5.3 The Definite Integral · 2017-01-22 · 5.3 The Definite Integral 343 The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over

5.3 The Definite Integral 343

The Definite Integral

In Section 5.2 we investigated the limit of a finite sum for a function defined over a closedinterval [a, b] using n subintervals of equal width (or length), In this sectionwe consider the limit of more general Riemann sums as the norm of the partitions of [a, b]approaches zero. For general Riemann sums the subintervals of the partitions need nothave equal widths. The limiting process then leads to the definition of the definite integralof a function over a closed interval [a, b].

Limits of Riemann Sums

The definition of the definite integral is based on the idea that for certain functions, as thenorm of the partitions of [a, b] approaches zero, the values of the corresponding Riemann

sb - ad>n .

5.3

sums approach a lim

iting value I. What w

e mean by this converging idea is that a R

iemann

sum w

ill be close to the number I

provided that the norm of its partition is sufficiently

small (so that all of its subintervals have thin enough w

idths). We introduce the sym

bol as a sm

all positive number that specifies how

close to Ithe R

iemann sum

must be, and the

symbol

as a second small positive num

ber that specifies how sm

all the norm of a parti-

tion must be in order for that to happen. H

ere is a precise formulation.

d

P

344Chapter 5: Integration

DEFINITION

The Definite Integral as a Limit of Riem

ann Sums

Let ƒ(x) be a function defined on a closed interval [a,b]. W

e say that a number I

is the definite integral of ƒover [a,

b]and that I

is the limit of the R

iemann

sums

if the following condition is satisfied:

Given any num

ber there is a corresponding num

ber such that

for every partition of [a,b] w

ith and any choice of

in w

e have

`a n

k=1 ƒsc

k d ¢x

k-

I `6

P.

[xk-

1 , xk ],

ck

7P 76d

P=5

x0 , x

1 ,Á, x

n 6d

70

P7

0g

nk=1 ƒsc

k d ¢x

k

Leibniz introduced a notation for the definite integral that captures its construction as

a limit of R

iemann sum

s. He envisioned the finite sum

s becom

ing an infi-nite sum

of function values ƒ(x) multiplied by “infinitesim

al” subinterval widths dx. T

hesum

symbol

is replaced in the limit by the integral sym

bol w

hose origin is in theletter “S

.” The function values

are replaced by a continuous selection of function val-ues ƒ(x). T

he subinterval widths

become the differential dx. It is as if w

e are summ

ingall products of the form

as x

goes from a

to b. While this notation captures the

process of constructing an integral, it is Riem

ann’s definition that gives a precise meaning

to the definite integral.

Notation and Existence of the Definite Integral

The sym

bol for the number I

in the definition of the definite integral is

which is read as “the integral from

ato b

of ƒof x

dee x” or sometim

es as “the integralfrom

ato b

of ƒof x

with respect to x.” T

he component parts in the integral sym

bol alsohave nam

es:

⌠⎮⎮⌡

⎧⎪⎪⎨⎪⎪⎩

The function is the integrand.

x is the variable of integration.

When you find the value

of the integral, you haveevaluated the integral.

Upper lim

it of integration

Integral sign

Low

er limit of integration

Integral of f from a to b

a bf(x) dx

Lb

aƒsxd dx

ƒsxd #dx ¢x

k

ƒsck d

1,

ag

nk=1 ƒsc

k d ¢x

k

When the definition is satisfied, w

e say the Riem

ann sums of ƒ

on [a,b] convergeto

the definite integral and that ƒ

is integrableover [a,b]. W

e have many

choices for a partition Pw

ith norm going to zero, and m

any choices of points for each

partition. The definite integral exists w

hen we alw

ays get the same lim

it I, no matter w

hatchoices are m

ade. When the lim

it exists we w

rite it as the definite integral

When each partition has n

equal subintervals, each of width

we w

illalso w

rite

The lim

it is always taken as the norm

of the partitions approaches zero and the number of

subintervals goes to infinity.T

he value of the definite integral of a function over any particular interval depends onthe function, not on the letter w

e choose to represent its independent variable. If we decide

to use tor uinstead of x, w

e simply w

rite the integral as

No m

atter how w

e write the integral, it is still the sam

e number, defined as a lim

it of Rie-

mann sum

s. Since it does not m

atter what letter w

e use, the variable of integration is calleda dum

my variable.

Since there are so m

any choices to be made in taking a lim

it of Riem

ann sums, it

might seem

difficult to show that such a lim

it exists. It turns out, however, that no m

atterw

hat choices are made, the R

iemann sum

s associated with a continuous

function convergeto the sam

e limit.

Lb

aƒstd dt

or L

b

aƒsud du

instead of L

b

aƒsxd dx.

limn:

q a n

k=1 ƒsc

k d ¢x

=I=L

b

aƒsxd dx. ¢

x=sb

-ad>n

,

limƒƒ P

ƒƒ :0 a n

k=1 ƒsc

k d ¢x

k=

I=L

b

aƒsxd dx.

ck

I=1

baƒsxd dx

5.3The Definite Integral

345

THEOREM

1The Existence of Definite Integrals

A continuous function is integrable. T

hat is, if a function ƒis continuous on an

interval [a,b], then its definite integral over [a,b] exists.

By the E

xtreme V

alue Theorem

(Theorem

1, Section 4.1), w

hen ƒis continuous w

ecan choose

so that gives the m

aximum

value of ƒon

giving an uppersum

. We can choose

to give the minim

um value of ƒ

on giving a low

er sum.

We can pick

to be the midpoint of

the rightmost point

or a random point.

We can take the partitions of equal or varying w

idths. In each case we get the sam

e limit

for as

The idea behind T

heorem 1 is that a R

iemann sum

asso-ciated w

ith a partition is no more than the upper sum

of that partition and no less than thelow

er sum. T

he upper and lower sum

s converge to the same value w

hen A

ll otherR

iemann sum

s lie between the upper and low

er sums and have the sam

e limit. A

proofof T

heorem 1 involves a careful analysis of functions, partitions, and lim

its along thisline of thinking and

is left to a more advanced text. A

n indication of this proof is given inE

xercises 80 and 81.

7P 7:0

.

7P 7:0

.g

nk=1 ƒsc

k d ¢x

k

xk ,

[xk-

1 , xk ],

ck

[xk-

1 , xk ],

ck

[xk-

1 , xk ],

ƒsck d

ck

Theorem

1 says nothing about how to calculate

definite integrals. A m

ethod of calcu-lation w

ill be developed in Section 5.4, through a connection to the process of taking anti-

derivatives.

Integrable and Nonintegrable Functions

Theorem

1 tells us that functions continuous over the interval [a,b] are integrable there.F

unctions that are not continuous may or m

ay not be integrable. Discontinuous functions

that are integrable include those that are increasing on [a,b] (Exercise 77), and the

piecewise-continuous functions

defined in the Additional E

xercises at the end of this chap-ter. (T

he latter are continuous except at a finite number of points in [a,b].) For integrabil-

ity to fail, a function needs to be sufficiently discontinuous so that the region between its

graph and the x-axis cannot be approximated w

ell by increasingly thin rectangles. Here is

an example of a function that is not integrable.

EXAMPLE 1

A Nonintegrable Function on [0, 1]

The function

has no Riem

ann integral over [0, 1]. Underlying this is the fact that betw

een any two num

-bers there is both a rational num

ber and an irrational number. T

hus the function jumps

upand dow

n too erratically over [0, 1] to allow the region beneath its graph and above the

x-axis to be approximated by rectangles, no m

atter how thin they are. W

e show, in fact, that

upper sum approxim

ations and lower sum

approximations converge to different lim

itingvalues.

If we pick a partition P

of [0, 1] and choose to be the m

aximum

value for ƒon

then the corresponding Riem

ann sum is

since each subinterval contains a rational num

ber where

Note that the

lengths of the intervals in the partition sum to 1,

So each such R

iemann

sum equals 1, and a lim

it of Riem

ann sums using these choices equals 1.

On the other hand, if w

e pick to be the m

inimum

value for ƒon

then theR

iemann sum

is

since each subinterval contains an irrational num

ber w

here T

helim

it of Riem

ann sums using these choices equals zero. S

ince the limit depends on the

choices of the function ƒ

is not integrable.

Properties of Definite Integrals

In defining as a lim

it of sums

we m

oved from left to right

across the interval [a,b]. What w

ould happen if we instead m

ove right to left, starting with

and ending at E

ach in the R

iemann sum

would change its sign, w

ithnow

negative instead of positive. With the sam

e choices of in each subinter-

val, the sign of any Riem

ann sum w

ould change, as would the sign of the lim

it, the integralc

kx

k-

xk-

1

¢x

kx

n=

a.

x0=

b

gnk=

1 ƒsck d ¢

xk ,

1baƒsxd dx

ck ,

ƒsck d

=0

.c

k[x

k-1 , x

k ]

L=a n

k=1 ƒsc

k d ¢x

k=a n

k=1 s0d ¢

xk=

0,

[xk-

1 , xk ],

ck

gnk=

1 ¢x

k=

1. ƒsc

k d=

1.

[xk-

1 , xk ]

U=a n

k=1 ƒsc

k d ¢x

k=a n

k=1 s1d ¢

xk=

1,

[xk-

1 , xk ]

ck

ƒsxd=e

1,if x is rational

0,if x is irrational


Since w

e have not previously given a meaning to integrating backw

ard, we are

led to define

Another extension of the integral is to an interval of zero w

idth, when

Since

is zero when the interval w

idth , w

e define

Theorem

2 states seven properties of integrals, given as rules that they satisfy, includ-ing the tw

o above. These rules becom

e very useful in the process of computing integrals.

We w

ill refer to them repeatedly to sim

plify our calculations.R

ules 2 through 7 have geometric interpretations, show

n in Figure5.11. T

he graphs inthese figures are of positive functions, but the rules apply to general integrable functions.

La

aƒsxd dx

=0

.

¢x

k=

0ƒsc

k d ¢x

k

a=

b.

La

bƒsxd dx

=-L

b

aƒsxd dx.

1abƒsxd dx.


347

THEOREM

2W

henƒ

andg

areintegrable,the

definiteintegralsatisfies

Rules

1to

7in

Table5.3.

TABLE 5.3Rules satisfied by definite integrals

1.O

rder of Integration:A

Definition

2.Z

ero Width Interval:

Also a D

efinition

3.C

onstant Multiple:

Any N

umber k

4.Sum

and Difference:

5.A

dditivity:

6.M

ax-Min Inequality:

If ƒhas m

aximum

value max

ƒand m

inimum

value m

in ƒon [a,b], then

7.D

omination:

(Special C

ase)ƒsxd

Ú0 on [a, b] Q

Lb

aƒsxd dx

Ú0

ƒsxdÚ

gsxd on [a, b] Q L

b

aƒsxd dx

ÚLb

agsxd dx

min ƒ #sb

-ad

…Lb

aƒsxd dx

… m

ax ƒ #sb-

ad.

