math 152.02 calculus with analytic geometry ii€¦ · lecture 6 - 1/14 sketching antiderivatives...

Math 152.02

Lecture 4 - 1/10

Integrals andarea

TheFundamentalTheorem ofCalculus

Interpretingintegrals

Average value ofa function

Lecture 5 - 1/12

Properties ofdefinite integrals

Bounding onintegrals

Antiderivatives

Lecture 6 - 1/14

Sketchingantiderivatives

Basic integrationrules

Math 152.02Calculus with Analytic Geometry II

January 14, 2011

Math 152.02

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Integrals and area

Integrals and area

If a < b then∫ b

a

f (x) dx

= (area under f above x-axis)− (area above f under x-axis)

Example 18

∫ 0

−3

√9− x2 dx = π·32

4 = 9π4

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Problem 19

Compute

∫ 6

−2

2x − 5 dx

by finding areas of regions between the graph of f and the x-axis.

Solution to Problem 19

x-intercept

0 = 2x − 5

x = 52

(area under f above x-axis)

A2 = 12 ·(6− 5

2

)(2 · 6− 5)

= 494

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Solution to Problem 19 (continued)

(area above f under x-axis)

A1 = 12 ·(

52 − (−2)

)(−2 · (−2) + 5)

= 12 ·(

92

)(9)

= 814∫ 6

−2

2x − 5 dx = A2 − A1 = 494 −

814 = −8

Example 20 (Area formulas seldom work)

Evaluate

∫ 3

−5

√25− x2 dx .

No area formula for this portion of a circle

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Theorem 21 (Fundamental Theorem of Calculus I)

If f is continuous on the interval [a, b] and F ′(t) = f (t) then∫ b

a

f (t) dt = F (b)− F (a).

Example 22d

dx

(sin 8x + x4

)= 8 cos 8x + 4x3 so by FTOC∫ 2

1

8 cos 8x + 4x3 dx = (sin(8 · 2) + 24)− (sin(8 · 1) + 14)

= sin 16− sin 8 + 15

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Notation

g(x)∣∣∣bx=a

= g(b)− g(a)

or if x is clearly the variable to plug a and b into can write

g(x)∣∣∣ba

= g(b)− g(a)

With this notation FTOC I (Fundamental Theorem of Calculus I) saysIf f is continuous on the interval [a, b] and F ′(t) = f (t) then∫ b

a

f (t) dt = F (t)∣∣∣bt=a

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FTOC says roughly

Computing areaunder f

≈ Finding a functionwith derivative f

Suprising! Why should area and slopes be related?

Could prove FTOC from definition of integral (Definition 9) anddefinition of the derivative, but we won’t.

Makes some sense visually

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Interpreting integrals

Note on units for integrals

• If x measured in units u1 and

• f (x) measured in units u2 then∫ b

a

f (x) dx

has units u1 · u2

Example 23

• If t is the time measured in hours

• P(t) number of people working in a factory at time t.∫ b

a

P(t) dt

is people · hours worked between time a and b

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Integral of a rate of change

• If F ′(t) is the rate of change of some quantity F (t)

• then FTOC I say that∫ b

a

F ′(t) dt = F (b)− F (a)

This is net change in F from time a to b

Example 24

As in Example 4 from last week

• t time (in h)

• v(t) velocity (rate of change of position) at time t (in km/h)

• then∫ 3

0

v(t) dt

is distance traveled (net change in position) in units(km/h) · h = km

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Example 25

• t time (in years)

• g(t) growth rate (in m/year) of person at age t

• then∫ 16

9

g(t) dt

is change in height (in m) of a person from age 9 to 16.

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Average value of a function

Definition 26 (Average value of a function)

The average value of f on the interval [a, b] is

fave =1

b − a

∫ b

a

f (t) dt.

Where does this formula come from?

Let’s derive it using

Philosophy of Calculus

• Estimate a quantity

• Figure out how to improve estimate

• Take limit

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Average temperature

Let f (t) be the temperature at time t.

Let’s estimate the average temperature between 2am and 10pm.

