

of the numerator is less than 4, which is the degree of the denominator, so there is no need for

any long division. One type of partial fraction decomposition is of the form

x2 − 2 x − 1

( x − 1)2( x2 + 1) =

α x + β

( x − 1)2 +

γ x + δ

x2 + 1

in which we take the most general numerator in each case, of degree less than the degree of the

denominator. This is not the most useful type of partial fraction decomposition, but we will

carry this step out and then improve on it. Multiplying both sides by ( x

−1)2( x2 + 1), we obtain

a polynomial equation

x2 − 2 x − 1 = (α x + β)( x2 + 1) + (γ x + δ)( x2 − 2 x + 1)

⇔ x2 − 2 x − 1 = (α + γ ) x3 + ( β − 2γ + δ) x2 + (α + γ − 2δ) x + ( β + δ)

Equating coefficients of corresponding powers of x yields equations corresponding to the terms

of degrees 3, 2, 1, 0:

α + γ = 0

β − 2γ + δ = 1

α + γ − 2δ = −2 β + δ = −1

which we proceed to solve, obtaining

(α,β,γ,δ) = (1, −2, −1, 1) ,

so the decomposition is

x2 − 2 x − 1

( x − 1)2( x2 + 1) =

x − 2

( x − 1)2 +

− x + 1

x2 + 1 .

(to be continued)

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Release Date: Wednesday, February 10th, 2010

C.18.1 §7.4 Integration of Rational Functions by Partial Fractions (conclusion)

The general procedure (continued).

Example C.56 (continued from Example2010:31 on page 3116 of these notes, cf. [1, Exercise

28, p. 482]) We shall see that we will be able to integrate the second summand immediately, by

breaking it into two parts. We can integrate the first summand if we first apply the substitution

u = x − 1:

x − 2( x − 1)2 dx =

u − 1

u2 du =

1u

− 1u2

du

= ln |u| + 1

u + C

= ln | x − 1| + 1

x − 1 + C

In practice we anticipate the results of this substitution by refining the partial fraction decom-

position: in place of a summand of the form

an−1 xn−1 + an−2 x

n−2 + . . . + a0

( x−

a)n

we repeatedly divide x − a into the numerator, so that we can express the numerator as a sum

of powers of x −a; then we decompose the fraction and divide excess powers of x −a, to obtain

a decomposition of the form

bn

( x − a)n +

bn−1

( x − a)n−1 . . . +

b1

x − a

in which we can integrate each of the summands at sight. This decomposition can be accom-

plished as a second phase of partial fraction decomposition, or immediately, by assuming a

decomposition of the form

x2 − 2 x − 1

( x − 1)2( x2 + 1) =

ζ

( x − 1)2 +

η

x − 1 +

γ x + δ

x2 + 1

which leads to the polynomial identity

x2 − 2 x − 1 = ζ ( x2 + 1) + η( x − 1)( x2 + 1) + (γ x + δ)( x − 1)2

⇔ x2 − 2 x − 1 = (η + γ ) x3 + (ζ − η − 2γ + δ) x2 + (η + γ − 2δ) x + (ζ − η + δ)

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which we proceed to solve, obtaining

(ζ ,η ,γ,δ) = (−1, 1, −1, 1) ,

x2 − 2 x − 1

( x − 1)2( x2 + 1) d x

=

−1

( x − 1)2 dx +

1

( x − 1)1 d x +

− x

x2 + 1 d x +

1

x2 + 1 d x

= 1

x − 1 + ln | x − 1| − 1

2 ln | x2 + 1| + arctan x + C

= 1

x − 1

+ ln

| x

−1

| − 1

2

ln( x2 + 1) + arctan x + C

= 1

x − 1 + ln

| x − 1|√ x2 + 1

+ arctan x + C

7.4 Exercises

[1, Example 19, p. 482] Evaluate the integral

1

( x + 5)2( x − 1) d x .

Solution: Since the degree of the numerator is 0, and that of the denominator is 2 + 1 =

3 > 0, we can skip the long division step. We have to decompose the fraction into a sum

of partial fractions, which we may take to be of the form

1

( x + 5)2( x − 1) =

A

( x + 5)2 +

B

( x + 5)1 +

C

( x − 1)1 .

Multiplying both sides by ( x + 5)2( x − 1), we obtain

1 = A( x − 1) + B( x − 1)( x + 5) + C ( x + 5)2 . (76)

i.e.,

0 x2 + 0 x1 + 1 x0 = A( x − 1) + B( x2 + 4 x − 5) + C ( x2 + 10 x + 25) . (77)

One way to obtain the values of A, B, C is simply to equate coefficients of like powers of

x:

0 = B + C

0 = A + 4 B + 10C

1 = − A − 5 B + 25C

which we may solve to show that ( A, B, C ) =−1

6, − 1

36, 1

36

.

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Another method of solution is to assign to x “convenient” values of x, and thereby obtain

more convenient equations to solve. Two values which re “convenient” are x = 1 and x = −5, and they yield from equation (76) the following equations:

x = 1 ⇒ 1 = 36C

and

x = −5 ⇒ 1 = −6 A

implying that A = −16

and C = 136

. To obtain the value of B we would need a third

equation. This could be obtained by equating coefficients, as in the earlier method, or by

simply choosing another value; e.g.,

x = −1 ⇒ 1 = −2 A − 8 B + 16C = 26 − 8 B + 16

36

which yields the same values as before. Integration is straightforward:

1

( x + 5)2( x − 1) d x =

−16

( x + 5)2 +

− 136

x + 5 +

136

x − 1

dx

= 1

6 · 1

x + 5 − 1

36 ln | x + 5| +

1

36 ln | x − 1| + K .

This indefinite integral could be further simplified, e.g., by combining the two logarith-

mic terms into the logarithm of a quotient of polynomials of degree 1.

Repeated irreducible quadratic factors.

Example C.57 [7, Exercise 38, p. 504] To integrate x4 + 1

x( x2 + 1)2 dx.

Solution: What distinguishes this example from those studied earlier is the multiplicity of the

irreducible quadratic factor x2 + 1 in the denominator; this is the first case we have seen where

that multiplicity exceeds 1.

Before I begin the application of the methods of this section, I will apply a substitution that

could simplify this problem. By setting u = x2, I find that du = 2 x dx, x2 = u − 1, so x4 + 1

x( x2 + 1)2 dx =

1

2

u2 + 1

u(u + 1)2 du .

We can seek a partial fraction decomposition:

u2 + 1

u(u + 1)2 =

A

u+

B

(u + 1)2 +

C

u + 1

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which implies identity of the numerators after the right side is taken to a common denominator:

u2 + 1 = A(u + 1)2 + Bu + Cu(u + 1)

in which I will take several “convenient” values of u to obtain equations that can be solved for

the coefficients;

u = −1 ⇒ 2 = 0 − B + 0 ⇒ B = −2

u = 0 ⇒ 1 = A

u = 1 ⇒ 2 = 4 A + B + 2C ⇒ C = 0

Hence x4 + 1

x( x2 + 1)2 dx =

1

2

1

u− 2

(u + 1)2

du

= 1

2 · ln |u| +

1

u + 1 + K

= ln | x| + 1

x2 + 1 + K .

The preceding was fortuitous, a consequence of the fact that a substitution was possible. Now

let’s use the problem as an example for the implementation of partial fraction decomposition.

We proceed analogously to the last phase of the preceding example, always taking the numer-

ator to be the most general polynomial whose degree is less than the degree of the irreducible

factor in the denominator, but now taking separate summands for the powers of that irreducible

factor. Thus we assume a decomposition of the form

x4 + 1

x( x2 + 1)2 =

α

x +

β x + γ x2 + 1

2 +

δ x + η

x2 + 1

and multiply through by the denominator on the left, to obtain the polynomial identity

x4 + 1 = α x2 + 1

2+ ( β x + γ ) x + (δ x + η)

x2 + 1

x

in which the identification of coefficients of like powers of x leads to the 5 equations

α + δ = 1

η = 0

2α + β + δ = 0

γ + η = 0

α = 1

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implying that

(α,β,γ,δ,η) = (1, −2, 0, 0, 0) .

x4 + 1

x( x2 + 1)2 =

1

x +

−2 x x2 + 1

2 .

x4 + 1

x( x2 + 1)2 d x =

1

xdx +

−2 x x2 + 1

2 dx

= ln | x| + 1

x2 + 1 + K ,

the same solution as was foruitously found earlier by using a substitution.

1 x2 + 1

n dx. Should one of the partial fractions be of the form

constant x2 + 1

n , where n is

an integer greater than 1, we can begin the integration by a substitution u = arctan x, which

implies that du = 1

x2 + 1 d x, and

1 x2 + 1

n d x =

1

sec2n−2 udu

=

cos2n−2 u du

=

1 + cos2u2

n

−1

du ,

which, in principle, we know how to integrate.

Example C.58 To evaluate the indefinite integral,

1 x2 + 1

2 dx.

Let x = tan u, i.e., u = arctan x. Then du = dx

1 + x2.

1

x2 + 12 dx =

1

sec2 udu

=

cos2 u du

=

1 + cos2u

2 du

= 1

2

u

2 +

sin 2u

4

+ C

8/15/2019 Brown's Notes 2010



= arctan x

2 +

sin u · cos u

2 + C

= arctan x

2 +

tan u

2sec2 u+ C

= arctan x

2 +

x

2tan2 u + 1

+ C

= arctan x

2 +

x

2 x2 + 1

+ C

The presence of the arctangent function in the integral indicates that this could not have been

evaluated without some step equivalent to this use of the arctangent function in a substitution.

Rationalizing Substitutions. In some integrals in which the integrand is not originally a

rational function, it can be transformed into a rational integrand by an appropriate substitution.

Example C.59 ([7, Exercise 40, p. 504]) “Make a substitution to express the integrand as a

rational function, and then evaluate the integral

1

x −√

x + 2dx.”

Solution: One substitution that suggests itself is u =√

x + 2. Under this substitution, u2 =

x + 2, d x = 2u du, 1

x −√

x + 2dx =

2u

(u − 2)(u + 1) du

=

2

3 2

u − 2 +

1

u + 1

du

= 2

3(2 ln |u − 2| + ln |u + 1|) + C

= 2

3 ln

√ x + 2 − 2

2 √

x + 2 + 1 + C

= 4

3 ln√

x + 2 − 2 +

2

3 ln√

x + 2 + 1

+ C ,

where I removed the absolute signs from the second term because√

x + 2 + 1 must be positive.

The arctangent substitution needs to be used when the quadratic, irreducible factors contain

a first degree term.

Example C.60 To integrate

dx

4 x2 + 4 x + 3.

Solution: Begin by dividing out the coefficient of the 2nd degree term, then completing the

square:

4 x2 + 4 x + 3 = 4

x2 + x +

3

4

= 4

x +

1

2

2

+ 1

2

= 2 + 4

x +

1

2

2

.

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It is clear that we can partially simplify the integral by a substitution of the form u = x + 1

2,

where du = dx. The integral becomes

du

2 + 4u2 . In order to interpret the integral as an

arctangent, we need to make some type of scale change. for example, we could interpret it as du

2 + 4u2 =

1

2

du

1 + 2u2 =

1

2

du

1 + (√

2 · u)2.

Thus an appropriate original substitution would be v =√

2 · x + 1

2

, where dv =

√ 2 dx, and

dx

4 x2 + 4 x + 3 =

1

4 dx

x2 + x + 3

4

= 1

4√

2

dv

v2

2 + 1

2

= 1

2√

2

dv

v2 + 1

= 1

2√

2arctan v + C

= 1

2√

2arctan

√ 2 x +

1√ 2

+ C ;

you should verify the correctness of this integration by diff erentiation.

8/15/2019 Brown's Notes 2010




Release Date: Friday, February 12th, 2010

C.19.1 §7.5 Strategy for Integration

The textbook suggests a strategy for solving integration problems:

1. Simplify the integrand if possible.

2. Look for an obvious substitution.

3. Classify the integrand according to its form.

4. Try again. The preceding instructions are vague, depend on your experience and yourintuition, and are occasionally not appropriate, as sometimes the best way to attach a

problem will be obscured by these methods. So be prepared to try again. As for experi-

ence, you need to work many problems to acquire it.

Table of integration formulæ Many of these formulæ are just recasts of familiar diff erenti-

ation formulæ that you already know. But you should remember the integrals of tan x, cot x,

sec x, csc x, even though you may know how to derive some of them.

Can We Integrate All Continuous Functions? Read this subsection. For most of the func-

tions you will meet in this course it will be possible to integrate them. If that is not the case,it could be that we are asking you indirectly to do something else than to integrate. You are

not expected to be able to detect which functions cannot be integrated in terms of elementary

functions: this is a difficult problem even for a trained mathematician. “You may be assured,

though, that the integrals in the...exercises are all elementary functions.”

7.5 Exercises

[1, Exercise 58, p. 489] Evaluate

x ln x√

x2 − 1dx.

Solution: A first impression is that the most complicated part of the integrand is the

logarithm, and I would like to dispose of it. One way to do that would be through asubstitution, but that could well render the denominator rather complicated. I propose to

try integration by parts, and to assign u and v so that the ln x term is part of u. But how

much of the integrand should be taken for u? I observe that if I take only the ln x term,

it leaves a function that is easy to integrate, so it is an ideal first step:

u = ln x ⇒ du = dx

x

8/15/2019 Brown's Notes 2010



dv = x√ x2

−1

dx

But what is v? If you don’t see, use the substitution U = x2: x√

x2 − 1dx =

1

2

1√

U − 1dU .

You should be able to evaluate this last integral by sight; but, if you can’t, try the substi-

tution V = U − 1:

v = 1

2

1√

U − 1dU =

1

2

1√

V dV =

√ V + C =

√ x2 − 1 + C .

Hence x ln x√

x2 − 1dx = (ln x) √ x2 − 1 −

√ x2 − 1 x

dx .

The problem isn’t solved yet, but at least the logarithm is gone. Now I propose to try a

trigonometric substitution to simplify the square root:

w = arcsec x ⇒ x2 − 1 = tan2 w, x dx = sec2 w tan w dw :

√ x2 − 1

xdx =

| tan w|sec w

· sec w tan w dw

= tan2 w dw

=

(sec2 w − 1) dw

= tan w − w + C

=√

sec2 w − 1 − arcsec x + C

=√

x2 − 1 − arcsec x + C

from which we may conclude that

x ln x√

x2

−1

dx = (ln x)√

x2 − 1 −√

x2 − 1 + arcsec x + C 1

This was not the only way to solve this problem, and it may not have been the best! But

I have written down the way I solved it first. One shorter, but less intuitive way would

be to use the substitution u =√

x2 − 1, which implies that du = x√ x2 − 1

dx. Then

x ln x√

x2 − 1dx =

ln

√ u2 + 1 du =

1

2

ln(u2 + 1) du .

8/15/2019 Brown's Notes 2010



Does this look simpler to you? It can be integrated by parts by taking u = ln(u2 + 1),

dv = du, so du = 2 du

u2 + 1

1

2

ln(u2 + 1) du =

1

2u ln(u2 + 1) −

u2

u2 + 1 du .

The point is that the last integrand is rational, and we know how to integrate all such

functions! (Indeed, division of the denominator into the numerator yields a quotient of

1 and a remainder of -1, so the integral can be evaluated as u2

u2 + 1 du =

1 du −

1

u2 + 1 du

= u

−arctan u + C

= √ x2 + 1 − arctan √ x2 + 1 + C

Another strategy would be to use the same trigonometric inverse substitution immedi-

ately: w = arcsec x ⇒ d x = sec w tan w dw ⇒ x ln x√

x2 − 1du =

sec2 w ln sec w

dw .

In the last integral you should recognize ln sec w as being an antiderivative of tan w. This

suggests using integration by parts with u = ln sec w and d v = sec2 w dw: d u = tan w dw,

v = tan w.

sec2 w ln sec w

dw = (ln sec w)(tan w) −

tan2 w dw

= (ln sec w)(tan w) −

(sec2 w − 1) dw

= (ln sec w)(tan w) − tan w + w + C

= (ln x)√

x2 − 1 −√

x2 − 1 + arcsec x + C

In several places I have casually suppressed absolute value signs; these steps can be

justified, and are consequences of the way in which we defined the inverse functions.

Exercise C.1 To integrate 2π

0

√ 1 + sin x dx.

Solution: This is an interesting integral which, at first glance, does not appear to fit into any of

the families of integrals we have been studying. However, it can easily be seen that, except for

the determination of a sign, the integral is not very difficult to evaluate. But the sign question

is delicate, and even the previous edition of a well known text book overlooked this di fficulty.

In the Student Solution Manual to that textbook the answer was given to be −2√

1 − sin t + C ,

and the following hint was given for integration:

8/15/2019 Brown's Notes 2010



“Multiply numerator and denominator of the integrand by√

1 − sin x.”

The hint is a good one, and does, indeed, lead to one way of solving the problem. Unfortu-

nately, the answer that was given in that textbook was correct only within certain intervals.

Solution following the suggestion

√ 1 + sin x dx =

√ 1 + sin x ·

√ 1 − sin x√ 1 − sin x

dx =

1 − sin2 x

1 − sin xdx

=

cos2 x

1 − sin xdx =

| cos x|√ 1

−sin x

dx

Had the absolute signs not been present, we could make the substitution u = sin x, subject to

which we would have du = cos x · dx cos x√

1 − sin xdx =

1√

1 − udu

= −2√

1 − u + C

= −2√

1 − sin x + C

Unfortunately, this function has derivative√

1 + sin x only when cos x is non-negative, i.e.,

only when 4n−12

·π

≤ x

≤ 4n+1

2

·π, where n is any integer. For example, the value of the definite

integral

2π

0

√ 1 + sin x dx is certainly not equal to

−2

√ 1 − sin x

2π

0= 0 , since the integrand

is positive for most x, so the area under the curve y =√

1 + sin x must be positive; the correct

value is −2

√ 1 − sin x

π2

0+2

√ 1 − sin x

3π2

π2

+−2

√ 1 − sin x

2π

3π2

=2

√ 1 − sin x

3π2

π2

+−2

√ 1 − sin x

π2

−π2

= 2√

2 + 2√

2 = 4√

2 0

Other ways to find the indefinite integral While we cannot remove the sign difficulties

in this problem, we can show that the problem does, in fact, lend itself to a more systematic

integration — i.e., the hint given above is not really necessary. One way to see this is to

remember that we have a trigonometric identity that expresses 1 + cos θ as a square. But, as

the trigonometric function given here is a sine, and not a cosine, one must first arrange for the

8/15/2019 Brown's Notes 2010



presence of a cosine. One way is as follows:

√ 1 + sin x dx =

1 + cos

π

2 − x

dx

=

2cos2

π

4 − x

2

dx

=√

2

cos

π

4 − x

2

dx

Another approach, suggested by a student in this course several years ago, is to observe that

√ 1 + sin x =

sin2 x2

+ cos2 x2

+ 2sin x2 · cos x

2

=

cos x

2 + sin

x

2

.

To integrate this we need to determine the sign of the function inside the absolute signs. This

can be done by observing that it is equal to√

2 sin

x

2 +

π

4

, essentially the same function as

determined just above.

A more systematic approach would have been to attempt to simplify the original integral

by the substitution u = sin x, which would imply that du = cos x dx, so

√ 1 + sin x dx =

√ 1 + u

cos x du

= ± √

1 + u√ 1 − u2

du = ±

1√ 1 − u

du

= ∓2√

1 − u + C = ∓2√

1 − sin x + C

where the sign still depends upon the interval in which x is located.

What, then, is one antiderivative of √

1 + sin x? We have found that, if we confine

ourselves to one interval of the form 4n+1

2 π < x < 4n+3

2 π, any antiderivative has the form−2√

1 − sin x + C ; and, if we confine ourselves to one interval of the form 4n+32

π < x <4n+5

2 π, any antiderivative has the form 2

√ 1 − sin x + C . By choosing the constants to make

the function continuous (indeed, diff erentiable) we can patch such subfunctions together to

form an antiderivative which is valid over an extended domain. The function f defined by the

8/15/2019 Brown's Notes 2010



following table is one such antiderivative:

x f ( x)

· · · · · ·−32

π ≤ x < −12

π −4√

2 + 2√

1 − sin x−12

π ≤ x < 12

π 0√

2 − 2√

1 − sin x12

π ≤ x < 32

π 0√

2 + 2√

1 − sin x32

π ≤ x < 52

π 4√

2 − 2√

1 − sin x52

π ≤ x < 72

π 4√

2 + 2√

1 − sin x72

π ≤ x < 92

π 8√

2 − 2√

1 − sin x92

π ≤ x < 112

π 8√

2 + 2√

1 − sin x11

2 π ≤ x <

13

2 π 12 √ 2 − 2 √ 1 − sin x· · · · · ·

We can verify that the value of the definite integral

2π

0

√ 1 + sin x dx is [ f ( x)]2π

0 = f (2π) −

f (0) =4

√ 2 − 2

√ 1 − sin 2π

−0

√ 2 − 2

√ 1 − sin0

= 4

√ 2.

C.19.2 §7.6 Integration Using Tables and Computer Algebra Systems (OMIT)

This section “is not examination material, but students are to try to solve the problems manu-

ally.”

C.19.3 §7.7 Approximate Integration (OMIT)

This section is not part of the syllabus of this course.

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Release Date: Monday, February 15th, 2010

C.20.1 §7.8 Improper Integrals

Piecewise continuous integrands Consider a function f that is continuous everywhere in an

interval [a, b], including continuity from the right at b and from the left at a. For such functions

we have developed the theory of the definite integral, and the Fundamental Theorem applies.

Now suppose that there is a point c such that a < c < b, where f has a jump discontinuity:

lim x→c−

f ( x) and lim x→c+

f ( x) both exist, but are diff erent. It is possible to define the integral of f as

follows: b

a f ( x) dx =

c

a f ( x) dx +

b

c f ( x) dx ,

and it can be shown that the familiar properties of the definite integral hold here, with the

exception of the Fundamental Theorem. We can proceed in the usual way with such integrals,

until we need to actually evaluate them; then we must split them up at the jump discontinuity

before we attempt to apply the Fundamental Theorem. More generally, this definition “works”

for functions with many jump discontinuities.46 47

Other types of generalizations In this section we wish to consider other types of general-

izations related to a condition of the original definition of the definite integral that fails to be

satisfied. We will follow the terminology of the textbook, calling two types of “impropriety”

Type 1 and Type48 2, but students should be aware that these terms are not in universal use. The

general spirit of these definitions, and of the preceding generalization to jump discontinuities,

is that the familiar properties of integrals proved for continuous functions should hold for these

broader classes wherever they make sense. We shall still have to keep away from any situation

that might lead us to attempt to, for example, add +∞ to −∞, and any other undefinable oper-

ations. Remember that, in this theory, you must rely on the definition, and not attempt to write

down what “makes sense”. The restrictions in some of these definitions are needed to avoid

paradoxes elsewhere.

46We will not explore here what limits — if any — exist on the number of discontinuities.47We have already seen that there may be other situations where the splitting of an integral into pieces for

diff erent parts of the domain may simplify the integration; but, where there is a jump discontinuity, the splittingis not by choice, it is by our definition of the meaning of the generalized symbol.

48Note that the terms Type 1 and Type 2 refer to a type of “impropriety” — not to a type of integral. One

improper integral could contain multiple instances of Type 2, and as many as 2 instances of an impropriety of

Type 1.

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Type 1: Infinite Intervals Our original definition of the definite integral was given for a

finite interval. If we wish to speak of an integral where either or both limits are infinite, weneed to define what these are to mean. The definition we give is one that is consistent with the

definition for finite intervals, and preserves those properties of the integral that are meaningful

when the limits before infinite. I repeat the boxed definition on [1, p. 509]:

Definition C.6 1.

∞ a

f ( x) dx = limt →∞

t a

f ( x) dx provided the integral on the right exists for

all t ≥ a, and the limit exists as a finite number. The “improper” integral on the left

is then said to converge or to be convergent . If the limit does not exist, the improper

integral is divergent .

2. An analogous definition holds when the upper limit is −∞, or when the lower limit iseither of ±∞. Read the textbook.

3. When both limits of the definite integral are infinite, we define the value to be the sum

of two integrals obtained by splitting the domain. It can be shown that it doesn’t matter

where the line is split, the following definition will always give the same value:

∞ −∞

f ( x) dx =

a −∞

f ( x) dx +

∞ a

f ( x) dx .

Of course, the integrand f must have the entire real line R as its domain!

4. It is ESSENTIAL to understand that, in the definition of

∞ −∞

, TWO limits have to be de-

termined independently. IT IS NOT CORRECT TO CONSIDER ONLY limt →∞

t −t

f ( x) dx.

In this case one has to add two finite numbers; if either of the limits does not exist as a fi-

nite number — and that includes being infinite — the sum is not finite, and the improper

integral does not exist .

Type 2: Discontinuous Integrands I have observed above how to cope with a finite, jumpdiscontinuity in the integrand. Here we are interested in other types of discontinuity, in partic-

ular, discontinuities where the function has a vertical asymptote. If the discontinuity occurs at

the left end-point of the interval, then the value of the “improper” integral is defined by

b a

f ( x) dx = limt →a+

b t

f ( x) dx .

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If the discontinuity is at the right end-point b, then the value of the “improper” integral is

defined by b a

f ( x) dx = limt →b−

t a

f ( x) dx .

And, if the discontinuity occurs at a point c between a and b, then the definition is based on

splitting the integral into two parts:

b a

f ( x) dx =

c a

f ( x) dx +

b c

f ( x) dx ,

where both of the summands on the right are defined to be limits as above, and the two limits

have to be evaluated independently. It is important to understand that in this case it is not acceptable that the two limits be linked so that one may a ff ect the other.

Example C.61 The improper integral

2 0

1

x − 1 d x does not converge, since at least one of the

limits

limt →1−

t 0

1

x − 1 d x , lim

t →1+

2 t

1

x − 1 d x

does not exist; indeed, neither of them exists! We say that this improper integral diverges.

Without this severe definition we might find that some of the properties we wish the integralto possess might not be present.

Example C.62 ([1, Example 4, p. 511]) “For what values of p is the integral

∞

1

1

x p d x con-

vergent?”

Solution: ∞

1

1

x p dx = lim

t →∞

t

1

1

x p dx

=

limt →∞

[ln x]t 1 when p = 1

limt →∞1

1 − p x

1

− pt

1 when p

1

=

limt →∞

[ln t − 0] = ∞ when p = 1

1

1 − plimt →∞

1

t p−1 − 1

=

1

p − 1 when p > 1

1

1 − plimt →∞

t 1− p − 1

= ∞ when p < 1

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This result will be needed in Chapter 11 in connection with the “ p-series test” [1, p. 700]. In

the special case p = 1, it states that the area under the hyperbola y = 1

x from x = 1 indefinitely

to the right is unbounded. By symmetry, that region should have the same area as the area

between x = 0 and x = 1 under the same curve, and over the line y = 1. That area is the value

of the improper integral 1

0

1

x d x = lim

t →0+

1

t

1

x d x

= limt →0+

[ln x]1t

= − limt →0+

ln t = +∞

Example C.63 Consider the improper integrals +∞

−∞

1

x2 + 1 d x ,

+∞

−∞

x

x2 + 1 d x .

In both of these cases we must split the interval of integration at a finite point, and consider

two improper integrals of Type 1 independently. In the first case an antiderivative is arctan x,

so +∞

−∞

1

x2 + 1 d x =

0

−∞

1

x2 + 1 d x +

+∞

0

1

x2 + 1 d x

= lima→−∞

0

a

1

x2 + 1 d x + lim

b→+∞

b

0

1

x2 + 1 d x

= lima→−∞

(0 − arctan a) + limb→+∞

(arctan b − 0)

= −−π

2

+

π

2 = π ,

and the improper integral is convergent. (We could have split the interval at any convenient

point other than 0.)

But, in the second case, we have

+∞

−∞

x

x2 + 1 d x = 0

−∞

x

x2 + 1 d x + +∞

0

x

x2 + 1 d x

= lima→−∞

0

a

x

x2 + 1 d x + lim

b→+∞

b

0

x

x2 + 1 d x

= lima→−∞

0 − ln( x2 + 1)

2

+ lim

b→+∞

ln( x2 + 1)

2 − 0

.

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Here we find that each of the limits is infinite: the integral is said to be divergent . In as-

signing a meaning to a doubly infinite integral we do not permit +∞ and −∞ to be ap-proached at the same rate. That is, it is not permitted to interpret this integral as being equal

to lima→∞

a −a

x

x2 + 1 d x. (Here the finite, symmetric integral is 0, so the limit is 0. But we do not

give that meaning to the original, improper integral, which we insist diverges.

Extended Definition of the Integral The original definition in the textbook — in terms of

Riemann sums — was for continuous functions f , where the interval of integration was a finite

interval [a.b]. This definition has been extended in several diff erent ways.

1. First extension: to functions with a finite number of finite jump discontinuities. The

same notation, viz.b

a

f ( x) dx

is used; but it now means the sum of integrals over the various (disjoint) subintervals

where the function is continuous. Remember that continuity over a closed interval [ a, b]

means

• continuity at every point x such that a < x < b,

• continuity from the right at x = a; and

• continuity from the left at x = b. (When a = b the integral is defined to be 0, with

no continuity restrictions.)

There is a need for an explicit definition here, since, where a function f defined on

a ≤ x ≤ b has a jump discontinuity at c, where a < c < b, then the function will usually

be continuous on one side of c, so, for the subinterval on the other side of c, there is still

the need for a limiting operation of some kind, since our original definition did require

continuity of the integrand over the full closed interval of the integral. If it happens that

the jump discontinuity is removable, then it can simply be ignored.)

2. Type 1 Improper Integral with one limit of integration equal to+

∞. We define

+∞ a

f ( x) dx = limb→+∞

b a

f ( x) dx ; and

a +∞

f ( x) dx = limb→+∞

a b

f ( x) dx .

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3. Type 1 Improper Integral with one limit of integration equal to−∞. We define

b −∞

f ( x) dx = lima→−∞

b a

f ( x) dx ; and

−∞ b

f ( x) dx = lima→−∞

a b

f ( x) dx .

4. Type 1 Improper Integral with one limit of integration equal to −∞ and one limit

of integration equal to +∞. We define

+∞ −∞

f ( x) dx =

a −∞

f ( x) dx +

+∞ a

f ( x) dx and,

−∞

+∞ f ( x) dx =

−∞

a

f ( x) dx +

a

+∞ f ( x) dx ,

where a is any convenient point chosen to break the line into two semi-infinite rays. In

practice one often chooses a = 0, but that is not required. It is not permitted to compute

the two limits together — each of them must exist as a finite limit, independent of the

other.

5. Type 2 Improper Integral where end-point a is a point of infinite discontinuity. Here

we “excise” the point a and define

b a

f ( x) dx = limc→a+

b c

f ( x) dx .

6. Type 2 Improper Integral where end-point b is a point of infinite discontinuity. Here

we “excise” the point b and define

b

a

f ( x) dx = limc→b−

c

a

f ( x) dx .

7. Type 2 Improper Integral where a point c such that a < c < b is a point of infinite

discontinuity. Here we “excise” the point c and define

b a

f ( x) dx =

c a

f ( x) dx +

b c

f ( x) dx .

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This means that the integral is the sum of two separate limits, where the “bad” point

has been excised at one end of each of the smaller intervals of integration. The twolimits must be computed separately, and both of them must exist (as finite limits) for the

integral to be said to converge. It is not permitted to compute the two limits at the same

time in a symmetric way.

With these extended definitions we can show that the familiar rules for definite integrals are

still operational, where they make sense. Because the definitions for improper integrals in-

volve limits, we use the terminology of convergence and divergence, the same terminology we

could use in connection with other limits, and which we will see again when we study infinite

sequences and series.

Comparison Test for Improper Integrals The definitions I have sketched concern limits of the values of certain integrals. Where we are unable to evaluate certain integrals directly, we

can still justify a comparison theorem similar to that we saw in connection with finite integrals:

Theorem C.64 (Comparison “Test” for Improper Integrals) Suppose that f and g are con-

tinuous functions such that, on the interval a ≤ x ≤ ∞, 0 ≤ g( x) ≤ f ( x). Then

∞ a

f ( x) dx converges ⇒∞

a

g( x) dx converges

∞

a

g( x) dx diverges ⇒∞

a

f ( x) dx diverges

Example C.65 Evaluate the definite integral

e3 1

dx

x√

ln x. Show that the integral is a conver-

gent, improper integral, and find its value.

Solution: The integrand is not defined at x = 1, since one factor in the denominator is expressed

in terms of ln x. For x > 1 the substitution u = ln x is valid. Hence

e3 1

dx

x√

ln x= lim

t →1+

e3 t

dx

x√

ln x

= limt →1+

3 ln t

1√ u

du

= limt →1+

2

√ u3

ln t

= limt →1+

2

√ 3 − 2

√ ln t

= 2

√ 3 − 2 lim

t →1+

√ ln t

= 2

√ 3 ,

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since, as t → 1+, ln t → 0, so√

ln t →√

0 = 0, by the continuity of the function√

t from the

right at t = 1.

7.8 Exercises

[1, Exercise 18, p. 515] Determine whether the integral

∞ 0

dz

z2 + 3 z + 2 is convergent or di-

vergent. Evaluate it if it is convergent.

Solution: Using standard methods of Partial Fractions we can show that

dz

z2 + 3 z + 2 =

1

z + 1 − 1

z + 2 dz

= ln | z + 1| − ln | z + 2| + C

= ln

z + 1

z + 2

+ C .

Hence

∞ 0

dz

z2 + 3 z + 2 = lim

a→∞

ln

z + 1

z + 2

a

0

= lima→∞ ln

1 + 1a

1 + 2a − ln

1

2= ln 1 − ln

1

2 = ln 2

Note that we could not have expressed

∞ 0

dz

z2 + 3 z + 2 as the diff erence of the improper

integrals

∞ 0

dz

z + 1 and

∞ 0

dz

z + 2, as both of these improper integrals are divergent and the

diff erence of their “values” is not defined.

[7, Exercise 52, p. 516] “Use the Comparison Theorem to determine whether the integral∞ 1

x√ 1 + x6

dx is convergent or divergent.”

Solution: At first glance, this integral suggests a substitution u = x2. While that would

simplify its form, it would not enable us to integrate it immediately, and it is not nec-

essary, since we can prove the convergence without this step. A simpler attack is to

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observe that 1 + x6 > x6, so, for positive x,√

1 + x6 > x3, 1

√ 1 + x6

< 1

x3. Hence we can

consider the limit limt →∞

t 1

x

x3 dx = lim

t →∞

−1

x

t

1

= 1 − limt →∞

1

t 1 = 1. The convergence of the

larger, improper integral implies the convergence of the given one.

[1, Exercise 75, p. 517] “Show that

∞ 0

x2e− x2

dx = 1

2

∞ 0

e− x2

dx .”

Solution: This is an interesting question, because we cannot express an antiderivative

of e− x2 in terms of elementary functions. So, at first glance, one wonders how it will be

possible to work with these integrals. Before doing that, I must prove that the integrals

are convergent — otherwise we don’t have any right to include them as numbers in an

equation.

I note that

x ≥ 1 ⇒ − x2 ≤ − x (multiplying the inequality by a negative number)

⇒ e− x2 ≤ e− x (exponential function is increasing)

⇒ for t ≥ 1

t

1

e− x2

dx ≤t

1

e− x dx = 1

e − 1

et

Hence, as t → ∞, the integral on the right approaches 1

e, i.e., the improper integral

∞ 1

e− xdx = 1

e .

By the [1, Comparison Theorem, p. 514], the improper integral

∞ 1

e− x2

dx is also con-

vergent, and is less than 1

e. But we were considering the integral from 0, not from 1!

The integral from 0 to 1 can be bounded in another way, since the reasoning given above

is valid only for x ≥ 1. For example,

− x2 < 0 ⇒ e− x2

< e0 = 1

⇒1

0

e− x2

dx <

1 0

1 dx = 1

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Hence

∞ 0

e− x2

dx =

1 0

e− x2

dx +∞

1

e− x2

dx < 1 + 1e

.

This is not the exact value of the integral; in fact, it can be shown that

∞ 0

e− x2

dx =

√ π

2 ,

but you are not expected to know this fact, nor how to prove it.

Now to prove the desired equality. Let us apply integration by parts with dv = xe− x2

dx

and u = x, so v = −12 e− x

2

, and du = d x:

t 0

x2e− x2

dx = −1

2 xe− x2

t

0

+ 1

2

t 0

e− x2

dx

= t

2et 2 +

1

2

t 0

e− x2

dx.

By l’Hospital’s Rule,

limt →∞t

et 2 = limt →∞1

2tet 2 = 0 .

Hence

∞ 0

x2e− x2

dx = limt →∞

t 0

x2e− x2

dx

= 1

2 limt →∞

y

0t e− x2

dx

= 1

2

∞

0

e− x2

dx .

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Release Date: Wednesday, February 17th, 2010

Textbook Chapter 8. FURTHER APPLICATIONS OF INTEGRATION.

C.21.1 §8.1 Arc Length

Just as with the earlier concepts of area, volume, and average, we are faced first with adopting

a definition that appears to have the properties that we associate with the concept, and, at the

same time, is workable in practice. The length of an arc will be defined to be the limit — if

there is a limit — of the sum of the lengths of the sides of an approximating polygon formed

by choosing points closer and closer together on the curve, and joining them by line segments.

Note that we haven’t even defined what we mean in general by a curve, so the definition we

give will apply at first only to the graph of a function.

Suppose that we wish to find the length of the arc of the graph of y = f ( x) between the

points (a, f (a)) and (b, f (b)). We can subdivide the interval [a, b] on the x-axis by intermediate

vertices, so that we have a sequence a = x0, x1, x2, . . . , xn = b of points on the x-axis. If we

define

∆ xi−1 = xi − xi−1, and

∆ f ( xi−1) = ∆ yi−1 = yi − yi−1 = f ( xi) − f ( xi−1)

then the distance between successive points Pi−1 = ( xi−1, f ( xi−1)) and Pi = ( xi, f ( xi)) is

|Pi−1Pi| =

( xi − xi−1)2 + ( yi − yi−1)2 =

(∆ xi)2 + (∆ yi)2 ,

which square root can be expressed as either of the following: 1 +

∆ f ( xi)

∆ xi

2

· ∆ xi =

1 +

∆ xi

∆ yi

2

· ∆ yi .

Note that the orientation of the increments in x and y is not relevant, as the increments appear

in these formulæ only as magnitudes. When we pass to the limit, as the “mesh” of points

selected on the x-axis become closer and closer together, the first of these expressions, gives

rise in the limit to the following integral representing the length of the arc:

b a

1 + ( f ( x))2 dx.

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If the curve is given by an equation in the form x = g( y), then we find the arc length from the

point (g(k ), k ) to (g( ), ) to be k

1 + (g( y))2 dy.

When the function whose graph is y = f ( x) is invertible, both formulæ are applicable, and they

give the same length.49

Evaluation of these integrals can often require an approximation method, as the integrands

tend to be of types for which a function expressible in terms of elementary functions is un-

available. For that reason the problems that one meets in calculus books are often confined to

a small set of functions for which antiderivatives can be found.

The Arc Length Function. If we fix a point on a curve, we can then define a function

that expresses distance along the curve from the fixed point. This distance is expressed as an

integral with a variable upper limit, and is signed , so that, in eff ect, we have parameterized the

curve with a variable — usually denoted by the symbol s — uniquely denoting the position of

a point on a path along the curve. This practice diff ers from that employed when we evaluate

the length of the arc between two points, where only the magnitude is of interest. I have written

“...on a path along the curve” rather than “...on the curve”, because we shall be generalizing

these ideas in [1, Chapter 10], to consider curves that are not the graphs of functions; in those

generalizations a curve may cross itself, and the same point could be traversed more than once

by a point whose path we are studying: in that case it is the length of the path that will be

given by the arc length function.

Example C.66 Circumference of a circle. What is the circumference of the circle x2+ y2 = R2

(where R > 0)?

Solution: Since the equation given is not the graph of a function (because the curve crosses

some vertical lines more than once), let’s find the length of the upper arc from x = − R to

x = + R, and double it. This is given by the function f ( x) =√

R2 − x2 = R

1 −

x

R

2

.

y = − x R

1 − x

R2

1 + y2

= 1

1 −

x R

2

49Passage between the two forms can be seen to result from the change of variable given by y = f ( x).

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Circumference =

2

R

− R

1 1 −

x R

2 dx

= 4

R 0

1 1 −

x R

2dx

since the integrand is even, and the interval symmetric around 0

= 4 limt → R−

t 0

1 1 −

x R

2

dx

since the integral has a Type 2 impropriety at x = R

= 4 R limt → R−

t R

0

1√ 1 − u2

du

under the substitution u = x

R

= 4 R lima→1−

a 0

1√ 1 − u2

du

under the substitution a = t

R

= 4 R lima→1−

arcsin a arcsin 0

cos v

| cos v| dv

under the substitution v = arcsin u

= 4 R lima→1−

arcsin a 0

1 dv

since cos v is non-negative for −π2 ≤ v ≤ π

2

= 4 R lima→1−

[v]arcsin a0

= 4 R arcsin 1 by continuity of arcsin= 4 R · π

2 = 2π R .

Of course, we didn’t need to apply this last substitution v = arcsin u because we know two

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antiderivatives of 1

√ 1 − u2

:

4 R limt → R−

t R

0

1√ 1 − u2

du = 4 R limt → R−

[arcsin u]t

R

0

= 4 R limt → R−

arcsin

t

R− 0

= 4 R [arcsin 1 − 0] by continuity of arcsin

= 4 R

π

2 − 0

= 2π R .

Example C.67 ([7, Exercise 8, p. 552]) Find the length of the arc y = x2

2 − ln x

4 , (2 ≤ x ≤ 4).

Solution: Note the way in which the information is presented: we need a description of the

underlying curve, here given by an equation, and specifications of the portion of the curve

whose length is to be determined, here given by an interval 2 ≤ x ≤ 4 or in the alternative

notation x ∈ [2, 4]. Only the absolute value of the length is of interest, so we need not be

careful about which of the end-points 2, 4 is in which limit of the integral; alternatively, it is

the absolute value of the integral that we seek.

y = x2

2 − ln x

4 ⇒ dy

dx= x − 1

4 x

⇒

1 + ( y)2 =

1 +

x2 − 1

2 +

1

16 x2

⇒

1 + ( y)2 =

x2 +

1

2 +

1

16 x2 =

x + 1

4 x

Hence the arc length between x = 2 and x = 4 (where the function

x + 1

4 x

is equal to x + 1

4 x

is

4

2

x +

1

4 x

dx =

4

2

x +

1

4 x

dx

=

x2

2 +

ln x

4

4

2

= 8 + 2 ln 2

4 − 2 − ln 2

4 = 6 +

ln 2

4 = 6 + ln

4√

2 .

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Example C.68 ([7, Exercise 12, p. 552]) Find the length of the curve y = ln x for 1 ≤ x ≤√

3.

Solution: Let’s attack this problem by integrating along either axis.Integrating along the x-axis:

y = ln x ⇒ dy

dx=

1

x

⇒

1 +

dy

dx

2

=

1 +

1

x2

Hence

arc length =

√ 3

1

1 + 1

x2 d x .

To complete the integration, one approach is to try the substitution u =√

x2 + 1 to

simplify the integral. Then

x dx = u du√ 3

1

1 +

1

x2 d x =

2 √

2

u2

u2 − 1 du

=

2

√ 21 +

12

u − 1 −

12

u+

1 du

=

u + ln

u − 1

u + 1

2

√ 2

= 2 −√

2 + ln 1√

3− ln

√ 2 − 1√ 2 + 1

= 2 −√

2 − ln 3

2 − ln

1√ 2 + 1

= 2

−

√ 2

− ln 3

2

+ ln(√

2 + 1) , etc.

Integrating along the y-axis: Here the curve can be rewritten as x = e y, whose length is to

be found for 0 ≤ y ≤ ln 32

.

arc length =

ln 32

0

√ 1 + e2 y dy

8/15/2019 Brown's Notes 2010



=

2

√

2

v2

v2 − 1 dv

under substitution v =√

1 + e2 y

=

2 √

2

1 +

12

v − 1 −

12

v + 1

dv

=

v +

1

2 ln

v − 1

v + 1

2

√ 2

= 2

−√

2

− ln 3

2 − 1

2

ln

√ 2 − 1

√ 2 + 1

= 2 −√

2 − ln 3

2 − 1

2 ln

√

2 − 1√ 2 + 1

·√

2 − 1√ 2 − 1

= 2 −

√ 2 − ln 3

2 − 1

2 ln√

2 − 12

= 2 −√

2 − ln 3

2 + ln(

√ 2 + 1) .

Example C.69 ([7, Exercise 14, p. 552]) Find the length of the curve y2 = 4 x, 0 ≤ y ≤ 2.

Solution: Since x =

y2

4 ,

dx

dy =

y

2 , and the arc length is the integral

2 0

1 +

y2

4 dy =

π4

0

sec v · 2sec2 v dv

under the substitution y = 2 tan v, i.e., v = arctan y

2

= [tan v · sec v + ln | sec v + tan v|]π4

0

=√

2 + ln

√ 2 + 1

.

(I have applied the reduction formula for sec3 v dv determined in [1, p. 458, Exercise 50,

§7.1].)

8.1 Exercises

[1, Exercise 12, p. 530] Find the length of the curve y = ln(cos x) for 0 ≤ x ≤ π

3.

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Solution: y =

−tan x

⇒ 1 + ( y)2 =

|sec x

|.

Arc length =

π3

0

| sec x| dx

=

π3

0

sec x dx

= [ln | sec x + tan x|]π3

0

= ln |2 +√

3| − ln |1 + 0| = ln(2 +√

3) .

C.21.2 §8.2 Area of a Surface of Revolution

I develop a formula for the area of a surface of revolution by observing that an element of arc

of length ∆ s will, when rotated about an axis whose distance from the element is r , generate

an element of surface whose area is approximately 2πr · ∆s. Remembering that

∆s =

1 +

∆ f ( x)

∆ x

2

· ∆ x =

1 +

∆ x

∆ y

2

· ∆ y .

we can integrate with respect to either x or y as the conditions of the problem demand (provided

the derivative exists).(to be continued)

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Release Date: Friday, February 19th, 2010; corrected 10 March, 2010

subject to further correction

C.22.1 §8.2 Area of a Surface of Revolution (conclusion)

8.2 Exercises

[1, Exercise 12, p. 537] Find the area of the surface obtained by rotating the curve x = 1 + 2 y2

(1 ≤ y ≤ 2) about the x-axis.

Solution:

x = 1 + 2 y2 ⇒ dx

dy = 4 y

⇒

1 +

dx

dy

2

=

1 + (4 y)2

⇒ Surface Area = 2π

2 1

y

1 + (4 y)2 dy

=

π

24

1 + 16 y2

32

2

1

=

π

24 · 65 √ 65 − 17 √ 17

[1, Exercise 22, p. 537] ...Find the area of the surface obtained by rotating the curve y =√ x2 + 1 (0 ≤ x ≤ 3) about the x-axis.

Solution:

y =√

x2 + 1

⇒ y = x√ x2 + 1

⇒ 1 + ( y)2 =

√ 1 + 2 x2

√ 1 + x2

⇒ Area of revolution =

3 0

2π√

1 + x2 ·√

1 + 2 x2

√ 1 + x2

dx

= 2π

3 0

√ 1 + 2 x2 dx


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=

2π

√ 2

arctan3√

2

0

sec3

θ d θ

under the substitution θ = arctan( x√

2)

= π√

2

arctan3√

2 0

sec θ d θ + π√

2[tan θ · sec θ ]arctan 3

√ 2

0

by the reduction formula [1, Exercise 50, p. 458]

= π√

2[tan θ · sec θ + ln | sec θ + tan θ |]arctan 3

√ 2

0

= π

3 √ 19 +

1

√ 2 ln( √ 19 + 3 √ 2)

.

Note that the textbook suggested the use of either a table of integrals or a computer

algebra system, but that neither was needed, as the solution of this problem is well

within the abilities of a student in this course (if she has time to do the calculations).

[1, Exercise 24, p. 537] Find the area of the surface obtained by rotating the curve y = ln( x+1)

(0 ≤ x ≤ 1) about the y-axis.

Solution:

y = ln( x + 1)

⇒ dy =

dx

x + 1

⇒

1 + ( y)2 =

1 +

1

( x + 1)2

Area of surface = 2π

1 0

x

1 +

1

( x + 1)2 dx

= 2π

2 1

(u − 1)

1 +

1

u2 du

under the substitution u =

x+

1

= 2π

2 1

√ 1 + u2 du − 2π

2 1

√ 1 + u2

udu

2 1

√ 1 + u2 du =

arctan2 π4

sec3 θ d θ

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under the substitution θ = arctan u

= 1

2 [tan θ · sec θ + ln | sec θ + tan θ |]arctan 2π4

by the reduction formula [1, Exercise 50, p. 458]

= 1

2

2

√ 5 + ln(

√ 5 + 2)

−√

2 + ln(√

2 + 1)

2 1

√ 1 + u2

u du =

arctan2 π4

(sec θ · tan θ + csc θ ) d θ

under the substitution θ = arctan u

= [sec θ + ln | csc θ − cot θ |]arctan2π4

=

√ 5 + ln

√ 5 − 1

2

−√

2 + ln(√

2 − 1)

etc.

Here again the integral does not require special software or tables, just patience.

[1, Exercise 4, p. 562] 1. The curve y = x2, (0 ≤ x ≤ 1), is rotated about the y-axis. Find

the area of the resulting surface.

2. Find the area of the surface obtained by rotating the curve in part 1 about the x-axis.

Solution:

1. Since dy

dx= 2 x,

Area = 2π

1

0

x√

1 + 4 x2 dx

= 2π

2

3 · 1

8 ·

1 + 4 x2 3

2

1

0

= π

6

5

32 − 1

.

2. Here Area = 2π

1 0

x2 √ 1 + 4 x2 dx . To simplify this integral I begin with the sub-

stitution 2 x = tan θ , i.e., θ = arctan 2 x. Then 2 dx = sec2 θ d θ , so

Area = 2π

arctan 2 0

tan2 θ

4 · sec θ · sec2 θ

2 d θ

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=

π

4

arctan2

0

tan

2

θ · sec

3

θ d θ

= π

4

arctan2 0

sec5 θ − sec3 θ

d θ .

Earlier, in Example C.42 on 3087 I had solved [1, Exercise 50, p. 458], proving

a reduction formula (Equation (62))for integrals of positive integer powers of the

secant:

secn x dx = 1

n − 1

secn−2 x

·tan x +

n − 2

n − 1 secn−2 x dx .

From this formula we see that sec3 θ d θ =

1

2 sec θ · tan θ +

1

2

sec θ d θ

= 1


1

2 ln | sec θ + tan θ | + C

sec5 θ d θ = 1

4 sec3 · tan θ +

3

4

sec3 θ d θ

= 1

4

sec3

·tan θ +

3

4

1

2

sec θ

·tan θ +

1

2

ln

|sec θ + tan θ

|+ C

= 1

4 sec3 · tan θ +

3


3

8 ln | sec θ + tan θ | + C

Note that the value of these antiderivatives (with C = 0) at θ = 0 is 0, and that

sec arctan 2 =√

5. Hence

Area = π

4

1

4 sec3 θ · tan θ − 1

8 sec θ · tan θ − 1

8 ln | sec θ + tan θ |

arctan2

0

= π

32

2 · 5

32 · 2 − 5

12 · 2 − ln

√

5 + 2

= π 9

16√ 5 − ln

√ 5 + 2

32

.

C.22.2 §8.3 Applications to Physics and Engineering (OMIT)

Omit this section. (But read it if you are majoring in Physics or Engineering!)

8/15/2019 Brown's Notes 2010



C.22.3 §8.4 Applications to Economics and Biology (OMIT)

Omit this section. (But read it if you are majoring into economics or biology.)

C.22.4 §8.5 Probability (OMIT)

Omit this section.

Textbook Chapter 9. DIFFERENTIAL EQUATIONS. (OMIT)

No parts of this chapter are included in the syllabus.

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C.23 Supplementary Notes for the Lecture of March 01st, 2010

Release Date: Monday, March 01st, 2010

Important Announcement

In earlier versions of the timetable for Section 1 of MATH 140 2010 01, the

times shown for Quizzes Q3, and Q4 were not correct. Quiz Q3 will be ad-

ministered during the week March 08-12, and Quiz Q3 will be administered

during the week March 22–26. The timetable has now been corrected.

Textbook Chapter 10. PARAMETRIC EQUATIONS AND POLARCOORDINATES.

C.23.1 §10.1 Curves Defined by Parametric Equations

A parameter is just a variable. When we call a variable by this term we usually are thinking

of a function or set of functions involving the variable as representing a family of objects. In

this first contact, the family will be a set of points. We will be taking the variable to be a

real variable, and so it is natural to consider not only the family of points on the graphs of a

function, but also the way in which the points are generated by the assignment of real numbers

to the parameter as representing a point moving along the curve. We can name the parameter

with any available symbol. Often we use the letter t ; and a common use for this representation

is to treat t as time, so that the curve can be thought of as the trajectory of (i.e., the path

traced out by) a moving point. If we adopt this point of view, and if the parameter values are

chosen from an interval a ≤ t ≤ b, then we can speak of the curve ( x(t ), y(t )), and can think

of ( x(a), y(a)) and ( x(b), y(b)) as, respectively, the initial and terminal points. Curves given

parametrically in this way need not be graphs of functions: a curve may cross vertical lines

more than once, and may even cross itself, possibly more than once.

Graphing Devices This subsection may be omitted, as, in this course, we shall not be con-

cerned with the use of graphing devices.

Can the graph of a function, given non-parametrically, be expressed in parametric form?

The curve y = f ( x) can be expressed in parametric form as, for example, x = t , y = f (t ). But

there are infinitely other ways of expressing it parametrically, for example

x = t 3, y = f (t 3),

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but not necessarily

x = t

2

, y = f (t

2

),since the latter would include only the points with non-negative abscissæ.

Parametric vs. nonparametric representation of a curve When we represent a curve in

parametric form, the parametrization sometimes contains information beyond what is available

in a non-parametric representation. Often we can see a dynamic way of actually tracing out

the curve by allowing the parameter to range through its domain. So, for example, the curve

x = cos t

y = sin t

can be thought of as being traced out by a point that moves around the unit circle centred at

the origin, starting at the point (1, 0) at time t = 0, counterclockwise at a rate of 1 radian per

unit time as t increases (and clockwise at a rate of 1 radian per unit time as t decreases). If we

compare the curve with

x = cos2t

y = sin2t

we see that both trace out the same circle, covering the curve infinitely often, but the second

curve moves twice as fast. If we wanted to eliminate the multiple covering, we could include

the inequalities 0 ≤ t < 0 in the first case, or 0 ≤ t < π

2 in the second.The standard way to transform from parametric to non-parametric equations is to eliminate

the parameter “between” the equations, which can be interpreted as solving one equation for

the parameter, and substituting that value into the second equation. However, this operation

sometimes “loses information”.

“Cartesian” representation of curves Where the textbook speaks of a Cartesian represen-

tation, I prefer to speak of a non-parametric representation. All the representations for curves

considered in these two sections are Cartesian, since they all refer to the system of represen-

tation associated with Rene Descartes. In the following two sections we will be considering

curves represented in another way — in the so called polar representation.

Example C.70 [7, Exercise 13, p. 656] asks you to find a Cartesian equation of the curve

x = cos2 θ , y = sin2 θ ,−π

2 < θ < π

2

. We can eliminate θ by solving one equation for θ and

substituting into the second; or, more elegantly, by adding the two equations together, obtaining

an equation that does not involve θ :

x + y = 1

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But this equation does not convey all of the information we started with. One type of infor-

mation that has been lost is the fact that both x and y, being squares, are non-negative. Thusit is not the whole line x + y = 1 that is equivalent, but only the line segment joining the

points (0, 1) and (1, 0). This feature is essential, since the doubly-infinite line is not the curve

represented by the parametric equations. Another type of information that is contained in the

parametrization is the way in which this line segment is traversed. The point is moving back

and forth between the point (1, 0) and the point (0, 1), covering the open segment in a parame-

ter interval of length π4

. Distance is not covered at a constant speed — the point moves fastest

in the middle of the segment; but, so far, we are not interested in how fast the point is moving.

There is more than one correct way to describe this curve, but giving the equation x + y = 1

is not enough.

Example C.71 [7, Exercise 17, p. 656]

(a) Eliminate the parameter to find a Cartesian equation of the curve.

(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as

the parameter increases:

x = cosh t y = sinh t

For every value of t a point with these coordinates can be seen to lie on the curve x2 − y2 = 1,

which is a hyperbola with two branches, one opening to the right, and passing through the point

(1, 0), the other opening to the left, and passing through the point (−1, 0); the curves are both

asymptotic to the lines y = ± x. But, since the hyperbolic cosine is positive, the parametrizationapplies only to the right branch of the hyperbola: the curve comes in from −∞ from below,

passes through (1, 0) at t = 0, and the moves off along the upper half as t → +∞. So one way

to describe the curve non-parametrically is

x2 − y2 = 1

x ≥ 0.

This curve may be parameterized in other ways. For example, we could represent it as

x = sec t

y = tan t

(cf. [7, Exercise 14, p. 656]) but this time we need to restrict the values the parameter may

take, for example by

−π

2 < θ <

π

2 .

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The Cycloid In this subsection the textbook describes the construction of this interesting

curve. You are not expected to remember specific properties of this curve, nor its history.However, you should be able to work with this curve if the parametric equations are given

to you, in the same way as you would be expected to be able to work with any reasonable

curve given parametrically. The particular parametrization developed in the textbook is x =

r (θ − sin θ ), y = r (1 − cos θ ).

Families of Parametric Curves Here the author considers a family of curves given para-

metrically, in which the various members of the family all have a similar equation obtained

by assigning diff erent values to a variable in the parametric equations. That variable is also

called a parameter , and the family of curves could be called a parametric family of curves

represented by parametric equations. Of course, the parameter that represents the family of

curves is not the same as the parameter that represents the family of points on any specific

curve. For example, x2 + y2 = a2 could be considered a parametric family of curves (circles)

given in non-parametric form; we could represent the same family in parametric form by

x = a cos t

y = a sin t

where the parameter t represents position on a specific curve, and the parameter a represents

the diff erent curves in the family. We could also interpret the equation by switching the roles

of the parameters: the curve

x = a cos t y = a sin t (78)

with t constant and a variable represents the points on the line through the origin inclined to

the positive x-axis by an angle of t radians. This point of view will become important in [1,

§10.3].

Here again, you are expected to be able to work with reasonable families of parametric

curves, but not to know specific properties of those families (with the exception of the obvious

parametric equations for familiar curves, like the circle).

Example C.72 Equation of a line in the plane. The line in the plane through the point

( x0, y0) and with slope m has non-parametric equation y = y0 + m( x − x0). It can be represented

parametrically in infinitely many ways. If we choose to relate the parameter to distance along

the line, one can show that the following equations represent the line:

x = x0 + t

y = y0 + mt

Check that the line segment joining any two points on this line has slope m

1 = m. (Take two

points with coordinates ( x0 + a, y0 + ma) and ( x0 + b, y0 + mb).)

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C.23.2 §10.2 Calculus with Parametric Curves

Tangents If a curve is given parametrically by ( x, y) = ( x(t ), y(t )), then (subject to certain

conditions) we can diff erentiate y with respect to x by passing through an intermediate variable

t . Recalling thatdx

dt · dt

dx=

dx

dx= 1 ,

we have

dy

dx=

dy

dt · dt

dx

= dy

dt · 1

dx

dt

so we can determine the first derivative dy

dx in terms of the derivatives of the functions f and

g. This calculation can be extended to the second derivative d 2 y

dx2, although the expressions are

not as pretty:

d 2 y

dx2 =

d

dx

dy

dx

= d

dt dy

dx ·

dt

dx

= d

dt

dy

dt dx

dt

· 1

dx

dt

=

d 2 y

dt 2 · dx

dt − d 2 x

dt 2 · dy

dt dx

dt

2

· 1

dx

dt

=

d 2 y

dt 2 · dx

dt − d 2 x

dt 2 · dy

dt dx

dt

3

Rather than substituting in this formula, memorized, you are advised to be able to carry out

this computation for a specific parameterized curve. It enables us to study the concavity of

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a parameterized curve, and to apply the 2nd derivative test if necessary in an optimization

problem.At this point in the lecture I began to discuss the curve which is featured in [1, Exercise 6,

p. 636]. The discussion will be continued at the next lecture.

