8.1concepts of indefinite integrals 8.2indefinite integration of functions 8.3integration by...

8.1 Concepts of Indefinite Integrals8.2 Indefinite Integration of Functions8.3 Integration by Substitution

Chapter Summary

Case Study

Indefinite Integrals8

8.4 Integration by Parts8.5 Applications of Indefinite Integrals

P. 2

By measuring the level of radioactivity with a counter, it is estimated that the number of radioactive particles, y, in the sample is decreasing at a rate of 1000e–0.046t per hour, where t is expressed in hours.

Case StudyCase Study

According to what we learnt in Section 7.5 (Rates of Change), we have

The process of finding a function from its derivative is called integration and will be discussed in this chapter.

I have already recorded the readings for the level of radioactivity.

Can you estimate the number of radioactive particles in the sample from your readings?

.0100 0.046– tedt

dy

In order to express y in terms of t, we need to find a function y of t such that its derivative is equal to –1000e–0.046t .

P. 3

In previous chapters, we learnt how to find the derivative of a given function.

8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals

Suppose we are given a function x2, by differentiation, we have

As 2x is the derivative of x2, we call x2 the primitive function (or antiderivative) of 2x.

Although x2 is a primitive function of 2x, it is not the unique primitive function.

A. A. Definition of Indefinite IntegralsDefinition of Indefinite Integrals

.2)( 2 xxdx

d

Definition 8.1

If , then F(x) is called a primitive function of f(x).

)()]([ xfxFdx

d

If we add an arbitrary constant C to x2 and differentiate it, we

have .202)()()( 22 xxCdx

dx

dx

dCx

dx

d

Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C.

Generally, for any differentiable function F(x), we have the following definition:

P. 4

In order to represent all the primitive functions of a function f (x), we introduce the concept of indefinite integral as below:

Note: 1. In the notation of , f (x) is called the integrand, and ‘ ’

is called the integral sign. The process of finding the primitive function is called integration.

Definition 8.2

If , then the indefinite integral of f(x), which is

denoted by , is given by

, where C is an arbitrary constant.

)()( xfxFdx

d

dxxf )(

CxFdxxf )()(

2. C is called the constant of integration (or integration constant).

dxxf )(


A. A. Definition of Indefinite IntegralsDefinition of Indefinite Integrals

P. 5

Note: 1. Formula 8.1 is also called the Power Rule for integration.

B. B. Basic Formulas of Indefinite IntegralsBasic Formulas of Indefinite Integrals


As integration is the reverse process of differentiation, the basic formulas for integrations can be derived from the differentiation formulas.

For example: Since , nn xnxdx

d)1()( 1

8.1 , for all real numbers n 1.Cxn

dxx nn

1

1

1

2. When n 0, the left hand side of the formula becomes .

For convenience we usually express as .

dxdxx 10

dxdx1

,1

1i.e., 1 nn xx

ndx

d

P. 6

B. B. Basic Formulas of Indefinite IntegralsBasic Formulas of Indefinite Integrals


In addition to the Power Rule, we can use the similar way to derive the following integration formulas:

8.2 , where k is a constant.

8.3 8.4

8.5 8.6

8.7 8.8

8.9 8.10 Cxxdxx csccotcsc

Cxxdx cotcsc2

Cxxdx sincos

Cedxe xx

Cxxdxx sectansec

Cxxdx tansec2

Cxxdx cossin

Cxdxx

ln1

Ckxkdx

P. 7

Theorem 8.1If k is a non-constant, then .