• (First estimate with 2 intervals)• total time = 10− 2• each interval is 10−2

2•

fave ≈f(2 + 1 · 10−2

2

)+ f

(2 + 2 · 10−2

2

)2

• (Better estimate with 3 intervals)• total time = 10− 2• each interval is 10−2

3•

fave ≈f(2 + 1 · 10−2

3

)+ f

(2 + 2 · 10−2

3

)+ f

(2 + 3 · 10−2

3

)3

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Average temperature (continued)

• (Better estimate with n intervals)• total time = 10− 2• each interval is 10−2

n•

fave ≈f(2 + 1 · 10−2

n

)+ · · ·+ f

(2 + n · 10−2

n

)n

=n∑

i=1

f

(2 + i · 10− 2

n

)· 1n

=1

10− 2

n∑i=1

f

(2 + i · 10− 2

n

)· 10− 2

n

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Average temperature (continued)

• (Take limit to get exact average)

fave = limn→∞

1

10− 2

n∑i=1

f

(2 + i · 10− 2

n

)· 10− 2

n

=1

10− 2lim

n→∞

n∑i=1

f

(2 + i · 10− 2

n

)· 10− 2

n

=1

10− 2

∫ 10

2

f (t) dt

We’ve derived the formula in Definition 26.

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Problem 27

Find the average value of the funtion

h(x) =√

4− x2

on the interval [−2, 2]


have =1

b − a

∫ b

a

h(x) dx

=1

2− (−2)

∫ 2

−2

√4− x2 dx

=1

4· π · 2

2

2

= π2

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Problem 28

• t is time measured in days since Jan. 1, 2003

• R(t) is the distance from the earth to the sun at time t

What does1

365− 0

∫ 365

0

R(t) dt

represent?


Average distance from earth to the sun in 2003

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Properties of definite integrals

Theorem 29

If f is integrable then∫ b

a

f (x) dx = −∫ a

b

f (x) dx

Theorem 30


a

f (x) dx +

∫ c

b

f (x) dx =

∫ c

a

f (x) dx

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Theorem 31


a

cf (x) dx = c

∫ b

a

f (x) dx

Proof. ∫ b

a

cf (x) dx = limn→∞

n∑i=1

cf (a + i∆x)∆x

= limn→∞

cn∑

i=1

f (a + i∆x)∆x

= c limn→∞

n∑i=1

f (a + i∆x)∆x

= c

∫ b

a

f (x) dx

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Theorem 32

If f and g are integrable then∫ b

a

f (x) + g(x) dx =

∫ b

a

f (x) dx +

∫ b

a

g(x) dx

Proof.∫ b

a

f (x) + g(x) dx = limn→∞

n∑i=1

(f (a + i∆x) + g(a + i∆x)) ∆x

= limn→∞

n∑i=1

(f (a + i∆x)∆x + g(a + i∆x)∆x)

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Proof of Theorem 32 (continued).

= limn→∞

n∑i=1

(f (a + i∆x)∆x + g(a + i∆x)∆x)

= limn→∞

(n∑

i=1

f (a + i∆x)∆x +n∑

i=1

g(a + i∆x)∆x

)

=

(lim

n→∞

n∑i=1

f (a + i∆x)∆x

)+

(lim

n→∞

n∑i=1

g(a + i∆x)∆x

)

=

∫ b

a

f (x) dx +

∫ b

a

g(x) dx

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Problem 33

Compute∫ 16

7t(v)− 7u(v) dv given

•∫ 16

7u(v) dv = −2

•∫ 16

7t(v) dv = 12

Solution to Problem 33∫ 16

7

t(v)− 7u(v) dv =

∫ 16

7

t(v) dv −∫ 16

7

7u(v) dv

=

∫ 16

7

t(v) dv − 7

∫ 16

7

u(v) dv

= 12− 7(−2)

= 26

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Warning

∫ b

a

f (x) · g(x) dx 6=∫ b

a

f (x) dx ·∫ b

a

g(x) dx

∫ b

a

f (x)

g(x)dx 6=

∫ b

af (x) dx∫ b

ag(x) dx

Example 34

f (x) =

{1, 0 ≤ x ≤ 10, 1 < x ≤ 2

and g(x) =

{0, 0 ≤ x ≤ 11, 1 < x ≤ 2

∫ 2

0

f (x)g(x) dx =

∫ 2

0

0 dx = 0

but

∫ 2

0

f (x) dx ·∫ 2

0

g(x) dx = 1 · 1 = 1

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Definition 35

• f is even if f (−x) = f (x)

• f is odd if f (−x) = −f (x)

Note: Most functions are neither. One function is both.