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C.24 Supplementary Notes for the Lecture of March 03rd, 2010

Release Date: Wednesday, March 03rd, 2010

C.24.1 §10.2 Calculus with Parametric Curves (continued)

Example C.73 (cf. [1, Exercise 6, p. 636]) (see Figure 7 on page 3158) Let’s investigate the

1

0

-2

0.5

-0.5

-1.5

-1

1.510.50-1 -0.5-1.5

Figure 7: The curve x = cos θ + sin 2θ , y = sin θ + cos2θ

curve given parametrically by

x = cos θ + sin2θ (79)

y = sin θ + cos2θ (80)

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Diff erentiating the parameterizing functions, we obtain

dxd θ

= − sin θ + 2cos2θ

= −4sin2 θ − sin θ + 2

= −4

sin2 θ +

1

4 sin θ − 1

2

= −4

sin θ + 1

8

2

− 33

64

= −4

sin θ + 1 +

√ 33

8

sin θ + 1 −

√ 33

8

dyd θ

= cos θ − 2sin2θ

= −4cos θ

sin θ − 1

4

The textbook asks us to find an equation for the tangent to the curve at the point with parameter

value θ = 0. We find that, when θ = 0,

dx

d θ , dy

d θ

= (2, 1), so the slope of the tangent at the

point ( x(0), y(0)) = (1, 1) is 12

, and an equation for the tangent is

y = 1 +

1

2 ( x − 1)

or x − 2 y + 1 = 0.

We can now use the same curve to illustrate some of the other theory of this section. For

example, we can determine where the curve is horizontal, by solving the equation dy

dx= 0,

which implies that dy

d θ = 0. We find that this happens when

cos θ = 0 or sin θ = 1

4 ,

which we could proceed to solve. We can also determine where the curve is vertical, by

determining where dxd θ

= 0; this happens when

sin θ = −1 ±

√ 33

8

All of these equations could be solved. Since the curve is expressed entirely in terms of

sin θ , cos θ , sin 2θ , cos 2θ , and these functions are periodic, repeating themselves after θ passes

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through an interval of length 2π, we can see the whole curve by confining θ to the interval

0 ≤ θ ≤ 2π. In that interval the cosine vanishes when θ = π

2 , 3π

2 , and sin θ = 1

4 when

θ = arcsin 1

4 and π − arcsin

1

4. These are the 4 points where the curve is horizontal; one of

those point is the origin, because the curve passes through the origin several times, one of those

times with a horizontal tangent. The curve has the shape of a “3-leafed rose” or a “trefoil” (3-

leafed clover) centred at the origin, with one petal bisected by the y-axis. We will see other

curves with this shape when we study polar coordinates in [1, §10.3].

For the same curve given parametrically by equations (79), (80), let’s consider the follow-

ing question, similar to [1, Exercise 25, p. 636]: “Show that the curve..has (several)...tangents

at ( x, y) = (0, 0), and find their equations.”

Solution: If we set the coordinates equal to zero and solve, we can simplify the resulting

equations, to obtain:

cos θ · (1 + 2sin θ ) = 0

(sin θ − 1) · (1 + 2sin θ ) = 0 .

The curve passes through the origin whenever both of the equations are satisfied. This means

that either

sin θ = −1

2 , (81)

or both50 of the following equations must hold:

cos θ = 0 (82)

sin θ = 1 . (83)

For θ between 0 and 2π, these last equations are satisfied when θ = π

2; equation (81) is satisfied

when θ = 7π

6 ,

11π

6 . The curve, which has the shape of a 3-leafed clover or a “3-leafed rose”,

passes through the origin 3 times, and we can find the slopes of the tangents in the usual way,

by taking the ratio

dy

d θ dx

d θ

.

50Note that the second equation implies the first, but the first does not imply the second; we could thus have

shown only the second equation.

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Areas. Consider the arc of the curve x = f (t ), y = g(t ) determined by points with parameter

value t between α ≤ t ≤ β. Since β

α

g(t ) · f (t ) dt =

f ( β)

f (α)

y · dx ,

the first integral represents the area between the arc and the x-axis. Remember, though, that

the curve need no longer be the graph of a function, so it could cross vertical lines more than

once. This means that the area could be “folded”, and there might be portions that could be

counted negatively and canceling the portions that you are interested in. For that reason you

should use this integral only where you are clear about the shape of the region whose area you

are finding, and there might be situations where the region should be broken up into parts and

the areas of the parts found separately.

Arc Length. In a similar way to the preceding, we can argue that the length of the arc

α ≤ t ≤ β of the curve x = u(t ), y = v(t ) between α ≤ t ≤ β is given by the integral

β

α

dx

dt

2

+

dy

dt

2

dt .

Note that, when we consider the parametrization

x = t y = f (t ) (a

≤ x

≤ b)

of the graph of the function y = f ( x) between the points (a, f (a)) and (b, f (b)), this reduces

to the formula we derived earlier. Here again, be careful that you are finding the length of the

curve that you intend. In this case there cannot be any cancellation, since the integrand is a

square root, which cannot be negative. If you obtain a negative length, it could simply be a

consequence of the direction in which you have parameterized the curve, which is harmless;

or of an error you have made in you calculations, which is serious.

Surface Area. We can also adapt, in the obvious ways, our previous formulæ for area of

surfaces of revolution.

Example C.74 Let’s determine a formula for the surface area of a doughnut. Suppose that the

doughnut is generated by the curve

x = R + r cos θ

y = 0 + r sin θ

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where 0 ≤ θ ≤ 2π, where this circle, centred at the point ( x, y) = ( R, 0), is revolved around the

y-axis. Assume R ≥ r .

Area of revolution =

2π

0

2π( R + r cos θ )

dx

d θ

2

+

dy

d θ

2

d θ

=

2π

0

2π( R + r cos θ )√

r 2 d θ

= 2πr [ Rθ + r sin θ ]2π0 = 2πr [ R · 2π − R · 0] = 4π2rR.

Volumes. The textbook appears to say little about this, but here again the earlier formulæ can

be adapted, to determine, for example, the volume of revolution generated by a given curve

about a given axis.

10.2 Exercises

[1, Exercise 54, p. 637] “Find the total length of the astroid x = a cos3 t , y = a sin3 t .”

Solution: This curve is generated over an interval of length 2π: we can take 0 ≤ t ≤ 2π.

the curve looks like a deformed circle, that has been pinched towards the centre at the

points away from where it crosses the coordinate axes.

Total length = 2π

0 dx

dt 2

+dy

dt 2

dt

=

2π

0

−3a cos2 · sin t 2

+3a sin2 t · cos t

2dt

=

2π

0

3|a| · | cos t · sin t | dt

By symmetry we can find the length of one quarter of the curve:

Total length = 4

π

2

0

3|a| · | cos t · sin t | dt

= 4|a|

π2

0

3cos t · sin t dt

= 6|a|sin2 t

π2

0= 6|a| .

[1, Exercise 48, p. 637] Find the length of the loop of the curve x = 3t − t 3, y = 3t 2.

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Solution: Where does this loop begin and end? We need to find all parameter values

t 1, t 2 where t 1

t 2 but

3t 1 − t 31 = 3t 2 − t 32

3t 21 = 3t 22

The equations can be simplified to

(t 1 − t 2)(3 − t 21 − t 1t 2 − t 22) = 0

(t 1 − t 2)(t 1 + t 2) = 0

When we know that t 1 t 2, the equations further simplify to the system

3 − t 21 − t 1t 2 − t 22 = 0

t 1 + t 2 = 0

which imply that t 1 = −t 2 = ±√

3. Thus these two curves intersect in the points that

have parameter values ±√

3.

Since dx

dt

2

+

dy

dt

2

=

3 − 3t 2

2

+ (6t )2

= 3(1 + t 2

) ,

the length of the arc of the loop is

√ 3

−√

3

3(1 + t 2) dt = 23t + t 3

√ 3

0= 12

√ 3 ,

since the integrand is an odd function, and the limits of integration are symmetrically

located around t = 0.

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C.25 Supplementary Notes for the Lecture of March 05th, 2010

Release Date: Friday, March 05th, 2010

C.25.1 §10.2 Calculus with Parametric Curves (conclusion)

One topic remaining, not discussed in the previous lectures is summarized in the following

subsection, which was included earlier in these notes on page 3160:

Areas. Consider the arc of the curve x = f (t ), y = g(t ) determined by points

with parameter value t between α ≤ t ≤ β. Since

β

α

g(t )

· f (t ) dt =

f ( β)

f (α)

y

·dx ,

the first integral represents the area between the arc and the x-axis. Remember,

though, that the curve need no longer be the graph of a function, so it could cross

vertical lines more than once. This means that the area could be “folded”, and there

might be portions that could be counted negatively and canceling the portions that

you are interested in. For that reason you should use this integral only where you

are clear about the shape of the region whose area you are finding, and there might

be situations where the region should be broken up into parts and the areas of the

parts found separately.

I illustrate this theory with the following example:

10.2 Exercises (conclusion)

[1, Exercise 34, p. 637] “Find the area of the region enclosed by the astroid x = a cos3 θ ,

y = a sin3 θ .” This is the same curve considered above in [1, Exercise 54, p. 637]. Let’s

consider the arch in the first quadrant. The area will be 4 times the area in the first

quadrant, which is

a 0

y dx =

π2

0

y(θ ) · dx(θ )

d θ d θ

=

π2

0

a sin3 θ ·−3a cos2 θ · (− sin θ )

d θ

.

You know one way to evaluate an integral of this type — by replacing sin 2 θ and cos2 θ

respectively by 1 − cos2θ

2 and

1 + cos2θ

2 . A variant of that method is as follows:

π2

0

a sin3 θ ·−3a cos2 θ · (− sin θ )

d θ

= 3a2

π2

0

a sin4 θ · cos2 θ d θ

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= 3a2

π2

0

1

−cos2θ

2 · sin2θ

22

d θ

= 3a2

8

π2

0

sin2 2θ − sin2 2θ · cos2θ

d θ

= 3a2

8

π2

0

1 − cos4θ

2 − sin2 2θ · cos2θ

d θ

= 3a2

8θ

2 −

sin4θ

8 −

sin3 2θ

6

π2

0

= 3πa2

32 ,

so the area of the interior of the astroid is 3πa2

8 .

Laboratory Project: Bezier Curves Omit this subsection.

C.25.2 §10.3 Polar Coordinates

Polar Curves. In the polar coordinate system we locate points in the plane by taking a special

point, called the pole, a special half-line or ray which emanates from the pole, called the polar

axis and a direction for measuring positive angles — usually taken to be the counter-clockwise

direction. Any point can be located if we know its distance from the pole, usually denoted by

r , and the angle that the line joining them makes with the polar axis in the positive direction,

usually denoted by θ . However, the angle is not unique, since angles of θ and θ + 2nπ will

give the same point for any integer n. So here we have one of the essential diff erences between

polar and cartesian coordinates:

Theorem C.75 The polar coordinates of a point are never unique.

In the case of the pole itself, the angle θ is totally undetermined: once we know that r = 0, any

angle θ will give the same point.

Convention permitting negative r . It is convenient to broaden the multiplicity of coordi-

nates by permitting the distance from the pole to be negative. We do this by agreeing that ( r , θ )

and (−r , θ + π) represent the same point. This convention permits for continuous representation

of certain curves, but causes complications at various stages: occasionally additional care is

required.

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Relations between polar and cartesian coordinates. In theory we can set up polar and

cartesian systems independently in the plane, placing the pole at any convenient place. Inpractice we often place the pole at the origin of a cartesian system, with the polar axis along the

positive x-axis. When the author of the textbook suggests that you are to consider two systems

at the same time, and gives no other information, this is what he expects you to do.51 When

the polar and cartesian systems are placed in this “standard” way, the following relationships

hold:

x = r · cos θ

y = r · sin θ

x2 + y2 = r 2

y

x

= tan θ

Note that, while it is possible to transform from polar to cartesian coordinates without ambi-

guity, it is not always possible to move painlessly in the other direction; this is because of the

non-uniqueness of polar coordinates, about which much more will be said. If you are given an

equation in cartesian coordinates, you can transform it to polar by substituting the appropriate

formulæ for x and y and simplifying — try [1, Problems 21-26, p. 648]; if you are given the

coordinates of a point in polar coordinates, you can transform to cartesian in the same way —

try [1, Exercises 3-4, p. 647].

Consequences of Non-uniqueness of Polar Coordinates. This is a difficult topic, and will

require considerable practice before you will become comfortable with it! Some of the prob-

lems you will have are related to understanding what is meant by an equation for a curve:

you need to understand that any curve can be represented in multiple ways, even when we use

cartesian coordinates. But, when the representation of the points themselves is not unique, the

results can be confusing.

We saw with parametric equations that the same point on a curve can appear more than once

on a “curve”. In that context there was a “natural” way of tracing out the curve, by following

increasing values of the parameter. When we come to study polar coordinates, the situation is

much more complicated, because there is no “natural” way of following the generation of the

curve, and no one set of coordinates for a point has preferential status with respect to others.

Polar curves can be expressed by any relationship between r and θ , although more often than

not the equation will be in the form r = f (θ ); however, you will see some equations in the formθ = f (r ) — for example, the equation θ =

π

4 represents the line through the pole inclined at an

angle of π

4 to the polar axis.

51However, there is an important application, involving conic sections — ellipses, parabolæ, hyperbolæ, where

we place the origin at a diff erent point; this topic is not on the syllabus of the present course, but you may read

about it in [1, §10.6].

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Symmetry When we studied symmetry in the context of cartesian coordinates, we could

have considered [1, pp. 19-20, 308]

• reflective symmetry in a vertical line, particularly in the y-axis;

• reflective symmetry in a horizontal line, particularly in the x-axis;

• rotational symmetry about a point, particularly about an angle of π around the origin

(which is equivalent to reflective symmetry in both the x-axis and the y-axis); and

• periodicity of a graph under a horizontal shift.

These are not the only types of symmetries that a graph can possess, so the study in the textbook

is selective, and we haven’t investigated thoroughly what principles were involved.Similarly, when we consider symmetry in the context of polar coordinates, we will not

attempt a thorough study, but will consider types of symmetries which the polar system is

particularly able to accommodate. Here are the types of symmetry that the textbook mentions:

• reflective symmetry in the polar axis — exhibited by the invariance of the equation of a

curve under the substitution θ → −θ .52

• rotational symmetry under a rotation through an angle of π around the pole — exhibited

by the invariance of the equation under the transformation r → −r or under θ → θ + π

• reflective symmetry under reflection in the line θ =

π

2

, exhibited by invariance under the

transformation θ → π − θ .

Again, these are not the full range of symmetries that can occur in the plane. Because of

the non-uniqueness of coordinates, curves can have the symmetries listed without exhibiting

invariance under the transformations listed. For example, the curve θ = π is the line that

extends the polar axis. It certainly has symmetry in the polar axis, but the equation is not

unchanged when we replace θ by −θ . You are not expected to be an expert in the subject of

symmetries.

Graphing Polar Curves with Graphing Devices. Omit this section — this is a device-free

course.52This description is incomplete. See the discussion below on page 3177 of Example C.78 ([7, Exercise 37, p.

678]).

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Where is the point with polar coordinates (r , θ )? The ambiguity in polar coordinates is not

in locating a point with given coordinates — it is only that all points possess multiple sets of coordinates. If you are given the coordinates (r , θ ), you may locate the point by:

• First, locating the ray obtained by turning the polar axis (the distinguished ray that em-

anates from the pole, relative to which we refer all coordinate angles) through an angle

of θ in the positive direction; and

• then

– if r ≥ 0, proceeding along that ray a distance of r ; or

– if r ≤ 0, proceeding along the extension of the polar axis beyond the pole a distance

of

−r .

Example C.76 ([7, Exercise 2, p. 677]) Plot the point whose polar coordinates are−1, −π

2

.

Then find two other pairs of polar coordinates of this point, one with r > 0, and one with r < 0.

Solution: The second coordinate tells us that the point is on the line inclined to the polar axis

at an angle of −π2

radians measured in the positive direction: this takes us to the ray which is

obtained by turning the polar axis in the clockwise direction through a right angle. But then

the negative first polar coordinate tells us to proceed along the opposite ray for 1 unit. If the

polar system is superimposed on a cartesian system in the usual way, the point is the unit point

on the positive y-axis. This point has the following other sets of polar coordinates:

with positive r : 1, π2

+ 2nπwith negative r :

−1, −π

2 + 2mπ

or

−1, 3π

2 + 2π

where n, m, and are any integers.

Example C.77 ([7, Exercise 12, p. 677]) “Sketch the region in the plane consisting of points

whose polar coordinates satisfy the conditions −1 ≤ r ≤ 1, π4 ≤ θ ≤ 3π

4 .”

Solution: Let’s separate the portion of the region described by positive r , zero r , and negative

r :

positive r . 0 < r ≤ 1, π4 ≤ θ ≤ 3π

4 describes a sector of the unit disk centred at the pole — all

points in the region bounded by two perpendicular radii bisecting the first and second

quadrants. The points on the radii and the bounding circle are included.

zero r . The region described by r = 0, π4 ≤ θ ≤ 3π

4 consists of the pole alone: the θ coordinate

is irrelevant for the pole.

negative r . −1 ≤ r < 0, π4 ≤ θ ≤ 3π

4 describes the region antipodal to the one described for

positive r — between radii at angles π + π4

and π + 3π4

.

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10.3 Exercises

[1, Exercise 14, p. 648] “Find a formula for the distance between the points with polar coor-

dinates (r 1, θ 1) and (r 2, θ 2).”

Solution: One way to solve this problem is to transform to cartesian coordinates, then

apply the usual distance formula, after which cancellations will occur. A more direct

way is to apply the Law of Cosines to a triangle whose vertices are the pole and the two

given points. That is easy if we assume that r 1 and r 2 are non-negative, so they represent

the lengths of the sides of the triangle. To solve the problem in this way we need to

consider several cases. The result is the same: the distance is

r 21

+ r 22

−2r 1r 2 cos(θ 1

−θ 2) .

[1, Exercise 18, p. 648] Identify the curve r = 2 cos θ + 2sin θ .

Solution: The textbook suggests finding a cartesian equation first. This is not always

easy, but is not impractical in the present problem. If we multiply the given equation by

r , we obtain the equation

r 2 − 2r cos θ − 2r sin θ = 0

which we see is equivalent to

x2 + y2 − 2 x − 2 y = 0

i.e., to

( x − 1)2 + ( y − 2)2 = 2

which is a circle with radius√

2 centred at the point (1, 1). However, we must note that,

in multiplying an equation by a factor — equivalently in multiplying the two equations

r = 2cos θ + 2sin θ (84)

r = 0 (85)

we were, in the second equation, possibly permitting new points to be included in the

“curve”. We must carefully analyze whether the equation r = 0 did that. But we know

that r = 0 represents only the pole! And the pole was already on the given curve: itappears there as

(r , θ ) =

0,

3π

4

.

So multiplying by r = 0 does not introduce any new points.53

53But what would happen if you were to multiply an equation like r = 1 by r = 0?

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[1, Exercise 30, p. 648] Describe the curve with the equation r 2 − 3r + 2 = 0.

Solution: The equation can be rewritten as

(r − 1)(r − 2) = 0 ,

which is satisfied by all points whose r -coordinate is either 1 or 2, i.e., by all points on

either of two concentric circles around the pole.

[1, Exercise 33, p. 648] Describe the curve with the equation r = 2(1 − sin θ ) (θ ≥ 0).

Solution: (see Figure 8 on page 3170) Under the transformation θ → π − θ the equation

0

-2

-1

-3

-4

210-1-2

Figure 8: The cardioid with equation r = 2(1 − sin θ )

is unchanged, so the curve has symmetry in the y-axis. This is a “heart-shaped” curve,

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called a cardioid . The textbook describes another example in [1, Example 9, pp. 645-

646], where the author shows, as we could show here, that the curve is tangent to the y-axis. We call the point of tangency (here, the pole) a cusp. (The condition θ ≥ 0 does

not restrict the curve in any way, since the function 2(1 − sin θ ) is periodic with period

2π: any part of the curve that might require a negative value θ 0 of θ to represent it, can

also be represented by the θ 0 + 2πn, where n is any integer; taking n sufficiently large

will make this value positive.)

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Release Date: Monday, March 08th, 2010

C.26.1 §10.3 Polar Coordinates (continued)

10.3 Exercises (conclusion)

[1, Exercise 34, p. 648] Describe the curve with equation r = 1 − 3 cos θ . (see Figure 9 on

page 3172)

2

0

1

-1

-2

0-1-2-3-4

Figure 9: The lima¸con r = 1 − 3cos θ ,

Solution: It is not practical to represent this curve in cartesian coordinates. We note

that, when θ is replaced by −θ , the equation is unchanged. This tells us that the curve is

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symmetric about the polar axis. Try tracing it out by starting with the point with θ = 0. If

we superimpose a cartesian system in the usual way, that point is the point with cartesiancoordinates (−2, 0). The curve passes through the pole first when θ = arccos 1

3, about

70 degrees. It crosses the y-axis at the point with cartesian coordinates ( x, y) = (0, 1),

and then moves to its maximum distance of 1 + 3 = 4 from the pole when θ = π. Then

it returns, crossing the negative y-axis at ( x, y) = (0, −1), and passing again through the

pole. The curve is one of the family called limacons, see them sketched in [1, Example

11, p. 647].

Tangents to Polar Curves I discuss how to determine the tangents to curves of the form

r = f (θ ). The discussion is based on the observation that, if we superimpose polar and cartesian

systems in the most usual way, we can express x and y in terms of θ by

x = f (θ ) · cos θ

y = f (θ ) · sin θ

and thereby interpret θ as a parameter in the parametric representation of a curve. We find that

dy

dx=

dy

d θ dx

d θ

=

dr

d θ · sin θ + r · cos θ

dr

d θ · cos θ − r · sin θ

. (86)

In the case of the cardioid r = 2(1 − sin θ ), we find that

dy

dx =

2 sin θ − 1

4cos θ

when cos θ 0. When θ = 0, at the unit point on the initial ray, the tangent has slope −14

. As θ

increases to π6

the tangent becomes vertical; then, as θ > π6

and θ → π2

−, the tangent approaches

the vertical, with positive slope. Similarly, as θ → π2

+, the tangent approaches the vertical, with

negative slope. We say that this curve has a cusp at the pole.

[1, Exercise 35, p. 648] Describe the curve with equation r = θ , with θ ≥ 0.

Solution: This curve is a spiral, turning around the pole. (see Figure 10 on page 3174)If, however, we were to ask about the curve with equation r = θ , with θ ≤ 0, it is also a

spiral around the pole. (see Figure 11 on page 3175) The superposition of the two curves

is shown in Figure 12 on page 3176.

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30

10

-30

30

20

20-20-30

-20

10

-10

-10

0

0

Figure 10: The spiral with equation r = θ , (θ ≥ 0)

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30

10

-30

30

20

20-20-30

-20

10

-10

-10

0

0

Figure 11: The spiral with equation r = θ , (θ ≤ 0)

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30

10

-30

20

0

3010-30

-20

-10

-20 -10 200

Figure 12: The full spiral with equation r = θ , −∞ < θ < +∞

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Example C.78 ([7, Exercises 37, 40, p. 678]) Describe the curves with equations r = sin 2θ ,

r = sin 5θ .Solution: First let’s consider the curve r = sin2θ . (see Figure 13 on page 3177) While the

0.6

0.2

-0.6

0.4

0

-0.4

-0.2

0.60.40.20-0.2-0.6 -0.4

Figure 13: The “4-leafed rose” with equation r = sin 2θ ,

equation does not remain unchanged when we replace θ by

−θ , it changes to r =

−sin 2θ ,

which contains the same points, since it can be rewritten as −r = sin 2(θ + π), which can beobtained from the original equation by the transformation

(r , θ ) → (−r , θ + π) .

Thus this curve is symmetric about the polar axis. It is also symmetric about the y-axis, since

the replacement of θ by π2 − θ leaves the equation unchanged. The curve is a “4-leafed rose”,

where each petal is tangent to the cartesian axes, if located in the usual way.

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But, when the multiplier of θ is an odd integer, the situation changes. (see Figure 14 on

page 3178) The curve r = sin 5θ is again a “rose”, passing through the pole every 36 degrees.

0.8

0

0.4

-0.4

-0.8

0.80 0.4-0.8 -0.4

Figure 14: The “5-leafed rose” with equation r = sin 5θ ,

It is not symmetric about the y-axis, nor about the x-axis. (It is symmetric under reflections in

other axes, and also under rotation through certain angles around the pole.)

Example C.79 ([7, Exercises 42, p. 678]) Sketch the curve with the polar equation r 2 = sin 2θ .

Note that there are no points on this curve for π2

< θ < π2

since, in that interval, sin 2θ < 0. The

entire curve is traced out for 0 ≤ θ ≤ π2

: every value of θ gives rise to two points on the graph.

(see Figure 15 on page 3179)

Example C.80 Where is the curve r = sin 2θ horizontal?

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0.8

0

0.4

-0.4

0.80.40-0.8 -0.4

-0.8

0.3

0.1

-0.3

0.2

0

-0.2

-1 0 0.5 1

-0.1

-0.5

Figure 15: The lemniscates r 2 = sin 2θ , r 2 = cos 2θ

Solution: We set dy

dx= 0 in (86):

dr

d θ · sin θ + r · cos θ = 0

⇔ 2cos2θ · sin θ + sin2θ · cos θ = 0

⇔ 2(1 − 2sin

2

θ ) · sin θ + 2sin θ · cos

2

θ = 0⇔ 2 sin θ (2 − 3sin2 θ ) = 0

so the tangents will be horizontal when

sin θ = 0, ±

2

3 ,

i.e., when

θ = 0, ± arcsin

2

3, π ± arcsin

2

3 .

C.26.2 §10.4 Areas and Lengths in Polar Coordinates

Areas To find the area bounded by a curve given in polar coordinates we express the area as

the limit of a sum of narrow triangles whose bases are along the bounding curve, and whose

upper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of the

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curve r = f (θ ) between θ = a and θ = b is then

b a

1

2 · ( f (θ ))2 d θ .

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Release Date: Wednesday, March 10th, 2010

C.27.1 §10.4 Areas and Lengths in Polar Coordinates (continued)

Areas To find the area bounded by a curve given in polar coordinates we express the area as

the limit of a sum of narrow triangles whose bases are along the bounding curve, and whose

upper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of the

curve r = f (θ ) between θ = a and θ = b is then

b

a

1

2 ·( f (θ ))2 d θ .

(This could be shown in two diff erent ways: by treating the element of area as a sector of a cir-

cle, or as a narrow, isosceles triangle.) As with all other formulæ involving polar coordinates,

one must use this formula with care. Be sure that you know precisely what the region looks

like; in case of doubt, break the region up into parts, and find the areas of the parts separately.

The limits must be chosen carefully, to be sure that, for example, you are not computing more

area than you intend. For example, the curve r = cos θ is a circle of radius 12

centred at the

point

12

, 0

, and passing through the pole. The curve is swept out as θ ranges from 0 to π, so

the area of the disk is

1

2

π 0

cos2 θ d θ = 1

4

π 0

(1 + cos2θ ) d θ = π

4 .

Compare this with the area of the disk r = 1, centred at the pole, where the curve is swept out

as θ ranges from 0 to 2π;

Area = 1

2

2π 0

1 d θ = π .

In that case, if we were to stop at π, we would obtain only the area of the upper half-disk.

Rather than attempting to memorize rules about limits for integrals, I suggest you carefullyanalyze each problem individually.

Finding the intersections of curves in polar coordinates The textbook [1, p. 651-652]

discusses the difficulties that of finding intersections which are a consequence of the multiple

sets of coordinates for points. Some textbooks suggest that finding the intersections can only

be done visually, and that is not true. In fact, intersections can be found algebraically, but one

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must be careful and thorough. I attach below a discussion I prepared for a class some years

ago, where another textbook had made a false statement that I felt obliged to correct; the erroris continued in the current edition of the same textbook.

Example C.81 In an example in another textbook [31, Example 8, p. 579], [30, Example 8, p.

635] the objective is to find the points where the curves

r = 1 + sin θ (87)

r 2 = 4 sin θ (88)

intersect. It was stated in the textbook solution that only one of the points of intersection

can be found algebraically, and that the others can be found only “when the equations are

graphed”. We show here all intersection points can be found algebraically! We never resort to

calculations on a sketch: all procedures can be justified theoretically — the sketch serves onlyto help visualize a situation that can be adequately described verbally and / or with mathematical

formulæ.

(cf. Figure 16, page 3182) Had the curves been given in cartesian coordinates, we could

–2

–1

0

1

2

–1 –0.5 0.5 1

Figure 16: Intersecting polar curves r = 1 + sin θ , r 2 = 4 sin θ

have found all intersections by solving the equations simultaneously. Why can’t we solve the

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polar equations in the same way? The difficulty derives from the fact that any point has in-

finitely many diff erent polar representations. More precisely, a point that can be representedby polar coordinates (r , θ ) also has coordinates ((−1)nr , θ + nπ), where n is any integer — pos-

itive or negative; moreover, the pole can be represented by (0, θ ), where θ is any real number.

To determine the points of intersection, one must consider the possibility that the same point

appears with di ff erent coordinates.

Solve the given equations algebraically: By eliminating sin θ between the two equations, we

obtain r 2 = 4(r −1), which implies that (r −2)2 = 0, so r = 2, and sin θ = 2−1 = 1. Hence

θ = π

2+2mπ, where m is any integer, and the points of intersection are

2,

π

2 + 2mπ

: but,

by the convention described above, these are representations of the same point, whose

“simplest” representation is 2, π

2.

Transform the equations in all possible ways and solve again: Apply to one of the equa-

tions the transformation

(r , θ ) → (−r , θ + π) (89)

and solve it with the original form of the other equation. Repeat this process until the

equations transform to a pair already solved. Equation (87) transforms to

−r = 1 + sin(θ + π) (90)

which is equivalent to

r = −

1 + sin θ (91)

which equation we solve with (88). Eliminating sin θ yields r = 2 ± 2√

2, so sin θ =

3 ± 2√

2. The upper sign is inadmissible, as a sine cannot exceed 1 in magnitude. Hence

r = 2 − 2√

2 and

sin θ = 3 − 2√

2 . (92)

The solutions to (92) are θ = sin−13 − 2

√ 2

+ 2mπ and θ = π − sin−13 − 2

√ 2

+ 2mπ;

we may take m = 0, as all other values of m give the same two points:2 − 2

√ 2, sin−1

3 − 2

√ 2

and2 − 2

√ 2, π − sin−1

3 − 2

√ 2

.

As the first coordinate in these cases is negative, we could equally well represent the

points as 2

√ 2 − 2, sin−1

3 − 2

√ 2

+ π

and 2

√ 2 − 2, − sin−1

3 − 2

√ 2

.

A second application of (89), to (90), restores the original equation; hence there are no

other intersection points, except possibly the pole.

8/15/2019 Brown's Notes 2010



Check whether the pole is on both curves: On (87) the pole appears as 0, 3π

2, etc.; on (88)

it appears as (0, 0), etc. Thus the pole is also a point of intersection. The reason we did

not find it when we solved pairs of equations is that it appears on the two curves only

with diff erent sets of coordinates, no two related by (89).

Example C.82 [7, Exercise 29, p., 683] “Find the area of the region that lies inside both

curves: r = sin θ , r = cos θ .”

Solution: (see Figure 17 on page 3184) The curves are circles through the pole. To see this,

0.75

0.5

0.0

0.25

−0.5

1.0

1.0

0.75

0.25

0.5

−0.25

0.0−0.25−0.5

Figure 17: Curves r = sin θ , r = cos θ

multiply the first by r = 0 (which, of course, could be bringing the pole into the curve). The

resulting equation is r 2 = r sin θ , which, in cartesian coordinates, would be x2 + y2 = y, a

8/15/2019 Brown's Notes 2010



circle with centre (in cartesian coordinates) 0, 12 and radius 1

2. The operation of multiplying

by r = 0 did not, however, alter the curve, since the pole was already on the curve, with polarcoordinates (r , θ ) = (0, 0). In the same way we can show that the second curve is a circle of

the same radius centred at the point (r , θ ) =

12

, 0

; polar coordinates for the centre of the first

circle are, for example,

12

, π4

.

Where do the curves meet? (This was [7, Exercise 37, p. 683], which should have pre-

ceded the present problem in the exercises!) We can begin this investigation by solving their

equations, which imply that tan θ = 1, so θ = π4

+ nπ, where n is any integer; that implies that

r = ± 1√ 2

: the + sign is associated with the cases where n is even, the − sign with those where

n is odd. But (r , θ ) =

1√ 2

, π4

+ 2mπ

and (r , θ ) =− 1√

2, π

4 + (2m + 1)π

are all the same point —

having cartesian coordinates 12

, 12.

But the curves also meet at the pole, even though this did not show up when we solved theequations. This is because the pole appears on the first curve with θ = nπ, and on the second

curve with θ = (n + 12

)π. The only practical way to determine whether curves intersect at the

pole is to examine each of them separately for that point.

Could it be that the curves meet in any other points? While a sketch does not suggest that,

we should never rely on a sketch! If we transform the first equation under the transformation

(r , θ ) → (−r , θ + π), we find that the equation does not change (except for a sign change on

both sides); the same applies to the second equation. Thus, in the present example, there are no

intersections other than the pole in which a point will appear on the two curves with di ff erent

coordinates. We can comfortably proceed to the integration part of the problem, confident that

we have not missed any points of intersection.

We can see that the region whose area is sought is symmetric about the line θ = π4

, so it

suffices to find half of it and to double it. The lower half of the area is subtended at the pole by

the arc 0 ≤ θ ≤ π4

of the circle r = sin θ ; hence

Area = 2 · 1

2

π4

0

sin2 θ d θ

= 1

2

π4

0

(1 − cos2θ ) d θ

= 1

2

θ − sin 2θ

2

π4

0

= π

8 − 1

4 .

(You could have solved this problem using Cartesian coordinates.)

10.4 Exercises

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[1, Exercise 22, p. 653] “Find the area enclosed by the loop of the strophoid r = 2cos θ −sec θ .”Solution: (see Figure 18 on page 3186) (The textbook did not expect a rigorous proof

0.2

-0.2

-0.6

10.80.60.40.2-0.2-0.4

0.6

0.4

0

-0.4

0

Figure 18: The strophoid r = 2 cos θ − sec θ

that the curve crosses itself only at the pole.) First we observe that the curve is entirelytraced out when θ passes through an interval of length 2π; so, without limiting generality,

we may confine ourselves to such an interval, say −π2

< θ < π2

, and thereby remove the

some possible ambiguity of coordinates. We must exclude the points where sec θ is

undefined. For convenience, let’s confine θ to the union of intervals

−π

2 < θ <

π

2 and

π

2 < θ <

3π

2 .

8/15/2019 Brown's Notes 2010



Consider a point on the second branch, say with θ = φ + π, where −π2

< φ < π2

. We find

that, for this point,

r = 2 cos(φ + π) − sec(φ + π)

= −2cos φ + sec φ

= −(2 cos φ − sec φ)

So we see that the point is the same as the point (−r , φ). Thus, in order to see the whole

curve, it suffices to investigate angles θ between − π2

and π2

; some of the points will,

however, appear with negative r -values.

We observe that the replacement of θ by −θ does not change the equation: this tells us

that the curve is symmetric about the polar axis and its extension.

For the problem to be meaningful, there should be just one loop. The curve passes

through the pole when 0 = 2 cos θ −sec θ , equivalently, when cos θ = ± 1√ 2

. In the interval

to which we have confined θ , this will occur when θ = ±π4

. If we follow the curve as θ

comes from −π2

, we find that r is negative, and the point is in the 2nd quadrant. It passes

through the pole first when θ = −π4

, and is then in the fourth quadrant, striking the polar

axis when θ = 0, at the point (r , θ ) = (1, 0) and it then moves into the first quadrant,

following the mirror image of the portion in the fourth quadrant, passing through the

pole again when θ = π4

, and moving into the 3rd quadrant. If we consider the cartesian

coordinates of the curve in parametric form, we have

x = 2cos2 θ − 1

y = sin2θ − tan θ

and see that, as cos θ → 0, x → −1, and y → ±∞, so the curve is asymptotic to the

vertical line x = −1. Our present problem is to determine the area of the loop from

θ = −π4

to θ = π4

, or, by symmetry,

2

1

2

π4

0

(2cos θ − sec θ )2 d θ =

π4

0

4cos2 θ + sec2 θ − 4

d θ

=

π4

0

2cos2θ + sec2 θ − 2

d θ

= [sin 2θ + tan θ − 2θ ]π4

0

=

1 + 1 − π

2

− 0 = 2 − π

2 .

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[1, Exercise 26, p. 653] Find the area of the region that lies inside the curve

r = 2 + sin θ (93)

and outside the curve

r = 3 sin θ . (94)

Solution: (see Figure 19 on page 3188) The first step is to determine where the curves

3

1

2

1 20-2 -1

-1

0

Figure 19: Curves r = 2 + sin θ , r = 3 sin θ

intersect. We can start by solving equations (93) and (94) algebraically. We find that

r = 3 and θ = π2

+ 2nπ, where n is any integer. The second curve is a circle with centre at

the point (r , θ ) =

32

, π2

. The first curve appears to be some sort of oval, and touches the

8/15/2019 Brown's Notes 2010



circle at its topmost point, which we shall call A, with coordinates (r , θ ) = 3, π2. Could

there be any other points of intersection? There don’t appear to be any, from the sketch,but we can resolve this question algebraically. The pole does not lie on first curve, since

that would entail that 0 = 2 + sin θ , i.e., that the sine of an angle exceeds 1 in magnitude.

Since that is not possible, the pole cannot lie on the curve. If there were to be any other

intersections, they would have to have diff erent sets of coordinates on the 2 curves. We

investigate what happens to (94) under the transformation

(r , θ ) → (−r , θ + π)

and find that the equation does not change significantly: it becomes −r = sin(θ + π) =

− sin θ , which is evidently equivalent to the original equation. Equation (93) changes to

−r = 2 + sin(θ + π) = 2 − sin θ , or

r = −2 + sin θ. (95)

When we solve this equation with (94), we find that r = −3, sin θ = −1: in the interval

0 ≤ θ ≤ 2π we find only the point (r , θ ) =−3, 3π

2

, which is the same point as (r , θ ) =

3, π2

, so we don’t find any new intersections. Thus there are no other intersections than

A.

A direct way to solve this problem is to find the area of the region bounded by the outer

curve, and then subtract from it the area of the disk inside. The outer curve is traced out

as θ ranges from 0 to 2π, so the area of the region it bounds will be

1

2

2π 0

(2 + sin θ )2 d θ = 1

2

2π 0

(4 + 4sin θ + sin2 θ ) d θ

= 1

2

2π 0

4 + 4sin θ +

1 − cos2θ

2

d θ

= 1

2

9θ

2 − 4cos θ − 1

4 sin 2θ

2π

0

= 9π

2

The inner curve bounds an area of

1

2

π 0

9sin2 θ d θ = 9

4

π 0

(1 − cos2θ ) d θ

= 9

4

θ − 1

2 sin 2θ

π

0

= 9

4 · π .

8/15/2019 Brown's Notes 2010



What would have happened if we had taken the upper limit of integration to be 2π? We

would have found twice the area, because the entire circle is swept out as θ ranges overan interval of length π.

Thus we see that the area of the region between the curves is

9π

2 − 9π

4 =

9π

4 .

Could we find the area by taking the diff erence of two squares under the integral sign,

as is suggested in [1, Example 2 and Figures 5 and 6, p. 651]? We could do this for the

portion of the area above the polar axis and its extension, where we would obtain

1

2

π 0

(2 + sin θ )2 − (3 sin θ )2

d θ

= 1

2

π 0

(4 + 4sin θ − 4(1 − cos2θ )) d θ

= 1

2 [−4cos θ + 2sin2θ ]π

0 = (2 + 0) − (−2 + 0) = 4

For the remainder of the area only curve (93) and the extended polar axis serve as a

boundary:

1

2

2π π

(2 + sin θ )2 d θ = 1

2

9θ

2 − 4cos θ − 1

4 sin 2θ

2π

π

= 1

2

(9π − 4 − 0) −

9π

2 − (−4) − 0

=

9π

2 − 4

which, when added to the area of the upper portion, gives the same area as before.

What would have happened if we had subtracted the square of 3 sin θ over the full range

of the integration? We would have been subtracting the area of the inside disk twice!

And what would have happened if we had taken (95) as the equation of the outer curve?

If we used it only to find the area of the (larger) region bounded, we would obtain thecorrect value there. But, if we had taken the diff erence of squares under the integral sign,

we would obtain values for regions that do not correspond to the one whose are we are

trying to find.

Before finding an area in polar coordinates, (or, indeed, in cartesian coordinates as well),

you are urged to make a sketch and study the element of area that you are summing in

the limit, to ensure that the integral represents the area that you are seeking.

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Example C.83 Another example of finding the intersections of curves.

Find the intersections of the lima¸con r = 1 − 2cos θ and the circle r = 1. [29]

Solution: (see Figure 20 on page 3191) When we solve the given equations, we find the inter-

1.5

0

1

0.5

-0.5

-1.5

10-1-2-3

-1

Figure 20: Intersections of the limacon r = 1

−2cos θ with the circle r = 1

section points (r , θ ) for which cos θ = 0, i.e.,1, ±π

2

. But a glance at the graphs shows that

there appears to be a third point of intersection. Under the transformation (r , θ ) → (−r , θ + π),

the given equation for the circle r = 1 does not change in a significant way. However, the

equation of the lima¸con changes to r = −1 − 2cos θ ; when we solve this latter equation with

the equation r = 1, we find that cos θ = −1, so the point (r , θ ) = (1, π) is another point of

intersection: i.e., (1, π) on the circle, (−1, 0) on the original equation for the limacon.

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To finish this investigation we should investigate whether the pole is on both curves, since

that may not be detected by this method of transforming the equations. We note, however, thatthe curve r = 1 cannot possibly contain the pole, since all of its points are 1 unit from the

pole. Thus the investigation is complete: there are three point of intersection, and we have

now found them all.

Example C.84 A problem from a recent examination:

[12 MARKS] Use polar coordinates — no other method will be accepted — to

find the area of the region bounded by the curve r = 2 and the line r = 1

cos θ , and

containing the pole.

Solution: (see Figure 21 on page 3193) The first step is to determine where the two curves

intersect. While the points of intersection can all be found algebraically, that procedure istedious, so it is best to make a sketch first to see whether the elaborate procedure is required.

You should be able to see immediately that the curve r = 2 is a circle of radius 2 centred at

the pole. But what about r = sec θ ? Since the secant is never less than 1 in magnitude, you

know this curve does not pass through the pole. You also know that, as θ ranges between −π2

and π2

, the function ranges from −∞ to +∞ monotonely, so the curve will be closest to the pole

when θ = 0 — when it is 1 unit from the pole. As θ increases to + π2

, r increases steadily —

monotonely — the curve crossing the circle r = 2 and moving a greater and greater distance

from the pole. What happens when r < 0? A point with coordinates (−a, θ ), where a > 0 will

still be a distance a from the pole! (It’s located on the ray obtained by turning the polar axis

through an angle of θ + π.) So, as r

→ −∞, the curve again crosses each of the circles centred

at the pole and moves a greater and distance from the pole. This is all you need to know inorder to solve the problem, but you should be able to see that this is a very simple curve: just

multiply both sides of r = 1

cos θ by cos θ , and you see that the equation is r cos θ = 1, or, in

cartesian coordinates, x = 1 — the curve is a straight line perpendicular to the polar axis!

We have seen that there will be just 2 points of intersection of the curves. Solving their

equations by eliminating r we obtain cos θ = 12

, so θ = ±π3

: the points of intersection are

(r , θ ) =

2, ±π

3

.

We saw that the element of area in the form of a thin triangle with apex at the pole (or a thin

sector of a disk with centre at the pole) has area of the form 1

2r 2 d θ . In this problem such an

element would have to be described in two diff erent ways:

−π

3 ≤ θ ≤ π

3: The area of this triangle is

1

2

π3

− π3

1

cos2 θ d θ =

1

2[tan θ ]

π3

− π3

=√

3 .

8/15/2019 Brown's Notes 2010



2

0

1

-1

20 1-2

-2

-1

Figure 21: Intersections of the curve r = sec θ with the circle r = 1

π

3 ≤ θ ≤ 5π

3 : The area of this portion of the disk is

1

2

5π3

π3

22

d θ =

1

2 [4θ ]

5π3

π3 =

8π

3 .

Thus the area of the region determined by the two curves, and containing the pole, is√

3 +8π

3 .

8/15/2019 Brown's Notes 2010



This problem could also have been solved by subtracting from the area of the disk, π22, the

area given by

1

2

π3

− π3

(22 − sec2 θ ) d θ = 1

2[4θ − tan θ ]

π3

− π3

= 4π

3 −

√ 3 .

8/15/2019 Brown's Notes 2010




Release Date: Friday, March 12th, 2010

C.28.1 §10.4 Areas and Lengths in Polar Coordinates (conclusion)

Arc Length To develop a formula for the arc length of a curve given in polar coordinates as

r = f (θ ), we can apply the theory of curves given in parametric form to the curve

x = f (θ ) · cos θ

y = f (θ ) · sin θ

We find that

dx

d θ 2

+dy

d θ 2

=dr

d θ 2

+ r 2

so the length is given by the integral

L =

b a

dr

d θ

2

+ r 2 d θ

where the limits θ = a and θ = b need to be determined from the parametrization of the curve.

As usual, you should be careful to determine the appropriate values of θ to define the portion

of the curve that interest you.

10.4 Exercises (continued)

[1, Exercise 46, p. 654] Find the exact length of the arc of the polar curve r = e2θ from θ = 0

to θ = 2π.

Solution:

Length =

2π 0

dr

d θ

2

+ r 2 d θ

=

2π

0

2e2θ 2+ e2θ r 2 d θ

=√

5

2π 0

e2θ d θ

=

√ 5

2

e2θ 2π

0=

√ 5e4π − 1

2

8/15/2019 Brown's Notes 2010



[1, Exercise 54, p. 654] Graph the curve r = cos2 θ 2

, and find its length.

Solution: The equation can be simplified to read

r =1 + cos

2 · θ

2

2

which we recognize to be a cardioid. As θ ranges over the interval 0 ≤ θ ≤ 2π the entire

curve is traced out once.

Length =

2π 0

dr

d θ

2

+ r 2 d θ

= 1

2

2π 0

(− sin θ )2 + (1 + cos θ )2 d θ

= 1

2

2π 0

2(1 + cos θ ) d θ

= 1

2

2π 0

2

2cos2

θ

2

d θ

=

2π 0

cos

θ 2

d θ

=

2sin

θ

2

π

0−

2sin θ

2

2π

π

= 2 − (−2) = 4 .

C.28.2 §10.5 Conic Sections

Omit this section.

Textbook Chapter 11. INFINITE SEQUENCES AND SERIES.

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C.28.3 §11.1 Sequences

A sequence is an ordered set of objects, usually labelled with either the non-negative integers

0, 1, . . . , n, . . . or the positive integers 1, 2, . . . , n, . . . . In this chapter we shall be interested in

sequences of real numbers or of real functions. Technically, such a sequence is a function that

maps either the non-negative integers or the positive integers on to the set of real numbers (or

to the set of real functions). We usually denote a sequence by a single letter, e.g. a, and show

the labelling by either a parenthesized value, as a(n), or a subscripted values, as an; when we

wish to talk about a sequence with specific terms, we may just list the terms with a general

term that shows the pattern we are describing, if there is one, as in

1, 1, 2, 3, 5, 8, 13, 21, 24, 55, 89, . . .

which is the Fibonacci sequence, in which every entry past the second is the sum of the twoentries immediately preceding it. Unless the general term is described unambiguously (as here,

where F n+2 = F n+1 + F n for all n ≥ 3), there will be more than one way to generalize a pattern

that appears to hold between a few terms at the beginning of the sequence.

Notation When a sequence consists of terms a0, a1, a2, . . . we may speak of the sequence

an, or perhaps the sequence ann=0,1,..., or sometimes simply the sequence an. Other notations

are also possible, and should be understandable from the context.

The limit of a sequence We define limn→∞

an = L in a way analogous to the definition of

lim x→∞ f ( x) = L; another way of writing the same definition is

an → L as n → ∞.

The precise definition is to be found as [1, Definition 2, p. 677], but is not on the syllabus —

you are not expected to be able to work with the precise definition of the limit of a sequence.

Note that we usually do not write the braces and when we speak of the limit of a sequence.

Since you bring to this course some understanding of limits of functions, we will occasion-

ally appeal to the following

11.1 Exercises

[1, Exercise 32, p. 685] Determine whether the sequence

ln nln 2n

converges.

Solution:

ln n

ln 2n=

ln n + ln 2 − ln 2

ln n + ln 2

= 1 − ln 2

ln 2n

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The subtracted fraction has a constant numerator, but the denominator is increasing, so

the fraction is decreasing as n increases; hence 1 minus the fraction is increasing; infact, we can see that the subtracted fraction is approaching 0, so the given fraction is

approaching 1.

C.28.4 Sketch of Solutions to Problems on the Final Examination in MATH 141 2005

01

(not for discussion at the lecture)

The best way to prepare for the examination in MATH 141 2010 01 is to study the textbook,

working problems, and verifying your solutions using the Student Solutions Manual: for sev-

eral reasons I don’t recommend your studying exclusively from old examinations. However

the old examinations do have some uses, and I am including one of them here, with solutions,

as a timely reminder that less than 1 month remains between now and the final examination in

this course. I am hoping to discuss last April’s examination briefly in a final lecture; there will

not be time for me to discuss the following examination in the lectures.

1. SHOW ALL YOUR WORK!

(a) [4 MARKS] Evaluate 3

0 | x

−1

|dx .

Solution:

3

0

| x − 1| dx =

1

0

| x − 1| dx +

3

1

| x − 1| dx

=

1

0

(1 − x) dx +

3

1

( x − 1) dx

=

x − x2

2

1

0

+

x2

2 − x

3

1

= 1

2 −0 + 3

2 − −1

2 =

5

2

(b) [3 MARKS] Evaluate d

dx

5

x

√ 4 + t 2 dt .

Solution: d

dx

5

x

√ 4 + t 2 dt = − d

dx

x

5

√ 4 + t 2 dt = −

√ 4 + x2 .

8/15/2019 Brown's Notes 2010



(c) [3 MARKS] Evaluate d

dx

x2

2π

sec t dt .

Solution: Let u = x2. By the Chain Rule

d

dx

x2

2π

sec t dt = d

du

u

2π

sec t dt · du

dx= sec u · 2 x = sec( x2) · 2 x .

(d) [3 MARKS] Evaluate

x5

3√

x3 + 1

dx .

Solution: Let u = x3 + 1

13

, so u3 = x3 + 1, 3u2 du = 3 x2 dx.

x5

3

√ x3 + 1

dx =

u3 − 1·u·u2 du = u

7

7 −u

4

4 +C =

x3 + 1 73

7 − x3 + 1 4

3

4 +C


For each of the following series you are expected to apply one or more tests for conver-

gence or divergence and determine whether the series is convergent. In each case you

must answer 3 questions:

• Name the test(s) that you are using.

• Explain why the test(s) you have chosen is / are applicable to the given series.

• Use the test(s) to conclude whether or not the series is convergent.

(a) [4 MARKS]

∞n=2

2 − cos n

n

Solution: This is a positive series. Since 0 < 1

n≤ 2 − cos n

n, for all n, the terms

are bounded below by the terms of the harmonic series, a positive series known

to diverge. We may apply the Comparison Test to such pairs of series, and may

conclude that the given series also diverges.

(b) [4 MARKS]

∞

n=0

n(−3)n

4n

Solution: As formulated in your textbook, we may apply the Ratio Test to the

sequence of absolute values of ratios of terms to their predecessors. Here

limn→∞

(n+1)(−3)n+1

4n+1

n(−3)n

4n

= limn→∞

1 +

1

n

· 3

4 =

3

4 < 1 ,

from which we may conclude that the given series is (absolutely) convergent.

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(c) [4 MARKS]

∞

n=2

1

n ln nSolution: Define f ( x) = 1

x ln x. Then f is a positive, continuous function, taking

on the values of the given sequence at the positive integer points. By the Integral

Test, the series and the following improper integral will either both converge or

both diverge. ∞

2

dx

x ln x= lim

b→∞

b

2

dx

x ln x= lim

b→∞[ln ln x]b

2 = limb→∞

lnln b − lnln2

which approaches +∞ as n → ∞. Hence the series diverges (to ∞).

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-

gral(s). It is not enough to quote a general formula: your integrals must have integrand

and limits specific to the given problems:

(a) [6 MARKS] The area of the region bounded by the parabola y = x2, the x-axis, and

the tangent to the parabola at the point (1, 1).

Solution: The tangent to the parabola at the point x, x2

has slope 2 x; at (1, 1) the

slope is 2. The equation of the tangent is y = 1 + 2( x − 1) = 2 x − 1, which line

intercepts the x-axis in the point

12

, 0.

If we evaluate the area by integration with respect to x, we have

12

0

x2 dx +

1

12

x2 − (2 x − 1)

dx =

x3

3

12

0

+

( x − 1)3

3

1

12

= 1

24 +

1

24

= 1

12

We can also integrate with respect to y: 1

0

− √

y + y + 1

2

dy =

−2

3 y

32 +

y2

4 +

y

2

1

0

= −2

3 +

1

4 +

1

2 =

1

12 .

(Note that you were not expected to evaluate the integrals: I did so in order to use

the opportunity of having 2 methods in order to verify my work.)

(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 4 the

region bounded by x = 0 and the curve x =

sin y (0 ≤ y ≤ π).

Solution: This problem is from your textbook [7, Exercise 25, p. 459], and is solved

in the Student Solution Manual, [9, p. 269]; it is one of the problems for which the

CD-Roms accompanying the textbook provide extra help.

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Integrating by the method of Cylindrical Shells centred on the line y = 4, we obtain

the volume to be 2π π

0(4 − y)

sin y dy.

(c) [3 MARKS] The area of the surface obtained by revolving about the y-axis the

curve y = e x, 1 ≤ y ≤ 2.

Solution: Integrating with respect to x gives the integral

ln 2

0

2π x√

1 + e2 x dx. Inte-

grating with respect to y gives the integral

2

1

2π(ln y)

1 +

1

y2 dy.

(d) [2 MARKS] The average value of the function 2 x

(1 + x2)2 over the interval 0 ≤ x ≤

2.

Solution: The answer expected was either 12

2

0

2 x(1+ x2)2 d x or

2

0

2 x(1+ x2)2 d x 2

0 1 dx

. The inte-

gral can be evaluated by using substitutions like u = x2, or u = x2 + 1.


[12 MARKS] Evaluate the indefinite integral 2 x3 + 3 x2 + 3

x2 + x − 12 dx .

Solution: Since the degree of the numerator of the rational function which is the inte-grand is not less than the degree of the denominator, the first step is to divide denominator

into numerator, obtaining

2 x3 + 3 x2 + 3

x2 + x − 12 = (2 x + 1) +

23 x + 15

x2 + x − 12 .

The polynomial quotient may be integrated to

(2 x + 1) dx = x2 + x + C . The excess

must be expanded using the method of Partial Fractions. Assuming an expansion of the

form23 x + 15

( x + 4)( x

−3)

= A

x + 4 +

B

x

−3

,

we multiply both sides by ( x+4)( x−3) to obtain the identity 23 x+15 = A( x−3)+ B( x+4).

We may obtain 2 equations for A, B by assigning to x the successive values x = 3 and

x = 4, obtaining B = 12 and A = 11. Hence 2 x3 + 3 x2 + 3

x2 + x − 12 dx =

2 x + 1 +

11

x + 4 +

12

x − 3

= x2 + x + 11ln | x + 4| − 12ln | x − 3| + C ,

8/15/2019 Brown's Notes 2010



which I then checked by diff erentiation.


(a) [9 MARKS] Use integration by parts to prove that, for integers

m ≥ 2, cosm x dx =

1

mcosm−1 x · sin x +

m − 1

m

cosm−2 x dx

Solution: (The instructions asked that the problem be solved using Integration by

Parts. Otherwise, the reduction formula could also have been proved by di ff erenti-

ating both sides and showing that the same derivative was obtained.)

Let u = cosm−1 x, dv = cos( x)

· dx, so du = (m

− 1) cosm−2 x

· (

−sin x) dx and

v = sin x. Then cosm x dx = cosm−1 x · sin x +

sin2 x(m − 1) cosm−2 x dx

= cosm−1 x · sin x + (m − 1)

cosm−2 x − cosm x

dx

⇒ m

m

x dx = cosm−1 x · sin x + (m − 1)

cosm−2 x dx

⇒

cosm x dx = 1


m − 1

m

cosm−2 x dx .

(b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate π2

0

cos6 x dx.

Solution:

cos6 x dx = 1

6 cos5 x · sin x +

5

6

cos4 x dx

= 1

6 cos5 x · sin x +

5

6

1

4 cos3 x · sin x +

3

4

cos2 x dx

= 1

6 cos5 x +

5

24 cos3 x sin x +

5

81

2 cos x

·sin x +

1

2 dx

=

1

6 cos5 x +

5

24 cos3 x +

5

16 cos x

sin x +

5

16 x + C .

Evaluating between 0 and π

2 we obtain 0 +

5π

32 =

5π

32.

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Consider the curve C defined by

x = 2cos t − cos2t

y = 2sin t − sin2t .

(a) [8 MARKS] Determine the points where the arc of the curve given by

π

4 ≤ t ≤ 7π

4

has a vertical tangent.

Solution:

dx

dt = −2sin t + 2sin2t

dy

dt = 2cos t − 2cos2t

Tangents are vertical where dy

dx is infinite, i.e., where dx

dt = 0 but

dy

dt 0.

dx

dt = 0 ⇔ −2sin t + 4sin t · cos t = 0

⇔ sin t

cos t − 1

2

= 0

⇔ sin t = 0 or cos t = 1

2

⇔ t = π, π

3, 5π

3

for t restricted to lie in the interval π

4 ≤ t ≤ 7π

4 . We check the values of

dy

dt at these

three points, and find that the respective values are −4, 2, 2, none of which is 0.

Hence the tangents will be vertical at the following points:

t = π ( x, y) = (−3, 0)

t = π

3 ( x, y) =

3

2,

√ 3

2

t =

5π

3 ( x, y) =

3

2, −−

√ 3

2

8/15/2019 Brown's Notes 2010



(b) [4 MARKS] Determine the length of the arc of the curve given by

0 ≤ t ≤ 2π .

Solution:

dx

dt

2

+

dy

dt

2

= (−2sin t + 2sin2t )2 + (2 cos t − 2cos2t )2

= 4 + 4 − 8(sin t · sin 2t + cos t · cos2t )

= 8 − 8 cos(2t − t ) = 8(1 − cos t ) .

The length of the arc is

√ 8

2π

0

√ 1 − cos t dt =

√ 8

2π

0

2sin2 t

2 dt

= 4

2π

0

sin t

2

dt

= 4

2π

0

sin

t

2

dt

= −8

cos

t

2

2π

0= −8(−1 − 1) = 16 .


(a) [5 MARKS] Determine whether the following integral is convergent; if it is con-

vergent, determine its value: 1

−1

dx√ 1 − x2

Solution: The integral is improper because the integrand is not defined — becomes

infinite — at both ends of the interval of integration. According to the definition,

we must integrate between points away from ±1, and allow the limit to be taken

independently. The safest way to do that is to split the improper integral into two,

and then to evaluate the two of them separately. 1

−1

dx√ 1 − x2

= lima→−1+

0

a

dx√ 1 − x2

+ limb→+1−

b

0

dx√ 1 − x2

= lima→−1+

(arcsin 0 − arcsin a) + limb→+1−

(arcsin b − arcsin 0)

= 0 −−π

2

+

π

2

− 0 = π .

8/15/2019 Brown's Notes 2010



(b) [5 MARKS] Determine whether the following series is conditionally convergent,

absolutely convergent, or divergent.∞

n=1

(−1)n n!

nn

Solution: I will apply the Ratio Test to the sequence of absolute values of the ratios

of terms to their predecessors:

(−1)n+1 (n+1)!

(n+1)n+1

(−1)n n!nn

= (n + 1)!

(n + 1)n+1 · nn

n! =

1 +

1

n

−n

→ 1

e

as n → ∞. As this limit is less than 1, the given series is absolutely convergent,

hence divergent.

(c) [3 MARKS] Determine whether the sequence an

= ln(n + 1)−

ln n is convergent;

if it is convergent, carefully determine its limit.

Solution: This is not a question about telescoping series! It is a problem about

sequences!

ln(n + 1) − ln n = ln

1 +

1

n

. As n → ∞, 1 + 1

n → 1, so ln

1 + 1

n

→ ln 1 = 0 (by

continuity of the logarithm function from the right at the point 1).


[12 MARKS] Find the area of the region bounded by the curves

r = 4 + 4sin θ r sin θ = 3

which does not contain the pole.

Solution: The problem is to determine the area cut o ff from the cardioid r = 4 + 4 sin θ

by the line y = 3. First we must determine where the curves cross. Solving the polar

equations by eliminating sin θ between them we obtain r 2 − 4r − 12 = 0, equivalent

to (r − 6)(r + 2) = 0. The values r = 6, −2 yield corresponding values sin θ = 12

, −32

.