C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals


dxxfkdxxkf )()(

Proof:

.)()(Let Cxgdxxf ).()]([Then xfxgdx

d

)()]([)]([ xkfxgdx

dkxkg

dx

d

kCxgkCxgkdxxfk )(])([)(

Since C and kC are arbitrary constants, the expressions kg(x) C and kg(x) kC represent the same family of primitive functions.

dxxfkdxxkf )()(

)()( xkgdxxkfBy definition, C , where C is an arbitrary constant. On the other hand,

P. 8



Theorem 8.2

dxxgdxxfdxxgxf )()()]()([

Let and , where C1 and C2 are

arbitrary constants. ).()]([ and )()]([Then xgxGdx

dxfxF

dx

d

)()()]()([ xgxfxGxFdx

d

21)()( CCxGxF Since C1 C2 is an arbitrary constant, the expressions F(x) G(x) C and F(x) G(x) C1 C2 represent the same family of primitive functions.

dxxgdxxfdxxgxf )()()]()([

1)()( CxFdxxf 2)()( CxGdxxg

.)()()]()([ CxGxFdxxgxf By definition,

])([])([)()( 21 CxGCxFdxxgdxxf On the other hand,

Proof:

P. 9

Example 8.1T

Solution:



Find .)543( 5 dxxx

dxxx )543( 5 dxdxxdxx 543 2

5

2

1

322

7

12

3

57

24

3

23 CxCxCx

Cxxx 57

82 73 Use C to express the sum

of all constants

P. 10

Example 8.2T



Solution:

dxx

x

1

13

dxx

xxx

1

1)()1(

3

3233

dxxx )1( 3

1

3

2

dxdxxdxx 3

1

3

2

Cxxx 3

4

3

5

4

3

5

3

.1

13

dx

x

x

Cancel the common factor

a3 b3 (a b)(a2 ab b2)

Find

P. 11

Example 8.3T



Solution:

. 2

23 dxx

x

dx

xx

223 dx

xdxxdx

1223

Cxxx ln222

32

Cxxx ln222

3 2

Find

P. 12

Example 8.4T



Solution:

dxe

xx x2

53 2

.25

3 2

dxe

xx x

dxedxx

dxx x21

53 2

Cexx x 2ln53

33

Cexx x 2ln53

Find

P. 13

Example 8.5T



Solution:

.)tan1( 2 dxx

dxx)tan1( 2 dxx 2sec

Cx tan

Find

P. 14

Example 8.6T



Let y ln x – ln (x 1).

(a) Find .

(b) Hence find .

Solution:

dx

dy

dxxx )1(

1

)1(1

11

x

dx

d

xxdx

dy(a)

1

11

xx

(b) By (a),

Cxxdx

xx1lnln

1

11

Cxxdxxx

xx1lnln

)1(

1

Cxxdxxx

1lnln)1(

1

P. 15

The integration formulas mentioned in Section 8.1 enable us to find the indefinite integrals of simple functions such as ex, sin x and cos x. But how about e2x, sin 4x and cos (7x + 5)?

8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions

A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)

Using the same method, we can obtain the following integration formulas:

.))(1()(Consider 1 nn baxnabaxdx

d

8.11 , where n –1 and a 0.Cna

baxdxbax

nn

)1(

)()(

1

Suppose a and b are real numbers with a 0.

8.12

8.13

8.14

8.15

Cbaxa

dxbax

ln

11

Cea

dxe baxbax 1

Cbaxa

dxbax )cos(1

)sin(

Cbaxa

dxbax )sin(1

)cos(

P. 16

Example 8.7T



Solution:

.2

93

dxx

dx

x3 2

9

dxx 3

1

)2(9

Cx

1

3

1

)2(1

3

19

Cx 3

2

)2(2

27

Find

P. 17

Example 8.8T



Solution:

.5232

1

dxxx

dx

xx 5232

1

dxxx

xx

xx 5232

5232

5232

1

dx

xx

xx

)52()32(

5232

dxxx )5232(8

1

Cxx

23

23

)52(1

21

28

1)32(

121

28

1

Cxx ])52()32[(24

1 2

3

2

3

Find

Rationalize the denominator

P. 18

Example 8.9T



Solution:

.456

432

dx

xx

x

dxxx

x

456

432

dxxx

x

)12)(43(

43

dx

x 12

1

Cx 12ln2

1

Find

Cancel the common factor

P. 19

Example 8.10T



.3

2

14

dte

eet

tt

dte

eet

tt

2

14 3

Cee tt

1

3

2

12

Cee tt 32

1 12

Solution:

Find

dtee tt )3( 12

dt

e

e

e

et

t

t

t

22

14

3

P. 20

Example 8.11T



Solution:

.)]23sin(2[ 2 dttt

dttt )]23sin(2[ 2

Ctt )23cos(31

32 3

Ctt

3

)23cos(

3

2 3

dttdtt )23sin(2 2

Find

P. 21

Example 8.12T


Solution:

.52 dxxx

dxxx 52 dxx10

dxe x )( 10ln

Ce x

10ln

)10(ln

Cx

10ln

10


If y ln 10, then ey 10 by definition,i.e., eln 10 10.