Lb

aƒsxd dx

+Lc

bƒsxd dx

=Lc

aƒsxd dx

Lb

asƒsxd

;gsxdd dx

=Lb

aƒsxd dx

;Lb

agsxd dx

k=

-1

Lb

a -

ƒsxd dx=

-Lb

aƒsxd dx

Lb

akƒsxd dx

=kL

b

aƒsxd dx

La

aƒsxd dx

=0

La

bƒsxd dx

=-L

b

aƒsxd dx

While R

ules 1 and 2 are definitions, Rules 3 to 7 of Table 5.3

must be proved. T

heproofs are based on the definition of the definite integral as a lim

it of Riem

ann sums. T

hefollow

ing is a proof of one of these rules. Sim

ilar proofs can be given to verify the otherproperties in Table 5.3.

Proof of Rule 6R

ule 6 says that the integral of ƒover [a,b] is never sm

aller than them

inimum

value of ƒtim

es the length of the interval and never larger than the maxim

umvalue of ƒ

times the length of the interval. T

he reason is that for every partition of [a,b]and for every choice of the points

Constant M

ultiple Rule

Constant M

ultiple Rule

=m

ax ƒ #sb-

ad.

=m

ax ƒ #a n

k=1 ¢

xk

ƒsck d

… m

ax f …a n

k=1 m

ax ƒ #¢x

k

min ƒ

…ƒsc

k d …a n

k=1 ƒsc

k d ¢x

k

=a n

k=1 m

in ƒ #¢x

k

a n

k=1 ¢

xk=

b-

a m

in ƒ #sb-

ad=

min ƒ #a n

k=1 ¢

xk

ck ,


x

y0a y �

f(x)

x

y0a

b y � f(x)

y � 2f(x)

x

y0a

b y � f(x)

y � f(x) �

g(x)

y � g(x)

x

y0a

cb

y � f(x)

b

a f(x) dxf(x) dx

Lc

bL

x

y0a

b y � f(x)

max f

min f

x

y0a

b

y � f(x)

y � g(x)

FIGURE 5.11

(a) Zero W

idth Interval:

(The area over a point is 0.)

La

aƒsxd dx

=0

.

(b) Constant M

ultiple:

(Show

n for )

k=

2.

Lb

a kƒsxd dx

=kL

b

a ƒsxd dx

.

(c) Sum:

(Areas add)

Lb

asƒsxd

+gsxdd dx

=Lb

aƒsxd dx

+Lb

agsxd dx

(d) Additivity for definite integrals:

Lb

aƒsxd dx

+Lc

bƒsxd dx

=Lc

aƒsxd dx

(e) Max-M

in Inequality:

…m

ax ƒ #sb

-ad

min

ƒ #sb-

ad…L

b

a ƒsxd dx

(f) Dom

ination:

QLb

a ƒsxd dx

ÚLb

a gsxd dx

ƒsxdÚ

gsxd on [a, b]

In short, all Riem

ann sums for ƒ

on [a,b] satisfy the inequality

Hence their lim

it, the integral, does too.

EXAMPLE 2

Using the Rules for Definite Integrals

Suppose that

Then

1.R

ule 1

2.R

ules 3 and 4

3.R

ule 5

EXAMPLE 3

Finding Bounds for an Integral

Show

that the value of is less than

.

SolutionT

heM

ax-Min

Inequalityfordefinite

integrals(R

ule6)says

that

is a lower bound

for the value of and that

is an upper bound.

The m

aximum

value of on [0, 1] is

so

Since

is bounded from above by

(which is 1.414

), the integralis less than

.

Area Under the Graph of a N

onnegative Function

We now

make precise the notion of the area of a region w

ith curved boundary, capturingthe idea of approxim

ating a region by increasingly many rectangles. T

he area under thegraph of a nonnegative continuous function is defined to be a definite integral.

3>2Á

22

110 2

1+

cos x dx

L1

0 21

+cos x dx

…22 #s1

-0d

=22

.

21

+1

=22

,2

1+

cos x

max ƒ #sb

-ad

1baƒsxd dx

min ƒ #sb

-ad

3>21

10 21

+cos x dx

L4

-1 ƒsxd dx

=L1

-1 ƒsxd dx

+L4

1ƒsxd dx

=5

+s-

2d=

3

=2s5d

+3s7d

=31

L1

-1 [2ƒsxd

+3hsxd] dx

=2L

1

-1 ƒsxd dx

+3L

1

-1 hsxd dx

L1

4ƒsxd dx

=-L

4

1ƒsxd dx

=-s-

2d=

2

L1

-1 ƒsxd dx

=5,

L4

1ƒsxd dx

=-

2, L

1

-1 hsxd dx

=7

.

min ƒ #sb

-ad

…a n

k=1 ƒsc

k d ¢x

k…

max ƒ #sb

-ad.


349

DEFINITION

Area Under a Curve as a Definite Integral

If is nonnegative and integrable over a closed interval [a,b], then the

area under the curve over [a,b] is the integral of ƒ

from a

to b,

A=L

b

aƒsxd dx.

y=

ƒsxdy

=ƒsxd

For the first time w

e have a rigorous definition for the area of a region whose bound-

ary is the graph of any continuous function. We now

apply this to a simple exam

ple, thearea under a straight line, w

here we can verify that our new

definition agrees with our pre-

vious notion of area.

EXAMPLE 4

Area Under the Line

Com

pute and find the area A

under over the interval [0,b],

SolutionT

heregion

ofinterestisa

triangle(Figure

5.12).We

compute

thearea

intw

o ways.

(a)To

compute

the definite

integral as

the lim

it of

Riem

ann sum

s, w

e calculate

for partitions whose norm

s go to zero. Theorem

1 tells us thatit does not m

atter how w

e choose the partitions or the points as long as the norm

sapproach zero. A

ll choices give the exact same lim

it. So w

e consider the partition Pthatsubdivides

theinterval[0,b]into

nsubintervals

ofequalwidth

and we choose

to be the right endpoint in each subinterval. The partition is

and S

o

Constant M

ultiple Rule

Sum

of First nIntegers

As

and this last expression on the right has the lim

it T

herefore,

(b)S

ince the area equals the definite integral for a nonnegative function, we can quickly

derive the definite integral by using the formula for the area of a triangle having base

length band height

The area is

Again w

e have

that

Exam

ple 4

can be

generalized to

integrate over

any closed

interval

Rule 5

Rule 1

Exam

ple 4 =

-a

22+

b22.

=-L

a

0x dx

+Lb

0x dx

Lb

ax dx

=L0

ax dx

+Lb

0x dx

[a, b], 06

a6

b.

ƒsxd=

x

1b0 x dx

=b

2>2.A

=s1>2d b #b

=b

2>2.

y=

b.

Lb

0x dx

=b

22.

b2>2.

7P 7:0

,n:

q

=b

22 s1

+1n d

=b

2

n2 # nsn

+1d

2

=b

2

n2 a n

k=1 k

=a n

k=1 kb

2

n2

ƒsck d

=c

k a n

k=1 ƒsc

k d ¢x

=a n

k=1 kbn

# bn

ck=

kbn.

P=e

0, bn, 2bn

, 3bn, Á

, nbn fc

kb>n

,¢

x=sb

-0d>n =

ck

limƒƒ Pƒƒ :

0 gnk=

1 ƒsck d ¢

xk

b7

0.

y=

xL

b

0x dx

y=

x


x

y0 b

b b

y � x

FIGURE 5.12

The region in

Exam

ple 4 is a triangle.

In conclusion,we have the follow

ing rule for integrating f(x)=

x:


351

(1)L

b

ax dx

=b

22-

a22,

a6

b

This com

putation gives the area of a trapezoid (Figure5.13). E

quation (1) remains valid

when a

and bare negative. W

hen the definite integral value

is anegative num

ber, the negative of the area of a trapezoid dropping down to the line

below the x-axis. W

hen and

Equation (1) is still valid and the definite inte-

gral gives the difference between tw

o areas, the area under the graph and above [0,b]m

inus the area below [a, 0] and over the graph.

The follow

ing results can also be established using a Riem

ann sum calculation sim

ilarto that in E

xample 4 (E

xercises 75 and 76).

b7

0,

a6

0y

=x

sb2-

a2d>2

a6

b6

0,

(2)

(3)L

b

ax

2 dx=

b33

-a

33,

a6

b

Lb

ac dx

=csb

-ad,

c any constant

x

y0 a

a

b

b

a

b

b � a

y � x

FIGURE 5.13

The area of

this trapezoidal region isA

=sb

2-

a2d>2

.

Average Value of a Continuous Function Revisited

In Section 5.1

we introduced inform

ally the average value of a nonnegative continuousfunction ƒ

over an interval [a,b], leading us to define this average as the area under the

graph of divided by

In integral notation we w

rite this as

We can use this form

ula to give a precise definition of the average value of any continuous(or integrable) function, w

hether positive, negative or both.A

lternately, we can use the follow

ing reasoning. We start w

ith the idea from arith-

metic that the average of n

numbers is their sum

divided by n. A continuous function ƒ

on[a,b] m

ay have infinitely many values, but w

e can still sample them

in an orderly way. W

edivide [a,b] into n

subintervals of equal width

and evaluate ƒat a point

in each (Figure5.14).T

he average of the nsam

pled values is

=1

b-

a a n

k=1 ƒsc

k d ¢x

¢x

=b

-a

n, so

1n=

¢x

b-

a =

¢x

b-

a a n

k=1 ƒsc

k d

ƒsc1 d

+ƒsc

2 d+

Á+

ƒscn d

n=

1n a n

k=1 ƒsc

k d

ck

¢x

=sb

-ad>n

Average

=1

b-

a L

b

aƒsxd dx.

b-

a.

y=

ƒsxd

x

y0

(ck , f(ck ))

y � f(x)

xn � b

ckx

0 � a

x1

FIGURE 5.14

A sam

ple of values of afunction on an interval [a, b].

The average is obtained by dividing a R

iemann sum

for ƒon [a,b] by

As w

eincrease the size of the sam

ple and let the norm of the partition approach zero, the average

approaches B

oth points of view lead us to the follow

ing definition.(1>(b

-a))1

baƒsxd dx.

sb-

ad.


DEFINITION

The Average or Mean Value of a Function

If ƒis integrable on [a,b], then its average value on

[a,b], also called its mean

value, is

avsƒd=

1b

-a

Lb

aƒsxd dx.

EXAMPLE 5

Finding an Average Value

Find the average value of on

SolutionW

e recognize as a function w

hose graph is the upper semi-

circle of radius 2 centered at the origin (Figure5.15).

The area betw

een the semicircle and the x-axis from

to 2 can be com

puted usingthe geom

etry formula

Because ƒ

is nonnegative, the area is also the value of the integral of ƒfrom

to 2,

Therefore, the average value of ƒ

is

avsƒd=

12

-s-

2d L2

-2 2

4-

x2 dx

=14

s2pd=p2

.

L2

-2 2

4-

x2 dx

=2p

.

-2

Area

=12

# pr

2=

12 # p

s2d 2=

2p

.