Of these the second is impossible, as a sine cannot be less than 1. We conclude that

sin θ = 12

and the values of θ in the interval 0 ≤ θ ≤ 2π are θ = π6

, 5π6

: the curves intersect

in (r , θ ) =

6,

π

6

and (r , θ ) =

6,

5π

6

. A naive way to solve the problem is to integratedirectly,

Area = 1

2

5π6

π6

(4 + 4sin θ )2 − (3 csc θ )2

d θ

= 1

2

5π6

π6

16

1 − cos2θ

2 + 2sin θ + 1

− 9csc2 θ

d θ

8/15/2019 Brown's Notes 2010



= 1

2 [24θ

−32 cos θ

−4sin2θ + 9cot θ ]

5π6

π6

= 8π + 12 √ 3 .

Another way to find the area is to observe that the triangle with base on y = 3 between

the points of intersection and its third vertex at the pole has area 12

(6√

3) ·3 = 9√

3. Then

one can simply subtract this area from 1

2

5π6

π6

(4 + 4sin θ )2 d θ .

8/15/2019 Brown's Notes 2010




Release Date: Monday, March 15th, 2010

C.29.1 §11.1 Sequences (conclusion)

Theorem C.85 [1, Theorem 3, p. 678] If f is a function, and an is such that f (n) = an for all

n, and if lim x→∞

f ( x) = L, then limn→∞

an = L.

We also generalize our definition of the limit of a sequence to permit limits to be ±∞; the

definition is again analogous to the corresponding definition for the meaning of lim x→∞

f ( x) =

±∞.

A sequence that has limit L is said to converge to L; if there is no limit, the sequence issaid to diverge. We do not permit L to be ±∞ in this usage, so a sequence whose limit is ±∞is said to diverge.

Limit Laws We can prove limit laws for sequences analogous to the limit laws we saw in

the previous course for functions. I shall not give the details in these notes. Remember the

usual restriction that we cannot divide by 0, so there must be a restriction on the quotient law.

The limit laws can be extended to infinite limits whenever the operations can be justified; so,

we can think of ∞ + ∞ = ∞, but we cannot attach a meaning to ∞ + (−∞), nor of 0 · ∞, as it is

not possible to assign to these expressions a meaning that will be consistent with the algebraic

operations on real numbers.

Increasing and Decreasing Sequences We defined the concepts of increasing and decreas-

ing in connection with real functions of a real variable (cf. [1, p. 20]. The identical defini-

tions apply for sequences, where we consider the domain of the function to be the positive

or non-negative integers. When a sequence is either increasing or decreasing we say that it is

monotonic or monotone. Sometimes we find it convenient to work with a slightly weaker prop-

erty than increasing, and may speak of a sequence as being non-decreasing, which permits the

function to remain constant for a while.

A function is bounded above if there exists a number M which is greater than all values of

the function. For example, the sine function is bounded above, since sin x ≤ 1 for all x in its

domain. We could also have observed that sin x < 10000, and conclude that such a statement justifies the conclusion that the function is bounded above: we don’t care how high above the

graph of the function the bounding line appears, only that such a line exists. But the function

tan x is not bounded above. In the same way we can define bounded below and bounded —

meaning bounded both above and below. An important theorem we shall need in this chapter

is

8/15/2019 Brown's Notes 2010



Theorem C.86 1. A sequence that is bounded above and monotonely increasing (or even

monotonely non-decreasing) is convergent.

2. A sequence that is bounded below and monotonely decreasing (or even monotonely non-

increasing) is convergent.

(Note that the theorem stated here is stronger than [1, Monotonic Sequence Theorem, p., 683].)

Example C.87 [1, Exercise 36, p. 685] Determine whether the sequence an = ln(n + 1) − ln n

converges.

Solution: We might be tempted to analyze a diff erence by using the diff erence law for limits.

But we find that each of the terms approaches +∞, and we cannot give a meaning to ∞ − ∞.

So that approach will not work.

However, if we observe that the diff erence of logarithms is

ln n + 1

n= ln

1 +

1

n

we can reason as follows. As n → ∞, 1n → 0, 1 + 1

n → 1 + 0 = 1. By continuity of the

logarithm, the sequence will approach ln 1 = 0. (I am using Theorem C.85 above.)

Sequences of powers For a fixed real number a we know the behavior of a function a x as

x → ∞, and how it depends on a; this permits us to study the behavior of an when the exponent

is restricted to integer values:

limn→∞

an = +∞

if a > 1

limn→∞

an = 1 if a = 1

limn→∞

an = 0 if −1 < a < 1

limn→∞

an does not exist if a ≤ −1

(Note that, when a limit is ±∞, as in the first case above, we still say that the limit does not

exist!)

Example C.88 Consider the sequence an, where an = n!

nn. Does it converge? If so, to what

value?

Solution: Consider the limit of the ratio of an+1 to an. It isan+1

an

= (n + 1)!

n! · nn

(n + 1)n+1 =

11 + 1

n

n → 1

e

as n → ∞. Thus, for sufficiently large n, every term is less than half of the one before it, which

is positive. Compared to the nth term, the (n + 10)th will be less than 11000

of the size; the

n + 20th will be less than one-millionth the size, etc. The sequence is thus approaching 0.

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8/15/2019 Brown's Notes 2010



definition, to permit us to speak of the sum of all the terms in a sequence. In this context we

no longer write the sequence with commas, as

a0, a1, . . . , an, . . . ,

but, instead, place plus signs between the terms, writing

a0 + a1 + . . . + an + . . . ,

or, more compactly, as∞

n=0

an ,

and we speak of a series, rather than a sequence. Note that the word series is both singularand plural in English: there is no word “serie” in English55. Our definition proceeds from the

partial sums, which we define to be the terms in the sequence

a0, a0 + a1, a0 + a1 + a2, . . . , a0 + a1 + . . . + an, . . .

or, more compactly, as

0n=0

an,

1n=0

an, . . . ,

mn=0

an, . . . ,

all of which are finite sums, and therefore well defined. (Note that we had to use a diff erentletter — m — for the general term in this sequence, because the letter n was “busy”56.) We say

that ∞n=0

an = L

if L is the limit of the sequence of partial sums of the series. The series is then said to converge,

or to be convergent ; if it does not converge, it diverges or is divergent . Here, as with series

and functions, we generalize to write that a sum = ∞ or = −∞, but still describe such series as

divergent. We will often consider series which we will know to be convergent, without being

able to specify the precise value of the limit.

“Geometric” series. A geometric series is one, each of whose terms after the first is a con-

stant multiple of its predecessor. We will often represent such series as

a + ar + ar 2 + . . . + ar n + . . .

55But the singular word for “series” in French is s ´ erie, and the plural is s´ eries.56Technically, we call n a bound variable in this case.

8/15/2019 Brown's Notes 2010



where a 0 and the “common ratio” is the ratio of each term to its predecessor, here denoted

by r . By “standard” methods, usually seen in high school, one can prove that the value of thepartial sum of m + 1 terms is

a + ar + . . . + ar m =

mn=0

ar n =

a (1 − r m)

1 − r if r 1

(m + 1)a if r = 1.

This ratio can be seen to have the following limit properties:

a + ar + . . . + ar m + . . . = limm

→∞ m

n=0

ar n =

= a

1 − r if −1 < r < 1

= +∞ if r ≥ 1 and a > 0

= −∞ if r ≥ 1 and a < 0diverges if r ≤ −1 and a 0

.

8/15/2019 Brown's Notes 2010




Release Date: Wednesday, March 17th, 2010

C.30.1 §11.2 Series (conclusion)

Divergence of the “harmonic” series. The harmonic series is

1 + 1

2 +

1

3 + . . . +

1

n+ . . . .

In the next section we will see a proof, using an improper integral, that this series diverges.

Here we shall see a simpler proof without integrals. We need only observe the following

inequalities:

12 ≥ 1

2

13

+ 14

> 14

+ 14

= 24

= 12

15

+ 16

+ 17

+ 18

> 18

+ 18

+ 18

+ 18

= 48

= 12

. . .

1

2n−1 + 1 +

1

2n−1 + 2 + . . . +

1

2n − 1 +

1

2n > 2n−1 · 1

2n =

1

2

Thus we can make the partial sums as large as we like by proceeding out sufficiently far in theseries. It follows that ∞n=1

1

n= +∞ ,

so the harmonic series diverges.

“Necessary” and “sufficient” conditions. Suppose that A and B are sentences that may be

true or false. If A cannot be true except when B is true, we say that the truth of A entails the

truth of B, or that

A implies B

and write this symbolically as

A ⇒ B .

We call A a su fficient condition for B. An example of such an implication where x and y are

members of the “universe” of real numbers is

x = y2 − 2 y + 4 ⇒ x ≥ 3

8/15/2019 Brown's Notes 2010



since y2 − 2 y + 4 = ( y − 1)2 + 3, and a square cannot be negative.

We can interpret the statement A ⇒ B

in another way, by observing that, when A is true, B cannot be false. We can say that B is a

necessary condition for A, since A cannot be true unless B is true.

Mathematicians usually search for conditions that are both necessary and sufficient, since

these can characterize a situation. But often we have to be satisfied with conditions that are

either of one type or the other.

A “necessary” condition for convergence of a series The implication that interests use here

is in the “universe” of series of real numbers. It states that

∞n=0

an is convergent ⇒ limn→∞

an = 0 .

This — once it has been proved — is a test that can be applied to a series to see whether

the series is convergent. It can never prove that a series is convergent! What it can prove,

sometimes, is that a certain series is not convergent, since no series that “fails the test” can

be convergent. Many textbooks call this test the “nth term test”; your textbook calls it “‘The’

test for divergence”. This is not a standard term, and you might wish to avoid using it, since

a listener who has not read Stewart’s books might not know what you are referring to. 57 In

practice you should “internalize” this test and always apply it, since you might otherwise

waste time trying to prove that a divergent series is convergent.

“Telescoping” series. Sometimes the partial sums can be interpreted as sums where there

is heavy cancellation of intermediate terms, leaving only a few at each end. An example is∞

n=1

1

n(n + 1) =

∞n=1

1

n− 1

n + 1

where the partial sum of the first N terms is simply 1 − 1

N + 1.

(cf. [1, Example 6, p. 691]).

Changes of variable in a sum. We have already seen the concept of changing a variable in

connection with definite integrals. We can carry out similar changes in sums — both finite and

infinite, although we will not investigate all the details. So, for example, we can change the

57You will meet other tests that could be given such a name, although this is the most likely one to distinguish

in this way. Stewart’s name for the test may be no more objectionable than the name in general use: it is not good

form to use the symbol n in the name of a test, since the test does not depend on giving that particular name to

the variable that indexes the members of the sequence; my objections to the name Stewart assigns are that (1) it is

not universally accepted; and (2) the definite article suggests uniqueness, and this is not the only test that exists.

8/15/2019 Brown's Notes 2010



name of the index of summation, and write

m

r =0

ar in place of the sum

m

n=0

an, by a “change of

variable” given by r = n. More generally, we could define s = n + 4, say, and change

mn=0

an

into

m+4s=4

as−4. Notice how the “limits of summation” have to be changed in the same way that

we changed the limits of integration in a definite integral.

Operations on series. Convergent series may be added term by term, or multiplied by a

constant: ∞n=0

(an + bn) =

∞n=0

an +

∞n=0

bn

∞n=0

can = c

∞n=0

an

But note well: while you may add convergent series term by term, you may not rearrange a

series; we shall see later that rearrangement of the terms can result in a series having a di ff erent

sum, or even in a convergent series being rendered divergent.

11.2 Exercises

[1, Exercise 36, p. 695] Determine whether the series

∞n=1

2n2 + 4n + 3

is convergent or diver-

gent by expressing the partial sum as a telescoping sum (as in [1, Example 6, p. 691]).

If it is convergent, find its sum.

Solution: Since

2

n2 + 4n + 3 =

2

(n + 1)(n + 3) =

1

n + 1 − 1

n + 3 ,

the N th partial sum is equal to

N n=1

2

n2 + 4n + 3 =

N n=1

1

n + 1 − 1

n + 3

=

N n=1

1

n + 1 −

N n=1

1

n + 3

=

N n=1

1

n + 1 −

N +2m=3

1

m + 1

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setting m = n + 2 in the second sum

=

N n=1

1n + 1

− N +2n=3

1n + 1

replacing the variable m by n

=

2n=1

1

n + 1 +

N n=3

1

n + 1 −

N n=3

1

n + 1 −

N +2n= N +1

1

n + 1

=

2n=1

1

n + 1 −

N +2n= N +1

1

n + 1

= 1

2 +

1

3 −

1

N + 2 −

1

N + 3 .

As N → ∞, this partial sum approaches 12

+ 13

= 56

. (In this type of problem we don’t

expect students to be write the proof out formally in this way; I have included this

solution so that you can see how the problem can be solved “properly”, without the use

of “dots”.


∞n=1

ln n

n + 1 is convergent or divergent

by expressing sn is a telescoping sum (as in [1, Example 6, p. 691]). If it is convergent,

find its sum.

Solution: In a similar way to that shown above in the solution to [1, Exercise 36, p. 695],

we can show that the partial sum of the first N terms of the series is ln 1 − ln N + 1. As

N → ∞ the partial sum approaches −∞: the series diverges.

[1, Exercise 44, p. 695] Express the “repeating decimal expansion” 6.254 as a ratio of inte-

gers.

Solution: By a “repeating decimal expansion”, we intend that the digits under the hori-

zontal line are to be repeated as a subsequence of the expansion indefinitely; thus

6.254 = 6.25454545454...,

which we can interpret as the sum of a series

6 + 2

10 +

54

1000 +

54

100000 +

54

10000000 + . . .

which can be seen as a constant added to a geometric series:6 +

2

10

+

54

1000

1 + 1

100 +

1

100

2

+

1

100

3

+ . . .

8/15/2019 Brown's Notes 2010



whose sum is

6 +

2

10 +

54

1000 · 1

1 − 1100

= 6 +

1

5 +

3

55 =

344

55 .

(In this way we can show that any “repeating decimal” is a rational number.)

[1, Exercise 50, p. 695] Find the values of x for which the series

∞n=0

( x + 3)n

2n converges. Find

the sum of the series for those values of x.

Solution: The given series can be rewritten as

∞n=0

( x + 3)

2

n

: it can be viewed as a

geometric series with initial term 1, and common ratio x + 3

2

. We know that a geometric

series converges if and only if its common ratio is less than 1 in magnitude. Thus the

given series converges iff

x + 3

2

< 1. Solving this inequality for x we obtain

x + 3

2

< 1 ⇔ −1 < x + 3

2 < +1

⇔ −2 < x + 3 < 2

⇔ −5 < x < −1 .

Thus the given series converges for and only for x ∈ (−5, −1). For x in this interval the

sum of the series is

first term

1 − common ratio =

1

1 − x+32

= − 2

1 + x .

C.30.2 §11.3 The Integral Test and Estimates of Sums

Thus far we have met only one “test” that can be applied to series to investigate whether the

series converges. That is the test Stewart calls “The Test for Divergence”, and it gives negative

information: it states that, if limn→∞

an does not exist, or if the limit does exist but its value is

not 0, then the series

∞

n=0

an is divergent; this test cannot be used to confirm a suspicion that a

series does converge. This is the only test we shall have that can be applied to “general” series,

where the signs of the terms do not follow a specific pattern.

The Integral Test. The idea of this test is to interpret the terms of a series

∞n=1

an as a sum

of areas. We can do this by interpreting an as the area of a rectangle whose width is 1, and

8/15/2019 Brown's Notes 2010



whose height is an. The test will be restricted to sequences where the terms are monotonely

decreasing. Suppose that we know of a function f defined on the interval [1, ∞), whose graphpasses through the left upper end-points of the rectangles. Then the area under the curve will

be less than the sum of the areas of the rectangles. If we can show that the area under the curve

is infinite, we will be able to conclude that the sum of the series is divergent. In a similar way,

if we can pass the graph of a function f through the right upper endpoints of the rectangles,

then the area under the curve will exceed the sum of the areas of the rectangles: if the area

under the curve is convergent, then the same can be said about the series. In these ways we

can shown that the series converges precisely when the integral representing the area under

the curve converges! We call this result The “Integral” Test.

8/15/2019 Brown's Notes 2010




Release Date: Friday, March 19th, 2010; corrected on 08 April, 2010; subject to further

correction

Review of the preceding lecture

• Theorem:

∞i=0

ai converges ⇒ limn→∞

an = 0

• The contrapositive of this theorem is “The” Test for Divergence: If limn→∞

an either does

not exist, or does exist but does not equal 0, then the series

∞

i=0

ai diverges.

• The harmonic series diverges — proved by grouping successive terms.

• “telescoping” series

• evaluating a “repeating” decimal expansion as a rational number

• The Integral Test — requiring the association with a positive series of a continuous,

decreasing, positive function f taking, at the positive integer points, values equal to the

terms of the series

C.31.1 §11.3 The Integral Test and Estimates of Sums (conclusion)

Theorem C.91 (The Integral Test) If f is a continuous, positive, decreasing 58 function de-

fined on the interval [1, ∞), and if the sequence an has the property that an = f (n). Then

∞n=1

an is convergent ⇔∞

1

f ( x) dx is convergent .

The convergence or divergence of both the series and the improper integral do not depend on

the lower limits of the sum and the definite integral. (However, when, below, we investigate

bounds for the actual value of the sum of the series, then our bounds will depend on the specificvalues we choose for the limits of sum and integral.)

58Actually, it suffices for the function to be non-increasing: we can permit the function values to stay at the

same level so long as they eventually decrease again and eventually approach 0.

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Tests of Positive Series for convergence. Most of the tests we shall meet require a specific

arrangement of signs for the terms. In fact, all but one of them will require that all the signsare the same. We call such series “positive” series, thinking of all the signs as being +; but

similar results hold when all signs are −. The first test of positive series will be discussed in

this section. I do not think of this as the most elementary of the tests, and would have preferred

to discuss the material of the next sections first; but I will grudgingly follow the order of topics

in the textbook.

“Ultimate” satisfaction of properties imposed on the terms of a series. In the statement

given for the a sum beginning with the 1st term, and compared it with the integral whose

lower bound was x = 1. Both of these lower bounds can be replaced by any integers you

like, provided the series and function are defined whenever you refer to them. Changing these

lower bounds does not change the truth of the logical equivalence. It could, however, change

the value of the sum of the series or the definite integral.

Application: the “ p-series”. The “ p-series” are the series 1

n p, where p is a positive

constant. The integral test shows that 1

n p is convergent if and only if p > 1 .

The case p = 1 is the harmonic series, which we have already shown to be divergent, using

a diff erent proof. (This application is sometimes called the “ p-series test”.) Since the tests

we will meet in [1, §11.4] involve comparing a given series with series that we know to beconvergent and divergent, the p-series are particularly important as they provide us with a

family of series that can be used for comparison purposes.

Estimating the Sum of a Series In the proof of the Integral Test we compared the sum of a

series of decreasing positive terms with the area under the graph of a function that is positive

and decreasing. This comparison can be refined, and gives rise to inequalities that bound the

value of the sum from below and above.

By associating with a decreasing, positive series

an a decreasing positive function f , we

are able to “trap” the value of the partial sums

N

n=1

an between the definite integrals giving the

areas for regions respectively contained in and containing the region represented by the partial

sum, when interpreted as the area of a region formed by rectangles of width 1 and respective

heights 1, 2, . . ., N . From the inequalities

N n=1

an ≤ a1 +

N

1

f ( x) dx

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N

n=1

an

≥ N +1

1

f ( x) dx

we are able to relate the convergence of the infinite sum to that of the improper integral ∞

1

f ( x) dx.

Let us, extending the notation used in the enunciation of the theorem, define

Rn =

∞m=n+1

am;

that is, Rn is the sum of the “tail” or Remainder of the series, starting with the (n + 1)st term.

Then ∞

n+1

f ( x) dx ≤ Rn ≤ ∞

n

f ( x) dx ,

or, alternatively, ∞

n+1

f ( x) dx ≤ Rn ≤ an+1 +

∞

n+1

f ( x) dx ,

or

0 ≤ Rn − ∞

n+1

f ( x) dx ≤ an+1 .

In particular, when n = 0, this gives the inequalities

0 ≤∞

n=1

an − ∞

1

f ( x) dx ≤ a1 .

While, for the purposes of testing convergence, it is su fficient to demand “ultimate” sat-

isfaction of the conditions of the Integral Test, this is not sufficient if we wish to determine

bounds for the sum. In that case the estimations discussed above depend on the comparison of

the area under a curve and the sum of the areas of the step function that represents the series;

these comparisons are valid only when the condition the the function be decreasing is satisfied.

It is possible to refine this type of bounding of partial sums and remainders, as in [1, Examples

5,6, pp. 701-702], but we will not go further in this course.

Example C.92 We proved earlier that the telescoping series∞

n=1

1

n(n + 1) converges, and we

found the sum. The integral test could be used to prove that it converges, but the test would

not give the exact value of the sum, although it would provide bounds which compare the sum

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to the improper integral ∞

1

1

x( x + 1) d x = lim

a→∞((ln a − ln 1) − (ln(a + 1) − ln 2))

= lima→∞

ln

a

a + 1 + ln 2

= ln 1 + ln 2 = 0 + ln 2 = ln 2 ,

so the sum of the series is bounded between ln 2 = 0.6931471806 and ln 2+ 12

= 1.1931471806,

which is a weaker result than the fact we already know that the sum is equal to 1.

11.3 Exercises

[1, Exercise 6, p. 703] Use the Integral Test to determine whether the series∞

n=1

1√ n + 4

is

convergent or divergent.

Solution: We begin by verifying that the Integral Test is applicable to this series:

• Since the terms are reciprocals of square roots, they are positive.

• The function f ( x) = 1√

x + 1is decreasing. This can be shown either by showing

that f ( x) = −12

( x + 1)− 32 < 0, or by the following reasoning:

x is an increasing function of x⇒ x + 1 is an increasing function of x

⇒√

x + 1 is an increasing function of x since√

preserves order

⇒ 1√ x + 1

is a decreasing function of x.

The Integral Test tells us that the series will converge iff the improper integral

∞ 1

dx√ x + 1

converges. But

∞ 1

dx√ x + 1

= lima→∞

a 1

dx√ x + 1

= lima→∞

2 √ x + 1

a

1= +∞ .

so the series must also diverge.

[1, Exercise 20, p. 704] To investigate the convergence of

∞n=1

1

n2 − 4n + 5.

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Solution: We know that 1

x2 − 4 x + 5 d x =

1

( x − 2)2 + 1 d x = arctan( x − 2) + C ,

and might be tempted to compare the integral

∞

1

1

x2 − 4 x + 5 d x with the series. This,

however, is not directly possible from the Integral Test, because the function f ( x) =1

x2 − 4 x + 5 is decreasing only for x ≥ 2: from x = 1 to x = 2 the function is increasing.

Thus we can apply the integral test to the series

∞n=2

1

x2 − 4 x + 5, provided we first verify

that the necessary conditions are satisfied by the function f whose values at the positive

integers coincide with the terms of the series:

1. Is f decreasing?

d

dx f ( x) =

−(2 x − 4) x2 − 4 x + 5

=

−2( x − 2)( x − 2)2 + 1

2 .

The numerator is negative for x > 2; the denominator is a non-zero square, always

positive. Thus the ratio is negative for x > 2, and so f is decreasing for x > 2.

2. Is f continuous? f is a rational function, and [1, Theorem 5(b), p. 122] all rational

functions are continuous on their domain. (The domain of f is all of R.)

Bounding the sum for n ≥ 2 from below yields

∞n=2

1

n2 − 4n + 5 ≥

∞

2

dx

x2 − 4 x + 5

= lima→∞

arctan(a − 2) − arctan 0

= π

2 − 0 =

π

2 .

Bounding the sum for n ≥ 3 from above yields

∞n=3

1

n2 − 4n + 5 ≤

∞

2

dx

x2 − 4 x + 5

= π

2 ,

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8/15/2019 Brown's Notes 2010



C.32 Supplementary Notes for the Lecture of March 22nd, 2010

Distribution Date: Monday, March 22nd, 2010, subject to further revision

Review of the last lecture

• After reviewing the previous lecture, I completed my treatment of the Integral Test.

• It time permits, I may return to further discussion of the use of integrals to bound series.

• Discussion included [1, Exercise 28, p. 704],

∞n=3

1

n(ln n) (lnln n) p , which can be shown

to behave “similarly” to p-series.

• There are often many, significantly diff erent choices of series with which to compare.

You need to work problems and check your solutions afterward in the manual to develop

the necessary experience.

C.32.1 §11.4 The Comparison Tests

We continue to study series in a “reverse-logical” order. That is, as we proceed, we will be

encountering properties that are more and more basic. In this section we will meet another

theorem that — like the Integral Test — applies only to series whose terms all have the same

sign, the so-called “positive series”, or “series of positive terms”. Finally, in [1,

§11.6] we will

see a theorem [1, Theorem 3, p., 715] that will justify our continued investigation of serieswith positive terms. When we have covered all the sections on the syllabus, I hope to return to

the various tests.

The comparison tests are designed so that we can infer the convergence or divergence of

a given positive series by comparison of its terms with those of another series that we know

to be convergent or divergent. So, in order to use this test, we need to have available as large

as possible a family of positive series whose convergence or divergence are known. Let’s

remember for which series we know such facts; these include

• positive geometric series of common ratio less than 1 (convergent)

• positive geometric series of common ration 1 or more (divergent)

• harmonic series (divergent)

• p-series (convergent if and only if p > 1)

• positive series whose terms do not approach 0 (divergent)

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8/15/2019 Brown's Notes 2010



Estimating Sums This question will be illustrated by the solution below of part of [1, Exer-

cise 32, p. 709].

Example C.95 ([7, Exercise 20, p. 735]) Determine whether the series

∞n=1

1 + 2n

1 + 3n converges

or diverges.

Solution: We could not compare the terms of this series naively with those of 2n

3n, which is a

convergent geometric series because the terms here are slightly larger. However a number of

other possibilities suggest themselves. For example, we have

1 + 2n

1 + 3n ≤ 1 + 2n

3n =

1

3n

+ 2

3n

< 2 2

3n

which is the general term of a geometric series with common ratio 23

< 1, which is conver-

gent. Other comparisons are possible, some of them complicated to prove. For the purpose of

proving only convergence (and not bounding the series) one tries to find the simplest possible

comparison.

This problem can also be solved by applying the Limit Comparison Test, again in compar-

ison with

23

n:

limn

→∞

1 + 2n

1 + 3n

23n = lim

n

→∞ 1 + 2n

2n · 3n

1 + 3n = lim

n→∞

2−n + 1 · 3−n + 1

= 1 × 1 = 1.

Since the limit is a positive real number, and since the geometric series

23

nconverges, the

given series must also converge.

Example C.96 ([7, Exercise 34, p. 735]) Estimate the error when the sum of the first 10 terms

of the following sum is used to approximate the sum of the series:

∞

n=1

1 + cos n

n5 .

Solution: The “tail” of the series will be estimated from above. Observe that

1 + cos n

n5 ≤ 2

n5 .

We know that1

n5 ≤ 1

x5 for n − 1 ≤ x ≤ n .

8/15/2019 Brown's Notes 2010



Comparing the sum of the terms on the left — interpreted as rectangles of width 1 lying under

the graph of f ( x) =

1

x5 — we have∞

n=11

1

n5 ≤

∞

10

dx

x5 =

1

104 .

It follows that the “tail” of the series — the sum beginning with term n = 11 — is no greater

than 2 × 10−4. This gives an upper bound for the error if the series is truncated at term n = 10.

This method will not give a useful lower bound for the error, however, since we know only that

1 + cos n

n5 ≥ 0

n5 .

This is certainly not the best possible lower bound, but we can’t do better in this course.

11.4 Exercises In some of the discussions below I give only hints about the solution.

[1, Exercise 12, p. 709] Determine whether the series converges or diverges:

∞n=0

1 + sin n

10n .

Solution: The Limit Comparison Test cannot be used, since the terms behave erratically

because of the sine function, and there is not likely to exist a limit, no matter what series

you choose to compare with. But

1 + sin n10n

< 210n

,

so the terms of the series are smaller than the terms of a geometric series with common

ratio 110

, which is less than 1, and must converge.

[1, Exercise 14, p. 709] Determine whether the series converges or diverges:

∞n=0

√ n

n − 1.

Solution: The terms can be made smaller by making the denominator larger, into n. That

yields a divergent p-series.

[1, Exercise 18, p. 709] Determine whether the series converges or diverges:∞

n=0

12n + 3

.

Solution: The terms can be made smaller by making the denominator into 2n + 4. The

new series is a multiple

12

of a harmonic series. The harmonic series diverges, and that

property doesn’t change if we discard 2 terms at the beginning, nor if we then multiply

by 12

.

8/15/2019 Brown's Notes 2010




∞

n=1

n!

nn

converges or diverges.

Solution: The general term is

n!

nn =

1

n· 2

n· 3

n· . . . · n − 1

n· n

n

≤ 1

n· 2

n· 1 =

2

n2

for n ≥ 3, which is the general term in a convergent p-series. Hence the given series

is convergent. It is impractical to use the limit comparison test here, because the limit

would be difficult to compute, and because it would be zero or infinite, depending on

which series is in the numerator; our version of this test does not permit the limit to be

either 0 or ∞.


∞n=1

sin 1

nconverges or diverges.

Solution: We know the value of limn→∞

sin 1n

1n

to be 1; so the Limit Comparison Test shows

this series diverges because the harmonic series diverges. Could we have used the Com-

parison Test? Unfortunately, the geometric argument that we used to prove the result

quoted earlier showed [1, p. 190-191] that

0 < sin 1n

< 1n

.

In its present form this inequality is not useful for proving the divergence of the given

series, since the series known to be divergent occupies the outer position rather than

lying between the terms of the given series and 0.


∞n=1

1

n1+ 1n

converges or diverges.

Solution: It appears that the series “resembles” the Harmonic Series, and we will try to

compare it in the limit. The guess is a fortunate one, since the limit does exist.

limn→∞

1n1+

1n

1n

= limn→∞

1

n1n

= limn→∞

1

eln n

n

= 1

elim

n→∞ln n

n

by continuity of exponential

8/15/2019 Brown's Notes 2010



= 1

e0 = 1 by L’Hospital’s Rule

We apply the Limit Comparison Test: since the Harmonic Series diverges and the ratio

of the terms of the given series to those of this divergent series approaches a non-zero

positive real number, the given series also diverges.

The same argument could have been used to prove that the series

∞n=1

1

n1+ 1ln n

is also divergent; the terms in this last series are smaller than those of the series we were

given.

C.32.2 Sketch of Solutions to Problems on the Final Examination in MATH 141 2006

01

(not for discussion at the lecture)

The best way to prepare for the examination in MATH 141 2009 01 is to study the textbook,

working problems, and verifying your solutions using the Student Solutions Manual: for sev-

eral reasons I don’t recommend your studying exclusively from old examinations. However

the old examinations do have some uses, and I am including a second one of them here, with

solutions. I am hoping to discuss last April’s examination briefly in a final lecture; there will

not be time for me to discuss the following examination in the lectures.








(a) [4 MARKS]

∞n=1

1

(tanh n)2 + 1

Solution: As n → ∞, tanh n = en−e−n

en+e−n = 1−e−2n

1+e−2n → 11

= 1 , so the terms summed in

this series have limit 112+1

= 12 0. By the “Test for Divergence” this series cannot

converge.

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(b) [4 MARKS]

∞

n=1

n2ne−n2

Solution:

nth root of nth term =n

n2ne−n2= n2e−n =

n2

en .

By l’Hospital’s Rule ( ∞∞ case) the limit of this nth root as n → ∞ is

limn→∞

n2

en = lim

n→∞2n

en = lim

n→∞2

en = 0 < 1 .

By the Root Test, the series is convergent.

(c) [4 MARKS]

∞

n=1

n2 − 85n + 12

n(n + 6)2

Solution: The ratio of the nth term of this series to the nth term of the Harmonic

series∞

n=1

1n

is

n2 − 85n + 12

(n + 6)2 =

1 − 85n

+ 12n2

1 + 6n

2 → 1 0 .

By the Limit Comparison Test the given series is divergent, since the Harmonic

series is divergent.


(a) [4 MARKS] Evaluate

2

−1

| x| dx .

Solution: 2

−1

| x| dx =

0

−1

| x| dx +

2

0

| x| dx

=

0

−1

(− x) dx +

2

0

x dx

= − x2

20

−1 + x2

22

0

= 1

2 + 2 =

5

2

(b) [3 MARKS] Evaluate

e3 1

dt

t √

1 + ln t .

8/15/2019 Brown's Notes 2010



Solution: Under the substitution u = ln t , du = dt t

and

e3 1

dt

t √

1 + ln t =

3

0

du√ 1 + u

= 2(1 + u)12

3

0= 2(

√ 4 −

√ 1) = 2(2 − 1) = 2


dx

x2

0

et 2 dt .

Solution: With u = x2,

d

dx

x2

0

et 2 dt = d

dx

u

0

et 2 dt = eu2

2 x = e( x2)2

2 x = 2 xe x4

,

by the Fundamental Theorem of Calculus.


limn→∞

1

n

0

n

7

+

1

n

7

+

2

n

7

+ . . . +

n − 1

n

7 .

Solution: You are asked to evaluate the limit of a Riemann sum for 1

0 x7 dx =

x8

8 1

0= 1

8, (where the values of the function are taken at the left end-point of each

subinterval). Accordingly the value is 18 .

SHOW ALL YOUR WORK!

3. For each of the following series you are expected to apply one or more tests for conver-






(a) [4 MARKS]

∞n=1

1

(tanh n)2 + 1

Solution: As n → ∞, tanh n = en−e−n

en+e−n = 1−e−2n

1+e−2n → 11

= 1 , so the terms summed in

this series have limit 112+1

= 12 0. By the “Test for Divergence” this series cannot

converge.

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(b) [4 MARKS]

∞

n=1

n2ne−n2

Solution:

nth root of nth term =n

n2ne−n2= n2e−n =

n2

en .

By l’Hospital’s Rule ( ∞∞ case) the limit of this nth root as n → ∞ is

limn→∞

n2

en = lim

n→∞2n

en = lim

n→∞2

en = 0 < 1 .

By the Root Test, the series is convergent.

(c) [4 MARKS]

∞n=1

n2

−85n + 12

n(n + 6)2

Solution: The ratio of the nth term of this series to the nth term of the Harmonic

series∞

n=1

1n

is

n2 − 85n + 12

(n + 6)2 =

1 − 85n

+ 12n2

1 + 6n

2 → 1 0 .

By the Limit Comparison Test the given series is divergent, since the Harmonic

series is divergent.

4. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-


and limits specific to the given problems, and should be simplified as much as possible,

except that you are not expected to evaluate the integrals.

(a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the region

bounded by the parabola y2 = 2 x + 6 and the line y = x − 1. An answer involving

integration along the y-axis will not be accepted.

DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)

8/15/2019 Brown's Notes 2010



Solution: −1

−3

2

2( x + 3) dx + 5

−1

2( x + 3) − ( x − 1)

dx


region bounded by the curves y = x3 and y = x2. For this question you are to use

only the method of “washers”.


Solution: Only the first integral was required.

π

1

0

1 − x3

2 −

1 − x22

dx

= π

1

0

−2 x3 + x6 + 2 x2 − x4

dx

= π −2 x4

4 +

x7

7 +

2 x3

3 − x5

5 1

0

= π

−2

4 +

1

7 +

2

3 − 1

5

=

23π

210 .

(c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 the


only the method of “cylindrical shells”.


Solution: Only the first integral was required.

2π

1

0

(1 − y) y

13 − y

12

dy

8/15/2019 Brown's Notes 2010



= 2π 1

0 y 1

3

− y

43

− y

12 + y

32 dy

= 2π

3

4 − 3

7 − 2

3 +

2

5

=

23π

210

(d) [3 MARKS] The length of the curve whose equation is

x2

4 +

y2

9 = 1 .


Solution: This is a special case of [7, Exercise 20, p. 553].

x2

4 +

y2

9 = 1 ⇔ 9 x2 + 4 y2 = 36

⇒ dy

dx= −9 x

4 y

⇒

1 +

dy

dx

2

=

81 x2 + 16 y2

4| y|

⇒ arc length = 2

2

−2

1

2

16 + 5 x2

4 − x2 dx

where the factor 2 is needed because the integral is the length of either the top or

the bottom arc of the ellipse; this integral is improper. A cleaner solution is found

by parameterizing the curve as x = 2cos θ , y = 3sin θ (0 ≤ θ ≤ 2π). Then the arc

length is

2π

0

4sin

2

θ + 9cos2

θ d θ .


[12 MARKS] Evaluate the indefinite integral x5 + x

x4 − 16 d x .

8/15/2019 Brown's Notes 2010



Solution: The first step is to divide denominator into numerator:

x5 + x = x( x4 − 16) + 17 x ⇒ x5 + x

x4 − 16 = x +

17 x

x4 − 16 .

Next we could separate the resulting remainder fraction, whose numerator has degree

less than that of its denominator, into partial fractions:

17 x

x4 − 16 =

Ax + B

x2 − 4 +

C x + D

x2 + 4

= E

x − 2 +

F

x + 2 +

C x + D

x2 + 4

⇒ 17 x = E ( x + 2)( x2 + 4) + F ( x

−2)( x2 + 4)

+C x( x − 2)( x + 2) + D( x − 2)( x + 2)

We can determine the constants by a combination of the standard methods: giving x

convenient values, and equating coefficients of like powers of x:

x = 2 ⇒ 34 = 32 E ⇒ E = 17

16

x = −2 ⇒ −34 = −32F ⇒ F = 17

16 x = 0 ⇒ 0 = 6 E − 6F − 8 D ⇒ D = 0

coeff x3 : ⇒

0 = E + F + C ⇒

C = −

17

8

Hence x5 + x

x4 − 16 = x +

17

16

1

x − 2 +

1

x + 2 − 2 x

x2 + 4

Hence

x5 + x

x4 − 16 d x =

x2

2 +

17

16 ln

x2 − 4

x2 + 4

+ C

A better tactic would have been to make a substitution u = x2 in the integral prior to

separation into partial fractions: 17 x

x4 − 16 d x =

17

2

du

(u − 4)(u + 4) =

17

16

1

u − 4 − 1

u + 4

du =

17

16 ln

u − 4

u + 4

+ C


Showing all your work, evaluate each of the following:

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(a) [4 MARKS] cos x

·cosh x dx

Solution: Using integration by parts,

u = cos x ⇒ du = −sin x dx

dv = cosh x dx ⇒ v = sinh x

⇒

cos x · cosh x dx = (cos x)(sinh x) +

sinh x · sin x dx .

A second integration by parts yields

U = sin x ⇒ dU = cos x dx

dV =

sinh x dx ⇒ V =

cosh x⇒

sin x · sinh x dx = (sin x)(cosh x) −

cosh x · cos x dx

⇒

cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) −

cos x · cosh x dx.

Now we can move the integral from the right side of the equation to join the same

integral with opposite sign on the left side, yielding

2

cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) + C

or

cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x)

2 + C .

(b) [5 MARKS]

1 −3

√ x2 + 2 x + 5 dx

Solution: We begin by completing the square:

1

−3

√ x2 + 2 x + 5 dx =

1

−3 ( x + 1)2 + 4 dx .

Then the integrand can be changed to the form of √

X 2 + 1:

1 −3

( x + 1)2 + 4 dx = 2

1 −3

x + 1

2

2

+ 1 dx .

8/15/2019 Brown's Notes 2010



Here an appropriate substitution would be to set x+12

equal to either tan θ or sinh θ ;

I have chosen to use the trigonometric substitution: x + 1

2 = tan θ ⇒ θ = arctan

x + 1

2

⇒ dx = 2 sec2 θ d θ

⇒1

−3

√ x2 + 2 x + 5 dx = 2

π4

− π4

| sec θ | · 2sec2 θ d θ

⇒1

−3

√ x2 + 2 x + 5 dx = 8

π4

0

sec3 θ d θ

Finally we have to find an antiderivative of sec3 θ , which can be integrated by

“standard” methods, e.g., by integration by parts with u = sec θ , dv = sec2 θ d θ :

du = sec θ tan θ d θ

v = tan θ sec3 θ d θ = sec θ tan θ −

sec θ tan2 θ d θ

= sec θ tan θ −

sec θ (sec2 θ − 1) d θ

= sec θ tan θ − sec

3

θ d θ +

sec θ d θ

= sec θ tan θ −

sec3 θ d θ + ln | sec θ + tan θ | + C

⇒ 2

sec3 θ d θ = sec θ tan θ + ln | sec θ + tan θ | + C

⇒

sec3 θ d θ = sec θ tan θ + ln | sec θ + tan θ |

2 + C .

Here1

−3

√ x2 + 2 x + 5 dx = 8

π4

0

sec3 θ d θ = 4√

2 + 4ln(√

2 + 1) .

(c) [4 MARKS]

sin2 x · cos2 x dx

Solution: One way to simplify this integral is to use the identity

sin x cos x = sin2 x

2 :

8/15/2019 Brown's Notes 2010



sin2 x

·cos2 x dx =

1

4 sin2 2 x dx

= 1

8

(1 − cos4 x) dx =

1

8

x − sin4 x

4

+ C

= x

8 − sin4 x

32 + C .

This could also have been simplified using the algorithm proposed in the textbook: sin2 x · cos2 x dx =

1

4

(1 − cos2 x)(1 + cos2 x) dx

= 1

4 (1

−cos2 2 x) dx

= 1

4

1 − 1 + cos4 x

2

dx

= 1

8

(1 − cos4 x) dx etc.



x = x(t ) = 10 − 3t 2

y = y(t ) = t 3

− 3t ,

where −∞ < t < +∞.

(a) [8 MARKS] Determine the value of d 2 y

dx2 at the points where the tangent is horizon-

tal.

Solution:

dx

dt = −6t

dy

dt = 3t 2

−3

dy

dx=

dy

dt dx

dt

= t 2 − 1

−2t = −1

2

t − 1

t

d 2 y

dx2 =

d

dx

dy

dx

8/15/2019 Brown's Notes 2010



= d

dt

dy

dx ·

dt

dx

=

d

dt

dy

dx

dx

dt

=

d

dt

−1

2

t − 1

t

dx

dt

= −1

2

d

dt t

− 1

t

−6t

= −1

2

1 + 1

t 2

−6t =

t 2 + 1

12t 3 .

The tangent is horizontal when dy

dx= 0, i.e., when t 2 − 1 = 0 or t = ±1. At such

parameter valuesd 2 y

dx2 =

2

±12 = ±1

6 .

(b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of thearc

( x(t ), y(t )) : −√

3 ≤ t ≤ 0

.

Solution: The area of the surface of revolution about the x-axis is

2π

0

−√

3

y(t )

dx

dt

2

+

dy

dt

2

dt = 2π

0

−√

3

(t 3 − 3t )

36t 2 + 9(t 2 − 1)2 dt

= 2π

0

−√

3

(t 3 − 3t )(3t 2 + 3) dt

= 2π

0

−√

3

(t 5 − 2t 3 − 3t ) dt

= 6π

t 6

6 − 2t 4

4 − 3t 2

2

0

−√

3

= 27π .


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(a) [5 MARKS] Showing detailed work, determine whether the following integral is

convergent; if it is convergent, determine its value: 0

−1

dx

x23

.

Solution: This integral is improper because the integrand is undefined, hence dis-

continuous at the right endpoint. 0

−1

dx

x23

= limb→0−

b

−1

dx

x23

by definition

= limb→0−

3 x13

b

−1= 3 lim

b→0−b

13 − (−3) = 0 + 3 = 3.

Since the limit exists, the improper integral is convergent.


absolutely convergent, or divergent.

∞n=1

(−1)n

n − ln n

Solution: Let f ( x) = 1

x − ln x. then

f ( x) =

−

1 − 1 x

( x − ln x)2

.

For x > 1, 1− 1 x

> 0, so f ( x) < 0. Thus the terms of the given series are monotonely

decreasing in magnitude, and alternating in sign.

We observe that

lim x→∞

ln x

x= lim

x→∞

1 x

1 = 0;

it follows that the limit of the ratio of corresponding terms of the given series and

the series

1n

is

limn→∞

1n−ln n

1

n

= limn→∞

1

1 − ln n

n

= 1

1 − limn→∞ln n

n

= 1 .

Since the Harmonic Series diverges, this implies (by the Limit Ratio Test) that the

series of absolute values of the given series is divergent; and the given series is

either conditionally convergent or divergent. The limit of the terms of the series is

limn→∞

1

n − ln n= lim

n→∞1

n· lim

n→∞1

1 − ln nn

= 0 · 1 = 0 .

8/15/2019 Brown's Notes 2010



This, combined with the property that the terms of the given series are alternating

in sign and monotonely decreasing in magnitude, permits us to conclude by theLeibniz Test that the series converges. Thus this alternating series is convergent,

but not absolutely convergent: it is conditionally convergent.

(c) [3 MARKS] Give an example of a sequence an with the property that limn→∞

an = 0

but

∞n=1

an = +∞. You are expected to give a formula for the general term an of

your sequence.

Solution: The Harmonic Series an = 1n

is one example of a series which has the

desired property.


[12 MARKS] The arc

r = 1 − cos θ (0 ≤ θ ≤ π)

divides the area bounded by the curve

r = 1 + sin θ (0 ≤ θ ≤ 2π)

into two parts. Showing all your work, carefully find the area of the part that contains


12

, π2

.

Solution: Since the functions defining the curves are both periodic with period 2π, we

can confine our attention to any parameter interval of length 2π. We need to determinewhere the curves intersect. Solving the two given equations algebraically gives tan θ =

−1, implying that θ = 3π4

, 7π4

in the interval 0 ≤ θ ≤ 2π. The only point whose parameter

value is in the intervals given for the two arcs is θ = 3π4

. However, we always need to be

alert to the possibility that there are intersections of the curves in which the parameter

values could be diff erent on the two curves. One place where precisely that occurs for

these two curves is the pole: on r = 1−cos θ the pole appears as (r , θ ) = (0, 0), which is in

the restricted interval 0 ≤ θ ≤ π; on the cardioid r = 1 + sin θ the pole appears as0, 3π

2

.

Could there be any other points of intersection? Any others could be found by solving

the variants of the equations obtainable by applying the transformation (r , θ ) = (−r , θ +π)

repeatedly. We can show that there are no such points that appear in this way.It follows that the region whose area we seek is bounded by two arcs:

• The arc of r = 1 − cos θ bounded by 0 ≤ θ ≤ 3π4

.

• One arc of r = 1 + sin θ : either −π2 ≤ θ ≤ 3π

4 or 3π

4 ≤ θ ≤ 3π

2 . (In the first of these

I have, for convenience, used the negative value θ = −π2

. You might wish to object

that the prescribed interval for θ was 0 ≤ θ ≤ 2π. I could, instead have described

8/15/2019 Brown's Notes 2010



2

1

1.5

0.5

0

1-1.5 -1 0.5-0.5-2 0

Figure 22: The curves with equations r = 1 − cos θ , (θ ≤ 0), and r = 1 + sin θ , and the point12

, π2

the first of these arcs as being made up of two pieces: one being 3π2 ≤ θ ≤ 2π, and

the second being 0 ≤ θ ≤ 3π4

.)

But the point

12

, π2

is only in the region whose boundary contains the arc 3π

4 ≤ θ ≤ 3π

2 of

the curve r = 1 − cos θ .

To find the area, think of a line segment being drawn from the pole to1 +

1

√ 2 ,

3π

4. The

area is then the sum of the area below the line segment and the area above it, i.e.,

1

2

3π4

0

(1 − cos θ )2 d θ + 1

2

3π2

3π4

(1 + sin θ )2 d θ

= 1

2

3π4

0

1 − 2cos θ +

1 + cos2θ

2

d θ +

1

2

3π2

3π4

1 + 2sin θ +

1 − cos2θ

2

d θ

8/15/2019 Brown's Notes 2010



= 1

23θ

2 −2 sin θ +

sin 2θ

4

3π4

0

+ 1

23θ

2 −2cos θ

− sin 2θ

4

3π2

3π4

= 9π

8 −

√ 2 − 1

4 .

8/15/2019 Brown's Notes 2010




Distribution Date: Wednesday, March 24th, 2010, subject to further revision

C.33.1 §11.5 Alternating Series

I continue to follow the syllabus in the order of topics in the textbook. This order is the reverse

of that which I would prefer, in that the most basic results for this topic will appear in the next

section, where the textbook considers theorems about positive series. As with the preceding

section, we are investigating a specialized topic concerning series prior to meeting a more

general theorem in [1, §11.6] at the next lecture. The present section will broader the class of

series that we can work with, albeit to a very special class of series.

An alternating series is one in which the signs alternate. Our main interest, as in the

preceding sections will be in three questions, listed in the order of our priorities:

1. Does the series converge?

2. If the series converges, what is its sum?

3. If we truncate the series at a specific partial sum, can we obtain a “good” upper bound

for the error — i.e., for the diff erence between the sum of the series and the given partial

sum.

Our interest in alternating series derives partly from some specific series that will arise later in

[1, Chapter 11], in the portion of the chapter assigned to Calculus III. There we would meetseveral important power series which represent some familiar functions; and many of those

series have alternating signs.

By the “Test for Divergence”, an alternating series cannot converge unless the limit of the

sequence of its terms is 0. For these series, and these series alone, we have a partial converse

of that test:

Theorem C.97 (Leibniz’s Alternating Series Test) An alternating series∞

n=1

(−1)n−1bn for which:

(i) 0 ≤ bn+1 ≤ bn for all n; and

(ii) limn→∞

bn = 0

converges.

8/15/2019 Brown's Notes 2010



Estimating Sums While the estimation of errors can be complicated for other series, the

following very simple bounds hold for alternating series:

Corollary C.98 (to the Leibniz Alternating Series Test) Continuing with the same notation

as the theorem, the remainder upon truncation of an alternating series

∞n=1

(−1)n−1bn (bn ≥ 0; n =

1, 2, . . .) at the N th term cannot exceed the very next term; more precisely,∞

n= N +1

(−1)n−1bn

< |b N +1| .

Example C.99 Determine whether the series

∞

n=0

(−

1)n 1

2n + 1 converges.

Solution: This is an alternating series. Since the denominators are increasing, while the nu-

merators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the terms

are approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course,

but it can be shown that this series converges to arctan 1, i.e., to π4

, and thereby provides a way

of approximating π to any desired accuracy. It is, however, not a good series for that purpose,

since it converges rather slowly.)

Example C.100 Determine whether the series

∞n=0

(−1)n 1

n + 1 converges.

Solution: This is an alternating series. Since the denominators are increasing, while the nu-merators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the terms

are approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course,

but this series can be proved to converge to ln(1 + 1), i.e., to ln 2.)

Example C.101 ([7, Exercise 26, p. 740]) How many terms of the series

∞n=1

(−1)nn

4n do we

need to add in order to find the sum to within an error of 0 .002?

Solution: As n → ∞ lim n4n = lim 1

4n ln n = 0. The terms are decreasing, since

n

4n

> n + 1

4n+1

⇔ 4n > n + 1

⇔ n >

1

3

.

Thus the Leibniz Theorem is applicable, and we apply the error estimate associated with that

Theorem. If we truncate the series at the N th term, then we know that the error will be less

than N + 1

4 N +1 . If we can find N satisfying the condition that

N + 1

4 N +1 < 0.002 , (96)

8/15/2019 Brown's Notes 2010



we will have the desired accuracy. Since

d dx

x4 x

= 1 · 4 x − x · 4 x · ln 4

(4 x)2 = 1 − x ln 4

4 x ,

which is negative provided x > 1ln 4

, which is certainly true if x ≥ 1. Thus we need only find the

first value of N for which inequality (96) holds. By repeated calculations we find the first value

is N = 5. That is, the series may be truncated by summing so that the last term we include is(−1)55

45 and the error will be within the prescribed tolerance.59

The student should understand that the Leibniz estimate for error is not necessarily as sharp

as it might be. It is often possible to obtain much better approximations than the Leibniz esti-

mate might guarantee; such investigations are beyond this course. The Leibniz theorem states

that an alternating series whose terms are non-decreasing in magnitude and are approaching0 is convergent. However, an alternating series whose terms are not decreasing, but whose

terms do approach 0, may still converge! For example, suppose that we modify the “alternat-

ing harmonic series”∞

n=1

(−1)n+1 1n

in the following way: term number 2n is the negative fraction

(−1)2n+1 12n . Let’s divide just these terms by 2. The resulting alternating series can be shown

to still be convergent; but it will converge to a sum which is less than the sum of the original

series; in fact, the reduction converges to a sum of

1

2 ·

12

1 − 12

= 1

2 .

But the condition of the Leibniz Theorem that requires that the terms be monotonely decreasing

in magnitude is not satisfied, so the convergence cannot be inferred from the Leibniz Theorem.

Nonetheless, the series does converge (since the partial sums are equal to the diff erence of the

partial sums of two series known to converge).

11.5 Exercises

[1, Exercise 14, p. 713] Test for convergence or divergence the series

∞n=1

(−1)n−1 ln n

n.

Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’s

Rule to be limn→∞1n = 0. To determine whether the terms are decreasing, we can consider

d

dx

ln x

x

=

1 − ln x

x2 < 0

59We can actually sum this infinite series, using methods analogous to the method used to sum a geometric

progression. The sum can be shown to be − 425

. The sum of the first 5 terms is approximately -0.161133, so the

error is approximately 0.001133, which is within the desired tolerance.

8/15/2019 Brown's Notes 2010



for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s not

important where the terms start to decrease — only that they ultimately decrease.60


∞n=1

(−1)n

√ n

1 + 2√

n.

Solution: The general term of the given series is,

(−1)n

√ n

1 + 2√

n= (−1)n 1

1√ n

+ 2

whose magnitude → 1

2 as n → ∞. Thus the sequence of terms of this series does not

have a limit as n → ∞, and the series cannot converge, by the “Test for Divergence”. Infact, the other condition of the Leibniz Theorem, that the terms decrease in magnitude,

also fails to hold! But here we can say more: not only can we state that the Leibniz Test

yields no information, but we can state that the Test for Divergence shows that the series

actually diverges.


∞n=1

(−1)n−1 ln n

n .

Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’s

Rule to be limn

→∞

1n

= 0. We have determined this limit for 2 reasons:

• as an application of “The” Test for Divergence — our result is inconclusive, and

gives no information about divergence;

• as a condition of the Leibniz Alternating Series Test — here our result is positive,

and, when combined with the yet unproved fact that the terms are decreasing in

magnitude, will show that the series does, indeed, converge.

To determine whether the terms are decreasing, we can consider

d

dx

ln x

x

=

1 − ln x

x2 < 0

for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s not

important where the terms start to decrease — only that they ultimately decrease.61

60However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease.61However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease.

8/15/2019 Brown's Notes 2010




∞

n=1

(

−1)n sin

π

n

.

Solution: This problem should be compared with [1, Exercise 31, p. 709] (cf. these

notes, p. 3229). In the earlier problem we proved the divergence of a series whose terms

resemble those of the present series, except that it was not alternating. But the present

series satisfies the conditions of the alternating series test, as the terms alternate in sign,

are monotonely decreasing (because the sine function is an increasing function to the

right of 0), and the limit of the terms is 0. Thus the series converges. When we have

covered [1, §11.6] we will be able to describe the present series as being Conditionally

Convergent — it is convergent, but loses that property if all terms are replaced by their

absolute value.

[1, Exercise 18, p. 713] Test for convergence or divergence the series∞

n=1

(−1)n cos πn

.

Solution: Since the magnitudes of the terms of this series approach 1 as n → ∞, the

terms of the series have no limit! By “The” Test for Divergence, this series diverges!


∞n=1

(−1)n nn

n! (cf. [1, Ex-

ercise 30, p. 709][1, Exercise 18, p. 720])

Solution: The terms may be factorized as

(−1)n nn

n! = (−1)n n

1 · n

2 · . . . n

n − 1 · n

n.

While this is an alternating series, the limit of the terms cannot be 0, since the fractions

into which I have factored the terms are, in magnitude, all greater than or equal to 1;

thus the series must diverge, by “The” Test for Divergence; this also shows that not all

conditions of the Leibniz Alternating Series Test are satisfied, but from that we can only

conclude that that test does not apply, not that the series diverges.


∞n=1

(−1)n

n

5

n

.

Solution: This series also diverges, because the terms to not have limit 0.

C.33.2 Solutions to Problems on the Final Examination in MATH 141 2007 01

Among the instructions for this examination was the following:

There are two kinds of problems on this examination, each clearly marked as

to its type.

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• Most of the questions on this paper require that you

SHOW ALL YOUR WORK!

Their solutions are to be written in the space provided on the page where

the question is printed. When that space is exhausted, you may write on the

facing page. Any solution may be continued on the last pages, or the back

cover of the booklet, but you must indicate any continuation clearly on the

page where the question is printed!

• Some of the questions on this paper require only

BRIEF SOLUTIONS

for these you are expected to write the correct answer in the box provided;

you are not asked to show your work, and you should not expect partial marksfor solutions that are not correct.

You are expected to simplify your answers wherever possible.

You are advised to spend the first few minutes scanning the problems. (Please

inform the invigilator if you find that your booklet is defective.)


Your answers must be simplified as much as possible.

(a) [2 MARKS] Evaluate 2

−1 | x

|2 dx.

Solution: 2

−1

| x|2 dx =

2

−1

x2 dx =

x3

3

2

−1

= 8 − (−1)

3 = 3 .


0 1

t 4dt √ t 5 + 1

.

Solution: Either by observation, or by applying a substitution like u = t 5, or u =

t 5 + 1, or u =√

t 5 + 1, we can show that

0 1

t 4dt √ t 5 + 1

= 2

5

√ t 5 + 1

0

1=

2

5(1 − √

2) = −2

5(√

2 − 1) .

(c) [3 MARKS] Determine the value of

1

n

0

n

3

+

1

n

3

+

2

n

3

+ . . . +

n − 1

n

3 .

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



which simplifies to 2 x · e x(2 x2); when x = 1, the value is 2e.



gence or divergence to determine whether the series is absolutely convergent, condition-

ally convergent, or divergent. All tests used must be named, and all statements must be

carefully justified.

(a) [4 MARKS]

∞n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n

Solution:∞

n=1

(−n − 2)

n

(n − 2)

n

(2n2 + 1)n =

∞n=1

(−1)n (n

2

− 4)

n

(2n2 + 1)n .

This is an alternating series. Applying the Root Test to the series of absolute values

of the terms of the given series, we have

n

(n2 − 4)n

(2n2 + 1)n =

n2 − 4

2n2 + 1

=1 − 4

n2

2 + 1n2

for n ≥ 2. As n → ∞, this ratio → 12

< 1 . Since the limit of the ratio is less

than 1, the original alternating series is absolutely convergent.

(b) [4 MARKS]∞

n=1

(−1)n+1 n!

n22n

Solution: If we define an = (−1)n+1 n!

n22n, we find, using the Ratio Test, that

an+1

an

= (n + 1)n2

(n + 1)21

= n2

2(n + 1) → ∞ > 1 .

Hence the original series diverges.

(c) [4 MARKS]

∞n=1

(−1)n sin 1

n

Solution: The series of absolute values may be compared, using the Limit Compar-

ison Test, with the Harmonic series, a positive series known to be divergent. Since

8/15/2019 Brown's Notes 2010



limn→∞

sin 1n

1n

= 1 0, the series of absolute values is also divergent, and the original

series is either (1) conditionally convergent, or (2) divergent. But

d

dx

sin

1

x

=

cos

1

x

·− 1

x2

< 0

as x > 1. Hence the given series of alternating terms are monotonely decreasing

in magnitude. As n → ∞, sin 1n → sin0 = 0, by the continuity of the sine func-

tion. Thus the conditions of the Alternating Series Test are satisfied, and the series

converges. We conclude that the given series is conditionally convergent.

3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum,

product, or quotient of several definite integrals, simplified as much as possible; you arenot expected to evaluate the integrals.

R is defined to be the region enclosed by the curves x + y = 6 and y = x2; C is the arc

y = 3 x (−1 ≤ x ≤ 2).

(a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum of

integrals whose value is the volume of the resulting solid.


Solution: By solving the equations of the line x + y = 6 and the parabola y = x2,

we find the points of intersection to be (−3, 9) and (2, 4). Either of the following

integrals or sums of integrals was acceptable. (Students were not expected to eval-

uate the integrals; I have done so here in order to verify my own work, and to show

you that the integrals weren’t really hard to evaluate.)

i. Using the method of “Washers”, we find the volume to be

π 2

−3(6 − x)

2

− ( x

2

)

2 dx = π

2

−3− x

4

+ x

2

− 12 x + 36

dx

which can be shown to be equal to 500π

3 .

ii. Using the method of cylindrical shells, we find the volume to be 4

0

(2π y) · (2√

y) dy +

9

4

(2π y) · (6 − y + √

y) dy

8/15/2019 Brown's Notes 2010



whose value can again be shown to equal 500π

3 .

(b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sum

of integrals whose value is the volume of the resulting solid.


Solution: Here again, students were expected only to state the integrals, but not to

evaluate them.

i. Using the method of cylindrical shells: We will be rotating about x = 5

vertical elements of area; the width can be taken to be ∆ x, as integration will

be with respect to x. The element extends from a point ( x, x2) to ( x, 6 − x), so

its height is 6 − x − x2. The volume is 2

−3

2π(5 − x) · (6 − x − x2) dx = 2π

x4

4 − 4 x3

3 − 11 x2

2 + 30 x

2

−3

= 1375π

6 .

ii. Using the method of washers:

π 4

0

(5 + √ y)

2

− (−5 + √ r )2

dy + π 9

4 ((5 + √ y)2

− (5 − (6 − y))2

) dy

= 20π

4

0

√ y dy + π

9

4

(24 + 10√

y + 3 y − y2) dy = 1375π

6 .