Theorem 36

If f is even then∫ a

0

f (x) dx =

∫ 0

−af (x) dx = 1

2

∫ a

−af (x) dx

If f is odd then ∫ a

0

f (x) dx = −∫ 0

−af (x) dx

and ∫ a

−af (x) dx = 0

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Problem 37

Suppose

•∫ 4

2Q(s) ds = 3

•∫ 20

4Q(s) ds = −9

Evaluate ∫ 20

2

Q(s) ds


∫ 20

2

Q(s) ds =

∫ 4

2

Q(s) ds +

∫ 20

4

Q(s) ds

= 3 + (−9)

= −6

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Problem 38

Compute∫ 8

2y(u) du given

• y is an odd function

•∫ 8

−2y(u) du = 17


2

y(u) du =

∫ 8

−2

y(u) du −∫ 2

−2

y(u) du

= 17− 0

= 17

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Problem 39

Compute∫ 7

3h(r) dr given

• h is an even function

•∫ 7

0h(r) dr = 4

•∫ 7

−3h(r) dr = −10


3

h(r) dr =

∫ 7

−3

h(r) dr −∫ 3

−3

h(r) dr

=

∫ 7

−3

h(r) dr − 2

∫ 0

−3

h(r) dr

=

∫ 7

−3

h(r) dr − 2

(∫ 7

−3

h(r) dr −∫ 7

0

h(r) dr

)= −

∫ 7

−3

h(r) dr + 2

∫ 7

0

h(r) dr

= −(−10) + 2(4) = 18

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Bounding on integrals

Theorem 40

If m ≤ f (x) ≤ M for x ∈ [a, b] then

m(b − a) ≤∫ b

a

f (x) dx ≤ M(b − a)

Theorem 41

If f (x) ≤ g(x) for x ∈ [a, b] then∫ b

a

f (x) dx ≤∫ b

a

g(x) dx

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Problem 42

Show that

−199 ≤∫ 200

1

cos(x) · sin(x3) dx ≤ 199


For all x ∈ [1, 200]

| cos(x) · sin(x3)| = | cos(x)| · | sin(x3)| ≤ 1

so−1 ≤ cos(x) · sin(x3) ≤ 1

By Theorem 40

−1 · (200− 1) ≤∫ 200

1

cos(x) · sin(x3) dx ≤ 1 · (200− 1)

−199 ≤∫ 200

1

cos(x) · sin(x3) dx ≤ 199

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Problem 43

For each pairs of integrals decide which is the larger

1∫ π

4

0cos(x) dx and

∫ π4

0sin(x) dx

2∫ π

2π4

cos(x) dx and∫ π

2π4

sin(x) dx


1 For all x ∈ [0, π4 ]cos(x) ≥ sin(x)

so ∫ π4

0

cos(x) dx ≥∫ π

4

0

sin(x) dx

2 For all x ∈ [π4 ,π2 ]

cos(x) ≤ sin(x)

so ∫ π2

π4

cos(x) dx ≤∫ π

2

π4

sin(x) dx

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Antiderivatives

Definition 44 (An antiderivative)

F (x) is an antiderivative of f (x) if F ′(x) = f (x).

Example 45

ddx

(x2 cos(4x + 3)

)= 2x cos(4x + 3) + 4x2 sin(4x + 3)

sox2 cos(4x + 3)

is an antiderivative of

2x cos(4x + 3) + 4x2 sin(4x + 3)

x2 cos(4x + 3) + 30 is another antiderivative.

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Motivation

Why do we care about finding antiderivatives?

FTOC I says that computing ∫ b

a

f (t) dt

is easy if we have an antiderivative F of f .

Example 46

From Example 45 above∫ 7

1

2x cos(4x + 3) + 4x2 sin(4x + 3) dx

= x2 cos(4x + 3) + 30∣∣∣71

= (72 cos(4 · 7 + 3) + 30)− (12 cos(4 · 1 + 3) + 30)

= 49 cos(31)− cos(7)

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Notice in Example 45 we could have added any constant tox2 cos(4x + 3) and we would have had another antiderivative of2x cos(4x + 3) + 4x2 sin(4x + 3)

We usually add an unspecified constant to remind us that there aremany antiderivatives.

Definition 47 (The antiderivative)

The antiderivative of f (x) is the set of all antiderivatives of f (x).

Theorem 48

If f is continuous and F ′(x) = f (x) then every antiderivative of f isof the form

F (x) + C

for some constant C .