-2

ƒsxd=2

4-

x2

[-2, 2].

ƒsxd=2

4-

x2

–2–1

12

1 2

x

yf(x) �

�4 �

x2y �

�2

FIGURE 5.15

The average value of

on is

(Exam

ple 5).p>2

[-2, 2]

ƒsxd=2

4-

x2

352 Chapter 5: Integration

EXERCISES 5.3

Expressing Limits as IntegralsExpress the limits in Exercises 1–8 as definite integrals.

1. where P is a partition of [0, 2]

2. where P is a partition of


4. where P is a partition of [1, 4]limƒ ƒP ƒ ƒ:0

an

k=1a 1

ckb ¢xk ,

[-7, 5]limƒ ƒP ƒ ƒ:0

an

k=1sck

2 - 3ckd ¢xk ,

[-1, 0]limƒ ƒP ƒ ƒ:0

an

k=12ck

3 ¢xk ,

limƒ ƒP ƒ ƒ:0

an

k=1ck

2 ¢xk ,




8. where P is a partition of [0, p>4]limƒ ƒP ƒ ƒ:0

an

k=1stan ckd ¢xk ,

[-p>4, 0]limƒ ƒP ƒ ƒ:0

an

k=1ssec ckd ¢xk ,

limƒ ƒP ƒ ƒ:0

an

k=124 - ck

2 ¢xk ,

limƒ ƒP ƒ ƒ:0

an

k=1

11 - ck

¢xk ,

Using Properties and Known Values to Find

OtherIntegrals

9.S

uppose that ƒand g

are integrable and that

Use the rules in Table 5.3

to find

a.b.

c.d.

e.f.

10.S

uppose that ƒand h

are integrable and that


to find

a.b.

c.d.

e.f.

11.S

uppose that Find

a.b.

c.d.

12.S

uppose that Find

a.b.

c.d.

13.S

uppose that ƒis integrable and that

and

Find

a.b.

14.S

uppose that his integrable and that

and

Find

a.b.

Using Area to Evaluate Definite IntegralsIn E

xercises 15–22, graph the integrands and use areas to evaluate theintegrals.

15.16.L

3/2

1/2s-

2x+

4d dxL

4

-2 a

x2+

3b dx

-L1

3hsud du

L3

1hsrd dr

13-1 hsrd dr

=6

.1

1-1 hsrd dr

=0

L3

4ƒstd dt

L4

3ƒszd dz

140

ƒszd dz=

7.

130 ƒszd dz

=3

L0

-3 gsrd

22

drL

0

-3 [-

gsxd] dx

L0

-3 gsud du

L-

3

0gstd dt 1

0-3 gstd dt

=22

.L

2

1[-

ƒsxd] dxL

1

2ƒstd dt

L2

1 23ƒszd dz

L2

1ƒsud du 1

21ƒsxd dx

=5

.L

7

9[hsxd

-ƒsxd] dx

L7

1ƒsxd dx

L1

9ƒsxd dx

L9

7[2ƒsxd

-3hsxd] dx

L9

7[ƒsxd

+hsxd] dx

L9

1 -

2ƒsxd dx

L9

1ƒsxd dx

=-

1, L9

7ƒsxd dx

=5, L

9

7hsxd dx

=4

.

L5

1[4ƒsxd

-gsxd] dx

L5

1[ƒsxd

-gsxd] dx

L5

2ƒsxd dx

L2

13ƒsxd dx

L1

5gsxd dx

L2

2gsxd dx

L2

1ƒsxd dx

=-

4, L5

1ƒsxd dx

=6, L

5

1gsxd dx

=8

.

17.18.

19.20.

21.22.

Use areas to evaluate the integrals in E

xercises 23–26.

23.24.

25.26.

EvaluationsU

se the results of Equations (1) and (3) to evaluate the integrals in

Exercises 27–38.

27.28.

29.

30.31.

32.

33.34.

35.

36.37.

38.


and Equations (1)–(3) to evaluate the inte-

grals in Exercises 39–50.

39.40.

41.42.

43.44.

45.46.

47.48.

49.50.

Finding AreaIn E

xercises 51–54 use a definite integral to find the area of the regionbetw

een the given curve and the x-axis on the interval [0, b].

51.52.

53.54.

y=

x2+

1y

=2x

y=p

x2

y=

3x2

L0

1s3x

2+

x-

5d dxL

2

0s3x

2+

x-

5d dx

L1

1/2 24u2 du

L2

13u

2 du

L0

3s2z

-3d dz

L1

2 a1+

z2 b dz

L2

2

0 A t-2

2B dtL

2

0s2t

-3d dt

L5

3 x8

dxL

2

05x dx

L-

2

02

2 dxL

1

37 dx

L3b

0x

2 dxL2

3b

0x

2 dxL2

3a

ax dx

L2a

ax dx

Lp

/2

0u

2 du

L1/2

0t 2 dt

L0.3

0s

2 dsL2

37

0x

2 dxL

522

22

r dr

L2p

p

u du

L2.5

0.5x dx

L2

2

1 x dx

Lb

a3t dt,

06

a6

bL

b

a2s ds,

06

a6

b

Lb

04x dx,

b7

0L

b

0 x2

dx, b

70

L1

-1 A 1

+21

-x

2B dxL

1

-1 s2

-ƒ xƒ d dx

L1

-1 s1

-ƒ xƒ d dx

L1

-2 ƒ xƒ dx

L0

-4 2

16-

x2 dx

L3

-3 2

9-

x2 dx


353

Average ValueIn E

xercises 55–62, graph the function and find its average value overthe given interval.

55.

56.on

[0, 3]57.

on[0, 1]

58.on

[0, 1]

59.on

[0, 3]

60.on

61.on

a.b.[1, 3], and c.

62.on

a.b.[0, 1], and c.

Theory and Examples

63.W

hat values of aand b

maxim

ize the value of

(Hint:

Where is the integrand positive?)

64.W

hat values of aand b

minim

ize the value of

65.U

se the Max-M

in Inequality to find upper and lower bounds for

the value of

66.(C

ontinuation of Exercise 65) U

se the Max-M

in Inequality tofind upper and low

er bounds for

Add these to arrive at an im

proved estimate of

67.S

how that the value of

cannot possibly be 2.

68.S

how that the value of

lies between

and 3.

69.Integrals of nonnegative functions

Use the M

ax-Min Inequal-

ity to show that if ƒ

is integrable then

70.Integralsofnonpositive

functionsShow

thatifƒis

integrablethen

71.U

se the inequality sin w

hich holds for to find an

upper bound for the value of 110 sin

x dx.

xÚ

0,

x…

x,

ƒsxd…

0 on

[a, b] Q

Lb

aƒsxd dx

…0

.

ƒsxdÚ

0 on

[a, b] Q

Lb

aƒsxd dx

Ú0

. 222

L2.8

101 2

x+

8 dx

110 sinsx

2d dx

L1

0

11

+x

2 dx.

L0.5

0

11

+x

2 dx and L

1

0.5 1

1+

x2 dx

.

L1

0

11

+x

2 dx.

Lb

asx

4-

2x2d dx

?

Lb

asx

-x

2d dx?

[-1, 1]

[-1, 0],

hsxd=

-ƒ xƒ

[-1, 3]

[-1, 1],

gsxd=

ƒ xƒ-

1

[-2, 1]

ƒstd=

t 2-

t

ƒstd=st

-1d 2

ƒsxd=

3x2-

3

ƒsxd=

-3x

2-

1ƒsxd

=-

x22

ƒsxd=

x2-

1 on C 0, 2

3D72.

The inequality sec

holds on U

se itto find a low

er bound for the value of

73.If av(ƒ) really is a typical value of the integrable function ƒ(x) on[a,b], then the num

ber av(ƒ) should have the same integral over

[a,b] that ƒdoes. D

oes it? That is, does

Give reasons for your answ

er.

74.It w

ould be nice if average values of integrable functions obeyedthe follow

ing rules on an interval [a,b].

a.b.c.Do these rules ever hold? G

ive reasons for your answers.

75.U

se limits of R

iemann sum

s as in Exam

ple 4a to establish Equa-

tion (2).

76.U

se limits of R

iemann sum

s as in Exam

ple 4a to establish Equa-

tion (3).

77.U

pper and lower sum

s for increasing functions

a.S

uppose the graph of a continuous function ƒ(x) rises steadilyas x

moves from

left to right across an interval [a,b]. Let P

bea partition of [a,b] into n

subintervals of length S

howby

referringto

the accompanying figure that

the difference between the upper and low

er sums for ƒ

on thispartition can be represented graphically as the area of arectangle R

whose dim

ensions are by

(Hint: T

he difference is the sum

of areas of rectanglesw

hosediagonals

liealong

thecurve. T

here is no overlapping when these rectangles are

shifted horizontally onto R.)

b.S

uppose that instead of being equal, the lengths of the

subintervals of the partition of [a,b] vary in size. Show

that

where

is the norm of P

, and hence that

x

y0x

0 � a

xn � b

x1

Q1

Q2

Q3x

2

y � f(x)

f(b)�

f(a)R

Δx

sU-

Ld=

0.

limƒƒ Pƒƒ :

0¢

xm

ax

U-

L…

ƒ ƒsbd-

ƒsadƒ ¢x

max ,

¢x

k

Q0

Q1 , Q

1 Q

2 ,Á, Q

n-1 Q

n

U-

L¢

x.

[ƒsbd-

ƒsad]

sb-

ad>n.

¢x =

avsƒd…

avsgd if

ƒsxd…

gsxd on

[a, b].

avskƒd=

k avsƒd sany num

ber kdavsƒ

+gd

=avsƒd

+avsgd

Lb

a avsƒd dx

=Lb

aƒsxd dx

?

110 sec x dx

.s-p>2, p>2d.

xÚ

1+sx

2>2d354

Chapter 5: Integration

78.U

pper and lower sum

s for decreasing functions(C

ontinuationof E

xercise 77)

a.D

raw a figure like the one in E

xercise 77 for a continuousfunction ƒ(x) w

hose values decrease steadily as xm

oves fromleft to right across the interval [a,b]. L

et Pbe a partition of

[a,b] into subintervals of equal length. Find an expression forthat is analogous to the one you found for

inE

xercise 77a.

b.S

uppose that instead of being equal, the lengths of the

subintervals of Pvary in size. S

how that the inequality

of Exercise 77b still holds and hence that

79.U

se the formula

to find the area under the curve from

to

in two steps:

a.Partition the interval

into nsubintervals of equal

length and calculate the corresponding upper sum U

; then

b.Find the lim

it of Uas

and

80.S

uppose that ƒis continuous and nonnegative over [a,b], as in the

figure at the right. By inserting points

asshow

n,divide[a,b]into

nsubintervals

oflengthsw

hich need not be equal.

a.If

explain theconnection betw

een the lower sum

and the shaded region in the first part of the figure.

b.If

explain theconnection betw

een the upper sum

and the shaded region in the second part of the figure.

c.E

xplain the connection between

and the shadedregions along the curve in the third part of the figure.