(c) [3 MARKS] Express in terms of integrals — which you need not evaluate — the

average length that R cuts off from the vertical lines which it meets.


Solution: The length of the segment cut off from the line x = a has been shown

above to be 6 − a − a2. The average of this function over the interval −3 ≤ a ≤ 2 is

the ratio 2

−3(6 − a − a2) da

2 − (−3) =

25

6 .

8/15/2019 Brown's Notes 2010



Students weren’t asked to evaluate the integral. I have done so here to verify my

work; in this case I can’t verify by comparing answers obtained in two diff erentways, but I can, at least, examine the magnitude of my answer, and decide whether

it is reasonable for the given data. For example, if my answer had been 25, I

would know something was wrong, by considering the sizes of the numbers being

averaged, and the maximum value obtained by the function on the given interval.

(d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluate

the integral.


Solution: When y = 3 x, y = 3 x ln 3. The arc length is 2

−1

1 + (3 x ln 3)2 dx .

(e) [3 MARKS] Given an integral whose value is the area of the surface generated by

rotating C about the line y = −1; you need not evaluate the integral.


Solution:

2π

2

−1

(3 x + 1)

1 + 32 x(ln 3)2 dx


[12 MARKS] Evaluate the indefinite integral x x2 − 4

( x − 2) + 4

x2 − 4

( x − 2) dx .

Solution: The numerator of the integrand has degree not less than that of the denomina-

tor. Accordingly, the first step is to divide denominator into numerator and to integrate

8/15/2019 Brown's Notes 2010



the quotient separately, and to factorize the denominator and group like factors together:

x x2 − 4

( x − 2) + 4

x2 − 4

( x − 2) dx =

x +

4

( x + 2)( x − 2)2

dx

= x2

2 +

4

( x + 2)( x − 2)2 d x .

Next we apply to the new integrand the method of partial fractions:

4

( x + 2)( x − 2)2 =

A

x + 2 +

B

x − 2 +

C

( x − 2)2

⇒ 4 = A( x − 2)2 + B( x + 2)( x − 2) + C ( x + 2)

⇒ 4 = 4C when x = 2; and4 = 16 A when x = 2

implying that A = 14

and C = 1. Comparing coefficients of x2 on both sides of the

identity yields 0 = A + B, implying that B = − 14

. Now we can continue integration of

the original function:

x x2 − 4

( x − 2) + 4

x2 − 4

( x − 2) dx =

x2

2 +

4

( x + 2)( x − 2)2 d x

= x2

2 +

1

4 1

x + 2 − 1

x

−2 dx +

dx

( x

−2)2

= x2

2 +

1

4 ln

x + 2

x − 2

− 1

x − 2 + C .



(a) [4 MARKS]

e− x · cos x dx

Solution: I will apply Integration by Parts twice. First, with u = e− x, dv = cos x dx

(so du = −e− x dx, v = sin x) yields e− x cos x dx = e− x sin x +

e− x sin x dx ,

and the second (applied to the integral resulting from the first application), with

U = e− x, dV = sin x dx (so dU = −e− x dx, V = −cos x) yields e− x cos x dx = e− x (sin x − cos x) −

e− x cos x dx .

8/15/2019 Brown's Notes 2010



By collecting both integral terms on one side, and dividing by 2, we obtain e− x cos x dx =

12

e− x(sin x − cos x) + C .

Another way to solve this problem is to assume — from experience — that the

indefinite integral is equal to a sum of the form e− x · cos x dx = e− x( A sin x + B cos x) + C ,

where A and B are “undetermined” constants, to be determined. One could then

diff erentiate this equation, top obtain

e− x cos x = −

e− x( A sin x + B cos x)e− x( A cos x−

B sin x) .

This is an identity – true for all values of x. If we assign “convenient” values, for

example x = 0 and x = π

2, we obtain equations

A − B = 1 ,

A + B = 0 ,

which we can solved, to obtain A = 12

, B = −12

, yielding the same solution as found

earlier.

(b) [5 MARKS]

52

− 1

2

x

√ 8 + 2 x − x2 dx

Solution:

52

− 12

x√ 8 + 2 x − x2

dx =

52

− 12

x 9 − ( x − 1)2

dx

= 1

3

52

−1

2

x

1 −( x−1

3 )2

dx .

I will use the substitution u = x − 1

3 , so du =

dx

3 .

1

3

52

− 12

x 1 − ( x−1

3 )2

dx = 1

3

12

− 12

3u + 1√ 1 − u2

.3 du

8/15/2019 Brown's Notes 2010



= −3

√ 1

−u2 + arcsin u

12

12

=

−3

3

4 + arcsin

1

2

−−3

3

4 − arcsin

1

2

= 2 arcsin

1

2 = 2 · π

6 =

π

3 .

(c) [4 MARKS]

cos2 x +

1

cos2 x

· tan2 x dx

Solution:

cos2 x + 1

cos2

x ·

tan2 x dx = sin2 x + tan2 x

·sec2 x dx

=

1 − cos2 x

2 + tan2 x · sec2 x

dx

= x

2 − sin 2 x

4 +

tan3 x

3 + C .


Consider the arc C defined by

x = x(t ) = cos t + t sin t

y = y(t ) = sin t −

t cos t ,

where 0 ≤ t ≤ π2

.

(a) [6 MARKS] Determine as a function of t the value of d 2 y

dx2.

Solution:

dx

dt = − sin t + sin t + t cos t = t cos t

dy

dt = cos t − cos t + t sin t = t sin t

dydx

=

dy

dt dxdt

= t sin t t cos t

= tan t when t 0, π2

,

d 2 y

dx2 =

d

dx

dy

dx

=

d

dt

dy

dx

· dt

dx

=

d dt

dy

dx

dxdt

=

d dt

tan t

t cos t =

sec2 t

t cos t =

1

t cos3 t ,

8/15/2019 Brown's Notes 2010



when t 0, π2

.

(b) [6 MARKS] Determine the area of the surface generated by revolving C about the y-axis.

Solution: The area of the surface generated is

2π

π2

0

x(t )

(t cos t )2 + (t sin t )2 dt

= 2π

π2

0

(cos t + t sin t )|t | dt = 2π

π2

0

(t cos t + t 2 sin t ) dt .

At this point I interrupt the calculations to integrate by parts. First, taking u = t ,

dv = cos t dt (so du = dt , v = sin t ), t cos t dt = t sin t −

sin t dt = t sin t + cos t + C 1 .

Then, taking u = t 2, dv = sin t dt (so du = 2t dt and v = −cos t ), t 2 sin t dt = −t 2 cos t + 2

t cos t dt = −t 2 cos t + 2(t sin t + cos t ) + C 2 .

Incorporating these results into the earlier integral yields

2π

π2

0 (t cos t +

t

2

sin t ) dt

= 2π(t sin t + cos t ) + (−t 2 cos t + 2t sin t + 2cos t )

π2

0

= 6π

π

2 − 1

.



convergent; if it is convergent, determine its value:

π

π2

sec x dx .

Solution: The integrand has an infinite discontinuity at x = π

2.

π

π2

sec x dx = lima→( π

2 )+

π

a

sec x dx

8/15/2019 Brown's Notes 2010



= lima

→( π

2 )+

[ln | sec x + tan x|]πa

= lima→( π

2 )+

(ln | − 1 + 0| − ln | sec a + tan a|)

= 0 − lima→( π

2 )+

ln | sec a + tan a| .

Here both sec a and tan a are negative, as the argument is in the second quadrant,

so they both approach −∞. This implies that their sum also approaches −∞, and

that ln | sec a + tan a| approaches +∞; hence the improper integral approaches −∞,

and is not convergent.

(b) [5 MARKS] Showing all your work, carefully determine whether the series

∞

n=3

4

n ln nis convergent.

Solution: Since both factors in the denominator are increasing, the terms of this

positive series are decreasing monotonely. We can apply the Integral Test with the

function f ( x) = 4 x ln x

. But ∞

3

dx

x ln x= lim

b→∞[ln ln x]b

3 = +∞

is a divergent, improper integral, so the series is also divergent.

(c) [3 MARKS] Showing all your work, determine whether the following sequence

converges; if it converges, find its limit:

a1 = 1.

a2 = 1.23

a3 = 1.2345

a4 = 1.234545

a5 = 1.23454545

a6 = 1.2345454545

etc., where each term after a2 is obtained from its predecessor by the addition on

the right of the decimal digits 45.Solution:

1.23 + 45(0.0001 + 0.000001 + . . .) = 1.23 + 45

10000

1 +

1

100 +

1

1002 + . . .

= 1.23 + 45

104 · 1

1 − 1100

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4

-4

0

4 80-8 -4

-12

-8

Figure 23: The cardioids with equations r = 2 + 2sin θ , r = 6 − 6 sin θ

and (98)

r = 6 − 6 sin θ

π

6 ≤ θ ≤ π

2

(99)

Technically the first of these curves is described incorrectly, since the values shown for

the angle are not in the given interval; so we should write, instead,

r = 2 + 2 sin θ

3π

2 ≤ θ ≤ 2π

, (100)

r = 2 + 2 sin θ

0 ≤ θ ≤ π

6

, (101)

and (102)

8/15/2019 Brown's Notes 2010



r = 6 − 6 sin θ π

6 ≤ θ ≤ π

2 . (103)

There are various ways in which the area can be calculated (see Figure 24 on page

3263)). One method is to first draw a line joining (0, 0) and3, π

6

. The area of the

1.5

0.5

1

2

0

2.51.50.5 10

-0.5

Figure 24: The region bounded by cardioids r = 2 + 2sin θ , r = 6 − 6sin θ and containing the

point (r , θ ) = (1, 0)

subregion above the line is 1

2

π2

π6

(6 − 6sin θ )2 dt ; the area of the region below the line is

1

2

π6

− π2

(2 + 2sin θ )2 d θ . To evaluate these definite integrals, observe that

(1 − sin θ )2 d θ =

(1 − 2sin θ + sin2 θ ) d θ

8/15/2019 Brown's Notes 2010



= 1

−2 sin θ +

1 − cos2θ

2 d θ

= 3θ

2 + 2cos θ − sin 2θ

4 + C .

In a similar fashion we can prove that (1 + sin θ )2 d θ =

3θ

2 − 2cos θ − sin2θ

4 + C .

Hence

1

2

π2

π6

(6 − 6sin θ )2

dt =

27θ + 36 cos θ − 9

sin2θ

2 π

2

π6

= 27π

2 −

9π + 36 ·√

3

2 − 9

2 ·

√ 3

2

= 9π − 63

√ 3

4

1

2

π6

− π2

(2 + 2sin θ )2 dt =

3θ + 4cos θ − sin 2θ

2

π6

− π2

=

π

2 + 4 · √ 3

2 − √ 3

2

−−−3π

4

= 2π + 7

√ 3

4 ,

so the total area is 11π − 18√

3. This area could have been computed in other ways. For

example, half the area of the smaller cardiod can be seen to be 3π, and from this one

could subtract the integral

1

2

π2

π6

(2 + 2sin θ )

2

− (6 − 6sin θ )22

d θ ,

which can be shown to equal 18√

3 − 8π. The integration of this diff erence of squares

depends on the fact that the two curves are “traced out” in the same direction; this should

be verified before using this “combined” method.

8/15/2019 Brown's Notes 2010




Distribution Date: Friday, March 26th, 2010, subject to further revision

C.34.1 §11.6 Absolute Convergence and the Ratio and Root Tests

Finally, after meeting, in [1, §§11.2–11.4] several tests for the convergence of positive series,

we see a reason for our interest in this type of series. To state the theorem we require two

definitions:

Definition C.7 1. A series

an is absolutely convergent if the series |an| is convergent.

2. A series an is conditionally convergent if an is convergent but not absolutely con-

vergent, i.e., if

an converges but |an| diverges.

Note that, thus far, you must not interpret the word absolutely as an adverb modifying the

participle convergent ; until the following theorem is available, you should treat absolutely

convergent as a two-word name for a property, nothing more. Now we state the theorem that

will permit a broader interpretation of the name.

Theorem C.102 A series which is absolutely convergent is convergent.

(I shall not prove the theorem in the lectures, but a short proof can be found in your textbook.)

Now that we have this theorem, we can interpret the word absolutely as a modifier —

if we drop the word from a statement, the resulting statement is still true; we didn’t need

such a reservation with the term conditionally convergent , since the definition stated that a

conditionally convergent series is convergent and . . . .

While investigation of the convergence of general series, i.e., series whose terms have both

plus and minus signs, can be difficult, we can begin by considering the series of absolute

values: if the resulting series converges, we can infer that the original series is convergent.

But, if the series of absolute values diverges, or if we are unable to say anything about it, then

we can make no inference at all.

The Ratio Test. The Ratio Test is a test for positive series which considers the limit of the

ratio of a term to its predecessor. The textbook states the test in terms of the absolute values of

terms of a general series, so we will present it in that variant:

Theorem C.103 (Ratio Test) 1. If limn→∞

|an+1||an|

= L < 1, then the series

an is absolutely

convergent.

2. If limn→∞

|an+1||an|

= L > 1, then the series

an is divergent.

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8/15/2019 Brown's Notes 2010



is absolutely convergent, conditionally convergent, or divergent.

Solution: I will apply the Root Test. The nth root of the absolute value of the nth term is1

arctan n, which approaches

2

π< 1. Hence the series is absolutely convergent.

11.6 Exercises

[1, Exercise 3, p. 719] I modify the problem: Determine whether the series

∞n=0

(−1)n 1, 000, 000n

n!


Solution: Apply the Ratio Test to the series of absolute values. The ratio of the (n + 1)th

term to the nth is 100000

n + 1 , and

100000

n + 1 → 0 as n → ∞. Thus, by the Ratio Test, the

series is absolutely convergent. This conclusion could also be proved in other ways.


∞n=0

sin4n

4n


Solution: Neither the Root Test nor the Ratio Test is useful here, because the limits do

not exist. Compare the series of absolute values with the geometric series

14n . Since

the latter converges, so does the former. Thus the given series is absolutely convergent.


∞n=1

3 − cos n

n23 − 2


Solution: For n > 1000 — (we don’t need the best bound here) — this is a positive

series; the numerator behaves erratically, and we will not be able to calculate a limit

using the Ratio or Root Tests. But since, for large n, the denominator is larger than

n23 , we would expect this series to diverge by comparison with the appropriate p-series.

More precisely, the divergence of the p-series with p = 23

implies the divergence of the

given series, since3 − cos n

n23 − 2

> 3 − 1

n23 − 2

> 3 − 1

n23

= 2

n23

.

8/15/2019 Brown's Notes 2010




Distribution Date: Monday, March 29th, 2010, subject to further revision

C.35.1 §11.6 Absolute Convergence and the Ratio and Root Tests (conclusion)

Review of the last lecture After reviewing the Leibniz Alternating Series Test, I worked

examples on these series, and mentioned the concept of conditional convergence, to be defined

in the next section. In §11.6 we finally saw some justification for our development of tests for

the convergence of positive series.

• definitions of absolute and conditional convergence

• Theorem: Absolute convergence implies convergence.

• The Ratio Test and Root Test are essentially tests for the convergence of positive series,

but they are formulated in your textbook for any series, since the limits involved are each

applied to the absolute values of roots and / or ratios.

• In each of these tests a limit is investigated, and that limit must be diff erent from 1 for

the test to yield any useful information. If the limit does not exist, or if the limit exists

and is equal to 1, the test fails to give any information.

• When the limit exceeds 1, that implies that the terms do not approach 0, and so diver-

gence follows from “The” Test for Divergence.

11.6 Exercises (continued)


∞n=1

(−1)n−1 2n

n4 is absolutely convergent,

conditionally convergent, or divergent.

Solution: This is an alternating series: the only test we have to apply directly to such

series is the Leibniz Alternating Series Test (and the Test for Divergence, which forms a

part of the Leibniz Test). To determine limn→∞

2n

n4, where both numerator and denominator

become infinite, we can use L’Hospital’s Rule. This will need to be applied 4 times, afterwhich we have lim

n→∞2n(ln 2)4

24 = +∞ . The terms of the original series do not approach

+∞, since they are alternating in sign; but the non-existence of a limit shows that, in

particular, the limit of terms is not 0, so the Test for Divergence tells us that this series

must diverge. We don’t even need to check the second condition of the Leibniz Test

(which could be shown to fail). Thus, even though we thought we were applying the

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8/15/2019 Brown's Notes 2010



is far beyond this course. The obvious test to try is the Ratio Test, since the terms

are products, and the ratio of successive terms has many factors which cancel. Wefind that the ratio of the (n + 1)st term to the nth is

2(n + 1)

n2 = 2

1

n +

1

n2

which

approaches 0 as n → ∞. Since this limit exists, and is less than 1, the Ratio Test

tells us that the series of absolute values is convergent — i.e., that the original series

is absolutely convergent.

(cf. [1, Exercise 17, p. 720]) Determine whether the series

∞n=0

(−1)n

n ln n


Solution: Here the ratio of the absolute value of the (n + 1)th term to that of the nth is1 +

1

n

· ln(n + 1)

ln n

in which the first factor approaches 1, and the second (by L’Hospital’s Rule) also ap-

proaches 1. Thus the Ratio Test fails to give useful information about this series. We are

not helpless, however. Let’s first consider the series of absolute values. The terms are

decreasing and approaching 0, so we may use the Integral Test to show that it is diver-

gent. On the other hand, the original series is alternating, and the terms are decreasingand approaching 0; hence, by the Leibniz Test, that series converges; consequently it is

conditionally convergent .

[1, Exercise 17, p. 720] is concerned with the series

∞n=0

(−1)n

n ln n. The preceding discussion

shows that a series with terms which are smaller in magnitude is not absolutely conver-

gent; hence, by the Comparison Test, the series of this exercise is surely not absolutely

convergent; we could also probe the absence of absolute convergence by comparing with

the harmonic series. But it is conditionally convergent, since the conditions of the Leib-

niz test are easily shown to be satisfied. (We could also probe the absence of absolute

convergence by comparing with the harmonic series.)


2

5 +

2 · 6

5 · 8 +

2 · 6 · 10

5 · 8 · 11 +

2 · 6 · 10 · 14

5 · 8 · 11 · 14 + . . .


8/15/2019 Brown's Notes 2010



Solution: The ratio of the (n + 1)st term to the nth is 4n + 2

3n + 5, which approaches 4

3 > 1.

By the Ratio test, this series diverges.

[1, Exercise 30, p. 720] The terms of a series

an are defined “recursively” by a1 = 1, an+1 =2 + cos n√

n· an (n ≥ 1). Determine whether

an converges or diverges.

Solution: The ratio of the (n + 1)st to the nth term is 2 + cos n√

n (n ≥ 1). As n → ∞ the

numerator is bounded between 1 and 3, while the denominator becomes infinitely large;

by the Squeeze Theorem, applied to

0 ≤ an+1

an

≤ 3√ n

,

we see that limn→∞

an+1

an

= 0. By the Ratio Test, this shows that the series converges abso-

lutely.

[1, Exercise 32, p. 720] For which positive integers k is the series

∞n=1

(n!)2

(kn)! convergent?

Solution: The ratio of the (n + 1)st term to the nth is

((n + 1)!)2

(k (n + 1))!

(n!)2

(kn)!

=

(n + 1)!(n + 1)!(kn)!

n!n!(kn + k )! =

(n + 1)2

(kn + k )(kn + k − 1) · . . . · (kn + 1)

in which the factors in the denominator are k in number, and all of them lie between

n + 1 and kn + k . As long as k > 2, this fraction has more 1st degree factors in the

denominator than in the numerator, so it will approach 0 as n → ∞, and we note that

0 < 1; when k = 2 the fraction is (n + 1)2

(2n + 2)(2n + 1), which approaches 1

4 < 1 as n → ∞;

when k = 1 the fraction is equal to n + 1, and approaches +∞. By the Ratio Test the

series will converge for k ≥ 2, and will diverge when k = 1.

8/15/2019 Brown's Notes 2010



C.36 Supplementary Notes for the Lecture of Wednesday, March 31st,

2010Distribution Date: Wednesday, March 31st, 2010

subject to correction

C.36.1 §11.7 Strategy for Testing Series

This section is concerned with improving your methods for attacking series problems. Plan-

ning strategies really needs some experience: after you have worked a substantial number of

questions from earlier sections in the chapter, you might wish to read the 8 points that the

textbook gives on [1, p. 721]. These points will be of much more use to you after you have

accumulated some experience in applying the individual tests.Exercise C.2 ([7, Problems 2 and 4, p. 748]) You are asked to test for convergence or diver-

gence the series

∞n=1

n − 1

n2 + nand

∞n=1

(−1)n−1 n − 1

n2 + n.

Solution:

1. Consider the given positive series.

(a) It is useful, where possible, to formulate a good conjecture (guess) about the out-

come of the investigation, in order to know whether to concentrate e ff orts on at-

tempting to prove convergence or divergence. I suggest that a valuable first strate-

gic step is to replace the general term by a “simpler” term: in this case I wouldlook at

n

n

n2. Since this gives the harmonic series, a first guess in this case would

be that the series diverges. I would thus try to use some test that relates the given

series to the harmonic series.

(b) The simplest method would be to use the Limit Comparison Test:

limn→∞

n − 1

n2 + n1

n

= limn→∞

n − 1

n + 1 = 1 0 .

Hence the given series and the Harmonic Series converge or diverge together. Weknow that the Harmonic Series diverges; hence the given series also diverges.

(c) We could also have attacked this problem naively by using the Integral Test, with

f ( x) = x − 1

x2 + x. Since

f ( x) = − x2 + 3 x

x2 + x2

= − x( x − 3) x2 + x

2 ,

8/15/2019 Brown's Notes 2010



the function is decreasing for n > 3; and

limn→∞

n − 1

n2 + n= lim

n→∞

1 − 1n

1 + n= 0 .

Then the series will converge or diverge according as the following improper inte-

gral converges or diverges: ∞

3

x − 1

x2 + xdx =

∞

3

−1

x+

1

x + 1

dx

expanding by partial fractions

= limb→∞

b

3−

1

x

+ 2

x+

1 dx

= limb→∞

[− ln x + 2 ln( x + 1)]b3

= limb→∞

ln

( x + 1)2

x

b

3

= limb→∞

ln

(b + 1)2

b − ln

16

3

= limb→∞

ln

b + 2 +

1

b

− ln

16

3

= ∞ .

Thus this integral diverges, and the given series must also diverge.

2. Now consider the series

∞n=1

(−1)n−1 n − 1

n2 + n. We know from the preceding discussion

that the given alternating series does not converge absolutely. However, the Alternating

Series Test may be applied to show that the alternating series does converge. Thus the

series converges conditionally. So the answer to the question as stated is that the given

series does converge.

11.7 Exercises


n=1

n2e−n3

.

Solution: The terms suggest a function with an obvious antiderivative. More precisely,

if f ( x) = x2e− x3

, then

f ( x) dx = −13

e− x3

+ C . Thus it would appear that we could use

the Integral Test here. But let’s check that the conditions for that test are satisfied:

• Is the given series positive: YES

8/15/2019 Brown's Notes 2010



• Does the function f match the series values at integer terms: YES

• Is f (continuous and) positive (not just at the integer points): YES

• Is f decreasing? NOT CHECKED YET!

f ( x) = x2 − 3 x3

e− x3

, a product of 3 factors. The first factor is positive for x > 0; the

last factor is an exponential, so it is always positive; and the middle factor is negative

for x >

23

13

, so it is certainly negative for x ≥ 1. Thus f < 0, and f is decreasing. The

Integral Test tells us that the series will converge or diverge according as the improper

integral ∞

1 f ( x) dx converges or diverges. But

a

0

x2e− x3

dx =

−1

3e−a

−e−1 →

1

3e

as a

→ ∞.

From the convergence of the integral we infer the convergence of the given series.

Another approach would have been to use the Ratio Test:

limn→∞

(n + 1)2e−(n+1)3

n2e−n3 =

1 +

1

n

2

· 1

e3n2+3n+1→ 12 · 0 = 0 < 1 as n → ∞.

The Root Test could also be used, although the calculation is more difficult:

n

n2 · e−n3= e

2 ln nn · e−n2 → e0 · 0 = 1 · 0 = 0 < 1 as n → ∞ ,

by l’Hospital’s Rule. In either of these cases the limit being less than 1 implies that the

series is absolutely convergent.


n=1

sin n.

Solution: Here is a case where the Test for Divergence is needed. None of the other

tests you know will be helpful. As n → ∞, sin n does not approach a limit. (However,

students in this course could not be expected to prove that fact rigorously without some

strong hint.) Thus the series cannot converge.


∞k =1

k + 55k .

Solution: The series looks at first like a geometric series; but the numerators are increas-

ing, so we can’t compare with a geometric series; the Limit Comparison Test will also

not be helpful, at least not in comparing with the Geometric Series 1

5k . But several

other possibilities suggest themselves.

8/15/2019 Brown's Notes 2010



Compare with a geometric series which converges more slowly: e.g., with25

k . Here

limk →∞

k + 5

5k 2

5

k = lim

k →∞k + 5

2k = lim

k →∞1

2k ln 2 = 0 .

Unfortunately, this limit being 0, we cannot use the form of the Limit Comparison

Test given in your book. We could use the version given in [1, Exercise 40(a), p.

709], but that was not studied in our course.

Ratio Test:

limk →∞

ak +1

ak

= 1

5

limk →∞

k + 6

k + 5

= 1

5

< 1

hence the given positive series converges.

Root Test:

limk →∞

k √

ak = 1

5 limk →∞

k √

k + 5 = 1

5 limk →∞

e

ln(k + 5)

k = 1

5e

limk →∞

ln(k + 5)

k .

It can be shown by l’Hospital’s Rule that the exponent approaches 0, so the k th root

approaches 15

, which is less than 1.

In fact one can determine the actual sum here, but it involves techniques slightly beyond

the course.


n=1

tan

1n

.

Solution: As n → ∞, 1n → 0, and tan 1

n → 0 (since it is the ratio of sin 1

n to cos 1

n, and

cos 1n → cos0 = 1, while sin 1

n → 0). Thus the Test for Divergence does not eliminate

the possibility that the series may converge, just as the same test did not eliminate the

possibility that the Harmonic Series might converge. If you remember that

limn→∞

tan 1n

1n

= limn→∞

sin 1n

1n · cos 1

n

= limn→∞

sin 1n

1n

· limn→∞

sec 1n

= 1 · 1 = 1

you can apply the Limit Comparison Test, comparing the given series with the Harmonic

Series. Since the limit is a non-zero real number, and since the Harmonic Series is known

to diverge, the given series must also diverge.

8/15/2019 Brown's Notes 2010




∞

n=2

1

(ln n)ln n

.

Solution: Observe that

1

(ln n)ln n =

1elnln n

ln n =

1

e(lnln n)(ln n) =

1eln n

lnln n =

1

nlnln n .

We can arrange for the exponent, i.e., ln ln n, to be greater than, for example, 2, by taking

n > ee2

= 1618.17799.... Thus we may compare the given series with the p-series

1n2 ,

which is known to converge, and infer that the given series also is convergent.


∞

n=1

n√

2

−1.

Solution: I start with the observation that the following factorization holds:

an − bn = (a − b)an−1 + an−2b + . . . + abn−2 + an−1

,

for any real numbers a and b, Taking a = n√

2 and b = 1, we have

n√

2 − 1 =

n√

2n − 1

n√

2

n−1

+

n√

2

n−2

+ . . . + n√

2 + 1.

The numerator is equal to 1. All of the n terms in the denominator are smaller than n√

2n ≤ 2, so the denominator is smaller than 2n. Thus the terms are greater than terms

in a multiple of the harmonic series, so the series must diverge, by the Comparison Test.

We could have used the Limit Comparison Test also, since we know that

limn→∞

n√

2 = limn→∞

eln 2

n = elimn→∞ ln 2n e0 = 1 .

Now apply l’Hospital’s Rule:

limn→∞

n√

2

−1

1

n

= limn→∞

− ln 2

n2

n√

2

− 1

n2

= limn→∞

(ln 2) n√

2 = ln 2 ,

since

limn→∞

n√

2 = limn→∞

eln 2

1n

= limn→∞

eln 2

n = elim

n→∞ln 2

n = e0 = 1 .

8/15/2019 Brown's Notes 2010



By the Limit Comparison Test, the divergence of the harmonic series implies the diver-

gence of the given series. (Note that, if we alternate the signs in this problem, we obtaina series whose terms are decreasing and approaching 0, so the Alternating Series Test

tells us that that series converges, i.e., that it is conditionally convergent.)

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



Students are advised to spend time checking their work; for that purpose you could verify

your answers by solving problems in more than one way. Remember that indefinite integralscan be checked by diff erentiation.

W. G. Brown, Examiner.

Instructions

1. Fill in the above clearly.

2. Do not tear pages from this book; all your writing — even rough work — must be handed in.

You may do rough work for this paper anywhere in the booklet.

3. Calculators are not permitted. This is a closed book examination. Regular and translation

dictionaries are permitted.

4. This examination booklet consists of this cover, Pages 1 through 9 containing questions; and

Pages 10, 11, and 12, which are blank. Your neighbour’s version of this test may be different

from yours.

5. There are two kinds of problems on this examination, each clearly marked as to its type.

• Most of the questions on this paper require that you SHOW ALL YOUR WORK!

Their solutions are to be written in the space provided on the page where the question

is printed; in some of these problems you are instructed to write the answer in a box,but a correct answer alone will not be sufficient unless it is substantiated by your work,

clearly displayed outside the box. When space provided for that work is exhausted, you

may write on the facing page . Any solution may be continued on the last pages, or the

back cover of the booklet, but you must indicate any continuation clearly on the page

where the question is printed!

• Some of the questions on this paper require only BRIEF SOLUTIONS ; for these you

are expected to write the correct answer in the box provided; you are not asked to show

your work, and you should not expect partial marks for solutions that are not correct.


You are advised to spend the first few minutes scanning the problems. (Please inform the

invigilator if you find that your booklet is defective.)

6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION.



8/15/2019 Brown's Notes 2010



(a) [2 MARKS] Evaluate 4 − 6 x

1 + x2 dx .

Solution: 4 − 6 x

1 + x2 dx = 4

dx

1 + x2 − 3

2 x

1 + x2 dx

= 4 arctan x − 3 ln(1 + x2) + C


2 0

y2

y3 + 1 dy .

Solution: I apply the change of variable u = y3 (or v = y3 + 1), under which

du = 3 y2 dy:

2 0

y2

y3 + 1 dy =

8

0

√ u + 1 · 1

3 du

= 1

3 · 2

3

(u + 1)

32

8

0

= 2

9

9

32 − 1

32

=

52

9 .

(c) [3 MARKS] Evaluate

sin(18 θ ) · cos(30 θ ) d θ .

Solution:

i. The easiest way to solve this problem is by using a trigonometric identity

which relates products to sums: sin(18 θ ) cos(30 θ ) d θ =

1

2

(sin 48θ + sin(−12θ )) d θ

= − 1

96 cos 48θ +

1

24 cos 12θ + C .

ii. The problem could also be solved laboriously by two applications of inte-

gration by parts: Take u = sin 18θ , dv = cos 30θ d θ , implying that du =

18cos18θ , v = 130

sin 30θ : sin(18 θ ) · cos(30 θ ) d θ =

1

30 sin 18θ · sin 30θ − 18

30

cos 18θ · sin 30θ d θ .

8/15/2019 Brown's Notes 2010



A second application of integration by parts, with U = cos 18θ , dV = sin 30θ d θ ,

implying that dU = −18sin18θ d θ , V = − 1

30 cos 30θ , yields sin(18 θ ) · cos(30 θ ) d θ

= 1

30 sin 18θ · sin 30θ − 18

30

− 1

30 cos 18θ · cos 30θ − 18

30

sin 18θ · cos30θ d θ

.

This equation may be solved for the desired indefinite integral, yielding sin(18 θ ) cos(30 θ ) d θ

= 1

1 − 1830

2 1

30

sin 18θ

·sin 30θ +

18

302

cos 18θ

·cos 30θ + C

= 1

96(5sin18θ · sin 30θ + 3cos18θ · cos30θ ) + C .

The validity of this solution can be demonstrated by diff erentiation.


(a) [3 MARKS] Simplifying your answer as much as possible, evaluate d

dx

√ 3

− x

earcsin z dz .

Solution:

d

dx

√ 3

− x

earcsin z dz = − d

dx

− x

√ 3

earcsin z dz

= −earcsin(− x) · d

dx(− x)

= earcsin(− x) = e− arcsin x

(The last simplification was not required.)

(b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for the

function f ( x) = 3 − x, where the sample points are the left end-point of each of n

subintervals of equal length.

Solution: The intervals have length 5 − 2n

= 3n

; f (2) = 1, xi = 2 + 3in

, f ( xi) =

3 −

2 + 3i

n

= 1 − 3i

n. The Riemann sum is

3

n

n−1i=0

1 − 3i

n

or

3

n

ni=1

1 − 3(i − 1)

n

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



the ratios of successive terms is 1, so that test is useless here. But we can apply the

Limit Comparison Test, comparing with the divergent p-series

n−12

:

limn→∞

√ 4n + 5

3n + 101

n12

= limn→∞

4 + 5

n

3 + 10n

= 2

3 > 0 .

Since this limit is positive, and since the p-series to which we have compared is

divergent, we may conclude that the series of absolute values is also divergent; and

hence that the original series we were given is not absolutely convergent. But that

still leaves open the question of whether that series is conditionally convergent or

divergent. I will check whether the conditions of the Leibniz Alternating SeriesTest are satisfied:

limn→∞

√ 4n + 5

3n + 10 = lim

n→∞

1√

n·

4 + 5

n

3

= 0

d

dx

√

4 x + 5

3 x + 10

=

1

2 · 2 · 1√

4 x + 5· (3 x + 10)) − (

√ 4 x + 5 · 3)

(3 x + 10)2

= −9 x − 5

(3 x + 10)2 √ 4 x + 5

< 0 ,

hence we may apply the Leibniz Test, and conclude that the given series is conver-

gent. Since it has been shown to not be absolutely convergent, it is conditionally

convergent.

(c) [4 MARKS]

∞n=1

cot−1

1

n + 1

− cot−1

1

n

Solution: This is a telescoping series:

N

n=0

cot−1 1

n + 1 −

cot−1 1

n =

−cot−1 1 + cot−1 1

N + 2

.

As N → ∞ this partial sum approaches − cot−1 1 + cot−1 0 = −π4

+ π2

= π4

, so this is

the value to which the series converges. Since the derivative of cot−1 x is − 1

1 + x2,

the function is monotonely decreasing. Hence the terms of the given series are all

positive: a positive series which is convergent is absolutely convergent.

8/15/2019 Brown's Notes 2010



(The problem did not require the student to find the precise value of the sum. If

one did not notice that the given series was telescopic, one could still have appliedthe integral test to prove that it is absolutely convergent. It can be seen that ∞

0

cot−1 1

x + 1 − cot−1 1

x

dx

=

∞

0

(arctan( x + 1) − arctan( x)) dx

= lima→∞

arctan( x + 1) − ln(1 + ( x + 1)2)

2

−

arctan x − ln(1 + x2)

2

a

1

= lima→∞ arctan( x + 1)

−arctan x +

1

2

ln 1 + x2

1 + ( x + 1)2

a

1

= 0

− 1

2

ln 2

5

.4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves 2 y = x, y = 2 x, and x2 + y2 = 5.

(a) [4 MARKS] The region R is rotated about the line x = −1. Give an integral or sum


Solution:

Using “Washers”:

π 1

0

(2 y + 1)2 − (

4

2 + 1)2

dy + π 2

1

5 − y2 + 1

2

− y2 + 1

2 dy

Using Cylindrical Shells:

2π

1

0

( x + 1)

2 x − x

2

dx + 2π

2

1

( x + 1)

√ 5 − x2 − x

2

dx

(b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies inside

R. Express in terms of integrals — which you need not evaluate — the average of

the positive lengths L(a).

Solution:

1

2

1

0

2 y − 4

2

dy +

2

1

5 − y2 − 4

2

dy

(c) [4 MARKS] Let C1 be the curve x(t ) = t , y(t ) = cosh t (0 ≤ t ≤ ln 2). Simplifying

your answer as much as possible, find the length of C1.

8/15/2019 Brown's Notes 2010



Solution:

dx

dt = 1

dy

dt = sinh t

dx

dt

2

+

dy

dt

2

=

1 + sinh2 t = |cosh t | = cosh t

Length =

ln 2

0

cosh t dt

= [sinh t ]ln 20

= sinhln2 − sinh 0

= 3ln 2 − e− ln 2

2 − 0

=

2 − 1

2

2 =

3

4 .


(a) [8 MARKS] Evaluate the indefinite integral

36

( x + 4)( x − 2)2 dx .

Solution: We observe that the degree of the numerator of the integrand is less than

the degree of the denominator; thus we do not need to divide denominator into

numerator and obtain a quotient and remainder. Next we need to find a Partial

Fraction decomposition of the integrand. Assuming a decomposition of the form

36

( x + 4)( x − 2)2 =

A

x + 4 +

B

x − 2 +

C

( x − 2)2

we obtain, by taking the right side to a common denominator,

36 = A( x − 2)2 + B( x + 4)( x − 2) + C ( x + 4) .

Students should know two methods to find the values of these constants: either,

by assigning “convenient” values to x and thereby obtaining enough equations that

can be solved for the coefficients, or by comparing coefficients of powers of x on

8/15/2019 Brown's Notes 2010



the two sides of the equation. These methods, or a combination of them, yield

A = 1, B = −1, C = 6. Hence 36

( x + 4)( x − 2)2 d x =

1

x + 4 − 1

x − 2 +

6

( x − 2)2

dx

= ln

x + 4

x − 2

− 6

x − 2 + C .

(b) [4 MARKS] Determine whether

∞ 3

36

( x + 4)( x − 2)2 dx converges. If it converges,

find its value.

a

3

36

( x + 4)( x − 2)2 dx =

ln

1 + 4

a

1 − 2

a

− 6

a − 2

− (ln7 − 6)

→ 0 − 0 − (ln7 − 6) = 6 − ln 7 as a → ∞ .

Since the limit exists, the integral converges to the limiting value, 6 − ln 7.



(a) [4 MARKS]

e √ x dx

Solution: This integral can be solved by a series of substitutions; for example, the

substitution u =

√ x

2 , which implies that x = 4u2, d x = 8u du yields

e

√ t dt =

e

√ x

2 dx = 8

u · eu du ,

which can then be integrated by parts, as follows: with u = u, dv = eu du, du = du,

v = eu, so

e √ t dt = 8

u · eu du

= 8

u · eu −

eu du

= 8 ((u − 1)eu) + C

=4

√ x − 8

e

√ x

2 + C .

8/15/2019 Brown's Notes 2010



(b) [5 MARKS]

0

− 1

2

x

√ 3 − 4 x − 4 x2 dx

Solution:

√ 3 − 4 x − 4 x2 =

4 − (1 + 2 x)2

= 2

1 −

1 + 2 x

2

2

.

0

−12

x√ 3

−4 x

−4 x2

dx = 1

2 0

− 12

x

1 − 1+2 x

22

dx

= 1

2

12

0

u − 12√

1 − u2du

= 1

2

−

√ 1 − u2 − 1

2 · arcsin u

12

0

= 1

2

−

√ 3

2 − 1

2 arcsin

1

2

− (−1 − 0)

=

1

2 −√

3

2 − 1

2 · π

6 + 1 =

1

2 − π

24 −

√ 3

4 .

(c) [4 MARKS]

π

0

sin2 t cos4 t dt .

Solution: To evaluate integrals of this type the method of the textbook is to replace

the square powers of sines and cosines by functions of the cosine of twice the angle.

The following is a variant of that method, using both the sine and the cosine of the

double angle. π

0

sin2 t cos4 t dt =

π

0

(2sin t · cos t )2

4 · cos2 t dt

= π

0

sin

2

2t 4

· 1 + cos2t 2

dt

=

π

0

sin2 2t

8 +

sin2 2t · cos2t

8

dt

=

π

0

sin2 2t

8 +

(2 sin 2t · cos2t )2

3 2

dt

8/15/2019 Brown's Notes 2010



= π

01 − cos4t

16

+ sin2 2t

32 dt

=

t

16 − sin 4t

64 +

sin3 2t

48

π

0

= π

16 .


Consider the curve C2 defined by x = x(t ) = 1 + e−t , y = y(t ) = t + t 2 .

(a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis.

Solution:

y = 0 ⇒ t + t 2 = 0 ⇒ t = 0, −1

t = 0 ⇒ ( x, y) = (2, 0)

t = −1 ⇒ ( x, y) + (1 + e, 0) ,

so the points of intersection with the x axis are (2, 0) and (1 + e, 0).

(b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent is

horizontal.

Solution:

dx

dt = −e−t

dy

dt = 1 + 2t

dy

dx= 0 ⇒

dy

dt

dxdt

= 0 ⇒ dy

dt = 0 ⇒ t = −1

2

t = −1

2 ⇒ ( x, y) =

1 +

√ e, −1

4

.

(c) [6 MARKS] Determine the area of the finite region bounded byC

2 and the x-axis.

Solution: The area is

2 1+ 1

e

y dx =

0

−1

y(t ) · dx

dt (t ) dt =

0

−1

t + t 2

−e−t dt

8/15/2019 Brown's Notes 2010



= e−t (t + 1) + e−t

t 2 + 2t + 20

−1

=e−t ·

t 2 + 3t + 3

0

−1

= 3 − e(1 − 3 + 3) = 3 − e .

The preceding evaluation required two indefinite integrals that could be obtained

by integration by parts:

• Taking u = t , v = e−t , which imply that u = 1, v = −e−t , t · e−t dt = −te−t +

e−t dt = −e−t (t + 1) + C

• Taking U =

t

2

, V =

e−t

, which imply that U =

2t , V =

−e−t

, t 2 · e−t dt = −t 2e−t +

2t · e−t dt = −(t 2 + 2t + 2)e−t + C .


(a) [5 MARKS] Showing all your work, determine whether the series

∞n=2

√ n√

n + 2 −√

n − 2

is convergent or divergent.

Solution: This is a telescoping series.

N n=2

√ n√

n + 2 −√

n − 2

=

N n=2

√ n ·

√ n + 2 −

N −2m=0

√ m ·

√ m + 2

= −√

3 +

N ( N + 2) +

( N − 1)( N + 1)

→ ∞ as N

→ ∞.

As the partial sums are approaching infinity, they do not have a finite limit, and the

series is, by definition, divergent.

Alternatively, one could argue that

√ n√

n + 2 −√

n − 2

=

√ n (n + 2 − n + 2)√ n + 2 +

√ n − 2

8/15/2019 Brown's Notes 2010



= 4

√ n

√ n + 2 + √ n − 2

= 4 1 + 2

n +

1 − 2

n

→ 2 0 as n → ∞.

By “The” Test for Divergence, the series diverges.

(b) [5 MARKS] Showing all your work, determine whether the following sequence


a1 = 3.

a2 = 3.14

a3 = 3.1414

a4 = 3.141414

a5 = 3.14141414

a6 = 3.1414141414


the right of the decimal digits 14.

an = 3 + 0.14

1 +

1

100 +

1

1002 + . . . +

1

100n−1

= 3 +0.14

1 − 1100n

1 − 1

100

→ 3 + 0.14

0.99 =

311

99 as n → ∞ .


Curves C3 and C4, respectively represented by polar equations

r = 4 + 2cos θ (0 ≤ θ ≤ 2π) (104)

and

r = 4cos θ + 5 (0 ≤ θ ≤ 2π) , (105)

divide the plane into several regions.

(a) [8 MARKS] Showing all your work, carefully find the area of the one region which

is bounded by C3 and C4 and contains the pole.

Solution: (cf. Figure 25 on page 3291 of these notes) Solving the given equations

yields θ = 2π

3 ,

4π

3 and r = 3. The points of intersection are

3,

2π

3

,

3,

4π

3

.

8/15/2019 Brown's Notes 2010



6

2

-6

4

0

-4

-2

8420-2 6

Figure 25: The curves with equations r = 4 + 2cos θ , r = 4 cos θ + 5

(Strictly speaking, the student should also solve using equations −r = 4 + cos(−θ )

and −r = 4 cos(−θ )+5, taking all 2×2 combinations of the equations, in case a point

of intersection appeared on the two given curves with diff erent sets of coordinates;

that would not have yielded any new points of intersection in this problem. It could

also have been necessary to check for the curves’ possibly passing through the pole— but that does not happen in this problem, since, when we set r = 0, we obtain

an equation for θ which cannot be solved.)

There are several diff erent ways of finding the area in question.

i. Find the area of the small oval (a limacon) and subtract that of the region

8/15/2019 Brown's Notes 2010



on the left. The area of the limacon is

1

2

2π

0

(2cos θ + 4)2 d θ = 1

2

2π

0

(2(1 + cos2θ ) + 16 cos θ + 16) d θ

= 1

2 [18θ + sin2θ + 16 sin θ ]2π

0 = 18π .

The area of the region to be subtracted is

2 · 1

2

π

2π3

(4 + 2cos θ )2 − (4 cos θ + 5)2

d θ

=

π

2π3

((16 + 16 cos θ + 2(1 + cos2θ )) − (8(1 + cos2θ ) + 40 cos θ + 25)) d θ

= [−15θ − 3si n2θ − 24 sin θ ]π2π3

= (−15π − 0 − 0) −−10π + 3 ·

√ 3

2 − 24 ·

√ 3

2

=

21√

3

2 − 5π ,

so the area of the region in question is 18π − 21√

3

2 + 5π = 23π − 21

√ 3

2 .

ii. Find the area of the larger outer oval and subtract the area of the region

to the right of the desired region. The area bounded by the larger oval is

1

2

2π

0

(4cos θ + 5)2 d θ = 1

2

2π

0

(8(1 + cos2θ ) + 40 cos θ + 25) d θ

= 1

2 [33θ + 4sin2θ + 40 sin θ ]2π

0 = 33π .

The area to be subtracted is

2 · 1

2

2π3

0

(4 cos θ + 5)2 − (4 + 2cos θ )2

d θ = [15θ + 3sin2θ + 24 sin θ ]

2π3

0

= 10π − 3√

3

2 +

24√

3

2 − 0

= 10π + 21

√ 3

2 .

Thus the net area is equal to

33π −10π +

21√

3

2

= 23π − 21√

3

2 .

8/15/2019 Brown's Notes 2010



iii. Join the pole by line segments to the points of intersection of the two

curves. Then compute the areas of the two subregions of the desired regionbounded by these line segments. The area to the right of the line segments is

21

2

2π3

0

(2 cos θ + 4)2 d θ =

2π3

0

(2(1 + cos2θ ) + 16 cos θ + 16) d θ

= [18θ + sin2θ + 16 sin θ ]2π3

0

=

12π −√

3

2 +

16√

3

2

− 0 = 12π + 15

√ 3

2 .

The area to the left of the line segments is

21

2

π

2π3

(4cos θ + 5)2 d θ

= [33θ + 4sin2θ + 40 sin θ ]π2π3

= 11π − 18√

3 .

Summing these two areas yields 23π − 21√

3

2 .

(b) [4 MARKS] Find another equation — call it (105*) — that also represents C4, and

has the property that there do not exist coordinates (r , θ ) which satisfy equations

(104) and (105*) simultaneously. You are expected to show that equations (104)and (105*) have no simultaneous solutions.

Solution: In my discussion of the preceding part of the problem I have shown (by

replacing (r , θ ) by (−r , θ + π)) that an alternative equation for the second curve is

−r = 4 cos(−θ )+5, equivalently r = −4cos θ −5. When this equation is solved with

r = 4 + 2cos θ , we obtain, as a consequence, that cos θ = −3

2, which is impossible.

C.37.2 Draft Solutions to the Final Examination in MATH 141 2009 01 (Version 4)

Instructions



2. Calculators are not permitted. This is a closed book examination. Regular and translation dictio-

naries are permitted.

3. . . . A TOTAL OF 75 MARKS ARE AVAILABLE ON THIS EXAMINATION.

8/15/2019 Brown's Notes 2010



4. You are expected to simplify all answers wherever possible.

• Most questions on this paper require that you SHOW ALL YOUR WORK!

. . . To be awarded partial marks on a part of a question a student’s answer for that part

must be deemed to be more than 50% correct.

• Some questions on this paper require only BRIEF SOLUTIONS ; . . . you are not asked to

show your work, and you should not expect partial marks for solutions that are not correct.



t 3 cos t 2 dt .

Solution: One substitution is u = t

2

, which implies that du = 2t dt , t 3 cos t 2 dt =

1

2

u cos u du .

Now we can integrate by parts, taking U = u, dV = cos u du, so dU = du, V =

sin u. t 3 cos t 2 dt =

1

2

u cos u du

= 1

2

u sin u −

sin u du

= 12

(u sin u + cos u) + C

= 1

2

t 2 sin

t 2

+ cost 2

+ C .

This problem could also have been attacked by an immediate application of inte-

gration by parts: u = t 2, dv = t cos t 2 dt ⇒ du = 2t dt , v = 12

sin t 2: t 3 cos t 2 dt = t 2 · 1

2 sin t 2 −

sin t 2

· 2t dt

= 1

2 t 2 · sin t 2 + cos t 2 + C .

Of course, whichever method you used, you should check your answer by di ff er-

entiation.

(b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivative

d

dt

t 2 0

tanh x2 dx .

8/15/2019 Brown's Notes 2010



Solution: Let u = t 2.

d

dt

t 2

0

tanh x2 dx = d

dt

u

0

tanh x2 dx

=

d

du

u

0

tanh x2 dx

· du

dt

=tanh u2

· du

dt =

tanh t 4· 2t .




12

1√ 2

dx√ 1 − x2 · arcsin x

.

Solution: Let u = arcsin x, so du = dx√

1 − x2. Then

12

1√ 2

dx√ 1 − x2 · arcsin x

=

arcsin 12

arcsin 1√ 2

du

u

=

π6

π4

du

u

= ln |u|]π6π4

= ln π

6 − ln

π

4 = ln

2

3.

It should be no surprise that this answer is negative, since 1√

2>

1

2, and the inte-

grand is positive in the interval 12

, 1√ 2.


2 y y2 − y + 1

dy

Solution: By completion of the square we find that

y2 − y + 1 =

y − 1

2

2

+ 3

4

8/15/2019 Brown's Notes 2010



= 3

41 + 2

√ 3 y − 1

2

2

=

3

4

1 +

2 y − 1√

3

2 .

We may thus substitute u = 2 y − 1√

3, obtaining du =

2√ 3

· dy, y = u

√ 3 + 1

2 .

2 y

y2 − y + 1

dy =

u√

3 + 1√

32

√ 1 + u2

·√

3

2

du

= √ 3 u

√ 1 + u2 du + du

√ 1 + u2

=√

3 ·√

1 + u2 +

du√

1 + u2

= 2

y2 − y + 1 +

du√

1 + u2.

TO BE CONTINUED

8/15/2019 Brown's Notes 2010



C.38 Supplementary Notes for the Lecture of Friday, April 09th, 2010

Distribution Date: Friday, April 09th, 2010, subject to correction

C.38.1 Final Examination in MATH 141 2009 01 (Version 4, continued)

2. (b) [4 MARKS] Evaluate

2 y y2 − y + 1

dy

Solution: By completion of the square we find that

y2 − y + 1 =

y − 1

2

2

+ 3

4

= 34

1 +

2√ 3

y − 1

2

2=

3

4

1 +

2 y − 1√

3

2 .

We may thus substitute u = 2 y − 1√

3, obtaining du =

2√ 3

· dy, y = u

√ 3 + 1

2 .

2 y

y2

− y + 1

dy =

u√

3 + 1√

32

√ 1 + u2

·√

3

2

du

=√

3

u√

1 + u2du +

du√

1 + u2

=√

3 ·√

1 + u2 +

du√

1 + u2

= 2

y2 − y + 1 +

du√

1 + u2.

Now we can apply a trigonometric (or hyperbolic) substitution, like tan θ = u =2 y − 1√

3, i.e., θ = arctan

2 y − 1√ 3

, so du = sec2 θ d θ . Thus

2 y y2 − y + 1

dy = 2

y2 − y + 1 +

sec θ d θ

= 2

y2 − y + 1 + ln | sec θ + tan θ | + C

= 2

y2 − y + 1 + ln

1 +

2 y − 1√

3

2

+ 2 y − 1√

3

+ C

8/15/2019 Brown's Notes 2010



= 2 y2

− y + 1 + ln 2 y

2 − y + 1 + 2 y − 1

√ 3 + C

= 2

y2 − y + 1 + ln2 y2 − y + 1 + 2 y − 1

+ C .

SHOW ALL YOUR WORK!

3. For each of the following series determine whether the series diverges, converges condi-

tionally, or converges absolutely. All of your work must be justified; prior to using any

test you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]

∞

n=1

(−

1)n+1 cos n

2n

Solution: We can’t use the nth root test here, as the sequence

cos n

2

n=1,2,...

has no

limit as n → ∞. But, if we first consider the positive series of the magnitudes of

the terms of the given series, we have

0 ≤

cos n

2

n ≤ 1

2n .

Since the series whose general term is 1

2n is a convergent (geometric) series, we

may apply the Comparison Test to conclude that the series of absolute values of

the terms of the original series is convergent; hence the original series is absolutely

convergent.

(b) [4 MARKS]

∞n=2

(−1)n 1

n√

ln n.

Solution: As x → ∞, x, ln x, and√

x are all increasing. Hence f ( x) = 1

x√

ln xis

decreasing. As it is a continuous function for sufficiently large x, we may apply the

Integral Test. The divergence of the improper integral

∞ 2

dx x

√ ln x

= lima→∞

2 √ ln x

a

2= +∞ ,

implies the divergence of the series of absolute values. Thus the series

∞n=2

(−1)n 1

n√

ln n

is at most conditionally convergent. Since x, ln x, and√

x are increasing functions

8/15/2019 Brown's Notes 2010



of x, 1 x ln x

is a decreasing function; and, as x → ∞, this last function approaches

0. Since the conditions of the Leibniz Alternating Series Theorem are satisfies, theoriginal series is convergent; but, as it is not absolutely convergent, is is therefore

conditionally convergent.

(c) [4 MARKS]

∞n=4

(−1)n 1

nln (3n + 1)

Solution: I apply the “ Test for Divergence”:

limn→∞

ln (3n + 1)

n= lim

n→∞

3n · ln 3

3n + 1

1

= (ln 3) · limn→∞

1

1 + 13n

= ln 3 > ln e = 1 .

Hence the limit limn→∞

(−1)n ln (3n + 1)

n

does not exist, and the given series diverges.

4. [9 MARKS] SHOW ALL YOUR WORK!

(a) [3 MARKS] Evaluate limn→∞

1

n ·

n

r =1

cos2

r π

n

. (Hint: This could be a Riemann

sum.)Solution: First consider the area under the graph of f ( x) = cos2 x between x = 0

and x = π. Divide the interval into n parts of equal width ∆ x = π

n. If we hang

rectangular elements of area from the curve by their upper right-hand corners, then

the area of the ith rectangle will be π

n· cos2 iπ

n. In order to be able to interpret the

given sum as a Riemann sum, we will need to scale the function by a constant. So

let’s use the function cos2 x

π. Thus the area will be

π 0

cos2 x

π dx =

π 0

1 + cos2 x

2π dx = 1

2π x +

1

2 sin 2 xπ

0=

1

2 .

(b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an =

arctan(−2n) as n → ∞.

Solution: As n → ∞, 2n → ∞, and arctan(−2n) → −π

2.

8/15/2019 Brown's Notes 2010



(c) [3 MARKS] Showing all your work, prove divergence, or find the value of

∞

n=1

∞

i=1

1

5i

n

.

Solution: Since∞

i=1

1

5i = lim

n→∞

15

1 −

15

n1 − 1

5

=

15

45

= 1

4 ,

∞n=1

∞i=1

1

5i

n

=

∞n=1

1

4

n

=

14

1 − 14

= 1

3 .

TO BE CONTINUED

8/15/2019 Brown's Notes 2010



D Problem Assignments from Previous Years

D.1 1998 / 1999

The problem numbers listed below refer to the textbook in use at that time, [31], [33]. For

many of the problems there are answers in the textbook or in the Student Solution Manual

[34].

D.1.1 Assignment 1

§5.2: 5, 11, 15, 21, 29

§5.3: 3, 9, 15, 35, 47

§5.4: none

§5.5: 17, 27, 33, 41

§5.6: 47, 55, 59, 65

§5.7: 21, 27, 33, 39, 45, 51, 57

§5.8: 33, 39, 45, 51, 57

D.1.2 Assignment 2

§6.1: none

§6.2: 3, 9, 15, 21, 27, 31, 35, 41

§6.3: 3, 9, 15, 21, 27, 31, 39, 43

§6.4: 3, 9, 15, 21, 27, 31, 35, 41

§3.8: none

Chapter 7: none

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D.1.3 Assignment 3

§8.2: 5, 13, 21, 29, 39, 45, 53

§9.2: 5, 13, 21, 29, 39

§9.3: 5, 13, 21, 29, 39, 41

§9.4: 5, 13, 21, 29, 39

§9.5: 5, 9, 17, 21, 29, 33

§9.6: 5, 9, 17, 21, 29, 33

D.1.4 Assignment 4

§9.7: 13, 17, 21, 25, 29, 33

§9.8: 21, 23, 29, 33, 39

§10.2: 39, 41, 43, 45, 47, 49, 51, 53, 57

§10.3: 9, 13, 17, 21, 23, 29, 33, 35

§10.4: 3, 5, 9, 13

D.1.5 Assignment 5

§11.2: 9, 17, 23, 33, 39

§11.3: 3, 9, 15, 21, 29, 35, 47

§11.4: 3, 9, 15, 21, 29, 35, 45, 47

§11.5: 3, 9, 15, 21, 23

§11.6: 3, 9, 15, 21, 29

§11.7: 3, 9, 15, 21, 27, 33

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D.2 1999 / 2000

(Students had access to brief solutions that were mounted on the web.)

D.2.1 Assignment 1

Before attempting problems on this assignment you are advised to try some “easy” problems

in the textbook. In most of the following problems there is a reference to a “similar” problem

in the textbook. You should always endeavour to show as much of your work as possible, and

to reduce your solution to “simplest terms”. Remember that the main reason for submitting

this assignment is to have an opportunity for your tutor to grade your work; the actual grade

obtained should be of lesser significance.

In Exercises 1-5 below, evaluate the indefinite integral, and verify by di ff erentiation:

1. (cf. [31, Exercise 5.2.5, p. 294])

3

x4 − 5 x

12 − x2 + 4 x−3

dx

2.

3

x− 2

1 + x2

dx

3. (cf. [31, Exercise 5.2.13, p. 294])

xe x2 − e4 x

dx

4. (cf. [31, Exercise 5.2.19, p. 294]) (1

−

√ x)(2 x + 3)2 dx

5. (cf. [31, Exercise 5.2.27, p. 294])

(4cos8 x − 2sin π x + cos2π x − (sin 2π) x) dx

6. (cf. [31, Example 5.2.8, p. 289]) Determine the diff erentiable function y( x) such thatdy

dx=

1√ 1 − x2

and y2− 1

2

=

π

2.

7. (This is [31, Exercise 5.2.51, p. 295] written in purely mathematical terminology.) Solve

the initial value problem: d

dx

dy

dx

= sin x, where y = 0 and

dy

dx= 0 when x = 0. [Hint:

First use one of the initial values to determine the general value of dy

dx from the given“diff erential equation”; then use the second initial value to determine y( x) completely.]

8. ([31, Exercise 5.3.4, p. 306]) Write the following in “expanded notation”, i.e. without

using the symbol

:

6 j=1

(2 j − 1).

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9. (cf. [31, Exercise 5.3.18, p. 306]) Write the following sum in “summation notation”:

x − x3

3 +

x5

5 − x7

7 + ... ± x999

999

where the signs are alternating +, −, +, −, ... The sign of the last term has not been given

— you should determine it.

10. (cf. [31, Example 5.3.6, p. 302]) Given that

ni=1

i = n(n + 1)

2 ,

ni=1

i2 = n(n + 1)(2n + 1)

6 ,

ni=1

i3 = n2(n + 1)2

4 ,

determine limn→∞(n + 1)3 + (n + 2)3 + ... + (2n)3

n4 .

D.2.2 Assignment 2

1. Evaluate the following integrals:

(a)

3

1

( x − 1)4 dx

(b)

1

0

(2e x − 1)2 dx

(c) π

0 sin 4 x dx.

2. Interpreting the following integral as the area of a region, evaluate it using known area

formulas: 6

0

√ 36 − x2 dx.

3. Use properties of integrals to establish the following inequality without evaluating the

integral: 1

0

1

1 +√

xdx ≤

1

0

1

1 + x3 dx.

4. Deduce the Second Comparison Property of integrals from the First Comparison Prop-

erty [31, p. 325, §5.5].