Find

P. 22

To find integrals where the integrand is the product or power of trigonometric functions, we can first usedouble angle formulas and product-to-sum formulas to express the integrand in the sum of trigonometric functions.


B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions

Product-to-sum Formulas

sin A cos B [sin (A B) sin (A B)]

cos A sin B [sin (A B) sin (A B)]

cos A cos B [cos (A B) cos (A B)]

sin A sin B [cos (A B) cos (A B)]

2

1

2

1

2

1

2

1

cos 2A 2 cos2 A 1 or cos 2A 1 2 sin2 A

P. 23

Example 8.13T


Solution:


.2

cos2

sin6 d

d

2cos

2sin6 dsin3

C cos3

sin 2A 2 sin A cos A

Find

P. 24

Example 8.14T


Solution:


dxxx

4cos

4sin 22

.4

cos4

sin 22 dxxx

dx

x 2

2sin

2

1

dx

x

2

cos1

4

1

dxx)cos1(8

1

Cxx )sin(8

1

Find

dxx

2sin

4

1 2

sin 2A 2 sin A cos A

cos 2A 1 2 sin2 A

P. 25

Example 8.15T


Find

Solution:


tdtt 7sin8sin

.7sin8sin tdtt

dttt ])78cos()78[cos(2

1

dttt )15cos(cos2

1

Ct

t

15

15sinsin

2

1

Ctt

30

15sin

2

sin

Product-to-sum formula

P. 26

Example 8.16T


Solution:


.12cos

cos2

d

d12cos

cos2

d

2

12coscos

d2sin

cos

d)csccot(

C csc

Find

cos 2A 1 2 sin2 A

cot A A

A

sin

cos

P. 27

8.8.33 Integration by SubstitutionIntegration by Substitution

A. Change of VariablesA. Change of Variables

In Section 8.2, we learnt some basic formulas to find the indefinite integrals of functions.

However, not all functions can be integrated directly using these formulas.

In this case, we have to use the method of integration by substitution.

The following shows the basic principle of this method.

Let and u g (x). CuFduuf )()(

dx

duuF

du

duF

dx

d )]([)]([ Since

f (u) g(x) f [g(x)] g(x)

By the definition of integration, . CuFdxxg'xgf )()()]([

duufdxxg'xgf )()()]([

P. 28


For an integral , we can transform it into

a simpler integral , by the following steps.

dxxg'xgf )()]([

duuf )(

Step 1: Separate the integrand into two parts: f [g(x)] and g(x)dx.

Step 2: Replace every occurrence of g(x) in the integrand by u.

Step 3: Replace the expression ‘g(x)dx’ by ‘du’.

Let us use this method to find the integral together. Note that , xdxxdxxx 2112 22

so we let u x2 + 1, such that 2x. dx

du

duuxdxx 212

Cu 2

3

3

2

Cx 23

2 )1(3

2


dxxx 12 2

P. 29

Example 8.17T



Solution:Let u 1 – x2. Then .

.12 3 2 dxxx

xdx

du2

dxxx 3 212 dxxx )2(13 2

duu3

Cu 3

4

4

3

Cx 3

42 )1(

4

3

Find

Express the answer in terms of x

P. 30

Example 8.18T



Solution:Let u x2 – 7. Then . x

dx

du2

.)7( 1023 dxxx

dxxx 1023 )7( xdxxx 22

1)7( 1022

duuu2

1)7( 10

duuu )7(2

1 1011

Cuu

1112

11

7

12

1

2

1

Cxx 112122 )7(22

7)7(

24

1

Find

Rewrite x2 as (u 7)

P. 31



With the method of integration by substitution, we can find the integrals of trigonometric functions other than sine and cosine, as shown in the following example.