81.W

e say ƒis uniform

ly continuouson [a,

b] if given any there is a

such that if are in [a,b] and

then It can be show

n that a continuousfunction on [a,b] is uniform

ly continuous. Use this and the figure

at the right to show that if ƒ

is continuous and is given, it is

possible to make

by making the largest of

the sufficiently sm

all.

82.If you average 30 m

ih on a 150-m

i trip and then return over thesam

e 150 mi at the rate of 50 m

ih, w

hat is your average speed forthe

trip? G

ive reasons

for your

answer.

(Source:

David

H.

>>

¢x

k ’sU

-L

…P #sb

-ad

P7

0

ƒ ƒsx1 d

-ƒsx

2 dƒ6

P.

ƒ x1-

x2 ƒ

6d

x1 , x

2d

70

P7

0

U-

L

U=

M1 ¢

x1+

M2 ¢

x2+

Á+

Mn ¢

xn

Mk=

max

5ƒsxd for x in the k

th subinterval6,

L=

m1 ¢

x1+

m2 ¢

x2+

Á+

mn ¢

xn

mk=

min 5

ƒsxd for x in the kth subinterval6

,

¢x

2=

x2-

x1 ,Á

, ¢x

n=

b-

xn-

1 ,¢

x1=

x1-

a,

x1 , x

2 ,Á, x

k-1 , x

k ,Á, x

n-1

¢x

=sb

-ad>n:

0.

n:q

[0, p>2]x

=p>2

x=

0y

=sin

x

=cos sh>2d

-cos ssm

+s1>2ddhd

2 sin

sh>2dsin

h+

sin 2h

+sin

3h+

Á+

sin m

h

sU-

Ld=

0.

limƒƒ Pƒƒ :

0

U-

L…

ƒ ƒsbd-

ƒsadƒ ¢x

max

¢x

k

U-

LU

-L

Pleacher, T

he Mathem

atics Teacher, Vol. 85, N

o. 6, pp. 445–446,S

eptember 1992.)

COMPU

TER EXPLORATIONS

Finding Riemann Sum

sIf your C

AS

can draw rectangles associated w

ith Riem

ann sums, use it

to draw rectangles associated w

ith Riem

ann sums that converge to the

integrals in Exercises 83–88. U

se 10, 20, and 50 subintervals

of equal length in each case.

83.84.L

1

0sx

2+

1d dx=

43L

1

0s1

-xd dx

=12

n=

4,


355x

y0a

bx

1x

2x

3xk�

1xn�

1xk

y � f(x)

x

y0a

bxk�

1xk

x

y0a

bxk�

1xk

b � a

�

85.86.

87.

88.(T

he integral’s value is about 0.693.)

Average ValueIn E

xercises 89–92, use a CA

S to perform

the following steps:

a.P

lot the functions over the given interval.

b.Partition the interval into

200, and 1000 subintervalsof equal length, and evaluate the function at the m

idpoint of eachsubinterval.

n=

100,

L2

1 1x

dx

L1

-1 ƒ xƒ dx

=1

Lp

/4

0 sec

2 x dx=

1Lp

-p

cos x dx=

0c.

Com

pute the average value of the function values generated inpart (b).

d.S

olve the equation for x

using theaverage value calculated in part (c) for the partitioning.

89.

90.

91.

92.ƒsxd

=x sin

2 1x on c p4

, pdƒsxd

=x sin

1x on c p4

, pdƒsxd

=sin

2 x on

[0, p]

ƒsxd=

sin x

on [0, p

]

n=

1000ƒsxd

=saverage valued



The Fundamental Theorem of Calculus

In this section we present the Fundamental Theorem of Calculus, which is the central the-orem of integral calculus. It connects integration and differentiation, enabling us to com-pute integrals using an antiderivative of the integrand function rather than by taking limitsof Riemann sums as we did in Section 5.3. Leibniz and Newton exploited this relationshipand started mathematical developments that fueled the scientific revolution for the next200 years.

Along the way, we present the integral version of the Mean Value Theorem, which is an-other important theorem of integral calculus and used to prove the Fundamental Theorem.

Mean Value Theorem for Definite Integrals

In the previous section, we defined the average value of a continuous function over aclosed interval [a, b] as the definite integral divided by the length or width

of the interval. The Mean Value Theorem for Definite Integrals asserts that this av-erage value is always taken on at least once by the function ƒ in the interval.

The graph in Figure 5.16 shows a positive continuous function defined overthe interval [a, b]. Geometrically, the Mean Value Theorem says that there is a number c in[a, b] such that the rectangle with height equal to the average value ƒ(c) of the functionand base width has exactly the same area as the region beneath the graph of ƒ froma to b.

b - a

y = ƒsxd

b - a1b

a ƒsxd dx

5.4

HISTORICAL BIOGRAPHY

Sir Isaac Newton(1642–1727)

y

xa b0 c

y � f (x)

f (c),

b � a

averageheight

FIGURE 5.16 The value ƒ(c) in theMean Value Theorem is, in a sense, theaverage (or mean) height of ƒ on [a, b].When the area of the rectangleis the area under the graph of ƒ from ato b,

ƒscdsb - ad = Lb

a ƒsxd dx .

ƒ Ú 0,

THEOREM 3 The Mean Value Theorem for Definite IntegralsIf ƒ is continuous on [a, b], then at some point c in [a, b],

ƒscd = 1b - a

Lb

aƒsxd dx .

ProofIf w

e divide both sides of the Max-M

in Inequality (Table 5.3, Rule 6) by

we obtain

Since ƒ

is continuous, the Intermediate V

alue Theorem

for Continuous F

unctions (Section

2.6) says that ƒm

ust assume every value betw

een min ƒ

and max ƒ. It m

ust therefore as-

sume the value

at some point c

in [a,b].

The continuity of ƒ

is important here. It is possible that a discontinuous function never

equals its average value (Figure5.17).

EXAMPLE 1

Applying the Mean Value Theorem

for Integrals

Find the average value of on [0, 3] and w

here ƒactually takes on this value

at some point in the given dom

ain.

Solution

Section 5.3, E

qs. (1) and (2)

The average value of

over [0, 3] is . T

he function assumes this value

when

or (Figure

5.18)

In Exam

ple 1, we actually found a point c

where ƒ

assumed its average value by set-

ting ƒ(x) equal to the calculated average value and solving for x. It’s not always possible to

solve easily for the value c. What else can w

e learn from the M

ean Value T

heorem for inte-

grals? Here’s an exam

ple.

EXAMPLE 2

Show

that if ƒis continuous on

and if

then at least once in [a,b].

SolutionT

he average value of ƒon [a,b] is

By the M

ean Value T

heorem, ƒ

assumes this value at som

e pointc H

[a, b].

avsƒd=

1b

-a

Lb

aƒsxd dx

=1

b-

a # 0

=0

.

ƒsxd=

0

Lb

aƒsxd dx

=0

,

[a, b], aZ

b,

x=

3>2.

4-

x=

5>25>2

ƒsxd=

4-

x

=4

-32

=52

.

=13

a4s3-

0d-a 3

22-

022 bb

=1

3-

0 L

3

0s4

-xd dx

=13

aL3

04 dx

-L3

0x dxb

avsƒd=

1b

-a

Lb

aƒsxd dx

ƒsxd=

4-

x

s1>sb-

add1baƒsxd dx

min

ƒ…

1b

-a

Lb

aƒsxd dx

…m

ax ƒ

.

sb-

ad,

5.4The Fundam

ental Theorem of Calculus

357

x

y

0 1

12 A

verage value 1/2not assum

ed

y � f(x)

12

FIGURE 5.17

A discontinuous function

need not assume its average value.

y � 4 �

x

52

32

y

x0

12

34

1 4

FIGURE 5.18

The area of the rectangle

with

base[0,3]and height

(the averagevalue of the function

) isequal to the area betw

een the graph of ƒand the x-axis from

0 to 3 (Exam

ple 1).

ƒsxd=

4-

x5>2

Fundamental Theorem

, Part 1

If ƒ(t) is an integrable function over a finite interval I, then the integral from any fixed

number

to another number

defines a new function F

whose value at x

is

(1)

For example, if ƒ

is nonnegative and xlies to the right of a, then F

(x) is the area under thegraph from

ato x

(Figure5.19). T

he variable xis the upper lim

it of integration of an inte-gral, but F

is just like any other real-valued function of a real variable. For each value ofthe input x, there is a w

ell-defined numerical output, in this case the definite integral of ƒ

from a

to x.E

quation (1) gives a way to define new

functions, but its importance now

is the con-nection it m

akes between integrals and derivatives. If ƒ

is any continuous function, thenthe F

undamental T

heorem asserts that F

is a differentiable function of xw

hose derivativeis ƒ

itself. At every value of x,

To gain some insight into w

hy this result holds, we look at the geom

etry behind it.If

on [a,b], then the computation of

from the definition of the derivative

means taking the lim

it as of the difference quotient

For the num

erator is obtained by subtracting two areas, so it is the area under the

graph of ƒfrom

xto

(Figure5.20). If h

is small, this area is approxim

ately equal tothe area of the rectangle of height ƒ(x) and w

idth h, which can be seen from

Figure 5.20.T

hat is,

Dividing both sides of this approxim

ation by hand letting

it is reasonable to expectthat

This result is true even if the function ƒ

is not positive, and it forms the first part of the

Fundam

ental Theorem

of Calculus.

F¿sxd=

limh:

0 Fsx+

hd-

Fsxdh

=ƒsxd.

h:0

,

Fsx+

hd-

FsxdL

hƒsxd.

x+

hh

70

,

Fsx+

hd-

Fsxdh

.

h:0

F¿sxdƒ

Ú0

ddx Fsxd

=ddxL

x

aƒstd dt

=ƒsxd.

Fsxd=L

x

aƒstd dt.

x H I

a H

I


t

y0a

xb

area � F

(x)y �

f(t)

FIGURE 5.19

The function F

(x) definedby E

quation (1) gives the area under thegraph of ƒ

from a

to xw

hen ƒis

nonnegative and x7

a.y �

f(t)

t

y0a

xx �

hb

f(x)

FIGURE 5.20

In Equation (1), F

(x) isthe area to the left of x. A

lso, is

the area to the left of T

hedifference quotient is then approxim

ately equal to ƒ(x), the

height of the rectangle shown here.

[Fsx+

hd-

Fsxd]>hx

+h

.Fsx

+hd

THEOREM

4The Fundam

ental Theorem of Calculus Part 1

If ƒis continuous on [a,

b] then is continuous on [a,

b] and

differentiable on and its derivative is

(2)F¿sxd

=ddxL

x

aƒstd dt

=ƒsxd.

ƒsxd;(a, b)

Fsxd=1

xa ƒstd dt

Before proving T

heorem 4, w

e look at several examples to gain a better understanding

of what it says.

EXAMPLE 3

Applying the Fundamental Theorem

Use the F

undamental T

heorem to find

(a)

(b)

(c)

(d)

Solution

(a)E

q. 2 with

(b)E

q. 2 with

(c)R

ule1

forintegralsin

Table5.3

ofSection

5.3sets

thisup

fortheF

undamentalT

heorem.