5. Apply the Fundamental Theorem of Calculus [31, p. 331, §5.6] to find the derivative of

the given function: x

−1

(t 2 + 2)15 dt .

8/15/2019 Brown's Notes 2010



6. Diff erentiate the functions

(a)

x3

0

cos t dt

(b)

3 x

1

sin t 2 dt .

7. Solve the initial value problem dy

dx=

√ 1 + x2 , y(1) = 5 . Express your answer in

terms of a definite integral (which you need not attempt to evaluate). This problem can

be solved using the methods of [31, Chapter 5].

8. Evaluate the indefinite integrals:

(a)

2 x

√ 3 − 2 x2 dx

(b)

x2 sin(3 x3) dx

(c)

x + 3

x2 + 6 x + 3 d x

9. Evaluate the definite integrals:

(a) 8

0

t √

t + 2 dt

(b)

π/2

0

(1 + 3sin η)3/2 cos η d η

(c)

π

0

sin2 2t dt .

10. Sketch the region bounded by the given curves, then find its area:

(a) x = 4 y2, x + 12 y + 5 = 0

(b) y = cos x, y = sin x, 0 ≤ x ≤ π

4 .

11. Prove that the area of the ellipse x2

a2 +

y2

b2 = 1 is A = πab. This problem can be

solved using the methods of [31, Chapter 5]. It is not necessary to use methods of [31,

Chapter 9].


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D.2.3 Assignment 3

In all of these problems you are expected to show all your work neatly. (This assignment is

only a sampling. Your are advised to try other problems from your textbook; solutions to some

can be found in the Student Solution Manual [32].)

1. [31, Exercise 6.1.6, p. 382] As n → ∞, the interval [2, 4] is to be subdivided into n

subintervals of equal length ∆ x by n − 1 equally spaced points x1, x2, ..., xn−1 (where

x0 = 2, xn = 4). Evaluate limn→∞

ni=1

1

xi

∆ x by computing the value of the appropriate

related integral.

2. (a) [31, Exercise 6.2.6, p. 391] Use the method of cross-sections to find the volume of

the solid that is generated by rotating the plane region bounded by y = 9 − x2 and

y = 0 about the x-axis.

(b) (cf. Problem 2a) Use the method of cylindrical shells to find the volume of the solid

that is generated by rotating the plane region bounded by y = 9− x2 and y = 0 about

the x-axis.

(c) Use the method of cross-sections to find the volume of the solid that is generated

by rotating the plane region bounded by y = 9 − x2 and y = 0 about the y-axis.

(d) (cf. Problem 2c) Use the method of cylindrical shells to find the volume of the solid

that is generated by rotating the plane region bounded by y = 9− x2 and y = 0 about

the y-axis.

3. (a) [31, Exercise 6.2.24, p. 392] Find the volume of the solid that is generated by

rotating around the line y = −1 the region bounded by y = 2e− x, y = 2, and x = 1.

(b) (cf. Problem 3a) Set up an integral that would be obtained if the method of cylin-

drical shells were used to represent the volume of the solid that is generated by

rotating around the line y = −1 the region bounded by y = 2e− x, y = 2, and x = 1.

YOU ARE NOT EXPECTED TO EVALUATE THE INTEGRAL.

4. (cf. [31, Exercise 6.2.40, p. 392]) The base of a certain solid is a circular disk with di-

ameter AB of length 2a. Find the volume of the solid if each cross section perpendicular

to AB is an equilateral triangle.

5. (a) [31, Exercise 6.3.26, p. 401] Use the method of cylindrical shells to find the volume

of the solid generated by rotating around the y-axis the region bounded by the

curves y = 1

1 + x2, y = 0, x = 0, x = 2.

8/15/2019 Brown's Notes 2010



(b) (cf. Problem 5a) Use the method of cross sections to find the volume of the solid

generated by rotating around the y-axis the region bounded by the curves y =1

1 + x2, y = 0, x = 0, x = 2.

6. (cf. [31, Exercise 7.3.69, p. 450]) Find the length of the arc of the curve y = e x + e− x

2between the points (0, 1) and (ln 2, 2).

7. (a) [31, Exercise 6.4.30, p. 411] Find the area of the surface of revolution generated

by revolving the arc of the curve y = x3 from x = 1 to x = 2 around the x-axis.

(b) (cf. 7a) Set up an integral for, BUT DO NOT EVALUATE, the area of the surface

of revolution generated by revolving the arc of the curve y = x3 from x = 1 to x = 2

around the y-axis.

8. [31, Exercise 7.2.44, p. 442] Evaluate the indefinite integral x + 1

x2 + 2 x + 3 d x

9. (cf. [31, Exercise 7.2.36, p. 442]) Determine the value of the function f ( x) =

x

−1

t 2

8 − t 3 dt

for any point x < 2.

10. (cf. [31, Exercise 7.3.70, p. 450]) Find the area of the surface generated by revolvingaround the x-axis the curve of Problem 6.

D.2.4 Assignment 4

1. Diff erentiate the functions:

(a) sin−1( x50)

(b) arcsin(tan x)

(c) cot−1 e x + tan−1 e− x

2. Showing all your work, evaluate the integrals:

(a)

dx√

1 − 4 x2

(b)

dx

2√

x(1 + x)

8/15/2019 Brown's Notes 2010



(c) e x

1 + e2 x dx

(d)

cot

√ y csc

√ y√

ydy

(e)

(ln t )8

t dt

(f)

tan4 2 x sec2 2 x dx

(g) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5. x2

√ 16 x2 + 9

dx

3. Use integration by parts to compute the following integrals. Show all your work.

(a)

t cos t dt

(b)

√ y ln y dy

(c) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5.

x2 arctan x dx

(d)

csc3

x dx

(e)

ln(1 + x2) dx

4. Showing all your work, evaluate the following integrals:

(a)

cos2 7 x dx

(b)

cos2 x sin3 x dx

(c)

sin

3

2 xcos2 2 x

dx

(d)

sec6 2t dt

8/15/2019 Brown's Notes 2010



D.2.5 Assignment 5

1. [31, Exercise 9.5.6, p. 540] Find

x3

x2 + x − 6 d x. (Your solution should be valid for x

in any one of the intervals x < 3, −3 < x < 2, x > 2.)

2. [31, Exercise 9.5.8, p. 540] Find

1

( x + 1)( x2 + 1) d x.

3. (a) [31, Exercise 9.5.23] Find

x2

( x + 2)3 dx.

(b) Find the volume of the solid of revolution generated by the region bounded by

y = x

( x + 2)

3

2

, y = 0, x = 1, and x = 2 about the x-axis.

(c) Find the volume of the solid of revolution generated by the region bounded by

y = x

( x + 2)32

, y = 0, x = 1, and x = 2 about the y-axis.

4. [31, Exercise 9.5.38, p. 540] Make a preliminary substitution before using the method

of partial fractions: cos θ

sin2 θ (sin θ − 6)d θ

5. [31, Exercise 9.6.6, p. 547] Use trigonometric substitutions to evaluate the integral

x2

√ 9 − 4 x2 dx.

6. [31, Exercise 9.6.26, p. 547] Use trigonometric substitutions to evaluate the integral 1

9 + 4 x2 dx.

7. [31, Exercise 9.6.35, p. 547] Use trigonometric substitutions to evaluate the integral √ x2 − 5

x2 dx.

8. [31, Exercise 9.7.14, p. 553] Evaluate the integral x√

8 + 2 x − x2 dx.

9. [31, Exercise 9.8.17, p. 561] Determine whether the following improper integral con-

verges; if it does converge, evaluate it:

∞

−∞

x

x2 + 4 d x.

10. [31, Exercise 9.8.27, p. 561] Determine whether the following improper integral con-

verges; if it does converge, evaluate it:

∞

0

cos x dx.

8/15/2019 Brown's Notes 2010



11. (cf. [31, Exercise 9.8.14, p. 561]) Determine whether the following improper integral

converges; if it does converge, evaluate it: +8

−8

1( x + 4)

23

dx.

12. [31, Exercise 10.2.2, p. 580] Find two polar coordinate representations, one with r ≥ 0,

and the other with r ≤ 0 for the points with the following rectangular coordinates:

(a) (−1, −1),

(b) (√

3, −1),

(c) (2, 2),

(d) (−1,√

3),

(e) ( √ 2, − √ 2),

(f) (−3,√

3).

13. For each of the following curves, determine — showing all your work — equations in

both rectangular and polar coordinates:

(a) [31, Exercise 10.2.20, p. 580] The horizontal line through (1, 3).

(b) [31, Exercise 10.2.26, p. 580] The circle with centre (3, 4) and radius 5.

14. (a) [31, Exercise 10.2.56, p. 581] Showing all your work, find all points of intersection

of the curves with polar equations r = 1 + cos θ and r = 10 sin θ .

(b) Showing all your work, find all points of intersection of the curves with polar equa-

tions r 2 = 4 sin θ and r 2 = −4sin θ .

[Note: The procedure sketched in the solution of [31, Example 10.2.8, p. 579] for finding

points of intersection is incomplete. Your instructor will discuss a systematic procedure

in the lectures.]

D.2.6 Assignment 6

1. Find the area bounded by each of the following curves.

(a) r = 2 cos θ,(b) r = 1 + cos θ.

2. Find the area bounded by one loop of the given curve.

(a) r = 2 cos 2θ,

(b) r 2 = 4 sin θ.

8/15/2019 Brown's Notes 2010



3. Find the area of the region described.

(a) Inside both r = cos θ and r = √ 3sin θ.

(b) Inside both r = 2 cos θ and r = 2 sin θ .

4. Eliminate the parameter and then sketch the curve.

(a) x = t + 1, y = 2t 2 − t − 1.

(b) x = et , y = 4e2t .

(c) x = sin 2πt , y = cos 2πt ; 0 ≤ t ≤ 1. Describe the motion of the point ( x(t ), y(t )) as

t varies in the given interval.

5. Find the area of the region that lies between the parametric curve x = cos t , y =sin2 t , 0 ≤ t ≤ π, and the x-axis.

6. Find the arc length of the curve x = sin t − cos t , y = sin t + cos t ; π/4 ≤ t ≤ π/2.

7. Determine whether the sequence an converges, and find its limit if it does converge.

(a) an = n2 − n + 7

2n3 + n2 ,

(b) an = 1 + (−1)n

√ n

(3/2)n ,

(c) an = n sin πn,

(d) an =

n − 1

n + 1

n

.

8. Determine, for each of the following infinite series, whether it converges or diverges. If

it converges, find its sum.

(a) 1 + 3 + 5 + 7 + . . . + (2n − 1) + . . . ,

(b) 4 + 43

+ . . . + 43n + . . . ,

(c)

∞

n=1

(5−n − 7−n),

(d)

∞n=1

e

π

n

.

9. Find the set of all those values of x for which the series

∞n=1

x

3

n

is a convergent geomet-

ric series, then express the sum of the series as a function of x.

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10. Find the Taylor polynomial in powers of x − a with remainder by using the given values

of a and n.

(a) f ( x) = sin x; a = π/6, n = 3.

(b) f ( x) = 1

( x − 4)2; a = 5, n = 5 .

11. Find the Maclaurin series of the function e−3 x by substitution in the series for e x.

12. Find the Taylor series for f ( x) = ln x at the point a = 1.

13. Use comparison tests to determine whether each of the following infinite series converge

or diverge.

(a)

∞n=1

1

1 + 3n,

(b)

∞n=1

√ n

n2 + n,

(c)

∞n=1

sin2(1/n)

n2 .

D.3 2000 / 2001

(In the winter of the year 2001 Assignments based on WeBWorK were used, although the

experiment had to be terminated in mid-term because of technical problems.)

D.4 2001 / 2002

This was the first time WeBWorK assignments were used exclusively in this course.

D.5 MATH 141 2003 01

WeBWorK assignments were used exclusively for assignments. The questions are not avail-

able for publication.

D.6 MATH 141 2004 01

WeBWorK assignments were used exclusively for assignments. The questions are not avail-

able for publication.

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D.7 MATH 141 2005 01

WeBWorK assignments were used for online assignments; the questions are not available for

publication. In addition, these written assignments were intended to provide students with in-

dividualized opportunities to work problems for which the textbook often provided examples;

at the time, appropriate materials available from WeBWorK for this purpose were limited. The

individualization was often based on the student number.

D.7.1 Written Assignment W 1

Your written assignments will usually be mounted on the WeBWorK site, and will usually be

individualized, that is, your problems will not be exactly the same as those of other students.

This first written assignment is based on your WeBWorK assignment R1. Subsequent written

assignments may be designed in other ways. Because of its general form, it was possible to

release this assignment in the document Information for Students in MATH 141 2005 01; some

of the other assignments may appear only on your WeBWorK site.

Your completed assignment must be submitted together with your solutions to quiz Q1,

inside your answer sheet for that quiz. No other method of submission is acceptable.

Purpose of the written assignments These assignments are designed to help you learn how

to write full solutions to problems. While they carry a very small weight in the computation of

your final grade, conscientious completion of the assignments should help you substantially in

learning the calculus, and help prepare you for your final examination.

Certificate Your assignment will not be graded unless you attach or include the following

completed certificate of originality:

I have read the information on the web page

http: // www.mcgill.ca / integrity / studentguide / ,

and assert that my work submitted for W 1 and R1 does not violate McGill’s

regulations concerning plagiarism.

Signature(required) Date(required)

The assignment questions Your assignment consists of the following problems on your ver-

sion of R1:

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##3, 4, 5, 6, 7, 8

(Teaching Assistants are not primarily checking for plagiarism; but, if they detect it, they may

be obliged to report any apparent violations to the Associate Deans.)

Complete solutions are required It is not enough to give the correct answer; in fact, the

numerical answer alone may be worth 0 marks. You should submit a full solution, similar to

solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should

look if you have doubts about the amount of detail required in a solution.

Use of calculators You are expected to complete the entire assignment without the use of

a calculator. In particular, you are expected to be familiar with the values of trigonometric

functions at “simple” multiples and submultiples of π.

Not all problems may be graded On all of the written assignments it is possible that the

Teaching Assistant will grade only a small number of the solutions you submit. The numbers

of the questions that will be graded will not be announced in advance, even to the tutor. For

that reason you are advised to devote equal attention to all of the problems.


Written assignment W 2 is based on your WeBWorK assignment R3, but some problems are

being modified. Your completed assignment must be submitted together with your solutions toquiz Q2, inside your answer sheet for that quiz. No other method of submission is acceptable.

Purpose of the written assignments These assignments are designed to help you learn how

to write full solutions to problems. While they carry a very small weight in the computation of

your final grade, conscientious completion of the assignments should help you substantially in

learning the calculus, and help prepare you for your final examination.


completed certificate of originality, signed in ink:

8/15/2019 Brown's Notes 2010





and assert that my work submitted for W 2 and R3 does not violate McGill’s

regulations concerning plagiarism.


The assignment questions Your assignment consists of the following problems on your ver-

sion of R1:

1. Problem 1 of R3, solved by integration with respect to x. Include in your solution a rough

sketch of the region, showing a typical element of area.

2. Problem 1 of R3 solved by integration with respect to y. This will require rewriting the

equations of the curves appropriately. You may assume without proof that ln x dx = x(ln x − 1) + C ,

a fact which you will see derived later in the course. Include in your solution a rough

sketch of the region, showing a typical element of area.

3. Problem 6 of R3, evaluated using the Method of Washers. Include in your solution a

rough sketch of the plane region which generates the solid, showing a typical element of

area which will generate a typical washer.

4. Problem 6 of R3, evaluated using the Method of Cylindrical Shells. Include in your

solution a rough sketch of the plane region which generates the solid, showing a typical

element of area which will generate a typical cylindrical shell.


numerical answer alone may be worth 0 marks. You should submit a full solution, similar to

solutions in Stewart’s textbook or the Student Solutions Manual, which are where you should

look if you have doubts about the amount of detail required in a solution.

Use of calculators You are expected to complete the entire assignment without the use of a

calculator.

8/15/2019 Brown's Notes 2010




Teaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. For

that reason you are advised to devote equal attention to all of the problems.


Unlike the preceding written assignments Written Assignment W 3 is not directly based on your

WeBWorK assignments, although some problems will be similar to WeBWorK assignment

problems. Your completed assignment must be submitted together with your solutions to quiz

Q3, inside your answer sheet for that quiz. No other method of submission is acceptable.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted statement of originality, signed in ink:



and assert that my work submitted for W 3 does not violate McGill’s regula-

tions concerning plagiarism.


The assignment questions The parameters in these problems are based on the digits of your

9-digit McGill student number, according to the following table:

Parameter name: A B C D E F G H J

Your student number:

Before starting to solve the problems below, determine the values of each of these integer

constants; then substitute them into the descriptions of the problems before you begin your

solution.

1. Showing all your work, systematically determine

Ax2 + Bx + C

e− x dx by repeated

integration by parts: no other method of solution will be accepted. Verify by diff erenti-

ation that your answer is correct.

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2. Showing all your work, use trigonometric or hyperbolic substitutions to evaluate

each of

du u2 − (F + 1)2

,

du u2 + (F + 1)2

,

du −u2 + (F + 1)2

.

Verify that your answers are correct by diff erentiation.

3. Let K =

J +42

=

J +42

.62 Showing all your work, develop for this integer K a reduction

formula of the following type that can be used to evaluate

I n( x) =

xn (sin(K x)) dx in terms of I n−2( x):

xn (sin K x) dx = L· xn cos K x + M

· xn−1 sin K x + N xn−2 (sin K x) dx

where n ≥ 2 and L, M , N are constants that you are expected to determine only by

integration by parts. Again showing all your work, use the reduction formula you have

just determined to evaluate

x2 sin K x d x, and test by diff erentiation the answer that it

gives — you should recover x2 sin K x.

4. Showing all your work, evaluate both of the integrals sinF +1 x · cos2 x dx and

sinF +2 x · cos2 x dx


final answer alone may be worth 0 marks. You should submit a full solution, similar to solu-

tions in Stewart’s textbook or the Student Solutions Manual, which are where you should look

if you have doubts about the amount of detail required in a solution.


Teaching Assistant will grade only a small number of the solutions you submit. The numbers

of the questions that will be graded will not be announced in advance, even to the tutor. If for

no other reason, you are advised to devote equal attention to all of the problems.


Your completed assignment must be submitted together with your solutions to quiz Q4, inside

your answer sheet for that quiz. No other method of submission is acceptable.

62Determine K from J using the greatest integer function, defined in your textbook, page 110. Your textbook

uses the notation

J +42

, but some authors write the function as

J +4

2

.

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completed statement of originality, signed in ink:






The assignment questions Some of the parameters in these problems are based on the digits

of your 9-digit McGill student number, according to the following table:

Parameter name: A B M D E F G H J




solution. It is not enough to give the correct answer; in fact, the final answer alone could be

worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or

the Student Solutions Manual, which are where you should look if you have doubts about the

amount of detail required in a solution. Not all problems may be graded.

1. Showing all your work, systematically determine the value of the following integral: x

( x + J )2( x − H − 1)1 dx . Verify your work by diff erentiation of your answer: you

should recover the integrand. (Systematically means that you to use the methods you

learned in this course for the treatment of problems of this type, even if you happen to

see some other method that could be used in this particular example.)

2. Showing all your work, use a substitution to transform the integrand into a rational

function, then integrate the particular integral that is assigned for your particular valueof G:

If G = 1,4,or 7:

1

x −√

x + 2dx

If G = 0, 2, 5, or 8:

cos2 x

sin2 2 x + sin 2 xdx

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If G = 3, 6, or 9: 1

x − 2 − 2 √ x + 1

dx

3. Problem 55, pages 504-505 of your textbook, describes a substitution discovered by Karl

Weierstrass (1815-1897) for the evaluation of rational functions of sin x and cos x into

an ordinary rational function. It states that if, for x such that −π < x < π, we define

t = tan x2

, then cos x = 1 − t 2

1 + t 2 and sin x =

2t

1 + t 2 and that, at a consequence, dx =

2

1 + t 2 dt . You are not asked to verify these facts. You are asked to use the substitution

to transform the following integral and then to evaluate it:

π

2

π3

1

1 + sin x − cos xdx

4. Use the trigonometric identities given in your textbook on page 487 to evaluate the fol-

lowing integral:

cos M x ·cos Dx ·sin x dx where M and D are the digits of your student

number, defined above.

5. Showing all your work, determine whether each of the following integrals is convergent

or divergent. If it is convergent, determine its value (again showing all your work):

(a)

4

0

1

2 + E

·F

dx

(b)

∞

0

x2 − E − F

x2 + E + F dx

(c)

π B+4

0

1

x sin((G + 2) x) d x


Your completed assignment must be submitted together with your solutions to quiz Q5, inside

your answer sheet for that quiz. No other method of submission is acceptable.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted statement of originality, signed in ink:

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The assignment questions Some of the parameters in these problems are based on the digits

of your 9-digit McGill student number, according to the following table:

Parameter name: A B D E F G H J K




solution. It is not enough to give the correct answer; in fact, the final answer alone could be

worth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook or

the Student Solutions Manual, which are where you should look if you have doubts about the

amount of detail required in a solution. Not all problems may be graded.

1. For the point with polar coordinates H , −

π

K 2 + 5

,

(a) find four other pairs of polar coordinates, two with r ≤ 0 and two with r > 0:

(b) Find the cartesian coordinates, assuming that the positive x-axis is along the polar

axis, the origin is at the pole, and the positive y-axis is obtained by turning the polar

axis through a positive angle of π2

.

2. For the point whose cartesian coordinates are (F 2+1, F 2−2), determine polar coordinates

(r , θ ) with the following properties:

(a) r > 0, 0 ≤ θ < 2π

(b) r < 0, 0 ≤ θ < 2π

(c) r > 0, 5π2 ≤ θ < 9π

2

3. For the following curve given in polar coordinates, determine, showing all your compu-

tations, the slope of the tangent at the point with θ = π

4:

r = A + cos( B + 2)θ .

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4. Showing all your work, find the area contained between the outer loop and the inner loop

of the curver = 1 + 2sin θ .

Explain carefully how you have established the limits for your definite integral.

5. The curve C is given by the parametric equations

x = 1 + (C + 2)t 2

y = t − ( E 2 − 2)t 3

Showing all your work, determine the value of d 2 y

dx2(t ).

6. Giving an explanation, determine whether or not the following sequences converge.

Where a sequence converges, find its limit.

(a) an = ln n − ln(n + A + 1)

(b) an = ln n

ln(( A + 2)n)

(c) an =√

n2 + A2 −√

n2 + A2 + 2

D.8 MATH 141 2006 01

D.8.1 Solution to Written Assignment W 1

Release Date: Friday, 27 January, 2006

The Assignment was posted to the class via WeBCT on January 2nd, 2006

Instructions to Students

1. Your solution to this assignment should be brought to your regular tutorial, during the

week 16-20 January, and folded inside your solution sheet to the quiz which will be

written at that time.

2. Your TA may make special arrangements for submission at other times, but, no solutionsare ever accepted after the end of the week.

3. Your solution must use the data on your own WeBWorK assignment, as described be-

low.

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The Assignment Question Problem 7 of WeBWorK assignment R1 requires that you eval-

uate a definite integral of the form C

D

( E x2 − Ax + B) dx,

where A, B, C , D, E are various combinations of constants, individualized for each student.

Your written assignment is to evaluate a simplified version of this integral as the limit of a

sequence of Riemann sums, using left endpoints63. You may simplify your problem only in

the following way: you may take the lower limit of the integral (here called D) to be 0; this

is purely to make the algebra a little easier. You should model your proof on that given in [7,

Example 2(b), pp. 383-385]. You are expected to provide a full solution for your version of the

problem — it is not enough to supply the correct answer, and you must solve the version of

the problem on your own WeBWorK assignment R1. In your solution you will need to use the

well known formulæ for the sums of the 1st and 2nd powers of the natural numbers from 1 to n;

these formulæ are in your textbook [7, Formulæ ##4, 5, p. 383], and you will not be expected

to prove them here.64 (This is a time-consuming exercise; the purpose of the assignment is to

ensure that you will have correctly solved a problem of this type during the term.)

The Solution (This solution has been composed with variables, so that it will produce valid

numerical solutions for all students. The particular solution for your version of the problem

should look much simpler, because, in place of all variables except n, there will be specific

integers.)

The length of each of n subintervals is

∆ x = C − D

n.

Thus xi = D + i∆ x = D + i

C − D

n

. Since we are using left endpoints we apply [7, top

formula, p. 381] C

D

E x2 − Ax + B

dx

= limn→∞

ni=1

E x

2

i−1 − Axi−1 +

B ·

∆ x

= limn→∞

ni=1

E

D2 +

2 D(C − D)

n · (i − 1) +

(C − D)2

n2 · (i − 1)2

63Note that the solution given in the textbook uses right endpoints.64Of course, these formulæ may be proved by induction, and students who took MATH 140 2005 09 should

know how to write up such proofs if they had to.

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− A D + C − D

n· (i − 1) + B · ∆ x

= limn→∞

( ED2 − AD + B)n

i=1

1 + (2 DE − A)(C − D)

n

ni=1

(i − 1) + E (C − D)2

n2

ni=1

(i − 1)2

· ∆ x

Nown

i=1

1 = sum of n 1’s

= nn

i=1

(i − 1) =

n−1 j=1

j

= (n

−1)n

2 by [7, Formula 4, p. 383]

ni=1

(i − 1)2 =

n−1 j=1

j2

= (n − 1)n(2n − 1)

6 by [7, Formula 5, p. 383].

Substituting the values of these sums yields C

D

E x2 − Ax + B

dx

= limn→∞C

− D

n · ( ED2

− AD + B)n +

(2 DE

− A)(C

− D)

n · (n

−1)n

2

+ E (C − D)2

n2 · (n − 1)n(2n − 1)

6

= limn→∞

( ED2 − AD + B)(C − D) +

(2 DE − A)(C − D)2

2

1 − 1

n

+ E (C − D)3

6 ·

1 − 1

n

2 − 1

n

= ( ED2 − AD + B)(C − D) + (2 DE − A)(C − D)2

2 · 1 +

E (C − D)3

6 · 1 · 2

= ( ED2

− AD + B)(C − D) +

(2 DE

− A)(C

− D)2

2 +

E (C

− D)3

3

= E C 3 − D3

3

− AC 2 − D2

2

+ B(C − D)

It was suggested that you simplify your problem by taking D = 0; in that case the value would

have worked out to be EC 3

3 − AC 2

2 + BC .

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The Grading Scheme The assignment was to be graded OUT OF A TOTAL OF 10 MARKS.

(In the version of this solution circulated to TA’s, there followed some technical details about the grad-ing.)

D.8.2 Solution to Written Assignment W 2

Release Date: Mounted on the Web on February 8th, 2006; corrected on March 16th, 2006

Solutions were to be submitted inside answer sheets to Quiz Q2, at tutorials during the period

January 30th – February 2nd, 2006.

The Problem. In [7, Example 6, p. 441] of your textbook, the author solves the following

problem: “Find the area enclosed by the line y = x

− 1 and the parabola y2 = 2 x + 6,” by

integrating with respect to y. The textbook then states the following: “We could have foundthe area in Example 6 by integrating with respect to x instead of y, but the calculation is

much more involved. It would have meant splitting the region in two, and computing the

areas labelled A1 and A2 in Figure 14. The method we used in Example 6 is much easier.” Your

assignment is to solve the problem by integrating with respect to x, and you may find the figure

in the textbook helpful. The computation is not really very hard, but it does involve working

with√

2. You should not approximate√

2; just write it that way and work with it carefully,

and the square roots will cancel by the time you finish your solution.

You know the numerical answer to this question; the purpose of the assignment is to nudge

you into solving problems in two ways, so that you can verify your work; and to show you that

there really is nothing to fear, even if you do happen to choose the “wrong” way to approach a

problem.

Your TA will be alert to the possibility that students might be copying their solutions from

others. Please write up your own solution, so that your TA will not have to waste everyone’s

time by sending exact copies to the disciplinary officer of the Faculty. You need to know how

to solve problems of this type, since you could be expected to demonstrate that ability at a

future quiz or on the examination.

The Solution. The integration must be carried out separately for two subregions. This is

because the area to the left of the line x = −1 is bounded above and below by the parabola;

while the area to the right of x =

−1 is bounded above by the parabola, and below by the line

y = x − 1. The method you know for finding the area between two curves requires that theequations of the curves be written as the graphs of functions of x. But the parabola y2 = 2 x + 6

is not the graph of a function, since it crosses some vertical lines twice. However, we can

factorize the equation in the form y −

2( x + 3)

y +

2( x + 3)

= 0 :

8/15/2019 Brown's Notes 2010



the parabola is the graph of two functions — the upper arm of the parabola is the graph of

y = √ 2( x + 3), while the lower arm is the graph of y = − √ 2( x + 3). To the left of x = −1the region A1 is bounded above by the graph of y =

√ 2( x + 3), and below by the graph of

y = −√ 2( x + 3); thus the height of the vertical element of area is the diff erence between these

two functions, i.e.,√

2( x + 3) − (− √ 2( x + 3)) = 2

√ 2( x + 3), and this will be the integrand for

the integral; the area to the right of x = −1 is bounded above by the parabola y =√

2( x + 3),

and below by the line y = x−1, so the element of area for the integral will be√

2( x + 3)−( x−1).

The total area is thus −1

−3

2

2( x + 3) dx +

5

−1

2( x + 3) − ( x − 1)

dx

= 2 √ 2

· 2

3 ·( x + 3)

32

−1

−3

+ √ 2

· 2

3 ·( x + 3)

32

− x2

2 − x

5

−1

= 4

√ 2

3

2

32 − 0

+

2

√ 2

3 · 8

32 −

25

2 − 5

−2

√ 2

3 · 2

32 −

1

2 + 1

= 16

3 +

56

3 − 6

=

16

3 +

38

3 = 18.

The Grading Scheme. The assignment was to be graded out of a maximum of 15 marks.

(In the version of this document prepared for TA’s there were additional details on the marking

scheme.)

D.8.3 Solutions to Written Assignment W 3

Release Date: March 25th, 2006

Solutions were to be submitted inside answer sheets to Quiz Q3, at tutorials during the period

13-16 February, 2006

The Problems

1. Use two integrations by parts to evaluate the integral

(sin x) · (sinh x) dx. You will

need, at the last step, to solve an equation. The solution should resemble the solution in

your textbook to [7, Example 4, p. 478], where the author evaluates

e x sin x dx.

2. Write your student number (9 digits) in the following chart.

A1 A2 A3 A4 A5 A6 A7 A8 A9

Student # :


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Some of the digits are to be used in solving the following problem.

(a) You are to evaluate the following integral by integration by parts: A7 x

2 + A8 x1 + A9

e A1 x dx

Your answer should be simplified as much as possible.

(b) Then you are to diff erentiate the function you have obtained, to verify that it is

indeed an antiderivative of the given integrand.

Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in

the Handbook on Student Rights and Responsibilities (PDF English version - French version). Article

15(a) of the Code, which is devoted to plagiarism, reads as follows:

No student shall, with intent to deceive, represent the work of another person as his or

her own in any academic writing, essay, thesis, research report, project or assignment

submitted in a course or program of study or represent as his or her own an entire essay or

work of another, whether the material so represented constitutes a part or the entirety of

the work submitted.

Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you really

need to learn how to solve these kinds of problems yourself.

The Solutions1. Let u = sin x, v = sin x. Then u = cos x, v = cosh x (or cosh x plus any real constant

— I have taken the constant to be 0, as all choices of constant here will lead to the same

solution.) (sin x) · (sinh x) dx = (cos x) · sinh x −

(cosh x) · (cos x) dx .

To evaluate

(cosh x) · (cos x) dx I will set U = cos x, V = cosh x, so U = − sin x,

V = sinh x. (cosh x) · (cos x) dx = (cos x) · sinh x +

(sinh x) · sin x dx .

Combining these results yields (sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) −

(sin x) · (sinh x) dx

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Combining our results yields A7 x

2 + A8 x1 + A9

e A1 x dx

=

A7

A1

x2 +

A8

A1

− 2 A7

A21

x1 +

A9

A1

− A8

A21

+ 2 A7

A31

· e A1 x + C

(b) Remember that you were to diff erentiate the preceding product of polynomial and

exponential by the Product Rule to show that, indeed, its derivative is the given

integrand.


Release Date: March 25th, 2006

Solutions were to be submitted at inside answer sheets to Quiz Q4, at tutorials during the

week 06 – 09 March, 2006

The Problems This assignment is based on your WeBWorK assignment R5. You are asked

to write out complete solutions to the following modifications of your versions of problems on

that assignment. Note that, in some cases, the question asks for more than was asked on the

assignment. You are not permitted to use a Table of Integrals.

1. Problem 10:

(a) Evaluate the following integral with the specific values of the constants given in

your own WeBWorK assignment: E x2 + E x + F

x3 + Gx2 + H x + J dx .

HINT: −C is a root of the denominator.

(b) No marks will be given unless you verify your integration by diff erentiating your

answer.

2. Problem 12:

(a) Evaluate the following integral with the specific values of the constants given in

your own WeBWorK assignment: Bx + C

( x2 + A2)2 d x .

8/15/2019 Brown's Notes 2010



(b) No marks will be given unless you verify your integration by diff erentiating your

answer.

3. Problem 16: Determine whether the following integral (with the constants given in your

own WeBWorK assignment) is divergent or convergent. If it is convergent, evaluate it.

If not, prove that fact. A

−∞

1

x2 + 1 d x .

In each case your solution should begin by your writing out the full problem with your data,

so your TA does not have to look up your data on the WeBWorK system.

While these problems were generated by WeBWorK, they now constitute a conventional

mathematics assignment, and it does not suffice to make unsubstantiated statements. You are

expected to prove everything you state. Thus, for example, in Problem 10, you have to show

how you use the fact that a certain number is stated to be a root of the denominator.Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in




her own in any academic writing, essay, thesis, research report, project or assignment

submitted in a course or program of study or represent as his or her own an entire essay or


the work submitted.



The Solutions

1. Under a change of variable of the form u = x

A, the integral reduces to one of the form

Ku + Lu2 + 1

2 du. The term

Ku

u2 + 1 du simplifies under a substitution v = u2 to a multiple

of dv

(v + 1)2. The term

L

u2 + 12 du simplifies under a substitution θ = arctan u,

eventually proving to be a multiple of θ + tan θ

2(tan2 θ + 1), etc.

2. An antiderivative is arctan x. This is evaluated between an upper limit of some constant

A, and a lower limit we may call B. We need to observe that lim B→−∞

arctan B = −π2

.

8/15/2019 Brown's Notes 2010




Release Date: Solutions were to be submitted inside answer sheets to Quiz Q5, at tutorials

during the week 20 – 23 March, 2006

The Problems This assignment will be graded out of a maximum of 20 MARKS.

Write your student number (9 digits) in the following chart.

A1 A2 A3 A4 A5 A6 A7 A8 A9

Student # :

Some of the digits are to be used in solving the following problems.

1. Consider the curve in the plane defined parametrically by

x(t ) = t 2 + 1

y(t ) = A7t 2 + A8t + A9

(a) [3 MARKS] Showing all your work, determine the slope of the tangent to the curve

at the point with parameter value t = 1.

(b) [6 MARKS] Showing all your work, determine the value of d 2 y

dx2 at the point with

general parameter value t (t 0). For this problem you must not substitute in any

formula from your class notes or any textbook; you are expected to determine the2nd derivative by di ff erentiation, for example in the manner similar to that done in

your textbook, Example 1, page 661.

(c) [1 MARK] Determine the range of values of t — if there are any such values —

where the curve is concave upward.

2. This problem is based on Problem 12 on your WeBWorK assignment R8. For the given

arc of the given curve,

(a) [8 MARKS] determine the area of the surface of revolution of that arc about the

x-axis;

(b) [2 MARKS] set up an integral for the area of the surface of revolution of that arcabout the y-axis, but, in this case only, do not evaluate the integral.

In each case you are expected to show all your work.

Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears in



8/15/2019 Brown's Notes 2010




her own in any academic writing, essay, thesis, research report, project or assignmentsubmitted in a course or program of study or represent as his or her own an entire essay or


the work submitted.



The Solutions

1. (a)

x(t ) = t 2 + 1

y(t ) = A7t 2 + A8t + A9

dx

dt = 2t

dy

dt = 2 A7t + A8

dy

dx=

dy

dt

dxdt

= 2 A7t + A8

2t

dy

dx

t =1

= A7 + A8

2

(b)

d 2 y

dx2 =

d

dx

2 A7t + A8

2t

= d

dt

2 A7t + A8

2t

· dt

dx

= d

dt

2 A7t + A8

2t

· 1

dxdt

= d

dt 2 A7t + A8

2t · 1

2t

= (2 A7)(2t ) − (2 A7t + A8)(2)

(2t )3

= − A8

4t 3

(c) If the student’s A8 is 0, the curve is flat. Otherwise it is concave upwards precisely

when t < 0.

8/15/2019 Brown's Notes 2010



2. CORRECTED ON 20 MARCH 2006. The following solution is valid only

when K , the upper limit of integration, is sufficiently small. The correct so-lution is much more complicated, as we need to ensure that the respective

factors cos θ + θ sin θ and sin θ − θ cos θ remain positive. I have corrected my

original version by inserting absolute signs. However, the evaluation of these

integrals is complicated when we attempt to break the integral up into parts

based on the signs of these factors. This is certainly more difficult than was

intended from students in this course. Please exercise good judgment in grad-

ing the two parts of this problem, and do not expect those students for whom

the upper limit K is large to obtain the correct area.

x(θ ) = a(cos θ + θ sin θ ) y(θ ) = a(sin θ − θ cos θ )

⇒ x = aθ cos θ

y = aθ sin θ

⇒

( x)2 + ( y)2 = |aθ |(a) about the x-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

K

0

2πa| sin θ − θ cos θ | · |θ | d θ = 2πa

K

0

θ sin θ d θ − K

0

θ 2 cos θ d θ

for positive K su fficiently small. We must integrate by parts.

θ sin θ d θ = −θ cos θ +

cos θ d θ

= −θ cos θ + sin θ + C θ 2 cos θ d θ = θ 2 sin θ − 2

θ sin θ d θ

= θ 2 sin θ + 2θ cos θ − 2sin θ + C (θ sin θ − θ 2 cos θ ) d θ = −3θ cos θ + 3sin θ − θ 2 sin θ + C

Hence the area of the surface of revolution is −3K cos K + 3 sin K − K 2 sin K .

(b) about the y-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

K

0

2πa| cos θ + θ sin θ | · |θ | d θ = 2πa

K

0

θ cos θ d θ +

K

0

θ 2 sin θ d θ

.

D.9 MATH 141 2007 01

The use of written assignments was discontinued in 2007.

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E Quizzes from Previous Years

E.1 MATH 141 2007 01

E.1.1 Draft Solutions to Quiz Q1

Distribution Date: Mounted on the Web on 4 February, 2007

corrected 19 January, 2010 — subject to further corrections

There were four diff erent types of quizzes, for the days when the tutorials are scheduled. Each

type of quiz was generated in multiple varieties for each of the tutorial sections. The order of

the problems in the varieties was also randomly assigned. All of the quizzes had a heading that

included the instructions

• Time = 30 minutes

• No calculators!

• Show all your work: marks are not given for answers alone.

• Enclose this question sheet in your folded answer sheet.

In the following I will either provide a generic solution for all varieties, or a solution to one

typical variety.

Monday version

1. [5 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative

of the function

g( x) =

a

x

b tan(t ) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluate

g( x), by first verifying carefully that ln | sec x| is an antiderivative of tan x.

Solution:

(a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as a

function of its upper index of integration. Here the variable is the lower index of integration.

d

dx

a

x

b tan(t ) dt

= d

dx

− x

a

b tan(t ) dt

8/15/2019 Brown's Notes 2010



=

−

d

dx

x

a

b tan(t ) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definite

integral with respect to the lower index, and this should be accepted if work has

been shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-

position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

d

dxln | sec x| =

d

duln |u| · d

dxsecx

= 1

u ·sec x tan x

= 1

sec x· sec x tan x

= tan x .

Hence

g( x) = b ln | sec t |a x = b ln

sec a

sec x

= b ln

cos x

cos a

.

2. [5 MARKS] Find an antiderivative of the integrand of the integral

b

ak + y + my2

dy,

(where a, b, k , , m are constants), and then use the Fundamental Theorem of Calculus to

evaluate the integral. You are not expected to simplify your numerical answer.

Solution:

(a) One antiderivative of ky0 + y1 + my2 is

k · y1

1 + · y2

2 + m · y3

3 .

(b) Hence

b a

k + y + my

2 dy

=

k · y1

1 + · y2

2 + m · y3

3

b

a

=

k · b1

1 + · b2

2 + m · b3

3

−

k · a1

1 + · a2

2 + m · a3

3

.

8/15/2019 Brown's Notes 2010




of the functiong( x) =

√ x

a

b cos t

t dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u( x) =√

x.

(b) Then

d

dxg( x) =

d

dx √

x

a

b cos t

t dt

= d

dx

u( x)

a

b cos t

t dt

= d

du

u( x)

a

b cos t

t dt · du( x)

dx

= b cos u

u· du( x)

dx

= b cos u

u · 1

2√

x

= b cos

√ x

√ x ·

1

2 √ x= b

cos√

x

2 x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y = x

0

1

1 + at + bt 2 dt is concave upward, where a, b are constants. (Each version of this

quiz contained specific values for the constants a, b.)

Solution:

(a) By Part 1 of the Fundamental Theorem,

y( x) = 1

1 + ax + bx2 .

(b) Diff erentiating a second time yields

y( x) = d

dx

1

1 + ax + bx2

8/15/2019 Brown's Notes 2010



=

− 1

1 + ax + bx22 ·

d

dx1 + ax + bx2

= − a + 2bx1 + ax + bx2

2 .

(c) The curve is concave upward where y > 0:

− a + 2bx1 + ax + bx2

2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive

⇔ 2bx < −a

⇔

x < −

a

2b when b > 0

x > − a2b

when b < 0

never concave upward when b = 0, a > 0

always concave upward when b = 0, a < 0

Tuesday version

1. [5 MARKS] If

c a

f ( x) dx = k and

c b

f ( x) dx = , find

b a

f ( x) dx. Show your work.

Solution:

(a)c

a

f ( x) dx =

b a

f ( x) dx +

c b

f ( x) dx .

(b) Hencec

b

f ( x) dx =

c a

f ( x) dx −b

a

f ( x) dx .

(c)

= k − .

2. [5 MARKS] Find an antiderivative of the integrand of the integral

a

0

√ x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expected

to simplify your numerical answer, but no marks will be given unless all your work is

clearly shown.


8/15/2019 Brown's Notes 2010



(a) One antiderivative of x12 is

112

+ 1· x 1

2+1 =

2

3 x

32 .

(b) a

0

√ x dx =

2

3 x

32

a

0

= 2

3

a

32 − 0

=

2

3a

32 .


of the function

x2

a

b√

1 + t c dt .

Solution:

(a) Let the upper index of the integral be denoted by u = x2. Then

(b)

d

dx

x2 a

b√

1 + t c dt = d

dx

u a

b√

1 + t c dt

= d

du

u

a

b√

1 + t c dt

· du

dx

= b√

1 + uc · du

dx

= b√

1 + uc · 2 x

= b√

1 + x2c · 2 x

4. [10 MARKS] If F ( x) =

x

1

f (t ) dt , where f (t ) =

t 2

1

a + ub

udu and a, b are constants,

find F (2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F ( x) = f ( x) =

x2

1

a + ub

udu .

8/15/2019 Brown's Notes 2010



(b) A second application of Part 1 of the Fundamental Theorem yields

F ( x) = f ( x) = d

dx

x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

d

dx

x2

1

a + ub

udu =

d

dx

v

1

a + ub

udu

= d

dv

v

1

a + ub

u

du

·

dv

dx

= a + vb

v· dv

dx

= a + vb

v· 2 x

= a + x2b

x2 · 2 x

=2

a + x2b

x.

Wednesday version

1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ

aπ

cos θ d θ , where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ .

(b)

bπ aπ

cos θ d θ = [sin θ ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiples

of π is 0, so the value of the definite integral is 0.

8/15/2019 Brown's Notes 2010



2. [5 MARKS] Express limn→∞

n

i=1

axi sin xi ∆ x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: c

b

ax sin x dx .


of the function

g( x) =

bx

ax

t 2 + c

t 2 − cdt ,

where a, b, c are positive integers.

Solution:

(a) The Fundamental Theorem gives the derivative of a definite integral with respect to

the upper limit of integration, when the lower limit is constant. The given integral

must be expressed in terms of such specialized definite integrals.

g( x) =

bx

ax

t 2 + c

t 2 − cdt =

0

ax

t 2 + c

t 2 − cdt +

bx

0

t 2 + c

t 2 − cdt

= − ax

0

t 2 + c

t 2 − cdt +

bx

0

t 2 + c

t 2 − cdt

(b) For the summand

bx 0

t 2 + c

t 2 − cdt , let u = bx. Then

d

dx

bx 0

t 2 + c

t 2 − cdt =

d

dx

u 0

t 2 + c

t 2 − cdt

= d

du

u 0

t 2 + c

t 2 − cdt · du

dx

= u2 + c

u2 − c· du

dx

= u2 + c

u2 − c· b

= (bx)2 + c

(bx)2 − c· b

8/15/2019 Brown's Notes 2010



(c) For the summand ax

0

t 2 + c

t 2

− c

dt , let u = ax. Then, analogously to the preceding

step,d

dx

ax

0

t 2 + c

t 2 − cdt =

(ax)2 + c

(ax)2 − c· a .

(d)

g( x) = (bx)2 + c

(bx)2 − c· b − (ax)2 + c

(ax)2 − c· a .

4. [10 MARKS] Find a function f ( x) such that

k + x

a

f (t )

t 2 dt =

√ x (106)

for x > 0 and for some real number a; k and are constants given in the question,

diff erent constants to diff erent students. (HINT: Diff erentiate the given equation.)

Solution: A more proper wording of the problem would have been “Find a function f ( x)

and a real number a such that...”.

(a) Assume that equation (106) holds. Then diff erentiation of both sides with respect

to x yields

0 + f ( x)

x2 =

1

2 · · 1√

x.

(b) We solve the preceding equation to obtain that f ( x) = 2 x 32 .

(c) Substitution into equation (106) yields

k +

2

x

a

t −12 dt =

√ x .

We know how to integrate powers of t :

k +

cot2 ·

√ t

xa

= √

x .

(d) The preceding reduces to k = √ a, which may be solved to obtain

a =

k

2

.

8/15/2019 Brown's Notes 2010



Thursday version

1. [5 MARKS] Evaluate the integral

πb πa

sin t dt .

Solution:

(a) An antiderivative of sin t is − cos t .

(b)πb

πa

sin t dt = [− cos t ]πbπa

(c) [1 MARK] Your answer should be simplified as much as possible.

2. [5 MARKS] Evaluate the Riemann sum for f ( z) = a− x2, (0 ≤ x ≤ 2) with 4 subintervals,

taking the sample points to be the right endpoints. It is not necessary to simplify the final

numerical answer.

Solution:

(a) The interval 0 ≤ x ≤ 2 is divided by 3 points into 4 subintervals of length ∆ x =24

= 12

.

(b) The point x∗i selected in the ith interval will always be the right end-point, i.e.,

xi = i

12

(i = 1, 2, 3, 4).

(c) The Riemann sum is4

i=1

f ( x∗i )∆ x =

1

2

4i=1

a − i2

4

.


of the function

bx cos x

cos (t c) dt , where a, b, c are real numbers.

Solution:




bx cos x

cos (t c) dt =

0 cos x

cos (t c) dt +

bx 0

cos (t c) dt

8/15/2019 Brown's Notes 2010



= −

cos x

0

cos (t

c

) dt +

bx

0

cos (t

c

) dt

(b) For the summand

bx 0

cos (t c) dt , let u = b x. Then

d

dx

bx 0

cos (t c) dt = d

dx

u 0

cos (t c) dt

=

d

du

u

0

cos (t c

) dt · du

dx

= cos (uc) · b

= cos ((bx)c) · b

(c) For the summand

cos x 0

cos (t c) dt , let v = cos x.

d

dx

cos x

0

cos (t c) dt = d

dx

v

0

cos (t c) dt

= d

dv

v 0

cos (t c) dt · dv

dx

= cos (vc) · (− sin x)

= cos (cosc x) · (− sin x)

(d)bx

cos x

cos (t c) dt = −cos (cosc x) · (− sin x) + cos ((bx)c) · b .

4. [10 MARKS] Let f ( x) =

0 if x < 0

x if 0 ≤ x ≤ a

2a − x if a < x < 2a

0 if x > 2a

and g( x) =

x

0

f (t ) dt , where a is

a positive constant. Showing all your work, find a formula for the value of g( x) when

a < x < 2a.

8/15/2019 Brown's Notes 2010



Solution:

(a) The interval where we seek a formula is the third interval into which the domain

has been broken. For x in this interval the integral can be decomposed into x

0

f (t ) dt =

a

0

f (t ) dt +

x

a

f (t ) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, since

we are not finding area under that portion of the curve; the same applies to the

portion of the definition for x > 2a.

(b)

a

0

f (t ) dt = a

0

t dt

=

t 2

2

t =a

t =0

= a2

2 .

(c) x

a

f (t ) dt =

x

a

(2a − t ) dt

= 2at − t 2

2 t = x

t =a

=

2ax − x2

2

−

2a2 − a2

2

.

(d)

g( x) = a2

2 +

2ax − x2

2

−

2a2 − a2

2

= 2ax − x2

2 − a2 .


Distribution Date: Posted on the Web on 28 February, 2007

Caveat lector! There could be misprints or errors in these draft solutions.






8/15/2019 Brown's Notes 2010



• No calculators!




typical variety.

Monday version


xb + a +

1

x2 + 1

dx, (where a and b are given

positive integers).

Solution:

(a)

xb dx =

xb+1

b + 1 + C 1,

(b)

a dx = ax + C 2

(c)

1

x2 + 1 d x = arctan x + C 3

(d) xb + a +

1

x2 + 1 dx = xb+1

b + 1 + ax + arctan x + C .

2. [5 MARKS] Use a substitution to evaluate the indefinite integral

t 2 cos

a − t 3

dt ,

(where a is a given real number).

Solution:

(a) Try the substitution u = t 3.

(b) du = 3t 2 dt ⇒ t 2 dt = 13

du.

(c)

t 2 cos a −t 3 dt = 1

3

cos(a

−u) du

= −1

3 sin(a − u) + C

= −1

3 sin(a − t 3) + C .

(Some students may wish to employ a second substitution v = a − u. Alternatively,

a better substitution for the problem would have been to take u = a − t 3.)

8/15/2019 Brown's Notes 2010



3. [10 MARKS] Find the area of the region bounded by the parabola y = x2, the tangent

line to this parabola at (a, a2

), and the x-axis, (where a is a given real number).Solution: This area can be computed by integrating either with respect to y or with

respect to x.

Integrating with respect to y: (a) Since y = 2 x, the tangent line through (a, a2) has

equation

y − a2 = 2a( x − a) ⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabola

and the line in the form

x = function of y .

The branch of the parabola to the right of the y-axis is x = √ y. The line hasequation x =

y

2a+

a

2.

(c) The area of the horizontal element of area at height y is

y + a2

2a− √

y

∆ y.

(d) The area is the value of the integral a2

0

y + a2

2a− √

y

dy .

(e) Integration yields

y2

4a+ ay

2 − 2

3 y

32

a2

0

=

14

+ 12 − 2

3

a3 = 1

12 a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Its

intercept with the x-axis is at x = a

2.

(b) The area of the vertical element of area at horizontal position x ≤ a

2 is x2 − 0

∆ x.

(c) The area of the vertical element of area at horizontal position x ≥ a2

is x2 − (2ax − a2)

( x − a)2 ∆ x.

(d) The area of the region is the sum a2

0

x2 dx +

a

a2

( x − a)2 dx .

(e) Integration yields x3

3

a2

0

+

−( x − a)3

3

a

a2

= a3

12 .

8/15/2019 Brown's Notes 2010



4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded by

the given curves about the line y = 1: y =

n

√ x, y = x, where n is a given positive integer.Solution: A favoured method of solution was not prescribed.

Using the method of “washers”: (a) Find the intersections of the curves bounding the

region. Solving the 2 equations yields the points ( x, y) = (0, 0), (1, 1).

(b) Find the inner and outer dimensions of the washer. Since the axis of revolution

is a horizontal line, the element of area being rotated is vertical. For arbitrary

x the lower point on the element is ( x, x); the upper point is ( x, n√

x). The

distances of these points from the axis are, respectively 1 − x and 1 − n√

x.

(c) The volume of the “washer” is, therefore,

π−(1 − x)2 + (1 − n√ x)2

∆ x .

(d) Correctly evaluate the integral:

π

1

0

−(1 − x)2 + (1 − n

√ x)2

dx

= π

1

0

−2 x + x2 + 2 x

1n − x

2n

dx

= π

− x2 +

x2

3 +

2n

n + 1 x

n+1n − n

n + 2 x

n+2n

1

0

= π

−1 +

1

3 +

2n

n + 1 − n

n + 2

=

(n − 1)(n + 4)π

3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) Find the intersections of the curves bound-

ing the region. Solving the 2 equations yields the points ( x, y) = (0, 0), (1, 1).

(b) Find the inner and outer dimensions of the washer. Since the axis of revolution

is a horizontal line, the element of area being rotated is also horizontal. For

arbitrary y the left endpoint on the element is ( yn, y); the right endpoint is ( y, y).

The length of the element is, therefore, y− yn; the distances of the element from

the axis of symmetry is 1 − y.

(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 − y) · ( y − yn) · ∆ y .


2π

1

0

(1 − y)( y − yn) dy

8/15/2019 Brown's Notes 2010



= 2π 1

0− yn + yn+1 + y

− y2 dy

= 2π

− 1

n + 1 yn+1 +

1

n + 2 yn+2 +

1

2 y2 − 1

3 y3

1

0

= 2π

− 1

n + 1 +

1

n + 2 +

1

2 − 1

3

− 0

= 2π

1

6 − 1

(n + 1)(n + 2)

=

(n − 1)(n + 4)π

3(n + 1)(n + 2)

Tuesday version


(a − t )(b + t 2) dt .

Solution:

(a) Expand the product in the integrand: (a − t )(b + t 2) dt =

ab − bt + at 2 − t 3

dt .

(b) Integrate term by term:

ab − bt + at

2

− t 3

dt = ab · t − b

2 · t 2

+

a

3 · t 3

− 1

4 · t 4

+ C .

2. [5 MARKS] Using a substitution, evaluate the indefinite integral

cosn x sin x dx, where

n is a fixed, positive integer.

Solution:

(a) Use new variable u, where du = −sin x dx; one solution is u = cos x.

(b)

cosn x sin x dx = − un du

= − un+1

n + 1 + C

= − 1

n + 1 cosn+1 x + C

8/15/2019 Brown's Notes 2010



3. [10 MARKS] Find the area of the region bounded by the parabola x = y2, the tangent

line to this parabola at (a2

, a), and the y-axis, where a is a fixed, positive real number.Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-

lem 3 of the Monday quiz.

4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded by

y = xn and x = yn about the line x = −1, where n is a given positive integer.

Solution:

Case I: n is even

Using the method of “washers”: (a) Find the intersections of the curves bound-


(b) Find the inner and outer dimensions of the washer. Since the axis of revo-

lution is a vertical line, the element of area being rotated is horizontal. For

arbitrary y the farther endpoint on the element is ( y1n , y); the nearer end-

point is ( yn, y). The distances of these points from the axis are, respectively

1 + n√

y and 1 + yn.


π−(1 + yn)2 + (1 + n

√ y)2

∆ y .


π 1

0

−(1 + y)2 + (1 + n√ y)2

dy

= π

1

0

2 y

1n + y

2n − 2 yn − y2n

dy

= π

2n

n + 1 y

n+1n +

n

n + 2 y

n+2n − 2

n + 1 yn+1 − 1

2n + 1 y2n+1

1

0

= 2(n − 1)(3n2 + 7n + 3)π

(n + 1)(n + 2)(2n + 1)

Using the method of cylindrical shells: (a) Find the intersections of the curves

bounding the region. Solving the 2 equations yields the points ( x, y) =(0, 0), (1, 1).

(b) Since the axis of revolution is a vertical line, the element of area being

rotated is also vertical. For arbitrary x the top endpoint on the element is

( x, x1n ); the lower endpoint is ( x, xn). The length of the element is, there-

fore, x1n − xn; the distance of the element from the axis of symmetry is

1 + x.

8/15/2019 Brown's Notes 2010




2π(1 + x) · x 1n − xn

.


2π

1

0

(1 + x)( x1n − xn) dx

= 2π

1

0

x

1n − xn + x

n+1n − xn+1

dx

= 2π

n

n + 1 − 1

n + 1 +

n

2n + 1 − 1

n + 2

= 2(n − 1)(3n2

+ 7n + 3)π(n + 1)(n + 2)(2n + 1)

Case II: n is odd


ing the region. Solving the 2 equations yields the points ( x, y) = (0, 0), (±1, ±1).

Here there is an issue of interpretation. The textbook usually permits the

word region to apply to one that may have more than one component;

some authors would not wish to apply the term in such a situation. I will

follow the textbook, and permit a region here to have two components.

(b) Find the inner and outer dimensions of the washer. Since the axis of rev-

olution is a vertical line, the element of area being rotated is horizontal.

But there are two kinds of elements, depending on whether y is positive

or negative. For arbitrary, positive y the farther endpoint on the element is

( y1n , y); the nearer endpoint is ( yn, y). The distances of these points from

the axis are, respectively 1 + n√

y and 1 + yn. For arbitrary, negative y the

nearer endpoint on the element is ( y1n , y); the farther endpoint is ( yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and

1 + yn (both of which are less than 1).


π−(1 + y

n

)

2

+ (1 +

n

√ y)

2∆ y .


π

1

−1

−(1 + y)2 + (1 + n√

y)2 dy

= π

1

0

2 y

1n + y

2n − 2 yn − y2n

dy

8/15/2019 Brown's Notes 2010



+π 0

−1−2 y

1n

− y

2n + 2 yn + y2n dy

= π

2n

n + 1 y

n+1n +

n

n + 2 y

n+2n − 2

n + 1 yn+1 − 1

2n + 1 y2n+1

1

0

+π

− 2n

n + 1 y

n+1n − n

n + 2 y

n+2n +

2

n + 1 yn+1 +

1

2n + 1 y2n+1

0

−1

= 2(n − 1)(3n2 + 7n + 3)π

(n + 1)(n + 2)(2n + 1) +

2(n − 1)(n2 + 3n + 1)π

(n + 1)(n + 2)(2n + 1)

= 4(n − 1)π

n + 1 .

Using the method of cylindrical shells: (a) Find the intersections of the curvesbounding the region. Solving the 2 equations yields the points ( x, y) =

(0, 0), (±1, ±1).


rotated is also vertical. For arbitrary, positive x the top endpoint on the el-

ement is ( x, x1n ); the lower endpoint is ( x, xn); for arbitrary, negative x the

bottom endpoint on the element is ( x, x1n ); the upper endpoint is ( x, xn).

The length of the element is, therefore, x 1

n − xn; the distance of the ele-

ment from the axis of rotation is 1 + x.


2π(1 + x) · x 1

n − xn .


2π

1

−1

(1 + x) x 1

n − xn dx

= 2π

1

−1

x 1n − xn + x

n+1n − xn+1

dx

= 2π n

n + 1 −

1

n + 1

+ n

2n + 1 −

1

n + 2

+2π

n

n + 1 − 1

n + 1 − n

2n + 1 +

1

n + 2

= 4(n − 1)π

n + 1

Wednesday version

8/15/2019 Brown's Notes 2010



1. [5 MARKS] Showing all your work, evaluate the indefinite integral sin2t

cos t dt .

Solution:

(a) Apply a “double angle” formula: sin2t

cos t dt =

2 sin t cos t

cos t dt = 2

sin t dt .

(b) Complete the integration:

2

sin t dt = −2cos t + C .

2. Using a substitution, evaluate the indefinite integral

e x

e x + adx, where a is a non-zero

real number.

Solution:

(a) [2 MARKS] Try the substitution u = e x + a, so du = e x dx.

(b) e x

e x + adx =

du

u= ln |u| + C = ln |e x + a| + C .

(If the constant a is positive, then the absolute signs are not required.)

3. [10 MARKS] Find the number b such that the line y = b divides the region bounded

by the curves y = ax2 and y = k into two regions with equal area, where a, k are given

positive constants.

Solution:

(a) Determine the range of values for integration by finding the intersections of the

bounding curves: solving the equations yields the points

∓

k

a, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin by

repeating the calculation of the preceding part: the corner points have coordinates∓

b

a, b

. The area is

√ ba

−√

ba

(b − ax2) dx = 2

bx − ax3

3

√ ba

0

= 4

3b

b

a .

8/15/2019 Brown's Notes 2010



(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is 43

k

k a

.

(d) The condition of the problem is that

4

3b

b

a=

1

2 · 4

3k

k

a

which is equivalent to 4b3 = k 3, and implies that the line should be placed where

b = 2− 23 k .