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What if f is not continuous?

The antiderivative of a noncontinuous function

Let

F (x) =

{ln x + 14, x > 0ln(−x) + 8, x < 0

Then

F ′(x) =

{1x , x > 0−1−x , x < 0

=

{1x , x > 01x , x < 0

=1

x

So F (x) is an antiderivative of 1x .

Any choice of constants (14 and 8 weren’t special) gives same result.

Thus the antiderivative of 1x is

F (x) =

{ln x + C1, x > 0ln(−x) + C2, x < 0

=

{ln |x |+ C1, x > 0ln |x |+ C2, x < 0

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On the other hand

• Main reason we care about antiderivatives is the FTOC.

• FTOC only applies if f is integrable on [a, b]

• 1x is not integrable on intervals containing 0

• so in applications we only use one of the two constants at a time

Example 49∫ −2

−3

1

xdx = ln |x |+C2

∣∣∣−2

−3= (ln |−2|+C2)−(ln |−3|+C2) = ln 2−ln 3

Example 50∫ 7

−4

1

xdx cannot be evaluated using FTOC

Notational warning

By convention we say that F (x) + C is the antiderivative of f (x)whenever F ′(x) = f (x) even when this is technically incorrect.

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Notation ∫f (x) dx = F (x) + C

means F (x) + C is the antiderivative of f (x)

Terminology

Since FTOC links antidifferentiation and integration we also callantiderivatives (indefinite) integrals.

The following statements all mean the same thing:

• f (x) = ddx F (x)

•∫

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

• f (x) is the derivative of F (x)

• F (x) + C is the antiderivative of f (x)

• F (x) + C is the indefinite integral of f (x)

• F (x) + C is the integral of f (x)

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Problem 51

Check the following integrals

1∫

(6x + 3ex) · cos(3x2 + 3ex) dx = sin(3x2 + 3ex) + C

2∫

sec x dx = ln | sec x + tan x |+ C


1d

dx sin(3x2 + 3ex) = (6x + 3ex) · cos(3x2 + 3ex) X

2 ddx ln | sec x + tan x | =

sec x tan x + sec2 x

sec x + tan x

= (sec x) · tan x + sec x

sec x + tan x

= sec x XNote: Similar to 1

x antiderivative of sec x should have a different

C for each interval [ (2n−1)π2 , (2n+1)π

2 ] but nobody does this.

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(See backboard)

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Basic integration rules

Each rule for differentiation gives us a rule for integration

Fromc d

dx F (x) = ddx

(cF (x)

)we get

Theorem 52 (Constant rule for integration)

∫cf (x) dx = c

∫f (x) dx

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Proof of Theorem 52.

Suppose ddx F (x) = f (x).

We have the derivative rule

c ddx F (x) = d

dx

(cF (x)

)Reinterpreting this rule as an antiderivative gives∫

c ddx F (x) dx = cF (x) + C .

Thus we may conclude∫cf (x) dx =

∫c d

dx F (x) dx

= cF (x) + C

= c(F (x) + C2)

= c

∫f (x) dx .

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From ddx F (x) + d

dx G (x) = ddx

(F (x) + G (x)

)we get

Theorem 53 (Sum rule for integration)

∫f (x) + g(x) dx =

∫f (x) dx +

∫g(x) dx

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Proof of Theorem 53.

Suppose ddx F (x) = f (x) and d

dx G (x) = g(x).

We have the derivative rule

ddx F (x) + d

dx G (x) = ddx

(F (x) + G (x)

)Reinterpreting this rule as an antiderivative gives∫

ddx F (x) + d

dx G (x) dx = F (x) + G (x) + C .

Thus we may conclude∫f (x) + g(x) dx =

∫d

dx F (x) + ddx G (x) dx

= F (x) + G (x) + C

=

∫f (x) dx +

∫g(x) dx

Note: We drop constants when we have integrals on both sides of anequation.