Rule 1

(d)T

he upper limit of integration is not x

but T

his makes y

a composite of the tw

ofunctions,

We m

ust therefore apply the Chain R

ule when finding

.

=2x cos x

2

=cossx

2d #2x =

cos u # dudx

=adduL

u

1 cos t dtb # dudx

dydx=

dydu # dudx

dy>dxy

=Lu

1 cos t dt

and u

=x

2.

x2.

=-

3x sin x

=-

ddxLx

53t sin

t dt

dydx=

ddx L

5

x3t sin

t dt=

ddx a-L

x

53t sin

t dtbƒstd

=1

1+

t 2

ddx L

x

0

11

+t 2 dt

=1

1+

x2

ƒ(t)=

cos tddx

Lx

a cos t dt

=cos x

dydx if

y=L

x2

1 cos t dt

dydx if

y=L

5

x3t sin

t dt

ddx L

x

0

11

+t 2 dt

ddx L

x

a cos t dt

5.4The Fundam


359

EXAMPLE 4

Constructing a Function with a Given Derivative and Value

Find a function on the dom

ain w

ith derivative

that satisfies the condition

SolutionT

he Fundam

ental Theorem

makes it easy to construct a function w

ith deriva-tive tan x

that equals 0 at

Since

we have only to add 5 to this function to construct one

with derivative tan x

whose value at

is 5:

Although the solution to the problem

in Exam

ple 4 satisfies the two required condi-

tions, you might ask w

hether it is in a useful form. T

he answer is yes, since today w

e havecom

puters and calculators that are capable of approximating integrals. In C

hapter 7w

ew

ill learn to write the solution in E

xample 4 exactly as

We

nowgive

aproofofthe

Fundam

entalTheorem

for an arbitrary continuous function.

Proof of Theorem 4

We prove the F

undamental T

heorem by applying the definition of

the derivative directly to the function F(x), w

hen xand

are in T

his means

writing out the difference quotient

(3)

and showing that its lim

it as is the num

ber ƒ(x) for each xin

.W

hen we replace

and F(x) by their defining integrals, the num

erator inE

quation (3) becomes

The A

dditivity Rule for integrals (Table 5.3, R

ule 5) simplifies the right side to

so that Equation (3) becom

es

(4) =

1hLx+

h

xƒstd dt.

Fsx+

hd-

Fsxdh

=1h

[Fsx+

hd-

Fsxd]

Lx+

h

xƒstd dt,

Fsx+

hd-

Fsxd=L

x+h

aƒstd dt

-Lx

aƒstd dt.

Fsx+

hd(a, b)

h:0

Fsx+

hd-

Fsxdh

(a, b).x

+h

y=

ln` cos 3cos x `

+5

.

ƒsxd=L

x

3 tan

t dt+

5.

x=

3

ys3d=L

3

3 tan

t dt=

0,

y=L

x

3 tan

t dt.

x=

3:

ƒs3d=

5.

dydx=

tan x

s-p>2, p>2d

y=

ƒsxd


According to the M

ean Value T

heorem for D

efinite Integrals, the value of the last ex-pression in E

quation (4) is one of the values taken on by ƒin the interval betw

een xand

That is, for som

e number c

in this interval,

(5)

As

approaches x, forcing cto approach x

also (because cis trapped betw

eenx

and ). S

ince ƒis continuous at x, ƒ(c) approaches ƒ(x):

(6)

Going back to the beginning, then, w

e have

Definition of derivative

Eq. (4)

Eq. (5)

Eq. (6)

If then the lim

it of Equation (3) is interpreted as a one-sided lim

it with

or , respectively. T

hen Theorem

1 in Section 3.1 show

s that Fis continuous for

every point of [a, b]. This concludes the proof.

Fundamental Theorem

, Part 2 (The Evaluation Theorem)

We now

come to the second part of the F

undamental T

heorem of C

alculus. This part

describes how to evaluate definite integrals w

ithout having to calculate limits of R

iemann

sums. Instead w

e find and evaluate an antiderivative at the upper and lower lim

its ofintegration.

h:0

-h:

0+

x=

a or b,

=ƒsxd.

=limh:

0 ƒscd

=limh:

0 1hLx+

h

xƒstd dt

dFdx=

limh:

0 Fsx+

hd-

Fsxdh

limh:

0 ƒscd=

ƒsxd.

x+

hh:

0, x+

h

1hLx+

h

xƒstd dt

=ƒscd.

x+

h.

5.4The Fundam


361

THEOREM

4 (Continued)The Fundam

ental Theorem of Calculus Part 2

If ƒis continuous at every point of [a,b] and F

is any antiderivative of ƒ on [a,b],then

Lb

aƒsxd dx

=Fsbd

-Fsad.

ProofPart 1 of the Fundam

ental Theorem

tells us that an antiderivative of ƒexists, nam

ely

Thus, if F

is anyantiderivative of ƒ, then

for some constant C

for(by C

orollary 2 of the Mean V

alue Theorem

for Derivatives, S

ection 4.2).S

ince both Fand G

are continuous on [a, b], we see that

also holdsw

hen and

by taking one-sided limits (as

and x:b

-d.x:

a+

x=

bx

=a

F(x)

=G

(x)+

Ca

6x

6b

Fsxd=

Gsxd

+C

Gsxd

=Lx

aƒstd dt.

Evaluating

we have

The theorem

says that to calculate the definite integral of ƒover [a,b] all w

e need todo is:

1.Find an antiderivative F

of ƒ, and

2.C

alculate the number

The usual notation for

is

depending on whether F

has one or more term

s.

EXAMPLE 5

Evaluating Integrals

(a)

(b)

(c)

The process used in E

xample 5 w

as much easier than a R

iemann sum

computation.

The conclusions of the F

undamental T

heorem tell us several things. E

quation (2) canbe rew

ritten as

which says that if you first integrate the function ƒ

and then differentiate the result, you getthe function ƒ

back again. Likew

ise, the equation

says that if you first differentiate the function Fand then integrate the result, you get the

function Fback (adjusted by an integration constant). In a sense, the processes of integra-

Lx

a dFdt

dt=L

x

aƒstd dt

=Fsxd

-Fsad

ddxLx

aƒstd dt

=dFdx

=ƒsxd,

=[8

+1]

-[5]

=4

.

=cs4d 3/2+

44 d-cs1d 3/2

+41 d

L4

1 a 32 1

x-

4x2 b

dx=cx

3/2+

4x d1 4

L0

-p>4 sec x tan

x dx=

sec xd-p

/4

0

=sec 0

-sec a-

p4 b=

1-2

2

Lp

0 cos x dx

=sin

xd0 p

=sin

p-

sin 0

=0

-0

=0

Fsxdda b or

cFsxdda b,

Fsbd-

Fsad1

ba ƒsxd dx

=Fsbd

-Fsad.

=Lb

aƒstd dt.

=Lb

aƒstd dt

-0

=Lb

aƒstd dt

-La

aƒstd dt

=Gsbd

-Gsad

Fsbd-

Fsad=

[Gsbd

+C

]-

[Gsad

+C

]

Fsbd-

Fsad,


tion and differentiation are “inverses” of each other. The F

undamental T

heorem also says

that every continuous function ƒhas an antiderivative F

. And it says that the differential

equation has a solution (nam

ely, the function )

for every continu-ous function ƒ.

Total Area

The R

iemann sum

contains terms such as

which give the area of a rectangle w

henis positive. W

hen is negative, then the product

is the negative of therectangle’s area. W

hen we add up such term

s for a negative function we get the negative of

the area between the curve and the x-axis. If w

e then take the absolute value, we obtain the

correct positive area.

EXAMPLE 6

Finding Area Using Antiderivatives

Calculate the area bounded by the x-axis and the parabola

SolutionW

e find where the curve crosses the x-axis by setting

which gives

The curve is sketched in Figure

5.21, and is nonnegative on T

he area is

The curve in Figure 5.21

is an arch of a parabola, and it is interesting to note that the areaunder such an arch is exactly equal to tw

o-thirds the base times the altitude:

To compute the area of the region bounded by the graph of a function

andthe x-axis requires m

ore care when the function takes on both positive and negative values.

We m

ust be careful to break up the interval [a,b] into subintervals on which the function

doesn’t change sign. Otherw

ise we m

ight get cancellation between positive and negative

signed areas, leading to an incorrect total. The correct total area is obtained by adding the

absolute value of the definite integral over each subinterval where ƒ(x) does not change

sign. The term

“area” will be taken to m

ean total area.

EXAMPLE 7

Canceling Areas

Figure5.22

shows the graph of the function

between

and C

ompute

(a)the definite integral of ƒ(x) over

(b)the area betw

een the graph of ƒ(x) and the x-axis over [0, 2p

].

[0, 2p

].

x=

2p

.x

=0

ƒsxd=

sin x

y=

ƒsxd

23 s5da 254 b=

1256=

20 56 .

=a12-

2-

83 b-a-

18-

92+

273 b=

20 56 .

L2

-3 s6

-x

-x

2d dx=c6x

-x

22-

x33 d-

3

2

[-3, 2].

x=

-3

or x

=2

.

y=

0=

6-

x-

x2=s3

+xds2

-xd,

y=

6-

x-

x2.

ƒsck d ¢

kƒsc

k dƒsc

k dƒsc

k d ¢k

y=

F(x)

dy>dx=

ƒsxd

5.4The Fundam


363

–3–2

–10

12

x

y

y � 6 �

x � x

2

254

FIGURE 5.21

The area of this

parabolic arch is calculated with a

definite integral (Exam

ple 6).

–1 0 1

x

y

�2�

y � sin x

Area �

2

Area �

�–2� � 2

FIGURE 5.22

The total area betw

eenand

the x-axis for is the sum

of the absolute values of two

integrals (Exam

ple 7).

0…

x…

2p

y=

sin x

SolutionT

he definite integral for is given by

The definite integral is zero because the portions of the graph above and below

the x-axism

ake canceling contributions.T

he area between the graph of ƒ(x) and the x-axis over

is calculated by break-ing up the dom

ain of sin xinto tw

o pieces: the interval over w

hich it is nonnegativeand the interval

over which it is nonpositive.

The second integral gives a negative value. T

he area between the graph and the axis is ob-

tained by adding the absolute values

Area

=ƒ 2

ƒ +ƒ -

2ƒ=

4.

L2p

p

sin x dx

=-

cos xdp 2p

=-

[cos 2p

-cos p

]=

-[1

-s-

1d]=

-2

.

Lp

0 sin

x dx=

-cos xd0 p

=-

[cos p-

cos 0]=

-[-

1-

1]=

2.

[p, 2p

][0, p

] [0, 2p

]

L2p

0 sin

x dx=

-cos xd0 2

p

=-

[cos 2p

-cos 0]

=-

[1-

1]=

0.