4. [10 MARKS] Use the method of cylindrical shells to find the volume generated by ro-

tating the region bounded by the given curves about the specified axis.

y = √ x − 1 , y = 0 , x = a

about y = b, where a, b are fixed positive constants, and b ≥√

a − 1 .

Solution:

(a) Solve equations to determine the limits of integration. Solving x = a with y =√ x − 1 yields the single point of intersection

a,

√ a − 1

.

(b) The horizontal element of area at height y which generates the cylindrical shell has

left endpoint (1 + y2, y) and right endpoint (a, y), so its length is a − (1 + y2).

(c) The distance of the horizontal element of area which generates the shell from the

axis of rotation is b

− y.

(d) Set up the integral for the volume by cylindrical shells:

2π

√ a−1

0

(b − y)a − (1 + y2)

dy .

(e) Evaluate the integral

2π

√ a−1

0

(b − y)

a − (1 + y2)

dy

= 2π

√ a−1

0 b(a − 1) − (a − 1) y − by2 + y3

dy

= 2π

b(a − 1) y − a − 12

y2 − b3

y3 + 14

y4√

a−1

0

= 2π

b(a − 1)

32 − a − 1

2 (a − 1) − b

3(a − 1)

32 +

1

4(a − 1)2

= 2π(a − 1)32

2

3 · b − 1

4

√ a − 1

.


8/15/2019 Brown's Notes 2010



Thursday version

1. [5 MARKS] Showing all your work, evaluate the integral by making a substitution: b

(1 + ax)3 d x ,

where a, b are non-zero constants.

Solution:

(a) A substitution which suggests itself is u = 1 + a x, implying that du = a dx, so

dx = 1a

du.

(b) b

(1 + ax)3 dx = b

a

du

u3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

2. [5 MARKS] Evaluate the indefinite integral

seca x tan x dx, where a is a constant,

positive integer.

Solution:

(a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x.

(b)

sec

a

x tan x dx =

u

a

−1

du =

ua

a + C =

seca x

a + C .

Some students may have integrated by sight.

3. [10 MARKS] Find the number b such that the line divides the region bounded by the

curves x = ay2 and x = k into two regions with equal area.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-

lem 3 of the Wednesday quiz.

4. [10 MARKS] The region bounded by the given curves is rotated about the axis x = −1.

Find the volume of the resulting solid by any method:

y = 5, y = x2 − ax + b

Solution: Because there are constraints on the constants, I will work just one variant,

with a = 3, b = 7.

Using the method of cylindrical shells: (a) To find the extremes of integration, we

solve the equations y = 5 and y = x2 − 3 x + 7, obtaining ( x, y) = (1, 5), (2, 5).

8/15/2019 Brown's Notes 2010



(b) The height of a vertical element of area which generates a cylindrical shell is,

at horizontal position x, 5 − ( x2

− 3 x + 7) = − x2

+ 3 x − 2.(c) The distance of that vertical element of area from the axis of revolution is 1+ x.

(d) The volume is given by the integral

2π

2

1

(1 + x)(− x2 + 3 x − 2) dx

(e) Evaluating the integral:

2π

2

1

(1 + x)(− x2 + 3 x − 2) dx

= 2π 2

1− x3 + 2 x2 + x

−2 dx

= 2π

−1

4 x4 +

2

3 x3 +

1

2 x2 − 2 x

2

1

= 2π

−4 +

16

3 + 2 − 4 +

1

4 − 2

3 − 1

2 + 2

= 5π

6

Using the method of “washers” (a) To find the lowest point on the parabola, we solve

x2 − 3 x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be3

2 ,

19

4

.(b) The horizontal element generating the “washer” at height y extends between

the solutions in x to the equation y = x2 − 3 x + 7; these are

x =3 ±

4 y − 19

2 .

(c) The volume of the “washer” at height y is, therefore,

π

1 +3 +

4 y − 19

2

2

−

1 +3 −

4 y − 19

2

2

∆ y

= 5π 4 y−

19 ∆ y

(d) The volume is given by the integral 5π

5

194

4 y − 19 dy.

(e) Evaluation of the integral:

5π

5

194

4 y − 19 dy =

5π · 2

3 · 1

4(4 y − 19)

32

5

194

= 5π

6 .

8/15/2019 Brown's Notes 2010




Distribution Date: Posted on the Web on 21 March, 2007







• No calculators!




typical variety.

Monday version

1. [6 MARKS] Evaluate the integral t 3eat dt , where a is a non-zero constant.

Solution: This problem requires several consecutive applications of integration by parts

to reduce the exponent of the power of t to 0:

u = t 3, dv = eat dt ⇒ du = 3t 2 dt , v = 1

aeat

⇒

t 3eat dt = 1

a· t 3eat − 3

a

t 2eat dt

U = t 2, dV = eat dt ⇒ dU = 2t dt , V = 1

aeat

⇒ t 3eat dt = 1

a ·t 3

−

3

a2

t 2 eat + 6

a2 teat dt

u = t , d v = eat dt ⇒ d u = dt , v = 1

aeat

⇒

t 3eat dt =

1

a· t 3 − 3

a2 · t 2 +

6

a3t

eat − 6

a3

eat dt

=

1

a· t 3 − 3

a2 · t 2 +

6

a3t − 6

a4

eat + C

8/15/2019 Brown's Notes 2010



the correctness of which integration may be verified by diff erentiation of the product on

the right.

2. [8 MARKS] Showing all your work, find a reduction formula for the indefinite integral cosn a x d x, where a is a non-zero constant, and n is an integer not less than 2.

Solution:

(a) Introduce a symbol for the general indefinite integral sought:

I n =

cosn a x d x

(b) Integration by parts:

u = cosn−1 ax ⇒ du = −a(n − 1) cosn−2 ax · sin a x d x

dv = cos a x d x ⇒ v = 1

a· sin ax

I n = 1

a· cosn−1 ax · sin ax + (n − 1)

cosn−2 ax · sin2 a x d x

= 1

a· cosn−1 ax · sin ax + (n − 1)

cosn−2 ax ·

1 − cos2 ax

dx

= 1

a ·cosn−1 ax

·sin ax + (n

−1) I n−2

−(n

−1) I n .

(c) Solve the last equation for I n

I n = 1

an· sin ax · cosn−1 ax +

n − 1

n I n−2 + C .

3. [4 MARKS] Showing all your work, evaluate the integral

1

x2√

x2 − a2dx , where a is

a non-zero constant.

Solution: The following solution uses a trigonometric substitution; it is also possible to

solve this problem using a hyperbolic substitution.

As is customary, I will proceed mechanically, taking square roots where necessary with-

out much attention to the sign choices; and then verify at the end by diff erentiation that

this process has produced a valid indefinite integral. This procedure can be made rigor-

ous by defining the new variable in terms of an inverse trigonometric function of x. This

is a useful exercise, but becomes extremely complicated in this case, because we would

have to work with either the inverse cosine or the inverse secant, and the textbook we

8/15/2019 Brown's Notes 2010



are using chooses diff erent domains for these two functions. So I will avoid the niceties

and proceed as described.I propose to use a substitution which provides that x = a sec θ . Then

dx = a sec θ · tan θ · d θ ,

and 1

x2√

x2 − a2dx =

1

a2

cos θ d θ

= 1

a2 sin θ + C

= 1

a2 tan θ · cos θ + C

= sec2 θ − 1

a2 sec θ + C

=

√ x2 − a2

a2 x+ C ,

the correctness of which integration may be verified by diff erentiation of the quotient on

the right.


k

x( x − a)( x − b) + c

( x−

a)( x−

b) dx,

where a, b are distinct constants, c is a non-zero constant, and the limits of integration

k , are also prescribed. (The integrand was not presented to students in factored form.)

Solution:

(a) Since the degree of the numerator is not less than the degree of the denominator,

begin by dividing the denominator into the numerator, obtaining a quotient and a

remainder: x( x − a)( x − b) + c

( x − a)( x − b) dx =

x +

c

( x − a)( x − b) d x .

(b) Expand the fraction into partial fractions. Assume that

c

( x − a)( x − b) =

A

x − a+

B

x − b,

take to a common denominator, and equate the resulting polynomials:

c = A · ( x − b) + B · ( x − a) .

8/15/2019 Brown's Notes 2010



(c) Now either equate coefficients of like powers of x, or, equivalently, give x succes-

sive values x = a and x = b:

c = A(a − b)

c = B(b − a) ⇒ A =

c

a − b= − B .

(d) The integration reduces to x( x − a)( x − b) + c

( x − a)( x − b) dx =

x +

c

a − b · 1

x − a − c

a − b · 1

x − b

dx .

(e) Complete the integration:

x( x − a)( x − b) + c

( x − a)( x − b) dx =

x2

2 +

c

a − b(ln | x − a| − ln | x − b|) + C

= x2

2 +

c

a − bln

x − a

x − b

+ C ,

the correctness of which integration may be verified by diff erentiation of the func-

tion on the right.

(f) Provided it is convergent the given definite integral can now be evaluated:

k

x( x − a)( x − b) + c

( x − a)( x − b)

dx = x2

2

+ c

a − b

ln x − a

x − b

k

= 2 − k 2

2 +

c

a − b

ln

− a

− b

− ln

k − a

k − b

= 2 − k 2

2 +

c

a − b

ln

( − a)(k − b)

( − b)(k − a)

(g) All of the preceding is based on the integral being convergent. In some of the

versions the integral was divergent. This was because at least one of the roots of

the polynomial which is the denominator of the integrand was contained in the

interval of integration. In such a case the integral can be seen to diverge.


1√ 3

−1

earctan y

1 + y2 dy . I have stated the problem with just one

pair of possible limits for the integral; the variations of the problem included several

possible limits in each case, for each of which students should have been familiar with

the arctangent.

8/15/2019 Brown's Notes 2010



Solution: For simplicity, I work a specific instance of this problem. Use the substitution

u = arctan y, so du = dy1 + y2 . Then an antiderivative can be obtained as follows

earctan y

1 + y2 dy =

eu du = eu + C = earctan y + C

so the definite integral is equal to

earctan y

1√ 3

−1= e

π6 − e− π

4 .

Alternatively, the substitution may be executed in the definite integral, replacing the

lower limit of

−1 by arctan(

−1) =

−π4

, and the upper limit of 1√ 3

by arctan 1√ 3

= π6

.

Tuesday version


x2 cos a x d x, where a is a

non-zero constant.

Solution: Two applications of integration by parts will be used to reduce the exponent

of the power of x to 0.

(a)

u = x2

⇒ du = 2 x dx

dv = cos a x d x ⇒ v = 1

asin ax

x2 cos a x d x = x2

a · sin ax − 2

a

x · sin a x d x

(b)

U = x ⇒ dU = d x

dV = sin a x d x ⇒ V = −1

a· cos ax

x2 cos a x d x = x

2

a· sin ax − 2

a

− xa

· cos ax + 1a

cos a x d x

=

x2

a· sin ax +

2

a2 x · cos ax − 2

a2

cos a x d x

= x2

a· sin ax +

2

a2 x · cos ax − 2

a3 sin ax + C

8/15/2019 Brown's Notes 2010



The integration can be checked by diff erentiation of the alleged antiderivative.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral

xneax dx,

where a is a non-zero constant.

Solution:

(a) Introduce a symbol for the general indefinite integral sought:

I n =

xneax dx .


u = xn ⇒ du = n xn−1 dx

dv = eax dx ⇒ v = 1

a· eax

I n = 1

a· eax − n

a

xn−1eax dx

= 1

a· eax − n

a I n−1.

which is the desired reduction formula.

3. [4 MARKS] Showing your work, evaluate the integral

sin

3

ax · cos

2

a x d x, where a isa non-zero constant.

Solution: This integral is easily evaluated by a substitution giving du = constant ×sin a x d x. So a convenient substitution is u = cos ax, which yields du = −a sin a x d x.

sin3 ax · cos2 a x d x =

sin2 ax · u2 · −du

a

=

1 − cos2 ax

· u2 · −du

a

= 1 −u2 ·

u2

·

−du

a=

1

a

u4 − u2

du

= 1

a

u5

5 − u3

3

+ C

= 1

a

cos5 ax

5 − cos3 ax

3

+ C ,

8/15/2019 Brown's Notes 2010



which integration may be checked by diff erentiation. (Of course, there are other, equiv-

alent ways of expressing this indefinite integral.)


x3

√ x2 + a2

dx, where a is a

given non-zero constant.

Solution: To simplify the surd in the denominator one may use either a trigonometric

or a hyperbolic substitution. For students in this course a trigonometric substitution is

usually a better choice. To arrange that x = a tan u, we use a substitution

u = arctan x

a, (107)

and dx = a sec

2

u du. We may assume that a > 0. The interval of validity for substitution(107) is −π2

< x < π2

, in which the secant is positive. x3

√ x2 + a2

dx =

a3 tan3 u

|a sec u| · a sec2 u du

= a3

tan2 u · sec u tan u du

= a3

sec2 u − 1

· d

dusec u du

= a3

sec3 u

3 − sec u + C

eff ectively by substitution U = sec u

= 1

3

(a tan u)2 − 2a2

(a tan u)2 + a2 + C

= 1

3

x2 − 2a2

√ x2 + a2 + C

which integration may be verified by diff erentiation.

5. [9 MARKS] (To simplify the exposition of the solution, I work a specific example here.)

Showing all your work, evaluate the integral

x + 21

( x + 9)( x − 5) d x.

Solution: Since the degree of the numerator is less than that of the denominator, we candispense with the first step of dividing denominator into numerator.

(a) We need to expand the integrand into a sum of partial fractions; fortunately the

factorization of the denominator has been given. Assuming there are constants

A, B such that x + 21

( x + 9)( x − 5) =

A

x + 9 +

B

x − 5

8/15/2019 Brown's Notes 2010



and transforming all fractions to have a common denominator, we find that

x + 21 = A( x − 5) + B( x + 9) .

(b) The values of A, B may be obtained either by comparing coefficients of like powers

of x, or by assigning to x successively the “convenient” values 5, -9: we obtain that

26 = 14 B ⇒ B = 13

7

12 = −14 A ⇒ A = −6

7.

(c) We may now complete the integration:

x + 21

( x + 9)( x − 5) d x =

1

7 − 6

x + 9 +

13

x − 5

dx

= −1

7(−6ln( x + 9) + 13 ln( x − 5)) + C

which can also be expressed in other, equivalent ways. This integration may be

verified by diff erentiation.

Wednesday version

1. [4 MARKS] Showing your work, evaluate

a

0

( x2 + 1)e− x dx , where a is a given constant.

Solution: I will integrate by parts twice, in order to reduce the degree of the polynomial

factor of the integrand.

(a) First integration by parts:

u = x2 + 1 ⇒ du = 2 x dx

dv = e− x dx ⇒ v = −e− x

a 0

( x2 + 1)e− x dx =( x2 + 1)

−e− xa

0+

a

0

2 x · e− x dx .

(b) Second integration by parts:

U = 2 x ⇒ dU = 2 dx

dV = e− x dx ⇒ V = −e− x

a 0

( x2 + 1)e− x dx =( x2 + 1 + 2 x)

−e− xa

0+

a

0

2 · e− x dx .

8/15/2019 Brown's Notes 2010



(c) Completion of the integration:

a 0

( x2 + 1)e− x dx =( x2 + 2 + 2 x)

−e− xa

0

= −(a2 + 2a + 3)e−a + 3.


π2

0

sinn a x d x,

where a is a given integer.

Solution: Assume n is an integer greater than 1.

(a) Introduce a symbol for the definite integral sought:

I n =

π2

0

sinn a x d x .


dv = sin a x d x ⇒ v = −cos ax

adx

u = sinn−1 ax ⇒ du = a(n − 1) sinn−2 ax · cos a x d x

I n = −1

a ·sinn−1 ax

·cos ax

π2

0

+ (n−

1) π

2

0

sinn−2 ax·

cos2 a x d x

(c) Decomposition of integral: π2

0

sinn−2 ax · cos2 a x d x =

π2

0

sinn−2 ax ·1 − sin2 ax

dx

=

π2

0

sinn−2 a x d x − π

2

0

sinn a x d x

= I n−2 − I n

(d) Solution of equation to obtain reduction formula

I n =

−1

a · sinn−1 ax · cos ax

π2

0

+ (n − 1) ( I n−2 − I n)

⇒ nI n =

−1

a· sinn−1 ax · cos ax

π2

0

+ (n − 1) I n−2

⇒ I n = 1

n(0 − 0) +

n − 1

n I n−2

8/15/2019 Brown's Notes 2010



Because of the choice of limits and the fact that a is an integer, the “net change” is

0. Thus we obtain a very simple relationship, which can be solved. Students werenot asked to complete this part of the solution. For example, it is possible to prove

by induction that, if n = 2m, an even, positive integer, then

I 2m = 2m − 1

2m · 2m − 3

2m − 2 · . . . · 3

2 I 0

= (2m)!

4mm!m! · π

2 .

(This is Exercise 44, page 481 in the textbook.)

3. [4 MARKS] Showing all your work, evaluate the integral x2a2 − x2

32

dx, where a is a

non-zero constant.

Solution: Without limiting generality we take a > 0. A trigonometric substitution can

simplify this integral. One such substitution would have x = a sin u; more precisely,

u = arcsin xa

, defined for − π2 ≤ x ≤ π

2, in which interval the cosine and secant are

positive. Then d x = a du√ 1−u2

.

x2

a2 − x2

32

=

a2 sin2 u

a3 cos3 u· a cos u du

=

tan2

u

du =

sec2

u − 1

du

= tan u − u + C

= sin u

cos u− u + C

= sin u

1 − sin2 u− u + C

=

xa

1 − x2

a2

− arcsin x

a+ C

= x

√ a2 − x2 −arcsin

x

a+ C

which may be verified by diff erentiation.


k

x( x − a)( x − b) + c

( x − a)( x − b) dx,

where a, b, k , are distinct constants such that the integrand is defined throughout the

8/15/2019 Brown's Notes 2010



given interval, and c is a non-zero constant. (The integrand was not presented to students

in factored form.)Solution:

(a) Since the degree of the numerator is not less than the degree of the denominator,

begin by dividing the denominator into the numerator, obtaining a quotient and a

remainder: x( x − a)( x − b) + c

( x − a)( x − b) dx =

x +

c

( x − a)( x − b) d x .

(b) Expand the fraction into partial fractions. Assume that

c

( x − a)( x − b) =

A

x − a+

B

x − b,

take to a common denominator, and equate the resulting polynomials:

c = A · ( x − b) + B · ( x − a) .

(c) Now either equate coefficients of like powers of x, or, equivalently, give x succes-

sive values x = a and x = b:

c = A(a − b)

c = B(b − a) ⇒ A =

c

a − b

=

− B .

(d) The integration reduces to x( x − a)( x − b) + c

( x − a)( x − b) dx =

x +

c

a − b · 1

x − a − c

a − b · c

x − b

dx .

(e) Complete the integration: x( x − a)( x − b) + c

( x − a)( x − b) dx =

x2

2 +

c

a − b(ln( x − a) − ln( x − b)) + C

= x2

2 + c

a − bln x − a

x − b+ C ,

the correctness of which integration may be verified by diff erentiation of the func-

tion on the right.

8/15/2019 Brown's Notes 2010



(f) Evaluate the definite integral

k

x( x − a)( x − b) + c

( x − a)( x − b) dx =

x2

2 +

c

a − b(ln | x − a| − ln | x − b|)

k

=

x2

2 +

c

a − bln

x − a

x − b

k

= 2 − k 2

2 +

c

a − bln

( − a)(k − b)

( − b)(k − a)

5. [4 MARKS] Showing all work, evaluate the integral

sin3 a x d x, where a is a positive

integer.

Solution: sin3 a x d x =

1 − cos2 ax

· sin a x d x

under substitution u = cos ax, where du = −a sin a x d x

=

(1 − u2)(−1

a) du

= −1

a

u − u3

3

+ C

= −

cos ax

a+

cos3 ax

3a+ C .

Thursday version


b

a

e√

t dt .

Solution: Begin with a substitution√

t = u, so t = u2, dt = 2u du. When t = a, u =√

a,

etc.: e

√ t dt =

2ueu du .

Now apply integration by parts:

U = u ⇒ dU = du

dV = 2eu du ⇒ V = 2eu 2ueu du = u · 2eu −

2eu du

= u · 2eu − 2eu + C = 2(u − 1)eu + C

= 2(√

t − 1)e√

t + C

8/15/2019 Brown's Notes 2010



The definite integral given is then equal to 2(√

t − 1)e√

t b

a.


(ln(ax + 1))n dx

where a is a given, positive constant.

Solution: This problem is a slight generalization of Exercise 45, p. 481 in the textbook,

an odd-numbered problem for which there is a solution in the Student Solutions Manual,

and also hints on one of the CD-Roms supplied with the textbook.

(a) Introduce a symbol for the definite integral sought:

I n = (ln(ax + 1))n dx .

(b) Change the variable (a step which is helpful, but not necessary)

u = a x + 1 ⇒ du = a dx

I n = 1

a

(ln u)n du

(c) Integration by parts:

U = (ln u)n ⇒ dU = n(ln u)n−1 · 1

udV = 1 du

⇒ V = u

I n = u(ln u)n − n

(ln u)n−1 du

= (ax + 1)(ln(ax + 1))n − na

(ln(ax + 1))n−1 dx

= (ax + 1)(ln(ax + 1))n − na · I n−1


cos4 at dt , where a is a

given non-zero constant.

Solution: We have to apply the following double angle identity twice:

cos2θ = 2 cos2 θ − 1 .

cos4 atdt =

1 + cos2at

2

2

dt

= 1

4

cos2 2atdt + 2

cos2atdt +

1 dt

8/15/2019 Brown's Notes 2010



= 1

4

1 + cos4at

2

dt + 2 cos2at dt + 1 dt =

1

8 · 1

4asin4at +

1

2 · 1

2asin2at +

3

8t + C

= 1

32asin4at +

1

4asin 2at +

3

8t + C

which may be verified by diff erentiation. (Of course, the integral may be expressed in

other ways under transformation by trigonometric identities.)

4. [4 MARKS] Showing your work, evaluate the integral

dx

x√

x2 + a, where a is a given

positive integer, not a perfect square.

Solution: The surd in the denominator may be simplified by either a trigonometric or ahyperbolic substitution. For students in this course the trigonometric substitutions are

usually easier.

x =√

a · tan θ ⇒ dx =√

a sec2 θ d θ dx

x√

x2 + a=

sec2 θ d θ

tan θ · √ a · | sec θ | .

The actual substitution is given by θ = arctan x√ a

, valid for −π2

< x < π2

. In that interval

the secant function is positive, so the absolute signs may be dropped. The integral is

equal to

1√ a

csc θ d θ = 1√

aln | csc θ − cot θ | + C

= 1√

aln

sec θ − 1

tan θ

+ C

= 1√

aln

√

tan2 θ − 1

tan θ

+ C

= 1√

aln

√ x2 + a − √

a

x

+ C

which can be verified by diff erentiation.


k

x( x − a)( x − b) + c

( x − a)( x − b) dx,

where a, b, k , are distinct constants such that a, b are not contained in the interval whose

end-points are k , , and c is a non-zero constant.

Solution: I will first determine an indefinite integral.

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010




Distribution Date: Posted on the Web on 06 April, 2007; corrected on 09 April, 2007.




the problems in the varieties was also randomly assigned. Each version of the quiz was graded

out of a maximum of 30 marks, but 2 of the versions had 5 problems and 2 had 4 problems.

All of the quizzes had a heading that included the instructions


• No calculators!




typical variety.

Monday version

1. [6 MARKS] Showing all of your work, find the length of the following curve for the

interval 0 < a ≤ u ≤ b : y = ln

eu + 1

eu − 1

.

Solution:

(a)

dy

dx=

eu − 1

eu + 1

eu(eu − 1) − (eu + 1)eu

(eu − 1)2

= − 2eu

e2u

−1

1 +

dy

dx

2

=

e2u + 1

e2u − 1

2

= coth2 u

(b)

arc length =

b

a

1 +

dy

dx

2

du =

b

a

| coth u| du

8/15/2019 Brown's Notes 2010



(c) Successful completion of the integration: b

a

| coth u| du =

b

a

coth u du

since a < b

[ln sinh u]ba

= ln sinh b

sinh a= ln

eb − e−b

ea − e−a

0.0

2.5

−1.0

0.5

1.51.00.5

3.0

2.0

−0.5

1.5

1.0

0.0

−1.5

Figure 26: The limacon r = 1 + 2sin θ

2. [10 MARKS] (see Figure 26 on page 5071) The graph of the following curve is given.

8/15/2019 Brown's Notes 2010



Showing detailed work, find the area that is enclosed between the inner and the outer

loops: r = a(1 + 2sin θ ), where a is a positive constant.Solution:

(a) Determination of the limits of integration: we need first to locate where the curve

crosses itself. Since its formula is in terms of sin θ , the curve is periodic with period

(at most) 2π. As θ ranges over the values from 0 to 2π, the values r (θ ) range over

uniquely determined values. How then can the curve cross itself? This can happen

either

i. at points (r (θ 1), θ 1) and (r (θ 1 + π), θ 1 + π) where r (θ 1 + π) = −r (θ 1); or

ii. at the pole, where r = 0 for two distinct values of θ .

The first possibility would, for the present curve, require that

1 + 2 sin(θ 1 + π) = − (1 + 2sin θ 1)

which is equivalent to

2 + 2sin θ 1 + 2 sin(θ 1 + π) = 0

which is equivalent to 2 = 0, a contradiction. Thus the present curve can cross

itself only at the pole. That occurs where 1 + 2 sin θ = 0, i.e., where sin θ = − 12

.

The values of θ satisfying this equation are 2nπ− π6

and (2n +1)π+ π6

, where n is any

integer. The outer loop of this lima¸con is traced out, for example, for

−π6

≤ θ

≤ 7π

6 .

The inner loop is traced for 7π6 ≤ θ ≤ 11π

6 .

(b) The area of the region bounded by the larger, outer loop is

a2

7π6

− π6

1

2 · (1 + 2sin θ )2 d θ

= a2

7π6

− π6

1

2 ·1 + 4sin θ + 4 sin2 θ

d θ

= a2

7π

6

−π6

1

2 · (1 + 4sin θ + 2 − 2cos2θ ) d θ

= a2

2

7π6

− π6

(3 + 4sin θ − 2cos2θ ) d θ

= a2

2 [3θ − 4cos θ − sin2θ ]

7π6

− π6

= a2

2

7π

2 − 4cos

7π

6 − sin

7π

3

− a2

2

−π

2 − 4cos

−π

6 − sin

−π

3

8/15/2019 Brown's Notes 2010



= a2 2π + 3

√ 3

2 .

(c) The area of the inner loop is

a2

11π6

7π6

1

2 · (1 + 2sin θ )2 d θ

= a2

2 [3θ − 4cos θ − sin2θ ]

11π6

7π6

= a2

2

11π

2 − 4cos

11π

6 − sin

11π

3

− a2

2

7π

2 − 4cos

7π

6 − sin

7π

3

= a2π −

3√

3

2 .

(d) The area of the region between the loops is the excess of the area inside the outer

loop over the area inside the inner loop, i.e.,

a2

2π +

3√

3

2

−π − 3

√ 3

2

= a2(π + 3

√ 3) .

Note that a cleaner way of solving this problem would have been to first integrate

from 0 to 2π, which would give the area between the loops plus twice the area

inside the smaller loop; and then to subtract twice the area inside the smaller loop.

This method is better because the first integral is very easy to evaluate, since theperiodic terms contribute nothing.

This curve is discussed in Exercise 10.4.21, on page 683 of the textbook, and is solved

in the Student Solutions Manual and also on one of the CD-Roms which accompany the

textbook.

3. [4 MARKS] Showing full details of your work, find the exact length of the curve x =

et + e−t , y = a − 2t , 0 ≤ t ≤ b, where a, b are constants.

Solution:

dxdt

2

+

dydt

2

=et − e−t

2 + 4

=et + e−t 2

arc length =

b

0

et + e−t dt

=et − e−t b

0 = eb − e−b = 2 sinh b .

8/15/2019 Brown's Notes 2010



4. [4 MARKS] Find the value of the limit for the sequence. If it diverges, prove that fact:arctan

3n

3n + 1

.

Solution: As n → ∞, 3n3n+1

= 1 − 13n+1

→ 1. Since the arctangent function is continuous

at the point 1, the limit of the sequence is the arctangent of 1, i.e., π4

.

5. [6 MARKS] The given curve is rotated about the y-axis. Find the area of the resulting

surface: x = 1

2√

2

y2 − ln y

, (1 ≤ y ≤ a), where a is a real constant greater than 1.

Solution:

(a)

dxdy

= 12

√ 2

2 y − 1

y

1 +

dx

dy

2

= 1 + 1

8

4 y2 +

1

y2 − 4

= 1

8

4 y2 +

1

y2 + 4

=

1

2√

2

2 y +

1

y

2

(b)

surface area =

a

1

2π x

1 +

dx

dy

2

dy

= π

4

a

1

y2 − ln y

2 y +

1

y

dy

= π

4

a

1

2 y3 + y − 2 y ln y − ln y

y

dy

= π

4 y4

2 +

y2

2 − 1

2(ln y)2

a

1

− π

2 a

1

y ln y dy

(c) One way to integrate

y ln y dy is by parts, with u = ln y, v = y: u = 1 y

, v = y2

2 ,

y ln y dy =

y2

2 · ln y −

y

2 dy

= y2

4(2 ln y − 1) + C .

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



(a)

dy

dx=

1

2

x − 1

x

1 +

dy

dx

2

= 1 + 1

4

x2 +

1

x2 − 2

= 1

4

x2 +

1

x2 + 2

=

1

2

x +

1

x

2

(b)

surface area =

b

a

2π y

1 +

dy

dx

2

dx

= π

b

a

x2

4 − ln x

2

x +

1

x

dx

= π

b

a

x3

4 +

x

4 − x ln x

2 − ln x

2 x

dx

= π 1

16

x4 + x2

8 − 1

4

(ln x)2b

a −π

b

a

x ln x dx

(c) One way to integrate

x ln x dx is by parts, with u = ln x, v = x: u = 1 x

, v = x2

2 ,

x ln x dx =

x2

2 · ln x −

x

2 d x

= x2

4(2 ln x − 1) + C .

Another way is to use the substitution w = x2, so dw = 2 x dx, ln w = 2 ln x:

x ln x dx =

ln w

2 · dw

2 = 1

4

ln w dw = 1

4w (ln w − 1) + C etc.

Thus the surface area is

π

x4

16 +

x3

8 − 1

4(ln x)2 − x2

4(2 ln x − 1)

b

a

8/15/2019 Brown's Notes 2010



3. [4 MARKS] Find the exact length of the curve given by

x = et cos t y = et sin t (0 ≤ t ≤ a)

where a is a positive constant.

Solution:

(a)

dx

dt = et (cos t − sin t )

dy

dt = et (sin t + cos t )

dx

dt

2

+

dy

dt

2

= e2t (2 cos2 t + 2sin2 t ) = 2e2t

(b) The arc length is a

0

√ 2et dt =

√ 2(ea − 1) .

4. [10 MARKS] Working only with polar coordinates, find the area of the region that lies

inside the first curve and outside the second curve: r = b sin θ , r = a, where a and b are

positive constants.

Solution:

(a) Both of these curves are circles; we need to determine the coordinates of the points

of intersection. Solving the equations yields

r = a sin θ = a

b .

One point of intersection will be (r , θ ) =a, arcsin a

b

. Another point of intersection

will be (r , θ ) =

a, π − arcsin ab

— remember that the values of the arcsine function

are in the interval −π2

, π2 . It appears from a drawing that we have all the points of

intersection. If we solve the equation r = −a with r = b sin θ we obtain preciselythe same points, albeit with diff erent coordinates. If we attempt to replace the

equation r = b sin θ with that obtained under the identification (r , θ ) → (−r , θ + π)

there is no change. This algebraic investigation discloses all possible points of

intersection except the pole, which must be checked separately. But the pole cannot

lie on r = a, since the pole has first coordinate 0 always. Thus we have, indeed,

found all the points of intersection.

8/15/2019 Brown's Notes 2010



(b) The area bounded by the arcs can be considered to consist of the disk r = b sin θ

diminished by a sector of the circle r = a for arcsin

a

b ≤ θ ≤ π − arcsin

a

b , togetherwith two small segments of the disk r = b sin θ bounded by the rays θ = arcsin a

b

and θ = π − arcsin ab

. What I have provided is one prescription for computing the

area. An easier way would be to take the integral

1

2

π−arcsin ab

arcsin ab

(b sin θ )2 − a2

d θ

=

π2

arcsin ab

(b sin θ )2 − a2

d θ

by symmetry around the line θ = π2

= π

2

arcsin ba

b2

1 − cos2θ 2

− a2

d θ

=

π2

arcsin ba

b2

2 − a2

− b2

2 cos2θ

d θ

=

b2

2 − a2

θ − b2

4 sin2θ

π2

arcsin ba

= b2 − 2a2

2

π

2 − arcsin

a

b

+

a

2

√ b2 − a2

5. [6 MARKS] Determine whether the series is convergent or divergent. If it is convergent,

find its sum. Otherwise prove that it is divergent:∞

n=1

b

n(n + a)

, where a, b are positive

integers.

Solution:

(a) Expand the general term into partial fractions: there exist constants A, B such that

b

n(n + a) =

A

n+

B

n + a.

To determine the coefficients A, B we can proceed in several ways. If we take the

fractions to a common denominator n(n + a), we obtain the polynomial identity

b = A · ( N + A) + b · a .

In this identity, if we set the variable n equal to −a, we obtain that b = B(−a), so

B = −ba

; and, setting n = 0, we obtain A = ba

, hence

b

n(n + a) =

b

a

1

n− 1

n + a

.

8/15/2019 Brown's Notes 2010



(b) For sufficiently large N the N th partial sum is equal to

N n=1

b

n(n + a)

= b

a

1

1 +

1

2 + . . . +

1

a − 1 − 1

N + 1 − 1

N + 2 − . . . − 1

N + a

→ b

a

1

1 +

1

2 + . . . +

1

a − 1 − 0 − 0 − . . . − 0

as N → ∞. Hence the series converges to the sum

b

a1

1 +

1

2 + . . . +

1

a − 1

.

Convergence could be proved in other ways, thereby earning the student part marks. For

example, using the Comparison or Limit Comparison Tests, or the Integral Test.

Wednesday version

1. [10 MARKS] Showing detailed work, find all points diff erent from the origin on the

following curve where the tangent is horizontal; a is a positive constant:

x = a(cos θ

−cos2 θ ), y = a(sin θ

−sin θ cos θ ) .

Solution:

(a)

dx

d θ = a (− sin θ − 2cos θ (− sin θ ))

= a(sin θ )(2 cos θ − 1)

dy

d θ = a

− cos θ + 2cos2 θ − 2sin2 θ

= a − cos θ + 2cos2 θ

−1

= a(2 cos θ + 1)(cos θ − 1)

Actually, you weren’t expected to find dxdt

.

(b) There will be a horizontal tangent at the point with parameter value t if dy

dx = 0,

i.e., if

dy

d θ

dxd θ

= 0, implying that dy

d θ must be 0, provided dx

d θ 0 at the same value of t .

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



⇒ dx

d θ

2

+ dy

d θ

2

= 9a2 sin2 θ

·cos2 θ

· cos2 θ + sin2 θ = 9a2 sin2 θ cos2 θ .

The curve is in the first quadrant when both coordinates are positive; as each of these is

a cube of a sine or cosine, this means that the portion of the curve in the first quadrant is

that given by 0 ≤ θ ≤ π2

. The length of the arc is

π2

0

9a2 sin2 θ cos2 θ d θ =

π2

0

3a sin θ · cos θ d θ = 3a

sin2 θ

2

π2

0

= 3a

2 .

3. [10 MARKS] Find the area of the region that lies inside both of the following curves

r = a + 2sin θ , r = a − 1, where a is a suitable positive constant.

Solution:

(a) Determination of the limits of integration: we need first to locate where the curves

cross. We begin by solving the two given equations, and find that

sin θ = −1

2 ⇒ θ =

7π

6 or

11π

6

or any angle obtained from these by adding an integer multiple of 2π. This yields

the points a − 1,

7π

6

,

a − 1,

11π

6

.

Students weren’t expected to pursue this question further. Strictly speaking, they

should then have solved r = a + 2sin θ , r = −(a − 1), which would have yielded no

points; then solved r = −a + 2sin θ , r = a − 1, which would again yield no points;

then r = −a + 2 sin θ , r = −(a − 1), which would have yielded the 2 points already

found.

(b) This problem could then be approached in several ways. To find the area “directly”

would require finding the sum of the integrals

1

2

7π6

− pi6

(a − 1)2 d θ

and1

2

11π6

7 pi6

(a + 2sin θ )2 d θ .

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



similarly, the second series sums to 1a−1

. Hence the given series sums to the sum of these

limits, i.e. 1

b − 1 +

1

a − 1 =

a + b − 2

(a − 1)(b − 1) .

Thursday version

1. [10 MARKS] The curve x = a(1 − 2cos2 t ), y = (tan t )(1 − 2cos2 t ), where a is a

given positive integer, crosses itself at some point ( x0, y0). Showing all your work, find

the point of crossing, and the equations of both tangents at the point. (In determining

the point of crossing you are expected to investigate the parametric functions: it is not

sufficient to simply plot a finite number of points on the curve.)

Solution:

(a) Since the functions are all periodic with period π, it suffices to take an interval of

this length for t , and that will reveal all aspects of the behavior of this curve. (More

precisely, the tangent function has period π, and, while the cosine function has

period 2π, its square has period π.) So, without limiting generality, let’s consider

−π2 ≤ t ≤ π

2: we have to exclude both end points of this interval, since the tangent

function is not defined at either of them.

Suppose that the curve crosses itself at the points with parameter values t = t 1 and

t = t 2; without limiting generality, we can assume that these parameter values have

been so labelled that t 1 < t 2. Since the x-coordinates will need to be the same,

a(1 − 2cos2 t 1) = a(1 − 2cos2 t 2) (108)

so

cos t 1 = ±cos t 2 . (109)

Since the y-coordinates must also coincide, we also have

(tan t 1)(1 − 2cos2 t 1) = (tan t 2)(1 − 2cos2 t 2) (110)

which implies that either

cos2

t 1 = cos2

t 2 =

1

2 , (111)

or

tan t 1 = tan t 2 . (112)

In the interval we have chosen for t , the cosines are always positive; the only solu-

tion to (111) is t 1 = −π4

, t 2 = + π4

. In that same interval for t there will be no solu-

tions to (112), since the tangent function is increasing there. Thus the only possible

8/15/2019 Brown's Notes 2010



crossing points are t = ± π4

, and the point of crossing is the origin, ( x, y) = (0, 0).

While it isn’t required in the solution, note that as t → ±π

2 , x → 1: this curve isasymptotic to the vertical line x = 1.

(b) Tangent at the point with parameter value t = −π4

:

dx

dt = 4a(cos t )(sin t ) = 2a sin 2t = −2a

y = (tan t )(1 − 2cos2 t ) = tan t − sin2t

dy

dt = sec2 t − 2cos2t = 2 − 0

dy

dx=

dy

dt

dx

dt

= − 2

2a= −1

a,

and the tangent has equation y = − xa

.

(c) Tangent at the point with parameter value t = π4

:

dx

dt = 4a(cos t )(sin t ) = 2a sin2t = 2a

dy

dt = sec2 t − 2cos2t = 2 − 0

dy

dx=

dy

dt

dxdt

= 2

2a=

1

a,

and the tangent has equation y =

x

a .

2. [5 MARKS] Showing detailed work, find the surface area generated by rotating the

following curve about the y-axis.

x = at 2 , y = bt 3 , 0 ≤ t ≤ 5.

Solution:

dx

dt = 2at

dy

dt = 3bt 2

dx

dt

2

+

dy

dt

2

= 4a2t 2 + 9b2t 4

area about y-axis = 2π

5

0

at 2√

4a2t 2 + 9b2t 4 dt

= 2πa

5

0

t 3√

4a2 + 9b2t 2 dt

8/15/2019 Brown's Notes 2010



Under the substitution u = 4a2 + 9b2t 2,

du = 18b2t dt

2πa

5

0

t 3√

4a2 + 9b2t 2 dt = aπ

81b4

4a2+225b2

4a2

(u32 − 4a2u

12 ) du

= πa

81b4

2

5u

52 − 8a2

3 u

32

4a2+225b2

4a2

= 2πa

(81)(15)b4

(4a2 + 225b2)

32 (2a2 + 675b2) − 16a5

= ...

3. [10 MARKS] There is a region in the first quadrant that is bounded by arcs of both of

the following curves. Showing your work in detail, find the area of the region:

r 2 = a sin 2θ r 2 = a cos2θ .

Solution:

(a) The given curves are expressed only in terms of sine and cosine of 2θ . The given

functions are periodic with period π. When, for either of these curves, we permit

θ to range over an interval of length π, we will trace out the entire curves. The

intersections of the curves in the first quadrant will be where sin 2θ = cos2θ is

positive: thus the only point we have found by this algebraic solution of the two

equations is at θ = π8 .

However there are other ways in which curves can intersect, since points have

infinitely many diff erent sets of polar coordinates. If we transform either of the

given equations under the substitution (r , θ ) → (−r , θ + π), we find that there is

no change in the equation. Thus we haven’t missed any points because of the

convention that permits the first coordinate to be negative.

But there is another situation that leads to multiple sets of coordinates; that is at the

pole, where the second — angular — coordinate is totally arbitrary; the pole can

lie on a curve simply because of the fact that its distance coordinate r = 0, with no

reference to θ . To determine whether the pole lies on a curve we must investigate

whether the equation is satisfied by r = 0 with any value of θ . We find the curver 2 = a sin2θ does contain the pole: when r = 0 the equation is satisfied by any θ

such that sin 2θ = 0; so two solutions are θ = 0 and θ = π2

. Similarly, the pole lies

on the curve r 2 = a cos2θ with θ = π4

. I have given only the coordinates in the first

quadrant. To summarize: there are 2 intersection points in the first quadrant:

(r , θ ) =

a

12 2− 1

4 , π

4

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010




type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that



• No calculators!



In the following I will either provide a generic solution for all varieties, or a solution to onetypical variety.

Monday version

1. [5 MARKS] If

c a

f ( x) dx = k and

c b

f ( x) dx = , find

b a

f ( x) dx. Show your work.

Solution:

(a)c

a

f ( x) dx =

b a

f ( x) dx +

c b

f ( x) dx .

(b) Hencec

b

f ( x) dx =

c a

f ( x) dx −b

a

f ( x) dx .

(c)

= k − .

2. [5 MARKS] Find an antiderivative of the integrand of the integral a

0√ x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expected

to simplify your numerical answer, but no marks will be given unless all your work is

clearly shown.

8/15/2019 Brown's Notes 2010



(a) One antiderivative of x12 is

112

+ 1· x 1

2+1 =

2

3 x

32 .

(b) a

0

√ x dx =

2

3 x

32

a

0

= 2

3

a

32 − 0

=

2

3a

32 .

3. [10 MARKS] Showing all your work, diff erentiate the function g( x) =

x4 tan x

1√ 2 + t 2

dt .

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

g( x) =

0 tan x

1√ 2 + t 2

dt +

x4 0

1√ 2 + t 2

dt .

(b) Then reverse the limits in the first summand and change its sign, so that the variable

limit is the upper one:

g( x) = −tan x 0

1√ 2 + t 2

dt +

x4 0

1√ 2 + t 2

dt .

(c) Denote the upper limit of the first integral by u = tan x. Then

d

dx

tan x 0

1√ 2 + t 2

dt = d

du

u 0

1√ 2 + t 2

dt · du

dx

= 1√

2 + u2· sec2 x

=

2sec x

·tan x

√ 2 + tan2 x =

2sec x

·tan x

√ 1 + sec2 x .

(d) Denote the upper limit of the second integral by v = x4. Then

d

dx

x4 0

1√ 2 + t 2

dt = d

dv

v 0

1√ 2 + t 2

dt · dv

dx

8/15/2019 Brown's Notes 2010



= 1

√ 2 + v2

·4 x3

= 4 x3

√ 2 + x8

.

(e) Henced

dxg( x) = −2sec x · tan x√

2 + tan2 x+

4 x3

√ 2 + x8

.

4. [10 MARKS] If F ( x) =

x

1

f (t ) dt , where f (t ) =

t 2

1

a + ub

udu and a, b are constants,

find F (2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F ( x) = f ( x) =

x2

1

a + ub

udu .

(b) A second application of Part 1 of the Fundamental Theorem yields

F ( x) = f ( x) = d

dx x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

d

dx

x2

1

a + ub

udu =

d

dx

v

1

a + ub

udu

= d

dv

v

1

a + ub

u du · dv

dx

= a + vb

v· dv

dx

= a + vb

v· 2 x

= a + x2b

x2 · 2 x

=2

a + x2b

x .

8/15/2019 Brown's Notes 2010



Tuesday version


of the function

g( x) =

a

x

b tan(t ) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluate

g( x), by first verifying carefully that ln | sec x| is an antiderivative of tan x.

Solution:

(a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as a

function of its upper index of integration. Here the variable is the lower index of

integration.

d

dx

a

x

b tan(t ) dt

= d

dx

− x

a

b tan(t ) dt

= − d

dx

x

a

b tan(t ) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definite

integral with respect to the lower index, and this should be accepted if work has

been shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-

position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

d

dxln | sec x| =

d

duln |u| · d

dxsecx

= 1

u· sec x tan x

= 1

sec x ·sec x tan x

= tan x .

Hence

g( x) = b ln | sec t |a x = b ln

sec a

sec x

= b ln

cos x

cos a

.

8/15/2019 Brown's Notes 2010



2. [5 MARKS] Evaluate the limit by first recognizing the sum as a Riemann sum for a

function defined on [0, 1]:

limn→∞

1

n

7

n+

14

n+

21

n+ . . . +

7n

n

.

Solution:

(a) We are told to take the interval of integration to be [0, 1]; when this is divided into

n equal parts, each has length ∆ x = 1

n. Such a factor has been explicitly written in

the sum.

(b) The typical summand is — aside from the common factor 1n — of the form 7i

n .

Since the distance of the left end-point of the ith subinterval from 0 is i∆ x = in

, we

may interpret

7i

n =

√ 7 x .

(c) Thus the limit must be equal to 1

0

√ 7 x dx =

√ 7 · 2

3 · x 3

2

1

0

= 2

3

√ 7.


of the function

g( x) =

√ x

a

b cos t

t dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u( x) =√

x.

(b) Then

d

dx

g( x) = d

dx

√ x

a

b cos t

t

dt

= d

dx

u( x)

a

b cos t

t dt

= d

du

u( x)

a

b cos t

t dt · du( x)

dx

= b cos u

u· du( x)

dx

8/15/2019 Brown's Notes 2010



= b cos u

u·

1

2 √ x= b

cos√

x√ x

· 1

2√

x

= b cos

√ x

2 x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y = x

0

1

1 + at + bt 2 dt is concave downward, where a, b are constants.

Solution:

(a) By Part 1 of the Fundamental Theorem,

y( x) = 1

1 + ax + bx2 .

(b) Diff erentiating a second time yields

y( x) = d

dx

1

1 + ax + bx2

= −

11 + ax + bx2

2 ·

d

dx1

+ax

+bx

2= − a + 2bx

1 + ax + bx22

.

(c) The curve is concave downward where y < 0:

− a + 2bx1 + ax + bx2

2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive

⇔ 2bx <

−a

⇔

x < − a2b when b > 0 x > − a

2b when b < 0

never concave upward when b = 0

Wednesday version

8/15/2019 Brown's Notes 2010




πb

πa

sin t dt .

Solution:

(a) An antiderivative of sin t is − cos t .

(b)πb

πa

sin t dt = [− cos t ]πbπa

(c) Your answer should be simplified as much as possible.

2. [5 MARKS] Evaluate the following limit by first recognizing the sum as a Riemann

sum for a function defined on [0, 1]: limn→∞

ni=1

i8

n9. A full solution is required — it is not

sufficient to write only the value of the limit.

Solution:

(a) We are told to take the interval of integration to be [0, 1]; when this is divided into

n equal parts, each has length ∆ x = 1

n.

(b) The typical summand is — aside from the common factor 1n

— of the form

in

8

.

Since the distance of the left end-point of the ith subinterval from 0 is i∆ x = i

n, we

may interpret

i

n

8

= x8 .

(c) Thus the limit must be equal to

1

0

x8 dx = 1

9 x9

1

0

= 1

9 .


of the function

bx cos x

cos (t c) dt , where a, b, c are real numbers.

Solution:

8/15/2019 Brown's Notes 2010




the upper limit of integration, when the lower limit is constant. The given integralmust be expressed in terms of such specialized definite integrals.

bx cos x

cos (t c) dt =

0 cos x

cos (t c) dt +

bx 0

cos (t c) dt

= −cos x 0

cos (t c) dt +

bx 0

cos (t c) dt

(b) For the summand

bx

0

cos (t

c

) dt , let u = b x. Then

d

dx

bx 0

cos (t c) dt = d

dx

u 0

cos (t c) dt

= d

du

u 0

cos (t c) dt · du

dx

= cos (uc) · b

= cos ((bx)c)

·b

(c) For the summand

cos x 0

cos (t c) dt , let v = cos x.

d

dx

cos x 0

cos (t c) dt = d

dx

v 0

cos (t c) dt

= d

dv

v 0

cos (t c) dt · dv

dx

= cos (vc

) · (− sin x)= cos (cosc x) · (− sin x)

(d)bx

cos x

cos (t c) dt = −cos (cosc x) · (− sin x) + cos ((bx)c) · b .

8/15/2019 Brown's Notes 2010



4. [10 MARKS] Let f ( x) =

0 if x < 0

x if 0 ≤ x ≤ a2a − x if a < x < 2a

0 if x > 2a

and g( x) = x

0 f (t ) dt , where a is

a positive constant. Showing all your work, find a formula for the value of g( x) when

a < x < 2a.

Solution:

(a) The interval where we seek a formula is the third interval into which the domain

has been broken. For x in this interval the integral can be decomposed into

x

0

f (t ) dt = a

0

f (t ) dt + x

a

f (t ) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, since

we are not finding area under that portion of the curve; the same applies to the

portion of the definition for x > 2a.

(b) a

0

f (t ) dt =

a

0

t dt

= t 2

2 t =a

t =0

= a2

2 .

(c) x

a

f (t ) dt =

x

a

(2a − t ) dt

=

2at − t 2

2

t = x

t =a

=

2ax − x2

2

−

2a2 − a2

2

.

(d)

g( x) = a2

2 +

2ax − x2

2

−

2a2 − a2

2

= 2ax − x2

2 − a2 .

Thursday version

8/15/2019 Brown's Notes 2010



1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ

aπ

cos θ d θ , where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ .

(b)bπ

aπ

cos θ d θ = [sin θ ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiples

of π is 0, so the value of the definite integral is 0.

2. [5 MARKS] Express limn→∞

ni=1

axi sin xi ∆ x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: c

b

ax sin x dx .

3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivativeof the function

g( x) =

bx

ax

t 2 + c

t 2 − cdt ,

where a, b, c are positive integers.

Solution:




g( x) =

bx

ax

t 2 + c

t 2 − c dt =

0

ax

t 2 + c

t 2 − c dt +

bx

0

t 2 + c

t 2 − c dt

= − ax

0

t 2 + c

t 2 − c dt +

bx

0

t 2 + c

t 2 − c dt

8/15/2019 Brown's Notes 2010



(b) For the summand

bx

0

t 2 + c

t 2 − c dt , let u = bx. Then

d

dx

bx 0

t 2 + c

t 2 − cdt =

d

dx

u 0

t 2 + c

t 2 − cdt

= d

du

u 0

t 2 + c

t 2 − cdt · du

dx

= u2 + c

u2

−c ·

du

dx

= u2 + c

u2 − c · b

= (bx)2 + c

(bx)2 − c· b

(c) For the summand

ax

0

t 2 + c

t 2 − cdt , let u = ax. Then, analogously to the preceding

step,d

dx

ax

0

t 2 + c

t 2 − cdt =

(ax)2 + c

(ax)2 − c· a .

(d)

g( x) = (bx)2 + c

(bx)2 − c· b − (ax)2 + c

(ax)2 − c· a .

4. [10 MARKS] Find the derivative of the function f ( x) =

x3 √

x

√ t cos t dt .

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

f ( x) =

0 √

x

√ t cos t dt +

x3 0

√ t cos t dt

8/15/2019 Brown's Notes 2010



(b) Then reverse the limits in the first summand and change its sign, so that the variable

limit is the upper one:

f ( x) = −

√ x

0

√ t cos t dt +

x3 0

√ t cos t dt .

(c) Denote the upper limit of the first integral by u =√

x. Then

d

dx

√ x

0

√ t cos t dt =

d

dx

u 0

√ t cos t dt

= d

du

u 0

√ t cos t dt · du

dx

=√

u cos u · du

dx

=

√ x cos

√ x · 1

2√

x

= cos

√ x

2 x14

.

(d) Denote the upper limit of the second integral by v = x3. Then

d

dx

x3 0

√ t cos t dt =

d

dx

v 0

√ t cos t dt

= d

dv

v 0

√ t cos t dt · dv

dx

=√

v cos v · dv

dx

= √ x3 cos x3 · 3 x2

= 3 x72 cos

x3

(e) Henced

dx f ( x) = −cos

√ x

2 x14

+ 3 x72 cos

x3

.

8/15/2019 Brown's Notes 2010




Distribution Date: Mounted on the Web on Monday, March 03rd, 2008







• No calculators!




typical variety.

Monday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotating

about the line y = 1 the region bounded by the curves y = n√

x and y = x, where n is a

given positive integer.

Solution: A favoured method of solution was not prescribed.

Using the method of “washers”: (a) The solution I am giving is for the case where n

is even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-

tions yields the points ( x, y) = (0, 0), (1, 1).

(c) It’s not clear from the wording of the problem whether it was intended, in the

case of odd n, to permit the second intersection point ( x, y) = (−1, −1); the

decision was left to the individual TA’s. The remainder of this solution covers

the case of even n; for odd n this solution does not consider the solid generatedby rotating the region with vertices ( x, y) = (−1, −1), (0, 0).

(d) Find the inner and outer dimensions of the washer. Since the axis of revolution

is a horizontal line, the element of area being rotated is vertical. For arbitrary

x the lower point on the element is ( x, x); the upper point is ( x, n√

x). The

distances of these points from the axis are, respectively 1 − x and 1 − n√

x.

8/15/2019 Brown's Notes 2010



(e) The volume of the “washer” is, therefore,

π−(1 − x)2 + (1 − n√ x)2

∆ x .

(f) Correctly evaluate the integral:

π

1

0

−(1 − x)2 + (1 − n

√ x)2

dx

= π

1

0

−2 x + x2 + 2 x

1n − x

2n

dx

= π − x2 + x2

3 +

2n

n + 1 x

n+1n − n

n + 2 x

n+2n

1

0

= π

−1 +

1

3 +

2n

n + 1 − n

n + 2

=

(n − 1)(n + 4)π

3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) The solution I am giving is for the case

where n is even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-

tions yields the points ( x, y) = (0, 0), (1, 1).

(c) It’s not clear from the wording of the problem whether it was intended, in the

case of odd n, to permit the second intersection point ( x, y) = (−1, −1); the

decision was left to the individual TA’s. The remainder of this solution covers

the case of even n; for odd n this solution does not consider the solid generated

by rotating the region with vertices ( x, y) = (−1, −1), (0, 0).

(d) Find the inner and outer dimensions of the washer. Since the axis of revolution

is a horizontal line, the element of area being rotated is also horizontal. For

arbitrary y the left endpoint on the element is ( yn, y); the right endpoint is ( y, y).

The length of the element is, therefore, y− yn; the distances of the element from

the axis of symmetry is 1 − y.

(e) The volume of the cylindrical shell element of volume is, therefore,

2π(1 − y) · ( y − yn) · ∆ y .

(f) Correctly evaluate the integral:

2π

1

0

(1 − y)( y − yn) dy

= 2π

1

0

− yn + yn+1 + y − y2

dy

8/15/2019 Brown's Notes 2010



= 2π − 1

n + 1

yn+1 + 1

n + 2

yn+2 + 1

2

y2

− 1

3

y31

0

= 2π

− 1

n + 1 +

1

n + 2 +

1

2 − 1

3

− 0

= 2π

1

6 − 1

(n + 1)(n + 2)

=

(n − 1)(n + 4)π

3(n + 1)(n + 2)


(a − t )(b + t 2) dt .

Solution:

(a) Expand the product in the integrand: (a − t )(b + t 2) dt =

ab − bt + at 2 − t 3

dt .

(b) Integrate term by term: ab − bt + at 2 − t 3

dt = ab · t − b

2 · t 2 +

a

3 · t 3 − 1

4 · t 4 + C .

3. [10 MARKS] Showing all your work, determine a number b such that the line x = b

divides into two regions of equal area the region bounded by the curves x = ay2 and

x = k .


lem 1 of the Tuesday quiz.

4. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral e x

e x + adx, where a is a non-zero real number.

Solution:

(a) Try the substitution u = e x + a, so du = e x dx.

(b)

e x

e x + adx =

du

u= ln |u| + C = ln |e x + a| + C .

(If the constant a is positive, then the absolute signs are not required.)

8/15/2019 Brown's Notes 2010



Tuesday version

1. [10 MARKS] Showing all your work, find a number b such that the line y = b divides

the region bounded by the curves y = ax2 and y = k into two regions with equal area,

where a, k are given positive constants.

Solution:

(a) Determine the range of values for integration by finding the intersections of the

bounding curves: solving the equations yields the points

∓

k

a, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin by

repeating the calculation of the preceding part: the corner points have coordinates∓

b

a, b

. The area is

√ ba

−√

ba

(b − ax2) dx = 2

bx − ax3

3

√ ba

0

= 4

3b

b

a.

(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is 4

3k

k

a.

(d) The condition of the problem is that

4

3b

b

a=

1

2 · 4

3k

k

a

which is equivalent to 4b3 = k 3, and implies that the line should be placed where

b = 2− 23 k .

2. [10 MARKS] The region bounded by the curves y = 5 and y = x2 − ax + b is rotated

about the axis x = −1. Showing all your work, find the volume of the resulting solid.

Solution: Because there are constraints on the constants, I will work just one variant,

with a = 3, b = 7.

Using the method of cylindrical shells: (a) To find the extremes of integration, we

solve the equations y = 5 and y = x2 − 3 x + 7, obtaining ( x, y) = (1, 5), (2, 5).

(b) The height of a vertical element of area which generates a cylindrical shell is,

at horizontal position x, 5 − ( x2 − 3 x + 7) = − x2 + 3 x − 2.

(c) The distance of that vertical element of area from the axis of revolution is 1+ x.

8/15/2019 Brown's Notes 2010



(d) The volume is given by the integral

2π

2

1

(1 + x)(− x2 + 3 x − 2) dx

(e) Evaluating the integral:

2π

2

1

(1 + x)(− x2 + 3 x − 2) dx

= 2π

2

1

− x3 + 2 x2 + x − 2

dx

= 2π −1

4

x4 + 2

3

x3 + 1

2

x2

−2 x

2

1

= 2π

−4 +

16

3 + 2 − 4 +

1

4 − 2

3 − 1

2 + 2

= 5π

6

Using the method of “washers” (a) To find the lowest point on the parabola, we solve

x2 − 3 x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be

3

2, 19

4

.

(b) The horizontal element generating the “washer” at height y extends between

the solutions in x to the equation y = x2 − 3 x + 7; these are

x =3 ±

4 y − 19

2 .

(c) The volume of the “washer” at height y is, therefore,

π

1 +

3 +

4 y − 19

2

2

−1 +

3 −

4 y − 19

2

2∆ y

= 5π 4 y − 19 ∆ y

(d) The volume is given by the integral 5π

5

194

4 y − 19 dy.

(e) Evaluation of the integral:

5π

5

194

4 y − 19 dy =

5π · 2

3 · 1

4(4 y − 19)

32

5

194

= 5π

6 .

8/15/2019 Brown's Notes 2010



3. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral cosn x sin x dx, where n is a fixed, positive integer.

Solution:

(a) Use new variable u, where du = −sin x dx; one solution is u = cos x.

(b) cosn x sin x dx = −

un du

= − un+1

n + 1 + C

= − 1n + 1

cosn+1 x + C


xb + a +

1

x2 + 1

dx, (where

a and b are given positive integers).

Solution:

(a)

xb dx =

xb+1

b + 1 + C 1,

(b)

a dx = ax + C 2

(c)

1

x2 + 1 d x = arctan x + C 3

(d)

xb + a +

1

x2 + 1

dx =

xb+1

b + 1 + ax + arctan x + C .

Wednesday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotating

about the line x =

−1 the region bounded by y = xn and x = yn, where n is a given

positive integer.

Solution:

Case I: n is even



8/15/2019 Brown's Notes 2010



(b) Find the inner and outer dimensions of the washer. Since the axis of revo-

lution is a vertical line, the element of area being rotated is horizontal. Forarbitrary y the farther endpoint on the element is ( y

1n , y); the nearer end-

point is ( yn, y). The distances of these points from the axis are, respectively

1 + n√

y and 1 + yn.