P. 32

Example 8.19T



Let u csc x – cot x.

.csc xdx

dxxx

xxx

cotcsc

cotcsccsc xdxcsc

dx

xx

xxx

cotcsc

cotcsccsc2

.cotcsccscThen 2 xxxdx

du

duu

xdx1

csc

Cu ln

Cxx cotcscln

Solution:

Find

Alternative Solution:xxx

dx

dcotcsccsc

xxdx

d 2csccotand

)cot(csccsc xxx )cot(csc xx

dx

d

Let u csc x – cot x.

duu

xdx1

cscCxx cotcscln

.cscThen xudx

du

P. 33



It is tedious to write u and du every time when finding the integrals by substitution, as shown in the previous examples.

After becoming familiar with the method of integration by substitution, the working steps can be simplified by omitting the use of the variable u.

Let us study the following example.

P. 34

Example 8.20T



dxx

x)sin(ln

.)sin(ln

dxx

x

dxx

x1

)sin(ln

)(ln)sin(ln xdx

Cx )cos(ln

Solution:

Find

P. 35

Strategies for finding integrals in the form

Sometimes we need to handle indefinite integrals that involvethe products of powers of trigonometric functions, such as

or , where m and n are integers.


B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions

xdxx nm cossin xdxx nm sectan

In the following discussion, we will see how to apply different strategies according to different values of m and n.

xdxx nm cossin

Case 1: m is an odd number. Use sin x dx –d(cos x) and express all the other sine terms as cosine terms.

Case 2: n is an odd number. Use cos x dx d(sin x) and express all the other cosine terms as sine terms.Case 3: both m and n are even numbers. Use the double-angle formula to reduce the powers of the functions.

P. 36

Example 8.21T



Find

Solution: xdxx 53 cossin

.cossin 53 xdxx

dxxx )sin(cossin 52

)(coscos)cos1( 52 xxdx

)(cos)cos(cos 57 xdxx

Cxx

6

cos

8

cos 68

P. 37

Example 8.22T



Find

Solution: xdxx 33 sincos

.sincos 33 xdxx

dxxxx )sin(sincos 23

)(cos)cos1(cos 23 xdxx

)(cos)(cos)(cos 3

1

3

7

xdxx

Cxx 3

4

3

10

)(cos4

3)(cos

10

3

P. 38

Example 8.23T



Find

Solution: xdxx 42 cossin

.cossin 42 xdxx

xdx

x 22

cos2

2sin

xdxx 22 cos2sin4

1

dxxx )2cos1)(4cos1(16

1

dxxxxx )4cos2cos4cos2cos1(16

1

dxxxxx 2cos

2

16cos

2

14cos2cos1

16

1

dxxxx 6cos

2

14cos2cos

2

11

16

1

Cxxxx 6sin

192

14sin

64

12sin

64

1

16

P. 39

Similarly, integrals in the form may be

found by using the method of integration by substitution.



xdxx nm sectan

Strategies for finding integrals in the form xdxx nm sectan

Case 1: m is an odd number. Use tan x sec x dx d(sec x) and then express all other tangent terms as secant terms.

Case 2: n is an even number. Use sec2x dx d(tan x) and then express all other secant terms as tangent terms.

P. 40

Example 8.24T



Find

Solution: xdxx 53 sectan

.sectan 53 xdxx

xdxxxx sectansectan 42

)(secsec)1(sec 42 xxdx

)(sec)sec(sec 46 xdxx

Cxx

5

sec

7

sec 57

P. 41

Example 8.25T



Find

Solution:

.csccot 42 xdxx

xdxx 42 csccot

dxxxx )csc(csccot 222

)(cot)1(cotcot 22 xdxx

)(cot)cotcot( 24 xdxx

Cxx

3

cot

5

cot 35

P. 42

In the above examples, the case that m is even while n is odd is not considered. This is because there is no standard technique and the method varies from case to case.