ƒsxd=

sin x


x

y02

–1

y � x

3 � x

2 � 2x

Area ��

83– ⎢⎢

⎢⎢83

Area �

512

FIGURE 5.23

The region betw

een thecurve

and the x-axis(E

xample 8).y

=x

3-

x2-

2x

Summ

ary:To find the area betw

een the graph of and the x-axis over the interval

[a,b], do the following:

1.S

ubdivide [a,b] at the zeros of ƒ.

2.Integrate ƒ

over each subinterval.

3.A

dd the absolute values of the integrals.

y=

ƒsxd

EXAMPLE 8

Finding Area Using Antiderivatives

Find the area of the region between the x-axis and the graph of

SolutionFirst find the zeros of ƒ. S

ince

the zeros are and 2 (Figure

5.23). The zeros subdivide

into two subin-

tervals: on w

hich and [0,2], on w

hich W

e integrate ƒover each

subinterval and add the absolute values of the calculated integrals.

The total enclosed area is obtained by adding the absolute values of the calculated integrals,

Total enclosed area

=512

+`-83 `

=3712

.

L2

0sx

3-

x2-

2xd dx=c x

44-

x33

-x

2d0 2

=c4-

83-

4d-

0=

-83

L0

-1 sx

3-

x2-

2xd dx=c x

44-

x33

-x

2d-1

0

=0

-c 14+

13-

1d=

512

ƒ…

0.

ƒÚ

0,

[-1, 0],

[-1, 2]

x=

0, -1

,

ƒsxd=

x3-

x2-

2x=

xsx2-

x-

2d=

xsx+

1dsx-

2d,

-1

…x

…2

.ƒ(x)

=x

3-

x2-

2x,

5.4The Fundam


365

EXERCISES 5.4

Evaluating IntegralsE

valuate the integrals in Exercises 1–26.

1.2.

3.4.

5.6.

7.8.

9.10.

11.12.

13.14.

15.16.

17.18.

19.20.

21.22.

23.24.

25.26.

Derivatives of IntegralsFind the derivatives in E

xercises 27–30

a.by evaluating the integral and differentiating the result.

b.by differentiating the integral directly.

27.28.

29.30.

Find in E

xercises 31–36.

31.32.

33.34.

y=L

x2

0 cos 1

t dty

=L0

1x sin

st 2d dt

y=L

x

1 1t dt,

x7

0y

=Lx

0 21

+t 2 dt

dy>dxdduL

tan u

0 sec

2 y dyddtL

t4

0 1u du

ddxL sin

x

13t 2 dt

ddxL1

x

0 cos t dt

Lp

0 12

scos x+

ƒ cos xƒ d dxL

4

-4 ƒ xƒ dx

L4

9 1

-2u

2u

duL2

2

1 s

2+2

s

s2

ds

L1

1>2 a1y3

-1y4 b

dy

L1

22 a u

72-

1u5 b

du

L2

3

-23 st

+1dst 2

+4d dt

L-

1

1sr

+1d 2 dr

L-p>4

-p>3 a4

sec2 t

+pt 2 b

dtLp>2-p>2 s8y

2+

sin yd dy

Lp>3-p>3 1

-cos 2t2

dtL

0

p>2 1+

cos 2t2

dt

Lp>30

4 sec u

tan u du

L3p>4

p>4 csc u cot u d

u

L5p>6

p>6 csc

2 x dxLp>30

2 sec

2 x dx

Lp

0s1

+cos xd dx

Lp

0 sin

x dx

L-

1

-2

2x2 dx

L32

1x-

6>5 dx

L5

0x

3>2 dxL

1

0 A x2+1

xB dxL

2

-2 sx

3-

2x+

3d dxL

4

0 a3x-

x34 b

dx

L4

-3 a5

-x2 b

dxL

0

-2 s2x

+5d dx

35.

36.

AreaIn E

xercises 37–42, find the total area between the region and the

x-axis.

37.

38.

39.

40.

41.

42.

Find the areas of the shaded regions in Exercises 43–46.

43.

44.

45.46.

t

y

�4–

01

1 2

y � sec

2 t

y � 1 �

t 2

�

y

–�2

�2

�4�4

–0

y � sec � tan �

y

x

1

�65�6

y � sin x

x

y0 2

�

y � 2

x � �

y � 1 �

cos x

y=

x1>3

-x,

-1

…x

…8

y=

x1>3,

-1

…x

…8

y=

x3-

4x, -

2…

x…

2

y=

x3-

3x2+

2x, 0

…x

…2

y=

3x2-

3, -

2…

x…

2

y=

-x

2-

2x, -

3…

x…

2

y=L

0

tan x

dt

1+

t 2

y=L

sin x

0

dt

21

-t 2 ,

ƒ xƒ6p2

Initial Value Problems

Each of the follow

ing functions solves one of the initial value prob-lem

s in Exercises 47–50. W

hich function solves which problem

? Give

brief reasons for your answers.

a.b.

c.d.

47.48.

49.50.

Express the solutions of the initial value problem

s in Exercises 51–54

in terms of integrals.

51.

52.

53.

54.

Applications55.

Archim

edes’area

formula

for parabolas

Archim

edes(287–212

B.C.), inventor, military engineer, physicist, and the

greatestmathem

aticianofclassicaltim

esin

theW

esternw

orld,dis-covered that the area under a parabolic arch is tw

o-thirds the basetim

es the height. Sketch the parabolic arch assum

ing that hand b

are positive. Then use

calculus to find the area of the region enclosed between the arch

and the x-axis.

56.R

evenue from m

arginal revenueS

uppose that a company’s

marginal revenue from

the manufacture and sale of egg beaters is

where r

is measured in thousands of dollars and x

in thousands ofunits. H

ow m

uch money should the com

pany expect from a pro-

duction run of thousand egg beaters? To find out, integrate

the marginal revenue from

to

57.C

ost from m

arginal costT

he marginal cost of printing a poster

when x

posters have been printed is

dollars. Find the cost of printing posters 2–100.

58.(C

ontinuation of Exercise 57.) Find

the cost ofprinting posters 101–400.

cs400d-

cs100d,cs100d

-cs1d,

dcdx=

121

x

x=

3.

x=

0x

=3 drdx

=2

-2>sx

+1d 2,

-b>2

…x

…b>2

,y

=h

-s4h>b

2dx2,

dydt

=gstd,

yst0 d=y

0

dsdt=

ƒstd, sst0 d

=s

0

dydx=2

1+

x2,

ys1d=

-2

dydx=

sec x, ys2d

=3

y¿=

1x , ys1d

=-

3y¿

=sec x,

ys0d=

4

y¿=

sec x, ys-

1d=

4dydx

=1x ,

yspd=

-3

y=L

x

p

1t dt-

3y

=Lx

-1 sec t dt

+4

y=L

x

0 sec t dt

+4

y=L

x

1 1t dt

-3

Drawing Conclusions About M

otion from Graphs

59.S

uppose that ƒis the differentiable function show

n in the accom-

panying graph and that the position at time t

(sec) of a particlem

oving along a coordinate axis is

meters. U

se the graph to answer the follow

ing questions. Give

reasons for your answers.

a.W

hat is the particle’s velocity at time

b.Is the acceleration of the particle at tim

e positive, or

negative?

c.W

hat is the particle’s position at time

d.A

t what tim

e during the first 9 sec does shave its largest

value?

e.A

pproximately w

hen is the acceleration zero?

f.W

hen is the particle moving tow

ard the origin? away from

theorigin?

g.O

n which side of the origin does the particle lie at tim

e

60.S

uppose that gis the differentiable function graphed here and that

the position at time t(sec) of a particle m

oving along a coordinateaxis is

meters. U

se the graph to answer the follow

ing questions. Give

reasons for your answers.

x

y

36

9

2 4 6 8–2

–4–6

y � g(x) (6, 6)

(7, 6.5)

s=L

t

0gsxd dx

t=

9?

t=

3?

t=

5

t=

5?

y

x0

12

34

56

78

9

1 2 3 4–1–2

(1, 1)

(2, 2)(5, 2)

(3, 3)

y � f(x)

s=L

t

0ƒsxd dx


a.W

hat is the particle’s velocity at

b.Is the acceleration at tim

e positive, or negative?

c.W

hat is the particle’s position at time

d.W

hen does the particle pass through the origin?

e.W

hen is the acceleration zero?

f.W

hen is the particle moving aw

ay from the origin? tow

ard theorigin?

g.O

n which side of the origin does the particle lie at

Theory and Examples

61.S

how that if k

is a positive constant, then the area between the

x-axis and one arch of the curve is

.

62.Find

63.S

uppose Find ƒ(x).

64.Find ƒ(4) if

65.Find the linearization of

at

66.Find the linearization of

at

67.S

uppose that ƒhas a positive derivative for all values of x

and thatW

hich of the following statem

ents must be true of the

function


ers.

a.g

is a differentiable function of x.

b.g

is a continuous function of x.

c.T

he graph of ghas a horizontal tangent at

d.g

has a local maxim

um at

e.g

has a local minim

um at

f.T

he graph of ghas an inflection point at

g.T

he graph of crosses the x-axis at

68.S

uppose that ƒhas a negative derivative for all values of x

andthat

Which of the follow

ing statements m

ust be true ofthe function

hsxd=L

x

0 ƒstd dt?

ƒs1d=

0.

x=

1.

dg>dxx

=1

.

x=

1.

x=

1.

x=

1.

gsxd=L

x

0 ƒstd dt?

ƒs1d=

0.

x=

-1

.

gsxd=

3+L

x2

1 sec st

-1d dt

x=

1.

ƒsxd=

2-L

x+1

2

91

+t dt

1x0 ƒstd dt

=x cos p

x.

1x1 ƒstd dt

=x

2-

2x+

1.

limx:

0 1x3 L

x

0

t 2

t 4+

1 dt.

2>ky

=sin

kx

t=

9?

t=

3?

t=

3

t=

3?


ers.

a.h

is a twice-differentiable function of x.

b.h

and are both continuous.

c.T

he graph of hhas a horizontal tangent at

d.h

has a local maxim

um at

e.h

has a local minim

um at

f.T

he graph of hhas an inflection point at

g.T

he graph of crosses the x-axis at

69.T

he Fundam

ental Theorem

If ƒis continuous, w

e expect

to equal ƒ(x), as in the proof of Part 1 of the Fundam

ental Theo-

rem. For instance, if

then

(7)

The right-hand side of E

quation (7) is the difference quotient forthe derivative of the sine, and w

e expect its limit as

to becos x.G

raph cos xfor

Then, in a different color if

possible, graph the right-hand side of Equation (7) as a function

of xfor

and 0.1. Watch how

the latter curves con-verge to the graph of the cosine as

70.R

epeat Exercise 69 for

What is

Graph

for T

hen graph the quotientas a function of x

for and 0.1.