π−(1 + yn)2 + (1 + n

√ y)2

∆ y .


π

1

0

−(1 + y)2 + (1 + n

√ y)2

dy

= π 1

0

2 y

1n + y

2n − 2 yn − y2n

dy

= π

2n

n + 1 y

n+1n +

n

n + 2 y

n+2n − 2

n + 1 yn+1 − 1

2n + 1 y2n+1

1

0

= 2(n − 1)(3n2 + 7n + 3)π

(n + 1)(n + 2)(2n + 1)


bounding the region. Solving the 2 equations yields the points ( x, y) =

(0, 0), (1, 1).


rotated is also vertical. For arbitrary x the top endpoint on the element is( x, x

1n ); the lower endpoint is ( x, xn). The length of the element is, there-

fore, x1n − xn; the distance of the element from the axis of symmetry is

1 + x.


2π(1 + x) · x

1n − xn

.


2π

1

0

(1 + x)( x1n − xn) dx

= 2π

1

0

x

1n − xn + x

n+1n − xn+1

dx

= 2π

n

n + 1 − 1

n + 1 +

n

2n + 1 − 1

n + 2

= 2(n − 1)(3n2 + 7n + 3)π

(n + 1)(n + 2)(2n + 1)

8/15/2019 Brown's Notes 2010



Case II: n is odd

Using the method of “washers”: (a) Find the intersections of the curves bound-ing the region. Solving the 2 equations yields the points ( x, y) = (0, 0), (±1, ±1).

Here there is an issue of interpretation. The textbook usually permits the

word region to apply to one that may have more than one component;

some authors would not wish to apply the term in such a situation. I will

follow the textbook, and permit a region here to have two components.

(b) Find the inner and outer dimensions of the washer. Since the axis of rev-

olution is a vertical line, the element of area being rotated is horizontal.

But there are two kinds of elements, depending on whether y is positive

or negative. For arbitrary, positive y the farther endpoint on the element is

( y1n , y); the nearer endpoint is ( yn, y). The distances of these points from

the axis are, respectively 1 + n√ y and 1 + yn. For arbitrary, negative y the

nearer endpoint on the element is ( y1n , y); the farther endpoint is ( yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and

1 + yn (both of which are less than 1).


π−(1 + yn)2 + (1 + n

√ y)2

∆ y .


π 1

−1−

(1 + y)2 + (1 + n√

y)2 dy

= π

1

0

2 y

1n + y

2n − 2 yn − y2n

dy

+π

0

−1

−2 y

1n − y

2n + 2 yn + y2n

dy

= π

2n

n + 1 y

n+1n +

n

n + 2 y

n+2n − 2

n + 1 yn+1 − 1

2n + 1 y2n+1

1

0

+π

− 2n

n + 1 y

n+1n − n

n + 2 y

n+2n +

2

n + 1 yn+1 +

1

2n + 1 y2n+1

0

−1

= 2(n − 1)(3n

2

+ 7n + 3)π(n + 1)(n + 2)(2n + 1)

+ 2(n − 1)(n

2

+ 3n + 1)π(n + 1)(n + 2)(2n + 1)

= 4(n − 1)π

n + 1 .


bounding the region. Solving the 2 equations yields the points ( x, y) =

(0, 0), (±1, ±1).

8/15/2019 Brown's Notes 2010




rotated is also vertical. For arbitrary, positive x the top endpoint on the el-ement is ( x, x

1n ); the lower endpoint is ( x, xn); for arbitrary, negative x the

bottom endpoint on the element is ( x, x1n ); the upper endpoint is ( x, xn).

The length of the element is, therefore, x 1

n − xn; the distance of the ele-

ment from the axis of rotation is 1 + x.


2π(1 + x) · x 1

n − xn .


2π 1

−1(1 + x)

x 1n − xn

dx

= 2π

1

−1

x 1n − xn + x

n+1n − xn+1

dx

= 2π

n

n + 1 − 1

n + 1 +

n

2n + 1 − 1

n + 2

+2π

n

n + 1 − 1

n + 1 − n

2n + 1 +

1

n + 2

= 4(n − 1)π

n + 1

2. [5 MARKS] Showing all your work, evaluate the indefinite integral

sin2t

cos t dt .

Solution:

(a) Apply a “double angle” formula: sin2t

cos t dt =

2 sin t cos t

cos t dt = 2

sin t dt .

(b) Complete the integration:

2

sin t dt = −2cos t + C .

3. [10 MARKS] Showing all your work, find the area of the region bounded by the parabola

y = x2, the tangent line to this parabola at (a, a2), and the x-axis, (where a is a given real

number).

Solution: This area can be computed by integrating either with respect to y or with

respect to x.

8/15/2019 Brown's Notes 2010



Integrating with respect to y: (a) Since y = 2 x, the tangent line through (a, a2) has

equation y − a2 = 2a( x − a) ⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabola

and the line in the form

x = function of y .

The branch of the parabola to the right of the y-axis is x = √

y. The line has

equation x = y

2a+

a

2.

(c) The area of the horizontal element of area at height y is

y + a2

2a− √

y

∆ y.

(d) The area is the value of the integral a2

0

y + a2

2a− √

y

dy .

(e) Integration yields y2

4a+

ay

2 − 2

3 y

32

a2

0

=

1

4 +

1

2 − 2

3

a3 =

1

12 a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Its

intercept with the x-axis is at x = a

2.

(b) The area of the vertical element of area at horizontal position x ≤ a

2 is x2 − 0

∆ x.

(c) The area of the vertical element of area at horizontal position x ≥ a2

is x2 − (2ax − a2)

( x − a)2 ∆ x.

(d) The area of the region is the sum a2

0

x2 dx +

a

a2

( x − a)2 dx .

(e) Integration yields x3

3

a

2

0

+−( x − a)3

3

a

a2

= a3

12 .


seca x tan x dx,

where a is a constant, positive integer.

Solution:

8/15/2019 Brown's Notes 2010



(a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x.

(b) seca x tan x dx =

ua−1 du =

ua

a+ C =

seca x

a+ C .

Some students may have integrated by sight.

Thursday version

1. [10 MARKS] Showing all your work, use the method of cylindrical shells to find the

volume generated by rotating about the axis x = b the region bounded by the curves

y =√

x − 1, y = 0, x = a, where a, b are fixed real constants. b ≥√

a − 1 .

Solution:

(a) Solve equations to determine the limits of integration. Solving x = a with y =√ x − 1 yields the single point of intersection

a,

√ a − 1

.

(b) The horizontal element of area at height y which generates the cylindrical shell has

left endpoint (1 + y2, y) and right endpoint (a, y), so its length is a − (1 + y2).

(c) The distance of the horizontal element of area which generates the shell from the

axis of rotation is b − y.

(d) Set up the integral for the volume by cylindrical shells:

√ a−

1

0

(b − y)

a − (1 + y2)

dy .

(e) Evaluate the integral

√ a−1

0

(b − y)a − (1 + y2)

dy

=

√ a−1

0

b(a − 1) − (a − 1) y − by2 + y3

dy

=

b(a − 1) y − a

−1

2 y2

− b

3 y3

+

1

4 y4

√ a−1

0

= b(a − 1)32 − a − 1

2 (a − 1) − b

3(a − 1)

32 +

1

4(a − 1)2

= (a − 1)32

2

3 · b − 1

4

√ a − 1

.

8/15/2019 Brown's Notes 2010



2. [5 MARKS] Showing all your work, use a substitution to evaluate the integral b

(1+ax)3 dx ,

where a, b are non-zero constants.Solution:

(a) A substitution which suggests itself is u = 1 + a x, implying that du = a dx, so

dx = 1a

du.

(b) b

(1 + ax)3 dx =

b

a

du

u3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

3. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral

t 2 cos a −t 3 dt , (where a is a given real number).

Solution:

(a) Try the substitution u = t 3.

(b) du = 3t 2 dt ⇒ t 2 dt = 13

du.

(c) t 2 cos

a − t 3

dt =

1

3 cos(a − u) du

=

−

1

3 sin(a

−u) + C

= −1

3 sin(a − t 3) + C .

(Some students may wish to employ a second substitution v = a − u. Alternatively,

a better substitution for the problem would have been to take u = a − t 3.)

4. [10 MARKS] Showing all your work, determine the area of the region bounded by the

parabola x = y2, the tangent line to this parabola at (a2, a), and the y-axis, where a is a

fixed, positive real number.


lem 3 of the Wednesday quiz.


Release Date: Mounted on the Web on 05 April, 2008

These draft solutions could contain errors, and they must be subject to correction. Caveat

lector!

8/15/2019 Brown's Notes 2010




type of quiz was generated in multiple varieties for each of the tutorial sections. The order of the problems in the varieties was also randomly assigned. All of the quizzes had a heading that



• No calculators!




typical variety.

Monday version


x5 ln(20 x) dx.

Solution:

(a) I will integrate by parts, setting u = ln(20 x), dv = x5 dx. Then du = dx

x, v =

x6

6 .

(b) x5 ln(20 x) dx = (ln(20 x)) ·

x6

6

− 1

6

x5 dx

= x6 ln(20 x)

6 − x6

36 + C .


1 − cot2 x

csc2 xdx.

Solution:

1 − cot2 x

csc2

x

dx = sin2 x − cos2 x

sin2

x · csc2

x

dx

=

sin2 x − cos2 x

dx

=

(− cos2 x) dx

= −1

2 sin 2 x + C .

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



= cot θ + θ + C

= cos θ

sin θ + θ + C

= cos θ √

1 − cos2 θ + arccos

x

2 + C

=

x2

1 − x2

4

+ arccos x

2 + C

= x√

4 − x2+ arccos

x

2 + C .

Tuesday version


e5 x cos(2 x) dx .

Solution:

(a) Use integration by parts. In this case the factors e5 x and cos 2 x are rendered neither

“more complicated” nor “simpler” under either integration or diff erentiation. Two

applications of integration by parts, with appropriate choices of functions, will

yield an equation that can be solved for the value of the indefinite integral. For the

first application take, for example, u = e5 x and dv = cos2 x. Then du = 5e5 x dx,

and v = 12

sin 2 x.

(b) e5 x cos(2 x) dx = e5 x · 1

2 sin 2 x − 5

2

e5 x sin 2 x dx . (113)

(c) Now we apply integration by parts to evaluate

e5 x sin 2 x dx, taking U = e5 x and

dV = sin 2 x. Then dU = 5e5 x dx, and V = −12

cos 2 x.

(d) e5 x sin(2 x) dx = e5 x ·

−1

2 cos 2 x

+

5

2

e5 x cos2 x dx . (114)

(e) Combining equations (113), (114) yields e5 x cos(2 x) dx = e5 x · 1

2 sin 2 x − 5

2

−e5 x · 1

2 · cos2 x +

5

2

e5 x sin2 x dx

= e5 x ·

1

2 · sin2 x +

5

4 cos 2 x

− 25

4

e5 x cos(2 x) dx .

8/15/2019 Brown's Notes 2010



(f) Solving the last equation yields e5 x cos(2 x) dx =

1

29e5 x · (2sin2 x + 5cos2 x) + C .

2. [10 MARKS] Showing all your work, use a substitution to change this integral into the

integral of a rational function; then integrate and express your solution in terms of x: 1 + 7e x

1 − e x dx .

Solution:

(a) Let u = e x, so du = e x dx.

(b) Then

1 + 7e x

1 − e x dx =

1 + 7u

1 − u· du

uand we proceed to expand the integrand into

a sum of partial fractions.

(c) Assuming that 1 + 7u

u(1 − u) =

A

u+

B

1 − u, and multiplying both sides by u(1 − u), we

obtain the identity in u 1 + 7u = A(1 − u) + Bu. Setting u = 0 and u = 1 yields the

equations 1 = A and 8 = B. We can now continue integration.

(d)

1

+7u

1 − udu =

1u

+ 81 − u

du

= ln |u| − 8 ln |1 − u| + C

= ln e x − 8 ln |1 − e x| + C

= x − 8 ln |1 − e x| + C

which solution can be checked by diff erentiation.


dx

x2 + 8 x + 17

2

d x.

Solution:

(a) Completing the square of the polynomial in the denominator, we obtain

x2 + 8 x + 17 =

x +

8

2

2

+

17 −

8

2

2 = ( x + 4)2 + 1 .

Accordingly, we can simplify the integral by taking u = x + 4, du = d x.

8/15/2019 Brown's Notes 2010



(b) The preceding substitution is not sufficient, however. All we obtain is dx

x2 + 8 x + 172

dx =

duu2 + 1

2 .

We can simplify this further by taking u = tan θ , i.e., by taking θ = arctan u, so

d θ = du

1 + u2.

(c) du

1 + u2

2

=

sec2 θ

sec4 θ d θ

=

cos2 θ d θ

=

1 + cos2θ

2 d θ =

θ

2 +

sin 2θ

4 + C

= 1

2 arctan u +

u

2(1 + u2) + C

= 1

2

arctan( x + 4) +

x + 4

x2 + 8 x + 17

+ C

which can be verified by diff erentiation. You should always verify this type of

integration by diff erentiation, in order to locate silly algebra mistakes (or worse).

Wednesday version


x cos(18 x) dx.

Solution:

(a) This can be solved using integration by parts. Define u = x, dv = cos(18 x) dx, so

du = d x and v = sin(18 x)

18 .

(b) x cos(18 x) dx = x · sin(18 x)

18 −

sin(18 x)

18

= x sin(18 x)

18 +

cos(18 x)

182 + C ,

which can be verified by diff erentiation.

8/15/2019 Brown's Notes 2010



2. [5 MARKS] Showing all your work, evaluate the integral 1 − sin x

cos xdx.

Solution: 1 − sin x

cos xdx =

(sec x − tan x) dx

= ln | sec x + tan x| + ln | cos x| + C

= ln |(sec x + tan x) · (cos x)| + C

= ln |1 + sin x| + C = ln(1 + sin x) + C .

Note that the absolute signs are not needed, since 1 + sin x cannot be negative.


π3

π6

ln tan x

(sin x) · (cos x) d x.

Solution:

(a) In view of the complicated nature of the integrand, one would be advised to seek

a substitution that could render it more amenable. But the integrand involves both

sines and cosines. However, note that

(sin x)(cos x) = (tan x)(cos2 x) = tan x

sec2 x=

tan x

tan2 x + 1.

Taking u = tan x, we have du = sec2 x dx. When x = π

3, u =

√ 3; when x =

π

6,

u = 1√

3.

(b)

π3

π6

ln tan x

(sin x) · (cos x) d x

=

√ 3

1√ 3

ln uu

u2+1

duu2 + 1

=

√ 3

1√ 3

ln u

u du

8/15/2019 Brown's Notes 2010



(c) Now the integral looks as though it could be simplified by a substitution v = ln u,

so dv = du

u . When u = √ 3, v = ln 32 ; when u = 1√ 3 , v = −ln 32 .

(d)

√ 3

1√ 3

ln u

udu =

ln 22

− ln 32

v dv

=

v2

2

ln 32

− ln 32

= 0.


et √

49 − e2t dt .

Solution:

(a) Clearly a substitution of the form u = et is indicated, in order to simplify the

integrand. We find that du = et dt .

(b) We obtain

et

√ 49 − e2t dt =

√ 49 − u2 du. Now a trigonometric substitution

is indicated. Take u = 7 sin θ — more precisely, θ = arcsin u7

(the inverse cosine

could also have been used), so du = 7 cos θ d θ : √ 49 − u2 du = 49

cos2 θ d θ.

(c)

49

cos2 θ d θ =

49

2

(1 + cos2θ ) d θ

= 49

2(θ + sin θ · cos θ ) + C

= 49

2 arcsin

u

7 +

u

2

√ 49 − u2 + C

= 49

2 arcsin

et

7 +

et √ 49 − e2t

2 + C .

Thursday version


sin3 9 x dx.

8/15/2019 Brown's Notes 2010



Solution: The integrand is an odd power of the sine function. I will substitute u = cos 9 x,

so du = −9sin9 x dx. sin3 9 x dx =

1 − u2

1

9

du

= −u

9 +

u3

27 + C

= −cos x

9 +

cos3 x

27 + C .


π2

π4

cot2 x dx. Your answer

should be simplified as much as possible; the instructors are aware that you do not have

the use of a calculator.

Solution:

(a) Recall that cot2 x = csc2 x − 1, and that d

dxcot x = −csc2 x.

(b)

π2

π4

cot2

x dx =

π2

π4

csc

2

x − 1

dx

= [− cot x − x]π4π2

.

(c) = −π

2 + 1 +

π

4 = 1 − π

4.


√ 9 − x2

xdx.

Solution:

(a) To simplify the square root, substitute x = 3cos θ , i.e., θ = arccos x3

. Then dx =

−3sin θ d θ .

(b)

√ 9 − x2

xdx = −3

sin2 θ

cos θ d θ

8/15/2019 Brown's Notes 2010



=

−3

1 − cos2 θ

cos θ

d θ

= −3

(sec θ − cos θ ) d θ

= −3 ln | sec θ + tan θ | + 3 sin θ + C

(c)

−3 ln | sec θ + tan θ | + 3sin θ + C = −3 ln

1 +

1 − x2

9

x3

+ 3

1 − x2

9 + C

= −3 ln3 +

√ 9

− x2

x + √ 9 − x2 + C

4. [10 MARKS] Showing all your work, determine whether the following integral is con-

vergent or divergent. Evaluate it if it is convergent; in such a case you are expected to

simplify your answer as much as is consistent with not having the use of a calculator: 1

0

6

ln 7 x√

x

dx

Solution:

(a) Let’s look first at the associated indefinite integral,

ln 7 x√

xdx. The integrand

can be expressed as a product, in which one of the factors “simplifies” upon dif-

ferentiation, while the other does not become significantly “more difficult” upon

integration. So we will integrate by parts, taking u = ln 7 x, and dv = x− 12 dx. Then

du = 1 x

, and we may take v = 2√

x.

(b)

ln 7 x√

xdx = (ln7 x)(2

√ x) − 2

dx√

x

= (ln7 x)(2 √ x) − 4 √ x + C .

(c) The integrand is not defined at x = 0 in the given interval of integration. By the

definition of an improper integral, we have 1

0

ln 7 x√ x

dx = lima→0+

1

a

ln 7 x√ x

dx

8/15/2019 Brown's Notes 2010



= lima

→0+ (ln7 x)(2

√ x) − 4

√ x

1

a

= lima→0+

(2ln7 − 4) − √ a · (2 ln(7a) − 4)

= 2 l n 7 − 4 − lim

a→0+(ln7a)(2

√ a)

(d) The limit can be expressed as that of a ratio where numerator and denominator both

become infinite. Thus l’Hospital’s Rule may be used:

lima→0+

(ln7a)(2√

a) = 2 lima→0+

ln 7a

a− 12

= 2 lima→0

+

1a

−12 a−3

2

= 2 lima→0+

(−2√

a) = 0.

Thus the original integral is convergent, and its value is 6(2 ln 7 − 4).

Problems not used

1. [10 MARKS] Make a substitution to express the integrand as a rational function, and

then evaluate the integral.

4

1

√ x

x − 16

d x .

Solution:

(a) Start by substituting u =√

x, so x = u2, dx = 2u du. When x = 1, u = 1; when

x = 4, u = 2.

(b) 4

1

√ x

x − 16 d x =

2

1

2u2

u2 − 36 du

= 2

12 +

72

u2

− 36 du

=

2

1

2 +

6

u − 6 − 6

u + 6

du

=u2 + 6 ln |u − 6| − 6 ln |u + 6|

2

1

= 3 + 6 ln 7

10 .

8/15/2019 Brown's Notes 2010




Release Date: Mounted on the Web on Wednesday, April 9th, 2008

These are draft solutions that were prepared when the quizzes were being designed. It was in-

tended that Teaching Assistants would consult these draft solutions when they graded their stu-

dents’ quizzes, and would report any errors or omissions. As the Teaching Assistants may be-

lieve that they are inhibited from communicating with the instructor who manages this course,

it is not clear that the solutions have been thoroughly checked. The solutions are being released

with the cautionary warning, Caveat lector! — Let the reader beware! Use them at your own

risk.

There were four diff erent types of quizzes, for the days when the tutorials are scheduled.

Each type of quiz was generated in multiple varieties for each of the tutorial sections. The

order of the problems in the varieties was also randomly assigned. All of the quizzes had aheading that included the instructions.


• No calculators!




typical variety.

Monday version

1. [10 MARKS] Showing all your work, find the length of the curve y = x5

30 +

1

2 x3 (1 ≤

x ≤ 2). Simplify your answer as much as possible; the instructors are aware that you do

not have the use of a calculator.

Solution:

dy

dx=

x4

6 − 3

2 x4

⇒ 1 +

dydx

2

=

1 + x8

36 + 9

4 x8 − 1

2

=

x8

36 +

9

4 x8 +

1

2 =

x4

6 +

3

2 x4

2

=

x4

6 +

3

2 x4

.

8/15/2019 Brown's Notes 2010



The absolute signs may be dropped, since the given square root is a sum of positive

multiples of even powers, and must be non-negative. The length is 2

1

x4

6 +

3

2 x4

dx =

x5

30 − 1

2 x3

2

1

= 353

240 .

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating

x = 1

2

y2 + 2

32

(7 ≤ y ≤ 10) about the x-axis. Simplify your answer as much as

possible; the instructors are aware that you do not have the use of a calculator.

Solution:dx

dy=

1

3 · 3

2 · y2 + 2

12 · 2 y = y

y2 + 2 .

Thus 1 +

dx

dy

2

=

1 + y2(2 + y2) = y2 + 2 .

The area of the surface is

10

7 1 + y2

· (2π y) dy = 2π y2

2 +

y4

4 10

7

= 7701π

2 .

3. [10 MARKS] Showing all your work, find the area enclosed by the curve (in polar

coordinates) r = 7 + 2sin6θ .

Solution: The curve surrounds the pole, and is periodic with period 2π. The area may be

expressed as an integral over an interval of length 2π; for example, as

1

2 2π

0

(7 + 2sin6θ )2 d θ = 1

2 2π

0

(49 + 28sin6θ + 4sin2 6θ ) d θ

= 1

2

2π

0

(49 + 28sin6θ + 2(1 − cos 12θ )) d θ

= 1

2

49θ − 28

6 cos6θ + 2θ − 2

12 sin 12θ

2π

0

= 1

2(51)(2π) = 51π .

8/15/2019 Brown's Notes 2010



4. [10 MARKS] Showing all your work, find the exact length of the curve x = 6 + 3t 2,

y = 2 + 2t 3

(0 ≤ t ≤ 1).Solution:

dx

dt = 6t

dy

dt = 6t 2

dy

dx=

dy

dt dx

dt

= t

⇒

1 +

dy

dx

2

=√

1 + t 2

Arc Length =

1

0

dx

dt

2

+

dy

dt

2

=

1

0

√ 36t 2 + 36t 4 dt

= 6

1

0

t √

1 + t 2 dt

= 6 · 1

2 · 2

3

1 + t 2 3

2 1

0= 2

1 + t

2 32 1

0

= 22

32 − 1

= 4

√ 2 − 2 .

5. [10 MARKS] Showing all your work, sum a series in order to express the following

number as a ratio of integers: 0.35 = 0.35353535 . . ..

Solution:

0.35 = 35

100

1 +

1

100 +

1

1002 +

1

1003 + . . .

=

35100

1 − 1100

= 35

99 .

Tuesday version

1. [10 MARKS] Showing all your work, find the length of the curve x = 1

3

√ y · ( y − 3)

(49 ≤ y ≤ 64). Simplify your answer as much as possible; the instructors are aware that

you do not have the use of a calculator.

8/15/2019 Brown's Notes 2010



Solution:

dx

dy=

1

3 · 3

2

√ y − 3

3 · 1

2 y− 1

2 = 1

2

√ y − 1√

y

⇒

1 +

dx

dy

2

=

1 +

1

4

y +

1

y− 2

=

1

4

y +

1

y+ 2

= 1

4 √

y + 1√

y2

= 1

2

√ y +

1√ y

.

The absolute signs may be dropped, since the square root is non-negative. The length is

1

2

64

49

√ y +

1√ y

dy =

1

2

2

3 · y 3

2 + 2√

y

64

49

=

1

3 · y 3

2 + √

y

64

49

= 172

3 .

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotatingthe curve x = 6 + 2 y2 (0 ≤ y ≤ 3) about the x-axis.

Solution:

dx

dy= 4 y

⇒

1 +

dx

dy

2

=

1 + 16 y2

⇒ Area = 3

0 1 + 16 y2 · 2π y dy

= π

24

(1 + 16 y2)

32

3

0

= π

24

(145)

32 − 1

.

3. [15 MARKS] Showing all your work, find the area of the region that lies inside the curve

r = 15 cos θ , and outside the curve r = 5 + 5cos θ .

8/15/2019 Brown's Notes 2010



Solution: The first curve is a circle; the second is a cardioid whose axis of symmetry

is the initial ray. If we solve the equations we find that the curves intersect at θ =± arccos 12

= ±π3

. They also intersect at the pole, which appears on the circle when

θ = π2

, etc., and on the cardioid when θ = π, etc. The region whose area we seek lies

between the two curves when −π3 ≤ θ ≤ π

3 and r is positive. Integration shows the area

to be

1

2

π3

− π3

(15 cos θ )2 − (5 + 5cos θ )2

d θ

= 25

2

π3

− π3

8cos2 θ − 2cos θ − 1

d θ

= 252

π3

− π3

(4(1 + cos2θ ) − 2cos θ − 1) d θ

= 25

2

π3

− π3

(3 + 4cos2θ − 2cos θ ) d θ

= 25[3θ + 2sin2θ − 2sin θ ]π3

0

= 25π +

√ 3 −

√ 3

= 25π .

4. [15 MARKS] Showing all your work, find the length of the loop of the curve x = 18t −6t 3, y = 18t 2.

Solution:

dx

dt = 18 − 18t 2

dy

dt = 36t

dx

dt

2

+

dy

dt

2

=

182

1 − t 2

2+ 362t 2 = 18

1 + t 2

.

We must determine where the curve crosses itself. The student was expected to show that

she knew how to find this crossing point systematically, not just by guessing or exam-ining a rough graph. The crossing point can be found by solving for distinct parameter

values t 1 t 2 the equations x = 18t 1 − 8t 31

= 18t 2 − 8t 32

, y = 18t 21

= 18t 2. Collecting

terms and factorizing yields the system of equations

(t 1 − t 2)18 − 6

t 21 + t 1t 2 + t 22

= 0 ,

18 (t 1 − t 2) (t 1 + t 2) = 0 .

8/15/2019 Brown's Notes 2010



Since we are looking for a solution where t 1 t 2, we may divide by t 1 − t 2, which cannot

equal 0, and obtain the system3 −

t 21 + t 1t 2 + t 22

= 0 ,

t 1 + t 2 = 0 .

From the second equation we see that t 2 = −t 1, and then the first equation yields 3 = t 21

,

so the solutions are t 1 = −t 2 = ±√

3; we may take the loop as beginning with parameter

value −√

3 and ending with parameter value +√

3. The length of the arc will be +√

3

−√

3

181 + t 2

dt

= 2 √

3

0

18

1 + t 2

dt

since the integrand is even and the interval is symmetric around 0

= 36

t +

1

3t 3√

3

0

= 72√

3 .

Wednesday version

1. [10 MARKS] Showing all your work, find the length of the curve y = ln sec x. Simplify

your answer as much as possible.

Solution:dy

dx=

1

sec x· (sec x tan x) = tan x

⇒

1 +

dy

dx

2

=

1 + tan2 x = | sec x| .

In the following integral I will drop the absolute signs because the secant is positive over

the entire interval of integration; the length is

π

4

−π4

| sec x| dx = π

4

−π4

sec x dx

= [ln | sec x + tan x|]π4

− π4

= ln(√

2 + 1) − ln(√

2 − 1)

= ln

√ 2 + 1√ 2 − 1

= ln (

√ 2 + 1)2

2 − 1

= ln

(√

2 + 1)2

= 2 ln(√

2 + 1) .

8/15/2019 Brown's Notes 2010



It would be sufficient for a student to obtain the diff erent of logarithms above. The

subsequent steps simplify the argument, and would be useful if the user did not have theuse of a calculator.

2. [10 MARKS] The curve y = 3√

x (1 ≤ y ≤ 3) is rotated about the y-axis. Showing all

your work, find the area of the resulting surface.

Solution: The data are given partly in terms of x and partly in terms of y, so some care is

needed. Since the limits are given in terms of y, I will integrate with respect to y; it will

be convenient to rewrite the equation of the curve as x = y3.

dx

dy= 3 y2

⇒

1 +

dxdy

2

=

1 + 9 y4

Area =

3

1

1 + 9 y4 · 2π y3 dy

=

1

9 · 1

4 · 2π · 2

3

1 + 9 y4

32

3

1

= π

27

730

32 − 10

32

.

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the inner

loop of the curve r = 9 + 18 sin θ .Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve is

traced out as θ passes through an interval of that length. If, for example, we consider the

interval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π6

and

at θ = 11π6

. Between these values the smaller loop is traced out; the larger loop is traced

out, for example, for −π6 ≤ θ ≤ 7π

6 . We can find the area of the small loop by integrating

between the appropriate limits; the area is

1

2

11π6

7π6

(9 + 18 sin θ )2 d θ

= 812

11π6

7π6

(1 + 4sin θ + 4 sin2 θ ) d θ

= 81

2

11π6

7π6

(1 + 4sin θ + 2(1 − cos2θ )) d θ

= 81

2

11π6

7π6

(3 + 4sin θ − 2cos2θ ) d θ

8/15/2019 Brown's Notes 2010



= 81

2 [3θ

−4cos θ

−sin2θ ]

11π6

7π6

= 81

2

11π

2 − 4cos

11π

6 − sin

11π

3

− 81

2

7π

2 − 4cos

7π

6 − sin

7π

3

= 81

2

11π

2 − 2

√ 3 − −

√ 3

2

− 81

2

7π

2 + 2

√ 3 −

√ 3

2

=

81

2 (2π − 3

√ 3) = 81π − 243

√ 3

2 .

4. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating

the curve x = 3t − t 3, y = 3t 2 (0 ≤ t ≤ 4) about the x-axis.

Solution:

dx

dt = 3 − 3t 2

dy

dt = 6t

dx

dt

2

+

dy

dt

2

=

32

1 − t 22

+ 62t 2 = 31 + t 2

.

The area of the surface of revolution about the x-axis will be 4

0

3

1 + t 2· 2π

3t 2

dt = 18π

4

0

t 2 + t 4

dt

= 18π

1

3t 3 +

1

5t 54

0

= 6 × 64 × 53

5 =

20352

5 .

5. [10 MARKS] Showing all your work, determine the value of c, if it is known that

∞n=2

(5+

c)−n = 2.

Solution: We are told that the geometric series converges; this implies that its common

ratio is less than 1 in magnitude, i.e., that |5 + c| < 1, which implies that −1 < 5 + c <1, equivalently, that −6 < c < −4. The sum of the geometric series on the left is

1(5+c)2 · 1

1− 15+c

= 1(5+c)(4+c)

. Equating this to 2, we obtain c2 + 9c + 18 = 0, implying that

(c + 3)(c + 6) = 0, so c = −3, −6. Of these two values, −6 lies outside of the admissible

interval, and would yield a divergent series. Thus c can be equal only to −3.

8/15/2019 Brown's Notes 2010



Thursday version

1. [10 MARKS] Showing all your work, find the exact length of the polar curve r = 7e4θ

(0 ≤ θ ≤ 2π).

Solution:

dr

d θ = 28e4θ

r 2 +

dr

d θ

2

=

72 + 282

e8θ

= 7√

17e4θ .

The length of the arc is then

7√

17

2π

0

e4θ d θ = 7

√ 17

4 e4θ

2π

0

= 7

√ 17

4

e8π − 1

.

2. [10 MARKS] The curve y = 4 − x2 (1 ≤ x ≤ 3) is rotated about the y-axis. Showing all

your work, find the area of the resulting surface.

Solution:

dy

dx= −2 x ⇒

1 +

dy

dx

2

=√

1 + 4 x2

Area =

3

1

√ 1 + 4 x2 · 2π x dx

= 2π · 1

4 · 1

2 · 2

3 ·

1 + 4 x2 3

2

3

1

= π

6 37

32 − 5

32

.

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the outer

loop of the curve r = 9 + 18 sin θ : this region will include the entire inner loop.

Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve is

traced out as θ passes through an interval of that length. If, for example, we consider the

interval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π6

and

at θ = 11π6

. Between these values the smaller loop is traced out; the larger loop is traced

8/15/2019 Brown's Notes 2010



out, for example, for −π6 ≤ θ ≤ 7π

6 . We can find the area of the large loop by integrating

between those limits, and the area will include the entire smaller loop. The area is

1

2

11π6

7π6

(9 + 18 sin θ )2 d θ

= 81

2

7π6

−π6

(1 + 4 sin θ + 4sin2 θ ) d θ

= 81

2

7π6

−π6

(1 + 4 sin θ + 2(1 − cos2θ )) d θ

= 81

2

7π6

−π6

(3 + 4 sin θ

−2cos2θ ) d θ

= 81

2 [3θ − 4cos θ − sin 2θ ]

7π6

− π6

= 81

2

7π

2 − 4cos

7π

6 − sin

7π

3

− 81

2

−π

2 − 4cos

π

6 + sin

π

3

= 81

2

7π

2 + 2

√ 3 −

√ 3

2

− 81

2

−π

2 − 2

√ 3 +

√ 3

2

= 162π +

243√

3

2 .

4. [10 MARKS] Showing all your work, sum a series in order to express the following

number as a ratio of integers: 4.645 = 4.645454545 . . ..

Solution:

4.645 = 4.6 + 1

10

45

100

1 +

1

100 +

1

1002 +

1

1003 + . . .

= 4.6 +

451000

1 − 1100

= 46

10

+ 1

10 ·

45

99

= 511

110

.

5. [10 MARKS] Showing all your work, find an equation for the tangent to the curve x =

cos θ + sin7θ , y = sin θ + cos2θ (−∞ < θ < +∞) at the point corresponding to θ = 0.

Solution: The slope of the tangent is

dx

d θ = − sin θ + 7cos7θ

8/15/2019 Brown's Notes 2010



dy

d θ = cos θ

−2 sin θ

dy

dx=

dy

d θ dx

d θ

= cos θ − 2sin θ

− sin θ + 7cos7θ =

1

7 when θ = 0 .

The line through ( x(0), y(0)) = (1, 1) with slope 17

has equation y − 1 = 1

7( x − 1), i.e.,

x − 7 y = −6.

Problems prepared but not used

1. The curve x =

c2 − y2 (0 ≤ y ≤ c2 ) is rotated about the y-axis. (c is a fixed real number.)

Showing all your work, find the area of the resulting surface.

Solution:

dx

dy=

1

2 · 1

c2 − y2· (−2 y) = − y

c2 − y2 1 +

dx

dy

2

=

1 +

y2

c2 − y2 =

|c| c2 − y2

Area =

c2

0 |c

| c2 − y2 · 2π

c

2

− y2

dy

= 2π|c| c

2

0

dy = πc2 .

2. Showing all your work, find the slope of the tangent line to the curve with equation in

polar coordinates r = 12

θ , at the point corresponding to θ = π.

Solution:

dr

d θ

=

−12

θ 2dy

dx=

sin θ · dr d θ

+ r · cos θ

cos θ · dr d θ − r · sin θ

= − sin θ + θ cos θ

− cos θ − θ sin θ .

At the point θ = π this ratio is equal to −π.

8/15/2019 Brown's Notes 2010



3. Showing all your work, find the area of one of the regions bounded by the line θ = π2

and

the closed curve r = 8 + 6sin θ .Solution: (The actual wording of the problem referred to a figure which it is not conve-

nient to include in these notes.) The region can be interpreted as being swept out by a

radius vector from the pole moving between −π2

and + π2

. The area is this

1

2

+ π2

− π2

(8 + 6sin θ )2 d θ =

+ π2

− π2

32 + 48 sin θ + 18 sin2 θ

d θ

=

+ π2

− π2

(32 + 48 sin θ + 9(1 − cos2θ )) d θ

= + π

2

− π2

(41 + 48 sin θ − 9cos2θ ) d θ

=

41θ − 48 cos θ − 9

2 sin 2θ

+ π2

− π2

= 41π .

4. Showing all your work, find the area enclosed by the curve (in polar coordinates) r =

9 + cos2θ .

Solution: The curve surrounds the pole, and is periodic with period 2π. The area may be

expressed as an integral over an interval of length 2π; for example, as

1

2 2π

0 (9 + cos2θ )

2

d θ =

1

2 2π

0 (81 + 18cos2θ + cos

2

2θ ) d θ

= 1

2

2π

0

(81 + 18cos2θ + 1 + cos4θ

2 ) d θ

= 1

2

81θ + 9sin2θ +

1

2θ +

1

8 sin 4θ

2π

0

= 163

4 · 2π =

163π

2 .

5. Showing all your work, find the exact length of the polar curve r = 4θ 2 (0 ≤ θ ≤ 2π).

Solution:

dr

d θ = 8θ

r 2 +

dr

d θ

2

=√

16θ 4 + 64θ 2

= 4|θ |√

4 + θ 2 .

8/15/2019 Brown's Notes 2010



Over the interval in question θ is positive, and the absolute signs may be dropped. The

length of the arc is 2π

0

4θ √

4 + θ s d θ = 4

3

(4 + θ 2)

32

2π

0

= 32

3

(1 + π2)

32 − 1

.

6. Showing all your work, find equations of the tangents to the curve x = 3t 2 + 4, y = 2t 3 + 3

that pass through the point (7, 5).

Solution: We might, in error think that we need first to determine the parameter value

associated with the given point. We would then solve the system of equations

3t 2 + 4 = 7

2t 3 + 3 = 5

to obtain t = +1. This would be an error. It happens that the given curve passes through

the point (7, 5), but that is fortuitous: we want the tangents to pass through the point, not

the curve! And we can’t find the points of contact of the tangents directly. So let’s first

determine the general tangent to the curve, at the point with parameter value t .

dx

dt = 6t

dydt

= 6t 2

dy

dx =

dy

dt dx

dt

= 6t 2

6t = t ,

so the slope of the tangent at the point (3t 2 + 4, 2t 3 + 3) on the curve is t ; the equation of

that tangent is y −2t 3 + 3

= t

x −

3t 2 + 4

, or

y = t x − t 3 − 4t + 3 . (115)

We now impose the condition that this line pass through the point ( x, y) = (7, 5), i.e., thatits equation be satisfied by ( x, y) = (7, 5), obtaining t 3 − 3t + 2 = 0, whose left member

factorizes to (t − 1)2(t + 2) = 0, so the points of contact of the tangents are t = 1 and

t = −2. The equations of the tangents through the given point are found by giving the

parameter t these two values in equation (115):

y = x − 2 and y = −2 x + 19 .

8/15/2019 Brown's Notes 2010



7. Showing all your work, use methods of polar coordinates to find the length of the polar

curve r = 15 sin θ 0 ≤ θ ≤

4π

15.

Solution:

dr

d θ = 15 cos θ

r 2 +

dr

d θ

2

= 15 .

The length of the arc is, therefore

4π15

015 d θ = 4π .

E.3 MATH 141 2009 01


Instructions to students









Monday Versions

1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find the

derivative of the function

f ( x) =

√ x

√ 3

sin(t )

t 5 dt

Solution:

8/15/2019 Brown's Notes 2010



• (This step may not be shown explicitly, but it underlies the successful implemen-

tation of the Chain Rule.) Introduction of an intermediate variable: If the new

variable / function is called u = u( x) =√

x, then f ( x) = d

du

u √

3

sin(t )

t 5 dt · du

dx

• application of the Fundamental Theorem

d

du

u √

3

sin(t )

t 5 dt =

sin(u)

u5 .

• completion

sin(√

x)

√ x

5 · 1

2 x−1

2 =

sin√

x

2 x3


(a) 0

−π/3

8sec2( x) dx

Solution: [5 MARKS TOTAL]

• state one antiderivative, e.g., 8 tan x

• indicate that the value of the integral is the net change, 8 tan x]0

−π

3• compute the final answer. Students should know the trigonometric functions

of simple submultiples of π.

(b) 1

0

√ x(5 x2 + 4 x − 5) dx


• state one antiderivative, here the obvious method is to express as a sum of

fractional powers and to integrate each separately:

5 x

5

2 + 4 x

3

2 − 5 x

1

2

dx = 5 · 2

7 x

7

2 + 4 · 2

5 x

5

2 − 5 · 2

3 x

3

2 + C

• indicate that the value of the integral is the net change,5 · 2

7 x

72 + 4 · 2

5 x

52 − 5 · 2

3 x

32

1

0

• compute the final answer correctly = 107

+ 85 − 10

3

8/15/2019 Brown's Notes 2010



Tuesday Versions



f ( x) =

e x

5

7 + ln6 t

t dt

Solution:



variable / function is called u = u( x) = e x

, then

f ( x) = d

du

u

5

7 + ln6 t

t dt · du

dx


d

du

u

5

7 + ln6 t

t dt =

7 + ln6 u

u.

• completion

7 + (ln(e x))6

e x · e x =

√ 7 + x6

e x · e x =

√ 7 + x6 .

While you may use some judgment about how much simplification you expect, I don’t

believe it would not be appropriate to accept a composition like ln(e x) not simplified.


(a)

0

−π/6

[8 sec( x) tan( x) + 7 cos( x)] dx


• state one antiderivative, e.g., 8 sec x + 7sin x

• indicate that the value of the integral is the net change in the antiderivative,

e.g., [8 sec x + 7sin x]0− π

6

8/15/2019 Brown's Notes 2010



• compute the final answer correctly.

(8 + 0) −

8 · 2√ 3

− 7 · 1

2

=

9

2 − 16√

3

(b) 4

1

−3 x−1 + 5 x + 3√ x

dx


• state one antiderivative, here the obvious method is to express as a sum of

fractional powers and to integrate each separately: −3 x−1 + 5 x + 3√ x

dx = −3 ·

−2

1

x− 1

2 + 5 · 2

3 x

32 + 3 · 2

1 x

12 + C

• indicate that the value of the integral is the net change,

−3 ·

−2

1

x− 1

2 + 5 · 2

3 x

32 + 3 · 2

1 x

12

4

1

• compute the final answer correctly

=

6

2 +

10

3 · 8 + 6 · 2

−

6 + 10

3 + 6

=

97

3

Wednesday Versions



f ( x) =

ln x

5

et √

1 + t 2 dt

Solution:



variable / function is called u = u( x) = ln x, then

f ( x) = d

du

u

5

et √

1 + t 2 dt · du

dx

8/15/2019 Brown's Notes 2010




d

du

u

5

et √ 1 + t 2 dt = eu √

1 + u2 .

• completion

eln x

1 + (ln x)2 · 1

x=

1 + (ln x)2

It is essential that eln x be simplified to x for full marks in this part.


π4

− π6

f ( x) dx, where

f ( x) =

4 sin x if x ≤ 05sec x tan x if 0 < x < π

2

Solution:

• decompose the interval into subintervals matching the intervals where the 2 parts

of the definition apply: π4

− π6

f ( x) dx =

0

− π6

f ( x) dx +

π4

0

f ( x) dx

• matching the diff erent functions to the appropriate subintervals: 0

− π6

f ( x) dx +

π4

0

f ( x) dx =

0

− π6

4 sin x dx +

π4

0

5sec x tan x dx

• shift the constants outside of the integration: 0

− π6

4 sin x dx +

π4

0

5sec x tan x dx = 4

0

− π6

sin x dx + 5

π4

0

sec x tan x dx

• find antiderivatives for both of the 2 integrands, e.g., − cos x and sec x

• indicate that the value of each integral is the net change,

4[− cos x]0− π

6+ 5[sec x]

π4

0

• correctly complete the computations

4

− cos0 + cos

π

6

+5

sec

π

4 − sec0

= 4

−1 +

√ 3

2

+5(√

2−1) = 2√

3+5√

2−9

8/15/2019 Brown's Notes 2010



Thursday Versions



f ( x) =

7 x

3 x

√ 8 + 9t 2

t dt

Solution:

• The integral must be split into two, at a convenient place, each integral with one

fixed and one variable limit; note that the point where the integral is split CANNOT

BE 0, since the integrand is undefined there:

f ( x) =

7 x

3 x

√ 8 + 9t 2

t dt =

1

3 x

√ 8 + 9t 2

t dt +

7 x

1

√ 8 + 9t 2

t dt

• one integral must be reversed so that the dependence on x is in the upper limit:

f ( x) = − 3 x

1

√ 8 + 9t 2

t dt +

7 x

1

√ 8 + 9t 2

t dt

• diff erentiate each of the integrals separately, using the Fundamental Theorem, and

multiply by the factor of the form dudx

from the Chain Rule (see problems on earlier

versions)

d

dx f ( x) = −

8 + 9(3 x)2

3 x· d (3 x)

dx+

8 + 9(7 x)2

7 x· d (7 x)

dx

• completion

−

8 + 9(3 x)2

3 x· 3 +

8 + 9(7 x)2

7 x· 7 =

−√

8 + 81 x2 +

8 + 9(49) x2

x

2. [10 MARKS] Compute √

3

0

f ( x) dx, where

f ( x) =

3 x if 0 ≤ x ≤ 1

61+ x2 if x > 1

Solution:

8/15/2019 Brown's Notes 2010



• for decomposing the interval into subintervals matching the intervals where the 2

parts of the definition apply: √ 3

0

f ( x) dx =

1

0

f ( x) dx +

√ 3

1

f ( x) dx

• for matching the diff erent functions to the appropriate subintervals:

√ 3

0

f ( x) dx =

1

0

3 x dx +

√ 3

1

6

1 + x2 dx

• for shifting constants outside of the integration:

1

0

3 x dx +

√ 3

1

6

1 + x2 dx =

3

2

1

0

2 x dx + 6

√ 3

1

1

1 + x2 d x

• for finding antiderivatives for both of the 2 integrands, e.g., x2 and arctan x

• for indicating that the value of each integral is the net change,

3

2[ x2]1

0 + 6[arctan x]√

31

• for correctly completing the computations

3

2[ x2]1

0+6[arctan x]√

31

= 3

2(1−0)+6

arctan

√ 3 − arctan 1

=

3

2+6

π

3 − π

4

=

3

2+

π

2






2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.

All submissions should carry your name and student number.



8/15/2019 Brown's Notes 2010



Monday Versions


(a) sec2 x

1 + 6 tan xdx

(b) 9/2

1/2

e√

2 x

√ 2 x

dx

Solution:

(a) [4 MARKS] for this indefinite integral

• [1 MARK] for stating the substitution

• [1 MARK] for rewriting the indefinite integral in terms of the new variable

• [1 MARK] for finding an antiderivative in terms of the new variable

• [1 MARK] for restating the antiderivative in terms of the original variable

(b) [6 MARKS] for this definite integral

• [1 MARK] for stating the substitution

• [3 MARKS] for transforming the definite integral, including the upper andlower limit

• [1 MARK] for finding an antiderivative

• [1 MARK] for the final answer

(Some students may, instead, find an antiderivative [4 MARKS] and then find the

net change [2 MARKS].)

2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtained

by revolving the region bounded by the x-axis, the lines x = π

6 and x =

π

3, and the curve

y =√

4cos x.

Solution: It was intended that students solve this problem using the “Method of Wash-

ers”. A solution using Cylindrical Shells would certainly be acceptable, but would be

more difficult, as students do not yet know how to integrate arccos y, and may not have

mastered integration by parts. If they complete part of such a solution, allocate the marks

similarly to the scheme for Washers.

• [4 MARKS] for the integrand

8/15/2019 Brown's Notes 2010



• [2 MARKS] for the limits of integration

• [2 MARKS] for finding an antiderivative

• [2 MARKS] for completing the integration.

3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = π4

and x = π2

, and the curve y = 9sin x. Compute the volume of the solid of revolution

obtained by revolving the region R about the y-axis. Hint: Use the method of cylindrical

shells.

Solution:

• [3 MARKS] for determining the integrand correctly

• [2 MARKS] for determining the limits of integration correctly• [4 MARKS] for applying integration by parts and correctly determining the full

antiderivative

• [1 MARK] for apparently completing the integration correctly

Tuesday Versions


(a)

sec x tan x−7 − 8sec x

dx

(b)

e4

e

cos(9 ln x)

x dx

Solution:

(a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions.

(b) [6 MARKS] Same scheme as for Problem 1(b) on Monday Versions.

2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtained

by revolving the region bounded by the lines y = 1, x = −π

4, x =

π

4, and the curve

y = 3 sec x.

Solution: Same scheme as for Problem 2 on Monday Versions.

8/15/2019 Brown's Notes 2010



3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = 1

and x = 5, and the curve y = ln(5 x). Compute the volume of the solid of revolutionobtained by revolving the region R about the y-axis. Hint: Use the method of cylindrical

shells.

Solution:

• [3 MARKS] for determining the integrand correctly

• [2 MARKS] for determining the limits of integration correctly

• [4 MARKS] for applying integration by parts and correctly determining the full

antiderivative

• [1 MARK] for apparently completing the integration correctly

Wednesday Versions


(a) 9 x + 2√

9 x2 + 4 xdx

(b)

1

0

e3 x

e6 x + 1 d x

Solution:

(a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions.


2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained

by revolving the region bounded by the y-axis, the lines y = ln3 and y = ln 4, and the

curve y = ln x

2.

Solution: Note that this is a solid of revolution about the y-axis. To use the Method

of Washers, which is intended, students will have to rewrite the equation of the curve

in the form x = 2e y. A correct solution using the Method of Cylindrical Shells would

certainly be acceptable. Follow the same grading scheme as shown above for Question

2 of Monday Versions.

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010




2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained

by revolving the region bounded by the lines x = 1 and y = 2 and the curve y = 3

x.

Solution: Note that this is a solid of revolution about the y-axis; moreover, it has a hole

in the middle. It was intended that students solve this problem using the “Method of

Washers”. However, the Method of Cylindrical Shells would be acceptable, and is not

difficult.

• [4 MARKS] for the integrand

• [2 MARKS] for the limits of integration; note that the data given are partly in terms

of x-coordinates and partly in terms of y-.• [2 MARKS] for finding an antiderivative

• [2 MARKS] for completing the integration.

3. [10 MARKS] Find the average value of the function

x√ 3 + x

on the interval [−2, 1].

Solution:

• [2 MARKS] for setting up the integral correctly, with correct integrand and limits

of integration

• [7 MARKS] for the evaluation of this integral — more than one method is feasible:

Integration by Parts: – [2 MARKS] for a correct selection of u and dv

– [2 MARKS] for determining du and v

– [2 MARKS] for applying integration by parts

– [1 MARK] for the integration of

v du.

Substitution: – [2 MARKS] for selection of an appropriate substitution u =

u( x)

– [2 MARKS] for transforming the integrand correctly into terms of u

– [2 MARKS] for correctly changing the limits of integration into terms of

the new variable

– [1 MARKS] for correctly evaluating the new definite integral

• [1 MARK] for dividing the weighted integral by the length of the interval, and

obtaining the final answer.

8/15/2019 Brown's Notes 2010













Monday Versions

1. [10 MARKS]

(a) Use integration by parts to compute the integral x3/5 ln x dx.

(b) Make a substitution and then use integration by parts to compute the integral

x3e x2

dx .

Solution:

(a) [4 MARKS]

• [2 MARKS] for a correct choice of u and dv and correctly determining du and

v

• [2 MARKS] for correctly implementing the selection of u and v and complet-

ing the integration correctly

(b) [6 MARKS]

• [1 MARK] for correctly implementing an appropriate substitution


v• [2 MARKS] for correctly implementing the selection of u and v and complet-

ing the integration in terms of the new variable

• [1 MARK] for expressing the final, correct answer in terms of the original

variable

8/15/2019 Brown's Notes 2010



2. [10 MARKS] Use a trigonometric substitution to compute 1

( √ 25 − x2

)3

dx . Verify

your answer by diff erentiating it!

Solution:

• [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do).

(Strictly speaking, the substitution should be expressed first in terms of an inverse

sine or inverse cosine, but it is common practice not to make that step explicit, so

one can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral in

terms of the the square of the secant or cosecant.

• [2 MARKS] for correctly integrating in terms of the new variable• [2 MARKS] for transforming the integral into terms of the original variable x.

• [2 MARKS] for correctly diff erentiating the antiderivative and thereby obtaining

the original integrand

item [10 MARKS] Find the arc length of the parameterized curve x(t ) = e2t + e−2t , y(t ) =

4t − 2 . for t between 0 and 1

2.

Solution: Grading instructions:

• [1 MARKS] for an integral of the correct form

• [4 MARKS] for correctly computing the derivatives of x and y

• [3 MARKS] for correctly finding an antiderivative

• [2 MARKS] for correctly completing the evaluation of the integral.

Tuesday Versions

1. [10 MARKS]

(a) Use integration by parts to compute the integral

x sec2(5 x) dx

(b) Make a substitution and then use integration by parts to compute the integral x3 cos( x2) dx

8/15/2019 Brown's Notes 2010



Solution:

(a) [4 MARKS]


v

• [2 MARKS] for correctly implementing the selection of u and v and complet-

ing the integration correctly

(b) [6 MARKS]

• [1 MARK] for correctly implementing an appropriate substitution


v

• [2 MARKS] for correctly implementing the selection of u and v and complet-ing the integration in terms of the new variable

• [1 MARK] for expressing the final, correct answer in terms of the original

variable

2. [10 MARKS] Use a trigonometric substitution to compute

x2

√ 4 − x2

dx . Verify your

answer by diff erentiating it!

Solution:

• [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do).

(Strictly speaking, the substitution should be expressed first in terms of an inversesine or inverse cosine, but it is common practice not to make that step explicit, so

one can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral in

terms of the the square of the secant or cosecant.

• [2 MARKS] for correctly integrating in terms of the new variable

• [2 MARKS] for transforming the integral into terms of the original variable x. This

antiderivative does not include any term with plus / or / minus: there are no ambigu-

ities of signs! If a student shows an ambiguity, this means he has not properly

completed the diff erentiation of the next part, since only one of the signs will yield

the correct derivative.

• [2 MARKS] for correctly diff erentiating the antiderivative and thereby obtaining

the original integrand

3. [10 MARKS] Find the arc length of the parameterized curve x(t ) = −3 + e2t cos t , y(t ) =

5 + e2t sin t for t between 0 and 1

2.

8/15/2019 Brown's Notes 2010



Solution: Grading instructions:





Wednesday Versions

1. [10 MARKS] Compute the integral

e9π

1

cos(ln( x)) dx. Hint: A solution by integration

by parts could begin from the observation that cos(ln( x)) = 1·

cos(ln( x)). You could also

apply integration by parts after making a substitution.

Solution: There appear to be several ways of attacking this problem, but the attacks will

require 2 applications of integration by parts, followed by the solving of an equation.

Applying Integration by Parts immediately: • [1 MARK] for a correct selection

of u and dv for the first integration by parts

• [1 MARKS] for correctly determining du and v

• [2 MARKS] for correctly applying integration by parts and expressing the

given integral as the value of uv] minus a second integral, which will then

require a second application of integration by parts

• [1 MARK] for a correct selection of U and dV for the second integration by

parts

• [1 MARKS] for correctly determining dU and V


original integral as a sum of [uv + UV ] minus the same integral

• [2 MARKS] for solving the equation for the desired integral and completing

all calculations apparently correctly

Preceding Integration by Parts by a Substitution: • [0 MARKS] for selecting a cor-

rect substitution, and implementing that substitution correctly both in the inte-

grand and the limits of integration, so that the integral is now written in a form

where the use of integration by parts is well indicated.

• [1 MARK] for a correct selection of u and dv for the first integration by parts

• [1 MARKS] for correctly determining du and v


given integral as the value of uv] minus a second integral, which will then

require a second application of integration by parts

8/15/2019 Brown's Notes 2010



• [1 MARK] for a correct selection of U and dV for the second integration by

parts• [1 MARKS] for correctly determining dU and V


original integral as a sum of [uv + UV ] minus the same integral

• [2 MARKS] for solving the equation for the desired integral and completing

all calculations apparently correctly


13

( x + 3)( x2 + 4) d x .

Solution:

• [2+2 MARKS] for correctly factorizing the denominator and expressing the needto expand the function into a sum of 2 partial fractions, one with a linear denomi-

nator, the other having a general numerator of degree 1 and denominator of degree

2. Reserve a full 2 MARKS for the numerator of the fraction with the quadratic

denominator.

• [3 MARKS] for determining correctly the 3 undetermined constants

• [3 MARKS] for completing the integration correctly

3. [10 MARKS] Find the surface area of the solid of revolution obtained by revolving the

graph of the parametric curve x(t ) = 2t − 3, y(t ) = t 2 − 3t − 1, 0 ≤ t ≤ 3/2 about the

y-axis.Solution: Grading instructions:





Thursday Versions

1. [10 MARKS] Use a substitution and then integration by parts to compute the integral 6

3

e3/ x

x3 dx .

Solution:

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



F Final Examinations from Previous Years

F.1 Final Examination in Mathematics 189-121B (1996 / 1997)

1. [4 MARKS] Find the derivative of the function F defined by

F ( x) =

x4

x2

sin√

t dt .

2. [4 MARKS] Evaluate

π

−π2

f ( x) dx , where

f ( x) =

cos x, −π

2 ≤ x ≤ π

33π x + 1, π

3 < x ≤ π

.


x sin3 x2 cos x2 dx .


( x5 + 4− x) dx .

5. [10 MARKS] Calculate the area of the region bounded by the curves x = y2 and

x − y = 2 .

6. [10 MARKS] The region bounded by f ( x) = 4 x − x2 and the x-axis, between

x = 1 and x = 4 , is rotated about the y-axis. Find the volume of the solid that is

generated.


x ln x dx .


sin2 x cos5 x dx .

9. [6 MARKS] Determine the partial fraction decomposition of the following ratio of poly-

nomials: x5 + 2

x2 − 1 .

10. [4 MARKS] Determine whether or not the following sequence converges as n → ∞ .

If it does, find the limit: 1 +

x

n

3n

.

8/15/2019 Brown's Notes 2010



11. [4 MARKS] Determine the following limit, if it exists:

lim x→0+

√ x√ x + sin

√ x

.

12. [6 MARKS] Determine whether the series

∞k =2

ke−k 2 converges or diverges.

13. [6 MARKS] Test the following series for

(a) absolute convergence,

(b) conditional convergence.

∞k =10

(−1)k

√ k (k + 1)

.

14. [10 MARKS] Find the area of the region that consists of all points that lie within the

circle r = 2 cos θ , but outside the circle r = 1 .

15. [10 MARKS] Determine the length of the curve

r = 5(1 − cos θ ) , (0 ≤ θ ≤ 2π) .


1. [10 MARKS]

(a) Sketch the region bounded by the curves

y = x2 and y = 3 + 5 x − x2 .

(b) Determine the area of the region.

2. [10 MARKS] The triangular region bounded by the lines

y = x , y = 3

2 − x

2 , and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolution which

is generated.

8/15/2019 Brown's Notes 2010



3. [10 MARKS] Find the length of the curve y = x2

2 − ln

4√

x from x = 1 to

x = 2 .

4. [5 MARKS] Determine, at x = 12

, the value of the function sin−1 x and the slope of its

graph.

5. [5 MARKS] Evaluate lim x→2

x3 − 8

x4 − 16 .

6. [5 MARKS] Showing all your work, evaluate lim x→0+

x x .

7. [5 MARKS] Evaluate x3e− x2

dx .


x3 − 1

x3 + xdx .


x3

√ 1 − x2

dx , where | x| < 1 .

10. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2cos θ

and outside the circle r = 2 .

11. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f ( x) = x

0e

t (1−

t )

dt at x = 0 .

12. [5 MARKS] Showing all your work, determine whether the following infinite series

converges or diverges. If it converges, find its sum.

∞n=0

3n − 2n

4n

13. [5 MARKS] Showing all your work, determine whether or not the following series con-

verges:∞

n=1

2

1n

n2

14. [5 MARKS] Showing all your work, determine whether the following series converges:

∞n=1

1

n · 2n

8/15/2019 Brown's Notes 2010



F.3 Supplemental / Deferred Examination in Mathematics 189-141B (1997 / 1998

1. [10 MARKS]

(a) Sketch the region bounded by the curves

y = 8

x + 2 and x + y = 4 .

(b) Determine the area of the region.

2. [10 MARKS] The triangular region bounded by the lines

y =

x , y =

3

2 − x

2 , and y =

0

is revolved around the line y = 0. Determine the volume of the solid of revolution which

is generated.

3. [10 MARKS] Find the area of the surface of revolution generated by revolving the curve

y = 1

2

e x + e− x (0 ≤ x ≤ 1)

about the x-axis.

4. [5 MARKS] Determine, at x =

1

2 , the value of the function cos−1

x and the slope of itsgraph.

5. [10 MARKS] Evaluate lim x→2

x − 2cos π x

x2 − 4 .