For example, to find (m 0 and n 1), we may follow the

method in Example 8.19.

xdxsec

In some other cases, such as (m 2 and n 1), we may

need to use the technique ‘integration by parts’, which will be discussed later in this chapter.

xdxx sectan2


P. 43

If an indefinite integral involves radicals in the form

, or , we can use the method of integration by substitution to eliminate the radicals.


C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution

The following three trigonometric identities are very useful for the elimination:

For example, if we substitute x a sin into the expression ,

we have

22 ax 22 xa 22 xa

22 xa

22222 sinaaxa

coscos22

aa

Then we can express the integrand in terms of .

After finding the indefinite integral in terms of (say, 3 + C), the final answer should be expressed in terms of the original variable, say, x.

cos2 1 sin2 , sec2 1 tan2 , tan2 sec2 1

P. 44

In order to express in terms of x, let us first introduce the following notations:



Inverse of Trigonometric FunctionsLet x be a real number. 1. sin–1x is defined as the angle such that sin x (where –1 x

1)

and .

2. cos–1x is defined as the angle such that cos x (where –1 x 1)and 0 .

3. tan–1x is defined as the angle such that tan x and .

2

π

2

π

2

π

2

π

P. 45

Example 8.26TFind

Solution:



.1 22 xx

dx

Let x sinThen dx cos d

2222 sin1sin

cos

1

d

xx

dx

cossin

cos2

d

d2csc

C cot

Since sin x,

1csccot 2

1sin

12

112

x

x

x21

Cx

x

xx

dx 2

22

1

1

P. 46

Example 8.27T

Solution:



Let x sinThen dx cosd.

Since sin x,

.1 22 dxxx

dxxx 22 1

d22 cossin

d

2

2

2sin

d2sin4

1 2

C

4

4sin

8

1

,1cos 2x

2sin2cos24sin )sincos2)(sin21(2 2

23 1)2(4 xxx

dxxx 22 1

Cxxx

x

4

1)2(4sin

8

1 231

Cxxxx ]sin1)2[(8

1 123

d2

4cos1

4

1

dcossin1sin 22

)1)(21(4 22 xxx

Find

P. 47

Example 8.28T

Solution:



Let x 3tanThen dx 3sec2 d.

.9 22 xx

dx

22 9 xx

dx

sectan27

sec32

2 d

dcot

sin

cos

cos

1

9

1

Since ,3

tanx

2cot1csc x

2tan

11

Cx

x

xx

dx

9

9

9

2

22

22

2

tan99tan9

sec3 d

d2tan

sec

9

1

dcotcsc9

1

C csc9

1 2

91

x

x

x 92

Find

P. 48

Example 8.29T

Solution:



.342

xx

dx

1)2(34 22 x

dx

xx

dx

Let x + 2 sec

342 xx

dx

tan

tansec d

d

tansec

tansecsec

tansec

)tan(secd

C tansecln

Since sec x + 2, 341)2(1sectan 222 xxx

Cxxxxx

dx

342ln34

2

2

1sec

tansec2

d

dsec

Find

1

x + 2 __________(x 2)2 1

Then dx sectan d.

P. 49

Some indefinite integrals such as , and

cannot be found by using the techniques we have

learnt so far.

8.8.44 Integration by PartsIntegration by Parts

xdxln xdxx sin

dxxe x

To evaluate them, we need to introduce another method called integration by parts.

Theorem 8.3 Integration by PartsIf u(x) and v(x) are two differentiable functions, then

In other words, .

. vu'dxuvuv'dx

vduuvudv

Proof:Suppose u and v are two differentiable functions.

Since (uv) uv + vu, by definition, .dx

dCuvdxvu'uv' )(

Cuvvu'dxuv'dx vu'dxuvuv'dx

P. 50

From Theorem 8.3, we can see that the problem of finding

can be transformed into the problem of finding instead.


If the integral is much simpler than , then the original integralcan be found easily.