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Basic Integrals

Each basic derivative gives us a basic integral

Table 1: Basic integrals to memorize

differentiation integrationrule rule

ddx x r+1 = (r + 1)x r

∫x r dx = 1

r+1 x r+1 + C if r 6= −1

ddx ln |x | = 1

x

∫1x dx = ln |x |+ C

ddx cos x = − sin x

∫sin x dx = − cos x + C

ddx sin x = cos x

∫cos x dx = sin x + C

ddx ex = ex

∫ex dx = ex + C

ddx arctan x = 1

1+x2

∫1

1+x2 dx = arctan x + C

ddx arcsin x = 1√

1−x2

∫1√

1−x2dx = arcsin x + C

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More basic integrals

You also know a few more derivative rules

Table 2: More basic integrals to memorize

differentiation integrationrule rule

ddx tan x = sec2 x

∫sec2 x dx = tan x + C

ddx cot x = − csc2 x

∫csc2 x dx = − cot x + C

ddx sec x = sec x tan x

∫sec x tan x dx = sec x + C

ddx csc x = − csc x cot x

∫csc x cot x dx = − csc x + C

ddx ax = (ln a)ax

∫ax dx = ax

ln a + C

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Techniques of integration

Advanced derivative rules give us techniques of integration

differentiation technique ofrule integration

chain rule u-substitution (§7.1)

product rule integration by parts (§7.2)

We will return to these integration techniques later.

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 54

Find a formula for∫

10ex + 7 sin x dx

Solution to Problem 54∫10ex + 7 sin x dx =

∫10ex dx +

∫7 sin x dx (Sum rule)

= 10

∫ex dx + 7

∫sin x dx (Constant rule)

= 10ex − 7 cos x + C (Table 1)

Check your answer!

ddx (10ex − 7 cos x) = 10ex + 7 sin x X

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 55

Find a formula for∫

18√1−t2− 8t22 dt

Solution to Problem 55∫18√

1− t2− 8t22 dt =

∫18 · 1√

1− t2dt −

∫8 · t22 dt (Sum rule)

= 18

∫1√

1− t2dt − 8

∫t22 dt (Const. rule)

= 18 arcsin t − 8 · t23

23 + C (Table 1)

Check your answer!

ddt

(18 arcsin t − 8

23 · t23)

=18√

1− t2− 8

23 ·23t22 =18√

1− t2−8t22 X

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 56

Find a formula for∫ arcsin(3−π2)√

2+ sec2 u du


∫arcsin(3− π2)√

2+ cos u du

=

∫arcsin(3− π2)√

2du +

∫sec2 u du (Sum rule)

= arcsin(3−π2)√2

· u + tan u + C (Tables 1 and 2)

Check your answer!

ddt

(arcsin(3− π2)√

2· u + tan u

)=

arcsin(3− π2)√2

+ sec2 u X

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 57

Compute∫ 2

1

3√y(1−6 6

√y)2

3√y dy


1

3√

y(1− 6 6√

y)2

3√

ydy =

∫ 2

1

3y12 (1− 12y

16 + 36y

26 )

y13

dy

=

∫ 2

1

3y12 − 36y

46 + 108y

56 )

y13

dy

=

∫ 2

1

3y16 − 36y

26 + 108y

36 dy

=

∫ 2

1

3y16 − 36y

13 + 108y

12 dy

= 3 · 67 y

76 − 36 · 3

4 y43 + 108 · 2

3 y32

∣∣∣21

= 187 y

76 − 27y

43 + 72y

32

∣∣∣21

=(

187 · 2

76 − 27 · 2 4

3 + 72 · 2 32

)−(

187 + 45

)

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 58

Find an antiderivative G (x) of g(x) = sin x + 7 satisfyingG (π) = −20.


G (x) =

∫sin x + 7 dx

= − cos x + 7x + C

Use fact that G (π) = −20 to solve for C .

−20 = G (π) = − cosπ + C

SoC = −20 + cosπ = −20 + (−1) = −21

G (x) = − cos x + 7x − 21

Math 152.02

Lecture 4 - 1/10

Integrals andarea




Lecture 5 - 1/12



Antiderivatives

Lecture 6 - 1/14



Problem 59

The average value of h(x) = x3 − 3x2 on [−a, a] is 8 solve for a.


have =1

a− (−a)

∫ a

−ax3 − 3x2 dx

=1

2a· ( 1

4 x4 − x3)∣∣∣a−a

=1

2a·(

[ 14 (−a)4 − (−a)3]− ( 1

4 a4 − a3))

=1

2a·(

14 a4 + a3 − 1

4 a4 + a3)

=1

2a·(2a3)

= a2

Use fact that have = 8 to solve for a.

8 = have = a2 so a = ±√

8

math 152.02 calculus with analytic geometry ii€¦ · lecture 6 - 1/14 sketching antiderivatives...

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