6. [5 MARKS] Evaluate lim x→∞

cos

1

x2

x4

.


e2 x

1 + e4 x d x .


x2 cos x dx .


x3 − 1

x3 + xdx .


√ a2 − u2 du , where |u| < a.

8/15/2019 Brown's Notes 2010



11. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2cos θ

and outside the circle r = 2 .

12. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f ( x) =

x

0

es(1−s)ds at x = 0 .


converges or diverges. If it converges, find its sum.

∞n=0

1 + 2n + 3n

5n

14. [5 MARKS] Showing all your work, determine whether or not the following series con-

verges.∞

n=1

ln n

n

15. [5 MARKS] Showing all your work, determine whether the following series convereges.

∞n=1

n2 + 1

en(n + 1)2


1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and ( y−1)2 = 5− x.

2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about the

line x = 1 the region bounded by the curve ( x − 1)2 = 5 − 4 y and the line y = 1 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0

the region bounded by the curves

y = sin x

y = −2 x = 0

and x = 2π .

4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = x2 (0 ≤ x ≤

√ 2) about the y-axis.

8/15/2019 Brown's Notes 2010



5. Define the function F by F ( x) =

x

0

et 3 dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the following

inequalities are true.

e < F (e) < ee3+1 .

(b) [4 MARKS] Determine the value of d

dxF ( x3) at each of the

following points:

i. at x = 0 .

ii. at x = 2 .

6. [4 MARKS] Showing all your work, evaluate sin3 π x dx .

7. [4 MARKS] Showing all your work, evaluate x2e− x dx .

8. [4 MARKS] Showing all your work, evaluate x − 1

x3 − x2 − 2 xdx .

9. [4 MARKS] Showing all your work, evaluate x3 + x2 + x − 1

x2 + 2 x + 2 dx .

10. [8 MARKS] Find the area of the region inside the curve r = 3 sin θ and outside the curve

r = 2 − cos θ .


or divergent:

(a) [4 MARKS]

∞ 0

sin x dx .

8/15/2019 Brown's Notes 2010



(b) [4 MARKS]

2

0

dx

1 − x2 .

12. Showing all your work, determine whether each of the following sequences is convergent

or divergent.

(a) [4 MARKS]

n sin

π

n

(b) [4 MARKS]

(2 n + 1) e−n

13. Showing all your work, determine whether each of the following infinite series is con-

vergent or divergent:

(a) [4 MARKS]∞

n=1

1

4n3 .

(b) [4 MARKS]

∞n=1

1

n+

1

n2

.

14. Showing all your work, determine whether each of the following series is convergent,

divergent, conditionally convergent and / or absolutely convergent.

(a) [4 MARKS]

∞

n=1

(−1)n n + 2

n(n + 1) .

(b) [4 MARKS]∞

n=1

(−1)n cos n

n2 .


1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and y = 6 − x.

2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about the

line x = 0 the region bounded by the curve y = 4 − x2 and the lines x = 0 and

y = 0 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0the region bounded by the curves

y = sin x

y = 2

x = 0

and x = 2π .

8/15/2019 Brown's Notes 2010



4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = − x2 (0 ≤ x ≤ √ 2) about the y-axis.

5. Define the function F by F ( x) =

x 0

sin10 t dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the following

inequalities are true.

0 < F (e) < e .

(b) [4 MARKS] Determine the value of d

dxF ( x) at each of the

following points:

i. at x = 0 .

ii. at x = π

2 .

6. [8 MARKS] Showing all your work, evaluate x5e− x2

dx .

7. [4 MARKS] Showing all your work, evaluate

x

3

− x2

+ x + 1 x2 − 2 x + 2

dx .

8. [8 MARKS] Find the area of the region inside the curve r = 6 sin θ and outside the curve

r = 4 − 2 sin θ .


or divergent:

(a) [4 MARKS]

∞

0

cos x dx .

(b) [4 MARKS]

4 0

dx

4 − x2 .

10. Showing all your work, determine whether each of the following sequences is convergent

or divergent.

8/15/2019 Brown's Notes 2010



(a) [4 MARKS] n sin π

n(b) [4 MARKS]

(2 n + 1) e−n

11. Showing all your work, determine whether each of the following infinite series is con-

vergent or divergent:

(a) [4 MARKS]

∞n=1

1

4n5 .

(b) [4 MARKS]

∞n=1

1

n− 1

n3

.


1. [11 MARKS] Find the area of the region bounded by the curves x = y2 and x =

− y2 + 12 y − 16 .

2. [11 MARKS] Let C denote the arc of the curve y = cosh x for −1 ≤ x ≤ 1. Find

the volume of the solid of revolution generated by revolving about the line x = −2

the region bounded by C and the line y = e2 + 1

2e.

3. (a) [5 MARKS] Showing all your work, evaluate

92

32

√ 6t − t 2 dt .

(b) [6 MARKS] Showing all your work, evaluate

3π4

π4

√ 1 − sin u du .

4. (a) [7 MARKS] Showing all your work, determine a reduction formula which ex-

presses, for any integer n not less than 2, the value of

xn sin2 x dx in

terms of

xn−2 sin 2 x dx.

(b) [4 MARKS] Use your reduction formula to determine the indefinite integral

x2 sin2 x d

8/15/2019 Brown's Notes 2010



5. [11 MARKS] Showing all your work, evaluate 8 x2 − 21 x + 6

( x − 2)2( x + 2) d x .

6. [11 MARKS] Find the area of the region inside the curve

r = 1 + cos θ and outside the curve r = 1 − cos θ .

7. [11 MARKS] Determine whether the following integral is convergent or divergent. If it

is convergent, find its value. Show all your work.

3

0

1

( x − 1)

4

5

dx

8. [11 MARKS] Showing all your work, determine whether the following infinite series is

convergent or divergent:

∞n=1

n! e−(n − 1)2

.

9. Showing all your work, determine whether each of the following series is convergent,

divergent, conditionally convergent and / or absolutely convergent.

(a) [6 MARKS]

∞

n=1

(−1)n

√

n + 2 − √ n

.

(b) [6 MARKS]

∞n=1

(−1)n n

lnn2 .


1. [11 MARKS] Determine the area of the region bounded by the curves y = x4 and

y = 2 − x2 .

2. [11 MARKS] Determine the volume of the solid generated by rotating the region bounded

by the curves y = 2 x2 and y2 = 4 x around the x-axis.

3. Evaluate the integrals:

(a) [5 MARKS]

x7

√ 1 − x4

dx .

(b) [6 MARKS]

x2

√ 4 − x2

dx .

8/15/2019 Brown's Notes 2010



4. [11 MARKS] Showing all your work, find π/2

0

e2 x sin3 x dx .

5. [11 MARKS] Determine

6 x3 − 18 x

( x2 − 1)( x2 − 4) d x .

6. [11 MARKS] Find the area of the region inside the curve r = 2 + 2sin θ and outside

r = 2 .

7. [11 MARKS] Determine whether the following improper integral converges:

1

0

ln x

x2 dx .


converges:

∞n=1

1√

15n3 + 3.

9. Showing all your work, determine, for each of the following series, whether it is conver-

gent, divergent, conditionally convergent and / or absolutely convergent.

(a) [6 MARKS]

∞n=1

(−1)n ln n

n.

(b) [6 MARKS]

∞n=1

cos nπ

n.


1. Showing all your work, determine, for each of the following infinite series, whether or

not it converges.

(a) [3 MARKS]

∞i=1

n

n3 + 1 .

(b) [3 MARKS]

∞n=1

ln

n

3n + 1

.

(c) [6 MARKS]∞

n=2

(−1)n(3n + 1)4

5n .

2. [12 MARKS] Determine the volume of the solid of revolution generated by revolving

about the y-axis the region bounded by the curves

y = e− x2

,

8/15/2019 Brown's Notes 2010



y = 0 ,

x = 0 , x = 1 .

3. [12 MARKS] Determine the area of the surface of revolution generated by revolving

about the x-axis the curve

y = cos x ,

0 ≤ x ≤ π

6

.

[Hint: You may wish to make use of the fact that

2

sec3

θ d θ = sec θ tan θ + ln | sec θ + tan θ | + C .]

4. [12 MARKS] Find the area that is inside the circle

r = 3 cos θ and outside the curve r = 2 − cos θ .

5. [14 MARKS] Evaluate the integral x

( x − 1)( x2 + 4) d x .

6. For the curve given parametrically by x = t 3 + t 2 + 1 , y = 1

−t 2 , determine

(a) [6 MARKS] The equation of the tangent line at the point

( x, y) = (1, 0) , written in the form y = mx + b , where m and b are con-

stants;

(b) [6 MARKS] the value of d 2 y

dx2 at the point ( x, y) = (1, 0) .

7. (a) [10 MARKS] Use integration by parts to determine the value of e x cos x dx .


0

−∞e x cos x dx .

8. [12 MARKS] Find the area of the region bounded by the curves y = x2 − 4 and

y = −2 x2 + 5 x − 2 .

8/15/2019 Brown's Notes 2010




1. (a) [6 MARKS] Showing all your work, find F (1) when

F (t ) =

2t

1

x

x3 + x + 7 dx .

(b) [6 MARKS] Showing all your work, evaluate 6

0 | x − 2| dx .

2. Showing all of your work, evaluate each of the following integrals:

(a) [4 MARKS]

x + 1√

9

− x2

dx;

(b) [4 MARKS]

12 x3 + x

dx;

(c) [4 MARKS]

sin2 2 x cos2 2 x dx;

(d) [4 MARKS]

ln x dx

3. [15 MARKS] Showing all your work, find the area of the region bounded below by the

line y = 1

2 , and above by the curve y =

1

1 + x2 .

4. [15 MARKS] Showing all your work, find the volume generated by revolving about the y-axis the smaller region bounded by the circle x2 + y2 = 25 and the line x = 4 .

5. Showing all your work,

(a) [2 MARKS] sketch the curve r = 1 − sin θ ;

(b) [6 MARKS] find the length of the portion of the curve that lies in the region given

by r ≥ 0 , −π

2 ≤ θ ≤ π

2 ;

(c) [5 MARKS] find the coordinates of the points on the curve where the tangent line

is parallel to the line θ = 0 .

6. For each of the following integrals, determine whether it is convergent or divergent; if itis convergent, you are expected to determine its value. Show all your work.

(a) [7 MARKS]

2

−1

1

x3 dx ;

(b) [7 MARKS]

∞

1

xe− x2

dx .

8/15/2019 Brown's Notes 2010



7. Showing all your work, determine, for each of the following series, whether or not it

converges:

(a) [5 MARKS]

∞n=2

1

n(ln n)2;

(b) [5 MARKS]

∞n=1

(−1)n

n2 − 1

n2 + 1

;

(c) [5 MARKS]

∞n=1

n + 1

3n .

F.10 Final Examination in Mathematics 189-141B (2001 / 2002)1. Showing all your work, evaluate each of the following indefinite integrals:

(a) [3 MARKS]

x3

√ 4 − x2

dx

(b) [3 MARKS]

1

y

ln ydy

(c) [3 MARKS]

sec u

1 + sec u· tan u du

(d) [3 MARKS]

e

t

1 + e2t dt

2. Let K denote the curve

y = x2 , (0 ≤ x ≤ 1) .

(a) [6 MARKS] Determine the area of the surface of revolution generated by revolving

K about the y-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolving

about the line y = 0 the region bounded by K and the lines x = 1 and

y = 0 .

3. Consider the arc C given by r = θ 2 (0 ≤ θ ≤ π).

(a) [4 MARKS] Express the length of C as a definite integral. Then evaluate the inte-

gral.

(b) [4 MARKS] Determine the area of the region subtended by C at the pole — i.e. of

the region bounded by the arc C and the line θ = 0.

8/15/2019 Brown's Notes 2010



(c) [4 MARKS] The given curve can be represented in cartesian coordinates paramet-

rically as x = θ 2

cos θ , y = θ 2

sin θ . Determine the slope of the tangent to this curveat the point ( x, y) =

0,

π2

2

.

4. [12 MARKS] Showing all your work, evaluate the integral 40 − 16 x21 − 4 x2

(1 + 2 x)

d x .


curves y = arctan x and 4 y = π x in the first quadrant.

6. (a) [4 MARKS] Showing all your work, determine the value of sin3 x cos2 x dx .

(b) [4 MARKS] Showing all your work, determine the value of tan4 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2

0

tan4 x dx .

7. [12 MARKS] Showing all your work, determine the value of

d 2

dx2

x

0

e2t

1

√ u + 1 du

dt

when x = 0.

8. Showing all your work, determine, for each of the following infinite series, whether it is

absolutely convergent, conditionally convergent, or divergent.

(a) [4 MARKS]

∞n=5

(−1)n n2 − 1

6n2 + 4 .

(b) [4 MARKS]

∞n=2

(

−1)n

n(ln n)2 .

(c) [4 MARKS]

∞n=2

(−1)n(n+1)

2

2n .

(d) [4 MARKS]

∞n=0

n + 5

2n .

8/15/2019 Brown's Notes 2010




1. Showing all your work, evaluate each of the following, always simplifying your answer

as much as possible:

(a) [3 MARKS]

e x sin x dx

(b) [3 MARKS]

12

0

sin−1 y 1 − y2

dy

equivalently,

12

0

arcsin y 1 − y2

dy

.

(c) [3 MARKS]

(u2 + 2u)e−u du

(d) [3 MARKS] 1 + cos t

sin t dt

2. Let K denote the curve

y =√

2 x − x2 ,

0 ≤ x ≤ 1

2

.

(a) [6 MARKS] Showing all your work, use an integral to determine the area of the

surface of revolution generated by revolving K about the x-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolving

about the line y =√

32

the region bounded by K and the lines x = 0 and

y =√

3

2 .

(You may assume that √ 2 x − x2 dx =

x − 1

2

√ 2 x − x2 +

1

2 arccos(1 − x) .)

3. A curve C in the plane is given by parametric equations

x = t 3 − 3t 2

y = t 3 − 3t .

(a) [6 MARKS] Showing all your work, determine all points ( x, y) on C where thetangent is horizontal.

(b) [6 MARKS] By determining the value of d 2 y

dx2 as a function of t , determine all

points ( x, y) on C at which the ordinate ( y-coordinate) is a (local) maximum, and

all points at which the ordinate is a (local) minimum.

8/15/2019 Brown's Notes 2010



4. [12 MARKS] Showing all your work, evaluate the indefinite integral 4 x3

x2 − 9

(3 x + 9) d x .


curves x − 2 y + 7 = 0 and y2 − 6 y − x = 0 .

6. (a) [6 MARKS] Showing all your work, evaluate sin4 x cos2 x dx .

(b) [4 MARKS] Showing all your work, evaluate tan5 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2

π4

tan5 θ d θ .

7. [12 MARKS] Showing all your work, determine the value of

d 2

dx2

x

0

π3

−2t

4 + sin(−2u) du

dt

when x = π4

. Your answer should be simplified, if possible.



(a) [4 MARKS]

∞n=5

(−1)n 1√ n + 1

.

(b) [4 MARKS]

∞

n=2

(

−1)2n

n(ln n)3 .

(c) [4 MARKS]

∞n=2

2n

1 + 5n

3n

.

(d) [4 MARKS]

∞n=1

sin

1n

cos

1n

· 1n

.

8/15/2019 Brown's Notes 2010



9. [10 MARKS] Prove or disprove the following statement: The point with polar coordi-

nates

r = 2(√

2 − 1)

θ = −π + arcsin((√

2 − 1)2)

lies on the intersection of the curves with polar equations

r 2 = 4sin θ,

r = 1 + sin θ .

You are expected to justify every statement you make, but you do not need to sketch the

curves.

F.12 Final Examination in MATH 141 2003 01

1. [10 MARKS] Find the area of the region bounded in the first quadrant by the curves

y = e x, y = e− x, y = e2 x−3 .

Simplify your answer as much as possible. (Your instructors are aware that you do not

have the use of a calculator.)

2. Showing all your work, evaluate each of the following indefinite integrals:

(a) [5 MARKS]

1

x2 + 2 x + 17 d x

(b) [5 MARKS]

ln(ln x)

xdx

3. [12 MARKS] For each of the following integrals,

(a) [2 MARKS] Explain why the integral is improper.

(b) [10 MARKS] Determine its value, or show that the integral does not converge.

Show all your work.

I 1 =

∞

2

2( x2 − x + 1)

( x − 1)( x2 + 1) d x , I 2 =

1

0

2( x2 − x + 1)

( x − 1)( x2 + 1) d x

4. Let Ω denote the region in the first quadrant bounded by the curves x =

16 + y2, y = 0,

x = 0, and y = 3.

8/15/2019 Brown's Notes 2010



(a) [5 MARKS] Showing all your work, determine the volume of the solid of revolu-

tion obtained by rotating Ω about the y-axis.(b) [7 MARKS] Showing all your work, determine the area of the surface of revolution

obtained by rotating the arc

x =

16 + y2, (0 ≤ y ≤ 3)

about the y-axis. You may assume that

d

d θ (sec θ tan θ + ln |sec θ + tan θ |) = 2 sec3 θ .

.

5. Consider the arc C given parametrically by x =

t

0

4(1 − cos θ )θ 2 d θ

y = cos t + t sin t

(−π ≤ t ≤ 2π) .

Showing all your work

(a) [4 MARKS] Find the slope of the tangent to C at the point with parameter value

t = 3π

2

.

(b) [6 MARKS] Find the length of C .

6. Give, for each of the following statements, a specific example to show that the statement

is not a theorem:

(a) [3 MARKS] If an∞n=0 is a sequence such that lim

n→∞an = 0, then

∞n=0

an converges.

(b) [3 MARKS] If the series∞

n=0

an and∞

n=0

bn are both divergent, then∞

n=0

(an + bn) is

divergent.

(c) [3 MARKS] If a series ∞n=0

an converges, then ∞n=0

a2n converges.



(a) [4 MARKS]

∞n=0

(−1)n

4n2 + 1 .

8/15/2019 Brown's Notes 2010



(b) [4 MARKS]

∞

n=2

n − 1

n

n2

(c) [4 MARKS]

∞n=2

1√ n(n + 1)

.

8. [10 MARKS] Showing all your work, determine the area of the part of one “leaf” of the

“4-leafed rose” r = 2 cos(2θ ) that is inside the circle r = 1.

F.13 Supplemental / Deferred Examination in MATH 141 2003 01

1. [10 MARKS] Find the area of the region bounded by the curves

y = e x − 1, y = x2 − x, x = 1.

2. Showing all your work, evaluate each of the following:

(a) [5 MARKS]

e

1 x

x2 dx

(b) [5 MARKS]

5

2

| x2 − 4 x| dx

3. [12 MARKS] For each of the following integrals,

(a) [2 MARKS] Explain precisely whether the integral is improper.

(b) [10 MARKS] Determine its value, simplifying as much as possible; or show that

the integral does not converge. (The examiners are aware that you do not have

access to a calculator.)

Show all your work.

I 1 =

1

0

2 x

( x − 1)( x2 + 1) d x , I 2 =

3

2

2 x

( x − 1)( x2 + 1) d x

4. Let Ω denote the region bounded by the curves y = sin x, y = 0, x = π2 , x = π.

(a) [6 MARKS] Showing all your work, determine the volume of the solid of revolu-

tion obtained by rotating Ω about the y-axis.

(b) [6 MARKS] Showing all your work, determine the volume of the solid of revolu-

tion obtained by rotating Ω about the x-axis.

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits, sums, inte-grals. If it does not exist write “DIVERGENT”.

(a)

∞n=1

1 + 2n−1

3n =

ANSWER ONLY

(b) ∞

−∞ xe x2

dx =

ANSWER ONLY

(c) The limit of the Riemann sum limn→∞

ni=1

2n

3 + 2i

n

2 − 6

3 + 2in

5

=

ANSWER ONLY

(d)

−∞

∞ xe− x2

dx =

ANSWER ONLY

(e)

∞n=3

4

(2n + 1)(2n + 3) =

ANSWER ONLY

2. BRIEF SOLUTIONS

[3 MARKS EACH] Simplify your answers as much as possible.

8/15/2019 Brown's Notes 2010



(a) For the point with polar coordinates 3, π7 give another set of polar coordinates

(r , θ ) in which r < 0 and θ > 2.

ANSWER ONLY

(b) Determine the length of the arc of the curve r = θ 2 from (0, 0) to (1, 1).

ANSWER ONLY

(c) A curve is given parametrically by x(t ) = t

0 e−u2

du, y(t ) = 4

t eu2

du. Find the

slope of the tangent to the curve at ( x(1), y(1)).

ANSWER ONLY

(d) Give a definite integral whose value is the area of the surface obtained by rotating

the curve x = y3

6 +

1

2 y

12 ≤ y ≤ 1

about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

(e) On the interval 0 ≤ x ≤ 4 the average value of the function

f ( x) =

√ 1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4 is

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



(e) sec3 x tan3 x dx =

ANSWER ONLY


Let R be the finite region bounded by the curves x = y2 and x = 4 − 3 y4.

(a) [5 MARKS] Find the area of R.

(b) [5 MARKS] Find the volume of the solid generated by revolving R about the y-axis.


[12 MARKS] Evaluate the definite integral

12

− 12

4 x

2 x2 − x − 2

( x2 + 1)( x2 − 1) dx .


(a) [4 MARKS] Show that, for any positive integer n, (ln x)2n dx = x(ln x)2n − 2n

(ln x)2n−1 dx

(b) [7 MARKS] Evaluate the integral 1

0

y 2 y − y2

dy .

7. SHOW ALL YOUR WORK![4 MARKS EACH] Determine for each of the following series whether it

• converges absolutely;

• converges conditionally; or

• diverges.

8/15/2019 Brown's Notes 2010



(a)

∞

n=1

1 + 1

n

2

e−n

(b)

∞n=10

(−1)n√

2n

1 + 2√

n

(c)

∞n=2

(−1)n ·√

n + 2 −√

n − 1

n

(d)

∞n=0

2π + cos n

6n

8. SHOW ALL YOUR WORK![6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ .

Another version

1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits, sums, inte-

grals. If it does not exist write “DIVERGENT”.

(a)

∞

n=1

1 + 3n−1

4n =

ANSWER ONLY

(b)

∞

−∞ ye y2

dy =

ANSWER ONLY


ni=1

2n

4 + 2i

n

2 − 7

4 + 2in

5

=

ANSWER ONLY

8/15/2019 Brown's Notes 2010



(d) −∞

∞ ye− y2

dy =

ANSWER ONLY

(e)

∞n=3

4

(2n − 1)(2n + 1) =

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) On the interval 0 ≤ x ≤ 4 the average value of the function

f ( x) =

√ 1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4 is

ANSWER ONLY

(b) For the point with polar coordinates

3, π5

give another set of polar coordinates

(r , θ ) in which r < 0 and θ > 2.

ANSWER ONLY

(c) Determine the length of the arc of the curve r = θ 2 from (0, 0) to (1, 1).

ANSWER ONLY

8/15/2019 Brown's Notes 2010



(d) A curve is given parametrically by x(t ) = t

0 e−v2

dv, y(t ) = 4

t ev2

dv. Find the

slope of the tangent to the curve at ( x(1), y(1)).

ANSWER ONLY

(e) Give a definite integral whose value is the area of the surface obtained by rotating

the curve x = y3

6 +

1

2 y

12 ≤ y ≤ 1

about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

3. BRIEF SOLUTIONS

[3 MARKS EACH] Give the value of each of the following indefinite integrals:

(a) sec3 x tan3 x dx =

ANSWER ONLY

(b)

x

5 x2 + 1 d x =

ANSWER ONLY

(c)

e x

√ 1 + e x dx =

8/15/2019 Brown's Notes 2010



ANSWER ONLY

(d)

(3 sin2 x − cos2 x) dx =

ANSWER ONLY

(e)

tan2 4 x dx =

ANSWER ONLY


Let S be the finite region bounded by the curves y = x2 and y = 4 − 3 x4.

(a) [5 MARKS] Find the area of S .

(b) [5 MARKS] Find the volume of the solid generated by revolving S about the x-

axis.


[12 MARKS] Evaluate the definite integral

1

2

−1

2

4 x

4 x2 − 2 x − 4

( x2 + 1)( x2

−1)

dx .


(a) [4 MARKS] Show that, for any positive integer m, (ln y)2m dy = y(ln y)2m − 2m

(ln y)2m−1 dy

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



(b) 0

−∞ xe x dx =

ANSWER ONLY


π

n

ni=1

sin2

iπ

n

=

ANSWER ONLY

(d)

∞n=0

4

(2n + 1)(2n + 3) =

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) For the point with polar coordinates (r , θ ) =−10π

3 , −π

6

give another set of polar

coordinates (r 1, θ 1) in which r 1 > 0 and θ 1 > 2 .

ANSWER ONLY

(b) Find all points — if there are any — where the curves

r = 1 − cos θ and r = −32

intersect.

8/15/2019 Brown's Notes 2010



ANSWER ONLY

(c) Find the exact length of the curve r = e2θ , (0 ≤ θ ≤ π) .

ANSWER ONLY

(d) On the interval ln 12 ≤ x ≤ π the average value of the function

f ( x) =

sinh x if ln 1

2 ≤ x ≤ 0

sin x if 0 < x ≤ π is

ANSWER ONLY

3. BRIEF SOLUTIONS

[3 MARKS EACH] Give the value of each of the following indefinite integrals:

(a)

14 x2 + 1

d x =

ANSWER ONLY

8/15/2019 Brown's Notes 2010



(b) dx

√ x2

− 25

=

ANSWER ONLY

(c)

(cos x + 1)(cos x − 2) dx =

ANSWER ONLY

(d)

tan3 x dx =

ANSWER ONLY


Let C be the arc x = 1

3

y2 + 2

3, (−

√ 2 ≤ y ≤ 0) .

(a) [6 MARKS] Find the area of the surface obtained by revolving C about the x-axis.

(b) [6 MARKS] Find the volume of the solid generated by revolving about the y-axis

the region bounded by C , the coordinate axes, and the line y =

−

√ 2.


[12 MARKS] Evaluate the indefinite integral x3 − 8 x − 1

( x2 − 1)( x + 1) d x .

8/15/2019 Brown's Notes 2010




Simplify your answers as much as possible.

(a) [6 MARKS] Evaluate the indefinite integral

√ 1 − 9 x2 dx .

(b) [6 MARKS] Evaluate the definite integral

−√

3

0

arctan x dx .


[4 MARKS EACH] Determine for each of the following series whether it

• converges absolutely;

• converges conditionally; or

• diverges.

(a)

∞n=2

(−1)n ln√

n

ln(n2)

(b)

∞n=1

1 · 3 · 5 · . . . · (2n − 1)

3nn!

(c)

∞

n=1

sin2n

1 + 2n

(d)

∞n=1

(−1)n

√ n

3n − 1


[12 MARKS] Use polar coordinates — no other method will be accepted — to find the

area of the region bounded by the curve r = 2 and the line r = 1

cos θ , and containing the

pole.

F.16 Final Examination in MATH 141 2005 01



3

0

| x − 1| dx .

8/15/2019 Brown's Notes 2010




dx

5

x

√ 4 + t 2 dt .


dx

x2

2π

sec t dt .


x5

3√

x3 + 1

dx .

SHOW ALL YOUR WORK!







(a) [4 MARKS]

∞n=2

2 − cos n

n

(b) [4 MARKS]

∞

n=0

n(−3)n

4n

(c) [4 MARKS]

∞n=2

1

n ln n

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral

or a sum, product, or quotient of several definite integrals, but do not evaluate the inte-


and limits specific to the given problems:

(a) [6 MARKS] The area of the region bounded by the parabola y = x2, the x-axis, and

the tangent to the parabola at the point (1, 1).


8/15/2019 Brown's Notes 2010




region bounded by x = 0 and the curve x =

sin y (0 ≤ y ≤ π).DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)

(c) [3 MARKS] The area of the surface obtained by revolving about the y-axis the

curve y = e x, 1 ≤ y ≤ 2.


(d) [2 MARKS] The average value of the function 2 x

(1 + x2)2 over the interval 0 ≤ x ≤

2.



[12 MARKS] Evaluate the indefinite integral

2 x3 + 3 x2 + 3

x2 + x − 12 dx .



m ≥ 2, cosm x dx =

1


m − 1

m

cosm−2 x dx

8/15/2019 Brown's Notes 2010




0

cos6 x dx.



x = 2cos t − cos2t

y = 2sin t − sin2t .

(a) [8 MARKS] Determine the points where the arc of the curve given by

π

4 ≤ t ≤ 7π

4

has a vertical tangent.

(b) [4 MARKS] Determine the length of the arc of the curve given by

0 ≤ t ≤ 2π .


(a) [5 MARKS] Determine whether the following integral is convergent; if it is con-

vergent, determine its value: 1

−1dx√

1 − x2



∞n=1

(−1)n n!

nn

(c) [3 MARKS] Determine whether the sequence an = ln(n + 1) − ln n is convergent;

if it is convergent, carefully determine its limit.


[12 MARKS] Find the area of the region bounded by the curves

r = 4 + 4sin θ

r sin θ = 3

which does not contain the pole.

8/15/2019 Brown's Notes 2010




Instructions




3. Calculators are not permitted.


Pages 8, 9, and 10, which are blank.




is printed. When that space is exhausted, you may write on the facing page . Any

solution may be continued on the last pages, or the back cover of the booklet, but you

must indicate any continuation clearly on the page where the question is printed!










0

−2

| x2 − 1| dx .


dx 4

x

et 2 dt .


dx

x2

1

dt

1 + t 5 .


x sin( x2) dx .

SHOW ALL YOUR WORK!

8/15/2019 Brown's Notes 2010




gence or divergence and determine whether the series is convergent. In each case youmust answer 3 questions:




(a) [4 MARKS]

∞n=2

2n

3n − 1

n

(b) [4 MARKS]

∞n=2

(

−1)n

√ n − 1

(c) [4 MARKS]

∞n=2

(−4)n

3n + 2n


– possibly improper — or a sum, product, or quotient of several such integrals, but do

not evaluate the integral(s). It is not enough to quote a general formula: your integrals

must have integrand and limits specific to the given problems:

(a) [6 MARKS] The area of the infinite region containing the point 0, 1

2 bounded by

the curve y = e x, the x-axis, and the tangent to the curve at the point (1 , e).


(b) [3 MARKS] The volume generated by rotating the region bounded by the curves

y =√

x

−1, y = 0, x = 5 about the line y = 3.


8/15/2019 Brown's Notes 2010



(c) [3 MARKS] The area of the surface obtained by revolving about the x-axis the

curve x = ln y, 1 ≤ x ≤ 3.DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)


[12 MARKS] Evaluate the indefinite integral x3 − x2 − 2 x − 2

x( x2 + x + 1) dx .



m 1,

secm x dx = 1

m

−1

secm−2 x · tan x + m − 2

m

−1 secm−2 x dx


0

sec3 x dx.


The curve C has equations x = t 3 + 4t , y = 6t 2.

(a) [8 MARKS] Determine the points on C where the tangent is parallel to the line

with equations x = −7t , y = 12t − 5.

(b) [4 MARKS] Determine a definite integral whose value is the length of the arc of C between the points with parameter values t = 1 and t = 2. YOU ARE NOT

EXPECTED TO EVALUATE THE INTEGRAL!


[12 MARKS] Find the area of the region between the inner loop and the outer loop of

the curve r = 1 − 2cos θ .

8/15/2019 Brown's Notes 2010



F.18 Final Examination in MATH 141 2006 01 (One version)

Instructions


2. Do not tear pages from this book; all your writing — even rough work — must be handed

in. You may do rough work for this paper anywhere in the booklet.

3. Calculators are not permitted. This is a closed book examination. Regular and transla-

tion dictionaries are permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions;

and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this test may be

di ff erent from yours.



Their solutions are to be written in the space provided on the page where the ques-

tion is printed. When that space is exhausted, you may write on the facing page.

Any solution may be continued on the last pages, or the back cover of the booklet,

but you must indicate any continuation clearly on the page where the question is

printed!

• Some of the questions on this paper require only BRIEF SOLUTIONS ; for theseyou are expected to write the correct answer in the box provided; you are not asked

to show your work, and you should not expect partial marks for solutions that are

not correct.


You are advised to spend the first few minutes scanning the problems. (Please inform

the invigilator if you find that your booklet is defective.)




2

−1

| x| dx .


e3 1

dt

t √

1 + ln t .

8/15/2019 Brown's Notes 2010




dx

x2

0

et 2 dt .


limn→∞

1

n

0

n

7

+

1

n

7

+

2

n

7

+ . . . +

n − 1

n

7 .

SHOW ALL YOUR WORK!







(a) [4 MARKS]

∞n=1

1

(tanh n)2 + 1

(b) [4 MARKS]

∞

n=1

n2ne−n2

(c) [4 MARKS]∞

n=1

n2 − 85n + 12

n(n + 6)2






(a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the region

bounded by the parabola y2

= 2 x + 6 and the line y = x − 1. An answer involvingintegration along the y-axis will not be accepted.



only the method of “washers”.

8/15/2019 Brown's Notes 2010




region bounded by the curves y = x3

and y = x2

. For this question you are to useonly the method of “cylindrical shells”.

(d) [3 MARKS] The length of the curve whose equation is

x2

4 +

y2

9 = 1 .



x5 + x

x4 − 16 d x .



(a) [4 MARKS]

cos x · cosh x dx

(b) [5 MARKS]

1 −

3

√ x2 + 2 x + 5 dx

(c) [4 MARKS]

sin2 x · cos2 x dx



x = x(t ) = 10 − 3t 2

y = y(t ) = t 3 − 3t ,

where

−∞ < t < +

∞.

(a) [8 MARKS] Determine the value of d 2 y

dx2 at the points where the tangent is horizon-

tal.

(b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of the

arc ( x(t ), y(t )) : −

√ 3 ≤ t ≤ 0

.

8/15/2019 Brown's Notes 2010





convergent; if it is convergent, determine its value: 0

−1

dx

x23

.



∞

n=1

(−1)n

n − ln n

(c) [3 MARKS] Give an example of a sequence an with the property that limn→∞

an = 0

but

∞n=1

an = +∞. You are expected to give a formula for the general term an of

your sequence.


[12 MARKS] The arc

r = 1 − cos θ (0 ≤ θ ≤ π)

divides the area bounded by the curve

r = 1 + sin θ (0 ≤ θ ≤ 2π)

into two parts. Showing all your work, carefully find the area of the part that contains


12

, π2

.


Instructions






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8/15/2019 Brown's Notes 2010



• [1 MARK] Explain why the test(s) you have chosen is / are applicable to the given

series.• [2 MARKS] Use the test(s) to conclude whether or not the series is convergent.

(a) [4 MARKS]

∞n=1

(−1)n n

n2 + 4

(b) [4 MARKS]

∞n=1

ln

n

3n + 2

(c) [4 MARKS]

∞

n=1

3n + 6

5n






(a) [4 MARKS] The length of the curve whose equation is

x = 1 + et , y = t 2 , (−3 ≤ t ≤ 3).

(b) [4 MARKS] The volume of the solid obtained by rotating about the x-axis theregion bounded by the curves y = x and y = x2. For this question you are expected

to use only the method of “cylindrical shells”.


region bounded by the curves y = x and y = x2. For this question you are expected

to use only the method of “washers”.



x

3

( x2 + 4)( x − 2) d x .


Showing all your work, evaluate each of the following. Simplify your answers as much

as possible.

8/15/2019 Brown's Notes 2010


8/15/2019 Brown's Notes 2010



F.20 Final Examination in MATH 141 2007 01 (One version)

Instructions







from yours.

















2

−1

| x|2 dx .


0 1

t 4dt √ t 5 + 1

.

(c) [3 MARKS] Determine the value of

1

n

0

n

3

+

1

n

3

+

2

n

3

+ . . . +

n − 1

n

3 .

8/15/2019 Brown's Notes 2010



(d) [3 MARKS] Suppose it is known that f ( x) = 4 cosh x for all x. Showing all your

work, determine the value of f (1)− f (−1), expressed in terms of the values of eitherexponentials or hyperbolic functions.

(e) [4 MARKS] Evaluate d

dx

x2

12

et t

dt when x = 1 .

SHOW ALL YOUR WORK!


gence or divergence to determine whether the series is absolutely convergent, condition-

ally convergent, or divergent. All tests used must be named, and all statements must be

carefully justified.

(a) [4 MARKS]

∞n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n

(b) [4 MARKS]

∞n=1

(−1)n+1 n!

n22n

(c) [4 MARKS]

∞n=1

(−1)n sin 1

n


product, or quotient of several definite integrals, simplified as much as possible; you arenot expected to evaluate the integrals.

R is defined to be the region enclosed by the curves x + y = 6 and y = x2; C is the arc

y = 3 x (−1 ≤ x ≤ 2).

(a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum of

integrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sum


8/15/2019 Brown's Notes 2010




(c) [3 MARKS] Express in terms of integrals — which you need not evaluate — the

average length that R cuts off from the vertical lines which it meets.



the integral.



rotating C about the line y = −1; you need not evaluate the integral.



[12 MARKS] Evaluate the indefinite integral x x2 − 4

( x − 2) + 4

x2 − 4

( x − 2) dx .



8/15/2019 Brown's Notes 2010



(a) [4 MARKS] e− x

·cos x dx

(b) [5 MARKS]

52

− 12

x√ 8 + 2 x − x2

dx

(c) [4 MARKS]

cos2 x +

1

cos2 x

· tan2 x dx


Consider the arc C defined by

x = x(t ) = cos t + t sin t

y = y(t ) = sin t − t cos t ,

where 0 ≤ t ≤ π2

.

(a) [6 MARKS] Determine as a function of t the value of d 2 y

dx2.

(b) [6 MARKS] Determine the area of the surface generated by revolving C about the

y-axis.



convergent; if it is convergent, determine its value: π

π2

sec x dx .


∞n=3

4

n ln n

is convergent.

(c) [3 MARKS] Showing all your work, determine whether the following sequence


a1 = 1.a2 = 1.23

a3 = 1.2345

a4 = 1.234545

a5 = 1.23454545

a6 = 1.2345454545

8/15/2019 Brown's Notes 2010






[10 MARKS] The polar curves

r = 2 + 2sin θ (0 ≤ θ ≤ 2π)

and

r = 6 − 6sin θ (0 ≤ θ ≤ 2π)

divide the plane into several regions. Showing all your work, carefully find the area of

the region bounded by these curves which contains the point (r , θ ) = (1, 0).

F.21 Supplemental / Deferred Examination in MATH 141 2007 01 (One

version)

Instructions




3. The use of calculators is not permitted. This is a closed book examination. Use of regularand translation dictionaries is permitted.


Pages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may be

different from yours.





solution may be continued on the last pages, or the back cover of the booklet, but youmust indicate any continuation clearly on the page where the question is printed!




8/15/2019 Brown's Notes 2010




You are advised to spend the first few minutes scanning the problems. (Please inform theinvigilator if you find that your booklet is defective.)




0

− 1√ 2

1√ 1 − x2

dx .

(b) [2 MARKS] Evaluate t 3 cosh t 4 dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral

x

− x

tet 2 dt .


dx

1

sin x

(ln | sec t + tan t |) dt when x = π

4 .

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determine

whether the series is convergent or divergent. All tests used must be named, and all

statements must be carefully justified.

(a) [4 MARKS]

∞n=3

1

(n + 2)(n − 2)

(b) [4 MARKS]

∞n=1

∞i=n

3−i

(c) [4 MARKS]

∞n=1

n + 1

n

n2


product, or quotient of several definite integrals, simplified as much as possible; you are

not expected to evaluate the integrals.

R is defined to be the region enclosed by the curves y − x = 9 and y = ( x + 3)2; C is the

arc x = t , y = e3t (−2 ≤ t ≤ 1).

8/15/2019 Brown's Notes 2010



(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum of

integrals whose value is the volume of the resulting solid.DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)

(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sum



(c) [3 MARKS] Let f (t ) denote the vertical distance of the pointt , e3t

from the x-

axis. Express in terms of integrals — which you need not evaluate — the average

value of f (t ) over the interval −2 ≤ t ≤ 1.



the integral.



rotating C about the line x = 1; you need not evaluate the integral.

8/15/2019 Brown's Notes 2010






x3 x4 − 4 x2

− 16

x4 − 4 x2 dx .


(a) [4 MARKS] Showing all your work, evaluate

sin2(3 x)cos2(3 x) dx .


1

x2√

9 x2 − 16dx .

(c) [5 MARKS] Assume that

f ( x) =

x

0

sec100 t dt

is known. Showing all your work, express the value of x

0

sec102 t dt in terms of

f ( x). (You are not expected to determine f ( x) explicitly.)


Consider the closed arc C defined by

x = x(t ) = 3t 2

y = y(t ) = t 3 − 3t ,

where

−

√ 3

≤ t

≤

√ 3.

(a) [3 MARKS] Determine the area bounded by C.

(b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter

value t = 1

2.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about the

y-axis.

8/15/2019 Brown's Notes 2010




convergent; if it is convergent, determine its value: ∞

−∞

x

x2 + 4 d x .


∞n=3

(−1)n

√ ln n

n

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of the

limacon r = 2 sin θ − 1.

F.22 Final Examination in MATH 141 2008 01 (one version)

This examination was written during a labour disruption, when the services of Teaching As-

sistants were not available for grading purposes. The following additional instructions were

distributed with the examination.

VERSION nMcGILL UNIVERSITY

FACULTY OF SCIENCE

FINAL EXAMINATION

IMPORTANT ADDITIONAL INSTRUCTIONS

MATHEMATICS 141 2008 01CALCULUS 2EXAMINER: Professor W. G. Brown DATE: Monday, April 14th, 2008

ASSOCIATE EXAMINER: Mr. S. Shahabi TIME: 09:00 – 12:00 hours

A. Part marks will not be awarded for any part of any question worth [4 MARKS] or less.

8/15/2019 Brown's Notes 2010



B. To be awarded part marks on a part of a question whose maximum value is 5 marks or

more, a student’s answer must be deemed to be more than 75% correct.

C. While there are 100 marks available on this examination 80 MARKS CONSTITUTE A

PERFECT PAPER. You may attempt as many problems as you wish.

All other instructions remain valid. Where a problem requires that all work be shown, that

remains the requirement; where a problem requires only that an answer be written in a box

without work being graded, that also remains the requirement.

Students are advised to spend time checking their work; for that purpose you could verify

your answers by solving problems in more than one way. Remember that indefinite integrals

can be checked by diff erentiation.

W. G. Brown, Examiner.

Instructions








from yours.




is printed; in some of these problems you are instructed to write the answer in a box,

but a correct answer alone will not be sufficient unless it is substantiated by your work,

clearly displayed outside the box. When space provided for that work is exhausted, youmay write on the facing page . Any solution may be continued on the last pages, or the

back cover of the booklet, but you must indicate any continuation clearly on the page

where the question is printed!




8/15/2019 Brown's Notes 2010









4 − 6 x

1 + x2 dx .

ANSWER (SHOW YOUR WORK OUTSIDE THE BOX)


2 0

y2

y3 + 1 dy .

8/15/2019 Brown's Notes 2010




(c) [3 MARKS] Evaluate

sin(18 θ ) cos(30 θ ) d θ .



(a) [3 MARKS] Simplifying your answer as much as possible, evaluate d

dx

√ 3

− x

earcsin z dz .

8/15/2019 Brown's Notes 2010




(b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for the

function f ( x) = 3 − x, where the sample points are the left end-point of each of n

subintervals of equal length.

ANSWER ONLY

(c) [4 MARKS] Determine the value of the preceding Riemann sum as a function of

n, simplifying your work as much as possible. (NOTE: You are being asked to

determine the value of the sum as a function of n, not the limit as n → ∞.)

8/15/2019 Brown's Notes 2010





For each of the following series determine whether the series diverges, converges condi-



(a) [4 MARKS]

∞

n=3

1

n √ ln n

(b) [4 MARKS]

∞n=1

(−1)n+1

√ 4n + 5

3n + 10

(c) [4 MARKS]

∞n=1

cot−1

1

n + 1

− cot−1

1

n

4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves 2 y = x, y = 2 x, and x2 + y2 = 5.

(a) [4 MARKS] The region

Ris rotated about the line x =

−1. Give an integral or sum



(b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies inside

R. Express in terms of integrals — which you need not evaluate — the average of

the positive lengths L(a).


8/15/2019 Brown's Notes 2010



(c) [4 MARKS] Let C1 be the curve x(t ) = t , y(t ) = cosh t (0 ≤ t ≤ ln 2). Simplifying

your answer as much as possible, find the length of C1.ANSWER ONLY


(a) [8 MARKS] Evaluate the indefinite integral 36

( x + 4)( x − 2)2 dx .

(b) [4 MARKS] Determine whether

∞ 3

36

( x + 4)( x − 2)2 d x converges.

If it converges, find its value.



(a) [4 MARKS]

e √ x dx


8/15/2019 Brown's Notes 2010



(b) [5 MARKS]

0

− 1

2

x

√ 3 − 4 x − 4 x2 dx


(c) [4 MARKS]

π

0

sin2 t cos4 t dt .



8/15/2019 Brown's Notes 2010



Consider the curve C2 defined by x = x(t ) = 1 + e−t , y = y(t ) = t + t 2 .

(a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis.


(b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent is

horizontal.


8/15/2019 Brown's Notes 2010



(c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis.



(a) [5 MARKS] Showing all your work, determine whether the series∞

n=2

√ n√

n + 2 −√

n − 2

is convergent or divergent.

(b) [5 MARKS] Showing all your work, determine whether the following sequence


a1 = 3.

a2 = 3.14

a3 = 3.1414a4 = 3.141414

a5 = 3.14141414

a6 = 3.1414141414



8/15/2019 Brown's Notes 2010




Curves C3 and C4, respectively represented by polar equations

r = 4 + 2cos θ (0 ≤ θ ≤ 2π) (116)

and

r = 4cos θ + 5 (0 ≤ θ ≤ 2π) , (117)

divide the plane into several regions.

(a) [8 MARKS] Showing all your work, carefully find the area of the one region which

is bounded by C3 and C4 and contains the pole.

(b) [4 MARKS] Find another equation — call it (117*) — that also represents

C4, and

has the property that there do not exist coordinates (r , θ ) which satisfy equations(116) and (117*) simultaneously. You are expected to show that equations (116)

and (117*) have no simultaneous solutions.

F.23 Supplemental / Deferred Examination in MATH 141 2008 01 (one

version)

Instructions


2. Do not tear pages from this book; all your writing — even rough work — must be handed in.You may do rough work for this paper anywhere in the booklet.

3. The use of calculators is not permitted. This is a closed book examination. Use of regular

and translation dictionaries is permitted.


Pages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may be

different from yours.



Their solutions are to be written in the space provided on the page where the questionis printed. When that space is exhausted, you may write on the facing page . Any






8/15/2019 Brown's Notes 2010








0

− 1√ 2

1√ 1 − x2

dx .

(b) [2 MARKS] Evaluate t 3 cosh t 4 dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral

x

− x

tet 2 dt .


dx

1

sin x

(ln | sec t + tan t |) dt when x = π

4 .

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determine

whether the series is convergent or divergent. All tests used must be named, and all

statements must be carefully justified.

(a) [4 MARKS]

∞n=3

1

(n + 2)(n − 2)

(b) [4 MARKS]

∞n=1

∞i=n

3−i

(c) [4 MARKS]

∞n=1

n + 1

n

n2


product, or quotient of several definite integrals, simplified as much as possible; you are

not expected to evaluate the integrals.

R is defined to be the region enclosed by the curves y − x = 9 and y = ( x + 3)2; C is the

arc x = t , y = e3t (−2 ≤ t ≤ 1).

8/15/2019 Brown's Notes 2010



(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum of

integrals whose value is the volume of the resulting solid.DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)

(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sum



(c) [3 MARKS] Let f (t ) denote the vertical distance of the pointt , e3t

from the x-

axis. Express in terms of integrals — which you need not evaluate — the average

value of f (t ) over the interval −2 ≤ t ≤ 1.



the integral.



rotating C about the line x = 1; you need not evaluate the integral.

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x3 x4 − 4 x2

− 16

x4 − 4 x2 dx .


(a) [4 MARKS] Showing all your work, evaluate

sin2(3 x)cos2(3 x) dx .


1

x2√

9 x2 − 16dx .

(c) [5 MARKS] Assume that

f ( x) =

x

0

sec100 t dt

is known. Showing all your work, express the value of x

0

sec102 t dt in terms of

f ( x). (You are not expected to determine f ( x) explicitly.)


Consider the closed arc C defined by

x = x(t ) = 3t 2

y = y(t ) = t 3 − 3t ,

where

−

√ 3

≤ t

≤

√ 3.

(a) [3 MARKS] Determine the area bounded by C.

(b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter

value t = 1

2.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about the

y-axis.

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convergent; if it is convergent, determine its value: ∞

−∞

x

x2 + 4 d x .


∞n=3

(−1)n

√ ln n

n

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of the

limacon r = 2 sin θ − 1.

F.24 Final Examination in MATH 141 2009 01 (one version)

Instructions






Pages 9, 10 and 11, which are blank. A TOTAL OF 75 MARKS ARE AVAILABLE ON THIS

EXAMINATION.

4. You are expected to simplify all answers wherever possible.

• Most questions on this paper require that you SHOW ALL YOUR WORK!

Solutions are to be begun on the page where the question is printed; a correct answer

alone will not be sufficient unless substantiated by your work. You may continue your

solution on the facing page , or on the last pages, or the back cover of the booklet, but

you must indicate any continuation clearly on the page where the question is printed!

To be awarded partial marks on a part of a question a student’s answer for that part

must be deemed to be more than 50% correct. Most of these questions will require

that the answer be written in a box provided on the page where the question is printed;

even if you continue your work elsewhere, the answer should be in the box provided.

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• Some questions on this paper require only BRIEF SOLUTIONS ; for these you must

write the correct answer in the box provided; you are not asked to show your work, andyou should not expect partial marks for solutions that are not correct. Check your work!



t 3 cos t 2 dt .


(b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivative

d

dt

t 2

0

tanh x2 dx .

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12

1

√ 2

dx√ 1

− x2

·arcsin x

.


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(b) [4 MARKS] Evaluate 2 y

y

2

− y + 1

dy


SHOW ALL YOUR WORK!

3. For each of the following series determine whether the series diverges, converges condi-



(a) [4 MARKS]

∞n=1

(−1)n+1

cos n

2

n

(b) [4 MARKS]

∞n=2

(−1)n 1

n√

ln n.

(c) [4 MARKS]

∞n=4

(−1)n 1

nln (3n + 1)

4. [9 MARKS] SHOW ALL YOUR WORK!

(a) [3 MARKS] Evaluate limn→∞

1

n·

nr =1

cos2

r π

n

. (Hint: This could be a Riemann

sum.)

(b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an =

arctan(−2n) as n → ∞.


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(c) [3 MARKS] Showing all your work, prove divergence, or find the value of

∞

n=1

∞

i=1

1

5i

n

.


(a) [8 MARKS] Evaluate the indefinite integral t + 1

2t 2 − t − 1 dt .

(b) [2 MARKS] Determine whether the following improper integral converges or di-

verges; if it converges, find its value:

∞ 4

t + 1

2t 2 − t − 1 dt .


Consider the curve C1 defined by x = 3t 2 , y = 2t 3 , (t ≥ 0).

(a) [7 MARKS] Showing all your work, determine the area of the surface generated

when the arc 0 ≤ t ≤ 1 of C1 is rotated about the y-axis.

(b) [2 MARKS] Showing all your work, determine all points — if any — where the

normal to the curve is parallel to the line x + y = 8.


Curves C3 and C4, are respectively represented by polar equations

r = 3 + 3cos θ (0 ≤ θ ≤ 2π) (118)

and

r = 9cos θ (0 ≤ θ ≤ π) . (119)

(a) [7 MARKS] Showing all your work, carefully find the area of the region lying

inside both of the curves.

(b) [3 MARKS] Determine the length of the curves which form the boundary of the

region whose area you have found.

8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves y = 8

x2, x = y, x = 1.

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(a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensional

solid, S 1. Give an integral or sum of integrals whose value is the volume of S 1;you are not asked to evaluate the integral(s).


(b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensional

solid, S 2. Give an integral or sum of integrals whose value is the volume of S 2obtained only by the Method of Cylindrical Shells; you are not asked to evaluate

the integral(s).


(c) [3 MARKS] Calculate the area of R.

ANSWER ONLY

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G WeBWorK

G.1 Frequently Asked Questions (FAQ)

G.1.1 Where is WeBWorK?

WeBWorK is located on Web servers of the Department of Mathematics and Statistics, and is

accessible at the following URL:

http: // msr02.math.mcgill.ca / webwork2 / MATH141 WINTER2010 /

If you access WeBWorK through WebCT, the link on your page will have been programmed

to take you to the correct WeBWorK server automatically.

G.1.2 Do I need a password to use WeBWorK?

You will need a user code and a password.

Your user code. Your user code will be your 9-digit student number.

Your password. The WeBWorK system is administered by the Mathematics and Statistics

Department, and is not accessible directly through the myMcGill Portal; your initial password

will be diff erent from your MINERVA password, but you could change it to that if you wish.

Your initial password will be your 9-digit student ID number. You will be able to change this

password after you sign on to WeBWorK.65

Your e-mail address. The WeBWorK system requires each user to have an e-mail address.

After signing on to WeBWorK, you should verify that the e-mail address shown is the one that

you prefer. You should endeavour to keep your e-mail address up to date, since the instructors

may send messages to the entire class through this route.

We suggest that you use either your UEA66 or your po-box address. You may be able to

forward your mail from these addresses to another convenient address, (cf. §4.)

G.1.3 Do I have to pay an additional fee to use WeBWorK?

WeBWorK is available to all students registered in the course at no additional charge.

65If you forget your password you will have to send a message to Professor Brown so that the system adminis-

trator may be instructed to reset the password at its initial value.66Uniform E-mail Address


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G.1.4 When will assignments be available on WeBWorK?

Each assignment will have a begin date and a due date. The assignment is available to you

after the begin date; solutions will be made available soon after the due date.

G.1.5 Do WeBWorK assignments cover the full range of problems that I should be able

to solve in this course?

The questions on the WeBWorK assignments ( A1 through A6) are a sampling of some types

of problem you should be able to solve after successfully completing this course. Some types

of calculus problems do not lend themselves to this kind of treatment, and may not appear on

the WeBWorK assignments. Use of WeBWorK does not replace studying the textbook —

including the worked examples, attending lectures and tutorials, and working exercises

from the textbook — using the Student Solutions Manual [3] to check your work. Students

are cautioned not to draw conclusions from the presence, absence, or relative frequencies of

problems of particular types, or from particular sections of the textbook. Certain sections

of the textbook remain examination material even though no problems are included in the

WeBWorK assignments.

G.1.6 May I assume that the distribution of topics on quizzes and final examinations

will parallel the distribution of topics in the WeBWorK assignments?

No! While the order of topics on WeBWorK assignments should conform to the order of the

lectures, there are some topics on the syllabus that will not appear in WeBWorK questions.Use WeBWorK for the areas it covers, and supplement it by working problems from your

textbook. Also, remember that WeBWorK — which checks answer only — cannot ascertain

whether you are using a correct method for solving problems. But, if you write out a solution

to an odd-numbered textbook problem, you can compare it with the solution in the Solutions

Manual; and, if in doubt, you can show your work to a Teaching Assistant at one of the many

office hours that they hold through the week.

G.1.7 WeBWorK provides for diff erent kinds of “Display Mode”. Which should I use?

“Display mode” is the mode that you enter when you first view a problem; and, later, when

you submit your answer. You may wish to experiment with the diff erent formats. The defaultis jsMath mode, which should look similar to the version that you print out (cf. next question).

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G.1.8 WeBWorK provides for printing assignments in “Portable Document Format”

(.pdf), “PostScript” (.ps) and “TEXSource” forms. Which should I use?Most newer home computers have already been loaded with the Acrobat Reader for .pdf files; if

the Reader has not been installed on your computer67, you will find instructions for download-

ing this (free) software in §1.6.5 of these notes. Most computers available to you on campus

should be capable of printing in .pdf format.

G.1.9 What is the relation between WeBWorK and WebCT?

There is none. WebCT is the proprietary system of Web Course Tools that has been imple-

mented by McGill University. You may access the web page for this course, and WeBWorK

through your WebCT account68, and WebCT will link you to the appropriate server for WeB-

WorK. If you follow this route to WeBWorK, you will still have to log in when you reach

the WeBWorK site. At the present time we will be using WebCT primarily for the posting of

grades, and as a convenient repository for links to notes and announcements in the course. We

are not planning to use the potential WebCT sites that exist for the tutorial sections: use only

the site for the lecture section in which you are registered.

G.1.10 What do I have to do on WeBWorK?

After you sign on to WeBWorK, and click on “Begin Problem Sets”, you will see a list of As-

signments, each with a due date. Since there is no limit to the number of attempts at problems

on P0 or the other “Practice” assignments, you may play with these assignments to learn howto use the WeBWorK software.

You may print out a copy of your assignment by clicking on “Get hard copy”. This is

your version of the assignment, and it will diff er from the assignments of other students in

the course. You should spend some time working on the assignment away from the computer.

When you are ready to submit your solutions, sign on again, and select the same assignment.

This time click on “Do problem set”. You can expect to become more comfortable with the

system as you attempt several problems; but, in the beginning, there are likely to be situations

where you cannot understand what the system finds wrong with some of your answers. It is

useful to click on the Preview Answers button to see how the system interprets an answer that

you have typed in. As the problems may become more difficult, you may have to refer to the

“Help” page, and also to the “List of functions” which appears on the page listing the problems.Don’t submit an answer until you are happy with the interpretation that the Preview Answers

button shows that the system will be taking of your answer.

67At the time these notes were written, the latest version of the Reader was 9.0, but recent, earlier versions

should also work properly.68http: // mycourses.mcgill.ca

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G.1.11 How can I learn how to use WeBWorK?

As soon as your instructor announces that the WeBWorK accounts are ready, sign on and try

assignment P0, which does not count. The system is self-instructive, so we will not burden

you with a long list of instructions.

You will need to learn how to enter algebraic expressions into WeBWorK as it is coded to

read what you type in a way that may diff erent from what you expect. For example, the symbol

ˆ is used for writing exponents (powers). If you type 2ˆ3, WeBWorK will interpret this as

23 = 8. However, if you type 2ˆ3+x, WeBWorKwill interpret it as 23 + x, i.e. as 8 + x; if you

wish to write 23+ x, you have to type 2ˆ(3+x). You may obtain more information from the List

of Available Functions, available online, or at

http: // webwork.maa.org / wiki / Mathematical notation recognized by WeBWorK

G.1.12 Where should I go if I have difficulties with WeBWorK ?

If you have difficulties signing on to WeBWorK, or with the viewing or printing functions on

WeBWorK, or with the specific problems on your version of an assignment, you may send

an e-mail distress message directly from WeBWorK by clicking on the Feedback button.

You may also report the problem to your instructor and / or your tutor, but the fastest way of

resolving your difficulty is usually the Feedback . Please give as much information as you

can. (All of the instructors and tutors are able to view from within WeBWorK the answers

that you have submitted to questions.)If your problem is mathematical, and you need help in solving a problem, you should

consult one of the tutors at their office hours; you may go to any tutor’s office hours, not only

to the hours of the tutor of the section in which you are registered.

G.1.13 Can the WeBWorK system ever break down or degrade?

Like all computer systems, WeBWorK can experience technical problems. The systems man-

ager is continually monitoring its performance. If you experience a difficulty when online,

please click on the Feedback button and report it. If that option is not available to you,

please communicate with either instructor by e-mail.

If you leave your WeBWorK assignment until the hours close to the due time on the duedate, you should not be surprised if the system is slow to respond. This is not a malfunction,

but is simply a reflection of the fact that other students have also been procrastinating! To

benefit from the speed that the system can deliver under normal conditions, do not delay your

WeBWorK until the last possible day! If a systems failure interferes with the due date of an

assignment, arrangements could be made to change that date, and an e-mail message could

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be broadcast to all users (to the e-mail addresses on record), or a message could be posted on

WeBCT or the WeBWorK sign-on screen.69

G.1.14 How many attempts may I make to solve a particular problem on WeBWorK?

Practice Assignments P1 — P6 are intended to prepare you for Assignments A1 — A6, and

permit unlimited numbers of attempts; your grades on these “Practice” do not count in your

term mark. For the problems on assignments A1 — A6 you will normally be permitted about 5

tries: read the instructions at the head of the assignment.

G.1.15 Will all WeBWorK assignments have the same length? the same value?

The numbers of problems on the various assignments may not be the same, and the individualproblems may vary in difficulty. Assignments A1 — A6 will count equally in the computation

of your grade.

G.1.16 Is WeBWorK a good indicator of examination performance?

A low grade on WeBWorK has often been followed by a low grade on the examination.

A high grade on WeBWorK does not necessarily indicate a likely high grade on the exam-

ination.

To summarize: WeBWorK alone is not enough to prepare this course; but students who

don’t do WeBWorK appear to have a poor likelihood of success in MATH 141: that is one

reason why we have made the WeBWorK assignments compulsory.

69But slowness of the system just before the due time will not normally be considered a systems failure.

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H Contents of the DVD disks for

brown's notes 2010

Documents