Watch how

the latter curves converge to the graph of as

COMPU

TER EXPLORATIONS

In Exercises 71–74, let

for the specified function ƒand interval [a,b]. U

se a CA

S to perform

the following steps and an-

swer the questions posed.

a.P

lot the functions ƒand F

together over [a,b].

b.S

olve the equation W

hat can you see to be true aboutthe graphs of ƒ

and Fat points w

here Is your

observation borne out by Part 1 of the Fundam

ental Theorem

coupled with inform

ation provided by the first derivative?E

xplain your answer.

c.O

ver what intervals (approxim

ately) is the function Fincreasing

and decreasing? What is true about ƒ

over those intervals?

d.C

alculate the derivative and plot it together w

ith F. W

hat canyou see to be true about the graph of F

at points where

Is your observation borne out by Part 1 of theF

undamental T

heorem? E

xplain your answer.

ƒ¿sxd=

0?

ƒ¿

F¿sxd=

0?

F¿sxd=

0.

Fsxd=1

xa ƒstd dt

h:0

.3x

2h

=1, 0.5, 0.2

,ssx

+hd 3

-x

3d>h-

1…

x…

1.

ƒsxd=

3x2

limh:

0 1h L

x+h

x 3t 2 dt

=limh:

0 sx+

hd 3-

x3

h?

ƒstd=

3t 2. h:0

.h

=2, 1, 0.5

, -p

…x

…2p

.

h:0

1h L

x+h

x cos t dt

=sin

sx+

hd-

sin x

h.

ƒstd=

cos t,

limh:

0 1h L

x+h

xƒstd dt

x=

1.

dh>dxx

=1

.

x=

1.

x=

1.

x=

1.

dh>dx 5.4The Fundam


367

TT

71.

72.

73.

74.

In Exercises 75–78, let

for the specified a,u, andƒ. U

se a CA

S to perform

the following steps and answ

er the questionsposed.

a.Find the dom

ain of F.

b.C

alculate and determ

ine its zeros. For what points in its

domain is F

increasing? decreasing?

c.C

alculate and determ

ine its zero. Identify the localextrem

a and the points of inflection of F.

F–sxd

F¿sxd

Fsxd=1

u(x)a

ƒstd dt

ƒsxd=

x cos px,

[0, 2p

]

ƒsxd=

sin 2x cos x3

, [0, 2p

]

ƒsxd=

2x4-

17x3+

46x2-

43x+

12, c0, 92 dƒsxd

=x

3-

4x2+

3x, [0, 4]

d.U

sing the information from

parts (a)–(c), draw a rough hand-

sketch of over its dom

ain. Then graph F

(x) on yourC

AS

to support your sketch.

75.

76.

77.

78.

In Exercises 79 and 80, assum

e that fis continuous and u(x) is tw

ice-differentiable.

79.C

alculate and check your answ

er using a CA

S.

80.C

alculate and check your answ

er using a CA

S.

d2

dx2L

usxd

aƒstd dt

ddxLusxd

aƒstd dt

a=

0, usxd

=1

-x

2, ƒsxd

=x

2-

2x-

3

a=

0, usxd

=1

-x,

ƒsxd=

x2-

2x-

3

a=

0, usxd

=x

2, ƒsxd

=21

-x

2

a=

1, usxd

=x

2, ƒsxd

=21

-x

2

y=

Fsxd



Indefinite Integrals and the Substitution Rule

A definite integral is a number defined by taking the limit of Riemann sums associatedwith partitions of a finite closed interval whose norms go to zero. The Fundamental Theo-rem of Calculus says that a definite integral of a continuous function can be computed eas-ily if we can find an antiderivative of the function. Antiderivatives generally turn out to bemore difficult to find than derivatives. However, it is well worth the effort to learn tech-niques for computing them.

Recall from Section 4.8 that the set of all antiderivatives of the function ƒ is called theindefinite integral of ƒ with respect to x, and is symbolized by

The connection between antiderivatives and the definite integral stated in the FundamentalTheorem now explains this notation. When finding the indefinite integral of a function ƒ,remember that it always includes an arbitrary constant C.

We must distinguish carefully between definite and indefinite integrals. A definite in-

tegral is a number. An indefinite integral is a function plus an arbi-trary constant C.

So far, we have only been able to find antiderivatives of functions that are clearly rec-ognizable as derivatives. In this section we begin to develop more general techniques forfinding antiderivatives. The first integration techniques we develop are obtained by invert-ing rules for finding derivatives, such as the Power Rule and the Chain Rule.

The Power Rule in Integral Form

If u is a differentiable function of x and n is a rational number different from theChain Rule tells us that

ddx

a un+1

n + 1b = un

dudx

.

-1,

1ƒsxd dx1ba ƒsxd dx

Lƒsxd dx .

5.5

From another point of view

, this same equation says that

is one of the anti-derivatives of the function

Therefore,

The integral on the left-hand side of this equation is usually w

ritten in the simpler “differ-

ential” form,

obtained by treating the dx’s as differentials that cancel. We are thus led to the follow

ingrule.

Lu

n du,

L aun dudx b

dx=

un+

1

n+

1+

C.

unsdu>dxd.

un+

1>sn+

1d

5.5Indefinite Integrals and the Substitution Rule

369

If uis any differentiable function, then

(1)L

un du

=u

n+1

n+

1+

C sn

Z-

1, n rationald.

Equation (1) actually holds for any real exponent

as we see in C

hapter 7.In deriving E

quation (1), we assum

ed uto be a differentiable function of the variable

x, but the name of the variable does not m

atter and does not appear in the final formula.

We could have represented the variable w

ith or any other letter. E

quation (1) saysthat w

henever we can cast an integral in the form

with u

a differentiable function and duits differential, w

e can evaluate the integral as

EXAMPLE 1

Using the Power Rule

Sim

pler form

Replace u

by 1+

y2.

=23

s1+

y2d 3>2

+C

=23

u3>2

+C

=us1>2d+

1

s1>2d+

1+

C

=Lu

1>2 du

L 21

+y

2#2y dy=L 1

u #a dudy b dy

[un+

1>sn+

1d]+

C.

Lu

n du, sn

Z-

1d,

u, t, y,

nZ

-1,

Let

du>dy=

2yu

=1

+y

2,

Integrate, using Eq. (1)

with n

=1>2

.

EXAMPLE 2

Adjusting the Integrand by a Constant

Sim

pler form

Replace u

by

Substitution: Running the Chain Rule Backwards

The substitutions in E

xamples 1 and 2 are instances of the follow

ing general rule.

4t-

1.

=16

s4t-

1d 3>2+

C

=16

u3>2

+C

=14

# u3>23>2

+C

=14L

u1>2 du

=14L 1

u #a dudt b dt

L 24t

-1 dt

=L 14 #2

4t-

1 #4 dt


Let

du>dt=

4.

u=

4t-

1,

With the

out front,the integral is now

instandard form

.

1>4Integrate, using E

q. (1)w

ith n=

1>2.

THEOREM

5The Substitution Rule

If is a differentiable function w

hose range is an interval Iand ƒ

is con-tinuous on I, then

Lƒsgsxddg¿sxd dx

=Lƒsud du

.

u=

gsxd

ProofT

he rule is true because, by the Chain R

ule, F(g

(x)) is an antiderivative ofw

henever Fis an antiderivative of ƒ:

Chain R

ule

Because

If we m

ake the substitution then

Fundam

ental Theorem

Fundam

ental Theorem

F¿=

ƒ =L

ƒsud du

=LF¿sud du

u=

gsxd =

Fsud+

C

=Fsgsxdd

+C

Lƒsgsxddg¿sxd dx

=L ddx

Fsgsxdd dx

u=

gsxd

F¿=

ƒ =

ƒsgsxdd #g¿sxd. ddx

Fsgsxdd=

F¿sgsxdd #g¿sxdƒsgsxdd #g¿sxd

The S

ubstitution Rule provides the follow

ing method to evaluate the integral

when ƒ

and are continuous functions:

1.S

ubstitute and

to obtain the integral

2.Integrate w

ith respect to u.

3.R

eplace uby g

(x) in the result.

EXAMPLE 3

Using Substitution

Replace u

by

We can verify this solution by differentiating and checking that w

e obtain the originalfunction

EXAMPLE 4

Using Substitution

Integrate with respect to u.

Replace u

by x3.

=-

13 cos sx

3d+

C

=13

s-cos ud

+C

=13L

sin u du

=L sin

u # 13 du

Lx

2 sin sx

3d dx=L

sin sx

3d #x2 dx

cos s7u

+5d.

7u

+5

. =

17 sin

s7u

+5d

+C

=17

sin u

+C

=17L

cos u du

L cos s7

u+

5d du

=L cos u # 17

du

Lƒsud du

.

du=

g¿sxd dxu

=gsxd

g¿

Lƒsgsxddg¿sxd dx,


371

s1>7d du=

du

.L

et u=

7u

+5, du

=7 du,

With the (

) out front, theintegral is now

in standard form.

1>7Integrate w

ith respect to u,Table 4.2.

s1>3d du=

x2 dx

.

du=

3x2 dx,

Let u

=x

3,

EXAMPLE 5

Using Identities and Substitution

The success of the substitution m

ethod depends on finding a substitution that changesan integral w

e cannot evaluate directly into one that we can. If the first substitution fails,

try to simplify the integrand further w

ith an additional substitution or two (see E

xercises49 and 50). A

lternatively, we can start fresh. T

here can be more than one good w

ay to start,as in the next exam

ple.

EXAMPLE 6

Using Different Substitutions

Evaluate

SolutionW

e can use the substitution method of integration as an exploratory tool: S

ub-stitute for the m

ost troublesome part of the integrand and see how

things work out. For the

integral here, we m

ight try or w

e might even press our luck and take u

to bethe entire cube root. H

ere is what happens in each case.

Solution 1:S

ubstitute

In the form


Replace u

by z2+

1.

=32

sz2+

1d 2>3+

C

=32

u2>3

+C

=u

2>32>3+

C

1u

n du =L

u-

1>3 du

L

2z dz

23

z2+

1=L

du

u1>3

u=

z2+

1.

u=

z2+

1 L

2z dz

23

z2+

1 .

u=

2x =

12 tan

2x+

C

ddu tan

u=

sec2 u

=12

tan u

+C

=12L

sec2 u du

=Lsec

2 u # 12 du

1cos 2x

=sec 2x

L

1cos

2 2x dx

=Lsec

2 2x dx


dx=s1>2d du

du=

2 dx,u

=2x,

Let

du=

2z dz.u

=z

2+

1,

Solution 2:S

ubstitute instead.


Replace u

by

The Integrals of and

Som

etimes w

e can use trigonometric identities to transform

integrals we do not know

howto evaluate into ones w

e can using the substitution rule. Here is an exam

ple giving the in-tegral form

ulas for and

which arise frequently in applications.

EXAMPLE 7

(a)

(b)

EXAMPLE 8

Area Beneath the Curve

Figure5.24

shows the graph of

over the interval Find

(a)the definite integral of

over

(b)the area betw

een the graph of the function and the x-axis over

Solution

(a)From

Exam

ple 7(a), the definite integral is

(b)T

hefunction

isnonnegative,so

thearea

isequalto

thedefinite

integral,orp

.sin

2 x

=[p

-0]

-[0

-0]

=p

.