If we want to apply the technique of integration by parts to an integral,

such as , we need to transform the integral into the form

first, such as or .

udv vdu

udvvdu

udv xdxe x sin

)()(sin xedx )cos( xde x

P. 51

Example 8.30TFind

Solution:


.ln xdxx

2lnln

2xxdxdxx

)(ln

22)(ln

22

xdxx

x

xdxxx

2

1

2

ln2

dxx

xxx 1

22

ln 22

Cxxx 42

ln 22

P. 52

Example 8.31T

Solution:


.log52 xdxx

dx

xx

xdxx

5ln

ln

log2

52

3ln

5ln

1 3xxd

)(ln

33)(ln

5ln

1 33

xdxx

x

dxx

xxx 1

3

1

3

ln

5ln

1 33

dxx

xx 23

3

1

3

ln

5ln

1

Cx

xx

9ln

35ln

1 33

Find

P. 53

In some cases, there may be more than one choice for u and v.


For example, we can transform into

However, if we choose the former, then

xdxx sin

)(sin

22)(sin

2sin

222

xdxx

xx

xd

xdxxxx

cos2

1

2

sin 22

As a result, we get an integrand x2 cos x which is more complicated than the original one x sin x.

Thus we should try sin x dx d(–cos x) instead.

2

sin2x

xd )(cos xxdor .

P. 54

Example 8.32T

Solution:


.csc2 xdxx

xdxx 2csc

)cot( xxd

xdxxx cot)cot(

dxx

xxx

sin

coscot

)(sinsin

1cot xd

xxx

Cxxx sinlncot

Find

P. 55

Example 8.33T

Solution:


(a) Show that

(b) Hence find .

.1cos

1

2cot

x

x

dx

d

dx

x

x

1cos

22csc

2cot 2 x

dx

dxx

dx

d

2csc

2

1 2 x

2sin2

1

2 x

2

cos12

1x

1cos

1

x

(a) (b)

2cot

1cos

xxddx

x

x

dxxx

x2

cot2

cot

22

sin

2cos

22

cotx

dx

xx

x

2sin

2sin

12

2cot

xd

xx

x

Cxx

x 2

sinln22

cot

P. 56

Example 8.34T

Solution:


.2 dxex x

)(2 xedx

])([ 22 xdeex xx

)2( 2 dxxeex xx

)(22 xx exdex

])2(2[2 xdexeex xxx

dxexeex xxx 222

Cexeex xxx 222

Cexx x )22( 2

dxex x2

Find

P. 57

Example 8.35T

Solution:


dxx)cos(ln

.)cos(ln dxx

)][cos(ln)][cos(ln xxdxx

dxx

xxxx1

)sin(ln)cos(ln

dxxxx )sin(ln)cos(ln

)][sin(ln)][sin(ln)cos(ln xxdxxxx

dxx

xxxxxx1

)cos(ln)sin(ln)cos(ln

dxxxxxx )cos(ln)sin(ln)cos(ln

Therefore, Cxxxxdxx )sin(ln)cos(ln)cos(ln2

Cxxx

dxx )]cos(ln)[sin(ln2

)cos(ln

Find

P. 58

In previous chapters, we learnt that of a curve y F(x) is the slope function of the curve.

8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals

Since integration is the reverse process of differentiation,

if we let f(x), then by the definition of integration, we have

where C is an arbitrary constant.

Thus we can see that the equation of a family of curves y F(x) + C

can be found by integration, providing that the slope function of the curve is known.

A. Geometrical ApplicationsA. Geometrical Applications

dx

dy

dx

dy

dxxfy )(

CxF )(

dx

dy

P. 59

Example 8.36TThe equation of the slope of a curve at the point (x, y) is given by

. If the curve passes through (2, 2), find the equation of

the curve. Solution:



6)32( xdx

dy

6)32( xdx

dy

dxxy 6)32( ,72

)32( 7

Cx

where C is an arbitrary constant

Cx 14

)32( 7

When x 2, y 2, we have

14

2714

)322(2

7

C

C

∴ The equation of the curve is .14

27

14

)32( 7

xy

P. 60

Example 8.37TIt is given that at each point (x, y) on a certain curve, . If

the curve passes through and , find the equation of that curve. Solution:



xdx

yd 22

2

sin2

xdx

yd 22

2

sin2

dxxdx

dy 2sin2

dxx)2cos1(

dxC

xxy 12

2sin

21

2

4

2cos

2CxC

xx

Since and lie on the curve, we have

4

5 ,0

4

5 ,

14

10

4

5

2

2

C

C

12

2sinC

xx

1

2

1

1

2

2

14

1

24

5

C

C

C

∴ The equation of the curve is

.124

2cos

2

2

xxxy

4

5 ,0

4

5 ,

P. 61

In Section 7.5 of Book 1, we learnt that for a particle moving along a straight line, its velocity v and acceleration a at time t are given by

where s is the displacement of the particle at time t.


Since integration is the reverse process of differentiation, we have

B. Applications in PhysicsB. Applications in Physics

,and2

2

dt

sd

dt

dva

dt

dsv

P. 62

Example 8.38TA particle moves along a straight line such that its velocity v at time t is given by . Find the displacement s of the particle at time t, given that s 6 when t 2.

Solution:



52 ttv

dttts 52 )5(52

1 22 tdt

Ct 2

32 )5(

3

2

2

1Ct 2

32 )5(

3

1

When t 2, s 6, we have

3

)52(3

16 2

32

C

C

3)5(3

1 2

32 ts

P. 63

Example 8.39TA particle moves along a straight line so that its acceleration a at time t is

given by a for t > 0. When t 1, the velocity of the particle is 8 and

its displacement is 24. Find the displacement of the particle at time t. Solution:



t

1

dtt

v1

1ln Ct

When t 1, v 8, we have

81ln8

1

1

C

C

8ln tv

ttdt

dtts

8ln

)8(ln

tdtt

ttt

tttdtt

81

ln

8)(lnln

27ln8ln

Cttttdttt

When t 1, s 24, we have

17)1(71ln124

2

2

C

C

177ln ttts

P. 64

8.1 Concepts of Indefinite Integrals

Chapter Chapter SummarySummary

)()]([ xfxFdx

d

CxFdxxf )()(

1. If , then the indefinite integral of f(x)

is defined by , where C is an arbitrary constant.

2. 1. where,1

1 1

nCxn

dxx nn

Cedxe xx

Cxdxx

ln1

constant. a is where, kCkxkdx 3.

4.

5.

P. 65

8.1 Concepts of Indefinite Integrals


6.

7.

Cxxdxx csccotcsc

Cxxdx cotcsc2

Cxxdx sincos

Cxxdxx sectansec

Cxxdx tansec2

Cxxdx cossin

8.

9.

10.

11.

12.

13. dxxgdxxfdxxgxf )()()()(

constant. zero-non a is where,)()( kdxxfkdxxkf

P. 66

8.2 Indefinite Integration of Functions


Let a and b be real numbers with a 0.

1.

2.

3.

4.

.1 where,)1(

)()(

1

nCna

baxdxbax

nn

Cbaxa

dxbax

ln

11

Cea

dxe baxbax 1

Cbaxa

dxbax )cos(1

)sin(

Cbaxa

dxbax )sin(1

)cos(5.

P. 67

1. Let u g(x) be a differentiable function. Then, 8.3 Integration by Substitution


.)()()]([ duufdxxg'xgf

2. If the integrated involves terms like , and , we can simplify the integrand by substituting x a sin , x a tan or x a sec respectively.

22 ax 22 xa 22 xa

P. 68

If u(x) and v(x) are two differentiable functions, then8.4 Integration by Parts


. vduuvudv

P. 69

1. If the slope of a curve at point (x, y) is f(x), then the equation of the family of curves is given by

where F (x) f(x).

8.5 Applications of Indefinite Integrals


CxFdxxf )()(

2. Let s, v and a be the displacement, velocity and acceleration of a particle moving along a straight line respectively, then

. and adtvvdts

8.1concepts of indefinite integrals 8.2indefinite integration of functions 8.3integration by...

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