L2p

0sin

2 x dx=c x2

-sin

2x4 d0 2

p

=c 2p2

-sin

4p

4 d-c 02

-sin

04 d

[0, 2p

].

[0, 2p

].gsxd

[0, 2p

].gsxd

=sin

2 x y=

sin2 x

=x2

+sin

2x4

+C

Lcos

2 x dx=L

1+

cos 2x2

dx

=12

x-

12 sin

2x2

+C

=x2

-sin

2x4

+C

=12L

s1-

cos 2xd dx=

12L dx

-12L

cos 2x dx

Lsin

2 x dx=L

1-

cos 2x2

dx

cos2 x

sin2 x

cos 2 xsin

2 x

sz2+

1d 1>3. =

32 sz

2+

1d 2>3+

C

=3 # u

22+

C

=3L

u du

L

2z dz

23

z2+

1=L

3u2 duu

u=2

3z

2+

1


373

Let

3u2 du

=2z dz.

u3=

z2+

1,

u=2

3z

2+

1,

sin2 x

=1

-cos 2x2

cos2 x

=1

+cos 2x2

As in part (a), but

with a sign change

02

��

�2

1

x

y

y � sin

2 x

12

FIGURE 5.24

The area beneath the

curve over

equals square units (E

xample 8).

p[0, 2p

]y

=sin

2 x

EXAMPLE 9

Household Electricity

We can m

odel the voltage in our home w

iring with the sine function

which expresses the voltage V

in volts as a function of time tin seconds. T

he function runsthrough 60 cycles each second (its frequency is 60 hertz, or 60 H

z). The positive constant

(“vee max”) is the peak voltage.

The average value of V

over the half-cycle from 0 to

sec (see Figure5.25) is

The average value of the voltage over a full cycle is zero, as w

e can see from Figure 5.25.

(Also see E

xercise 63.) If we m

easured the voltage with a standard m

oving-coil gal-vanom

eter, the meter w

ould read zero.To m

easure the voltage effectively, we use an instrum

ent that measures the square root

of the average value of the square of the voltage, namely

The subscript “rm

s” (read the letters separately) stands for “root mean square.” S

ince theaverage value of

over a cycle is

(Exercise 63, part c), the rm

s voltage is

The

valuesgiven

forhouseholdcurrents

andvoltages

arealw

aysrm

svalues.T

hus,“115 voltsac” m

eans that the rms voltage is 115. T

he peak voltage, obtained from the last equation, is

which is considerably higher.

Vm

ax=2

2 Vrm

s=2

2 #115L

163 volts,

Vrm

s=B

sVm

ax d 2

2=

Vm

ax

22

.

sV2dav

=1

s1>60d-

0 L

1>600

sVm

ax d 2 sin2 120

pt dt

=sV

max d 2

2,

V2=sV

max d 2 sin

2 120p

t

Vrm

s=2

sV2dav

.

=2V

maxp

.

=V

maxp

[-cos p

+cos 0]

=120V

max c-

1120p

cos 120p

td0 1>120 V

av=

1s1>120d

-0

L1>120

0 V

max sin

120p

t dt

1>120V

max

V=

Vm

ax sin 120p

t,


t

V0

V �

Vm

ax sin 120�

tV

max

Vav �

2Vm

ax�

1120

160

FIGURE 5.25

The graph of the voltage

over a full cycle. Itsaverage value over a half-cycle is Its average value over a full cycle is zero(E

xample 9).

2Vm

ax >p.

V=

Vm

ax sin 120p

t


EXERCISES 5.5

Evaluating IntegralsEvaluate the indefinite integrals in Exercises 1–12 by using the givensubstitutions to reduce the integrals to standard form.

1. 2. Lx sin s2x2d dx, u = 2x2

L sin 3x dx, u = 3x

3.

4. L a1 - cos t2b2

sin t2

dt, u = 1 - cos t2

L sec 2t tan 2t dt, u = 2t

5.6.7.8.9.

10.

11.

a.U

sing b.

Using

12.

a.U

sing b.

Using

Evaluate the integrals in E

xercises 13–48.

13.14.

15.16.

17.18.

19.20.

21.22.

23.24.

25.26.

27.28.

29.30.

31.32.

33.

34.

35.36.L

6

cos ts2

+sin

td 3 dtL

sin s2t

+1d

cos2 s2t

+1d dt

Lcsc a

y-p

2 b cot a

y-p

2 b dy

L sec a

y+p2 b

tan ay

+p2 b

dy L

x1>3 sin

sx4>3

-8d dx

Lx

1>2 sin sx

3>2+

1d dx

Lr

4 a7-

r5

10 b3 dr

Lr

2 ar

3

18-

1b5 dr

Ltan

7 x2 sec

2 x2 dx

Lsin

5 x3 cos x3

dx

Ltan

2 x sec2 x dx

Lsec

2 s3x+

2d dx

Lsin

s8z-

5d dzL

cos s3z+

4d dz

L s1

+1xd 3

1x

dxL

1

1x s1

+1xd 2 dx

L

4y dy

22y

2+

1L

3y27

-3y

2 dy

L8u2

3u

2-

1 du

Lu2

41

-u

2 du

L

3 dx

s2-

xd 2L

1

25s

+4

ds

Ls2x

+1d 3 dx

L 23

-2s ds

u=2

5x+

8u

=5x

+8

L

dx

25x

+8

u=

csc 2u

u=

cot 2u

Lcsc

2 2u cot 2

u du

L 1x

2 cos2 a 1x b

dx, u

=-

1x

L 1x sin

2 sx3>2

-1d dx,

u=

x3>2

-1

L12sy

4+

4y2+

1d 2sy3+

2yd dy, u

=y

4+

4y2+

1

L

9r2 dr

21

-r

3 , u

=1

-r

3

Lx

3sx4-

1d 2 dx, u

=x

4-

1

L28s7x

-2d -

5 dx, u

=7x

-2

37.38.

39.40.

41.42.

43.

44.

45.46.

47.48.

Simplifying Integrals Step by Step

If you do not know w

hat substitution to make, try reducing the integral

step by step, using a trial substitution to simplify the integral a bit and

then another to simplify it som

e more. Y

ou will see w

hat we m

ean ifyou try the sequences of substitutions in E

xercises 49 and 50.

49.

a.follow

ed by then by

b.follow

ed by

c.

50.

a.follow

ed by then by

b.follow

ed by

c.

Evaluate the integrals in E

xercises 51 and 52.

51.

52.


Solve the initial value problem

s in Exercises 53–58.

53.

54.

55.

56.drdu

=3

cos2 ap4

-ub

, rs0d

=p8

dsdt=

8 sin

2 at+p12 b

, ss0d

=8

dydx=

4x sx2+

8d -1>3,

ys0d=

0

dsdt=

12t s3t 2-

1d 3, ss1d

=3

L

sin 2u

2u

cos3 1u

du

L s2r

-1d cos 2

3s2r-

1d 2+

6

23s2r

-1d 2

+6

dr

u=

1+

sin2 sx

-1d

y=

1+

u2

u=

sin sx

-1d,

w=

1+y

2y

=sin

u,

u=

x-

1,

L 21

+sin

2 sx-

1d sin sx

-1d cos sx

-1d dx

u=

2+

tan3 x

y=

2+

uu

=tan

3 x,

w=

2+y

y=

u3,

u=

tan x

,L

18 tan

2 x sec2 x

s2+

tan3 xd 2

dx

L3x

52x

3+

1 dxL

x32

x2+

1 dx

L Ax

-1

x5

dxL

t 3s1+

t 4d 3 dt

Lsu

4-

2u

2+

8u

-2dsu

3-u

+2d du

Lss

3+

2s2-

5s+

5ds3s2+

4s-

5d ds

L

cos 2u

2u

sin2 2u

du

L 1u

2 sin 1u

cos 1u du

L

11t coss1

t+

3d dtL

1t 2 cos a 1t-

1b dt

L sec z tan

z

2sec z

dzL 2

cot y csc2 y dy


375

57.

58.

59.T

he velocity of a particle moving back and forth on a line is

for all t. If w

hen find

the value of sw

hen

60.T

he acceleration of a particle moving back and forth on a line is

for all

t. If

and w

hen find s

when

Theory and Examples

61.It looks as if w

e can integrate 2 sin xcos x

with respect to x

inthree different w

ays:

a.b.c.

=-

cos 2x2

+C

3 . 2 sin

x cos x=

sin 2x

L2

sin x cos x dx

=L sin

2x dx

=-

u2+

C2=

-cos

2 x+

C2

u=

cos x,

L2

sin x cos x dx

=L-

2u du

=u

2+

C1=

sin2 x

+C

1

u=

sin x

, L

2 sin

x cos x dx=L

2u du

t=

1 sec

.t=

0,

8 m/sec

y =s

=0

a=

d2s>dt 2

=p

2 cos pt m>sec

2

t=p>2

sec.

t=

0,

s=

0y

=ds>dt

=6

sin 2t m>sec

d2y

dx2

=4 sec

2 2x tan 2x,

y¿s0d=

4, ys0d

=-

1

d2s

dt 2=

-4

sin a2t

-p2 b

, s¿s0d

=100,

ss0d=

0C

an all three integrations be correct? Give reasons for your an-

swer.

62.T

he substitution gives

The substitution

gives

Can both integrations be correct? G

ive reasons for your answer.

63.(C

ontinuation of Exam

ple 9.)

a.S

how by evaluating the integral in the expression

that the average value of over a full cycle

is zero.

b.T

he circuit that runs your electric stove is rated 240 volts rms.

What is the peak value of the allow

able voltage?

c.S

how that

L1>60

0sV

max d 2 sin

2 120 p

t dt=sV

max d 2

120.

V=

Vm

ax sin 120

pt

1s1>60d

-0

L1>60

0V

max sin

120 p

t dt

Lsec

2 x tan x dx

=Lu du

=u

22+

C=

sec2 x

2+

C.

u=

sec x

Lsec

2 x tan x dx

=Lu du

=u

22+

C=

tan2 x

2+

C.

u=

tan x



Substitution and Area Between Curves

There are two methods for evaluating a definite integral by substitution. The first methodis to find an antiderivative using substitution, and then to evaluate the definite integral byapplying the Fundamental Theorem. We used this method in Examples 8 and 9 of the pre-ceding section. The second method extends the process of substitution directly to definiteintegrals. We apply the new formula introduced here to the problem of computing the areabetween two curves.

Substitution Formula

In the following formula, the limits of integration change when the variable of integrationis changed by substitution.

5.6

THEOREM 6 Substitution in Definite IntegralsIf is continuous on the interval [a, b] and ƒ is continuous on the range of g, then

Lb

aƒsg sxdd # g¿sxd dx = L

gsbd

gsadƒsud du

g¿

ProofL

et Fdenote any antiderivative of ƒ. T

hen,

To use the formula, m

ake the same u-substitution

and you

would use to evaluate the corresponding indefinite integral. T

hen integrate the trans-form

ed integral with respect to u

from the value g

(a) (the value of uat

) to the valueg

(b) (the value of uat

).

EXAMPLE 1

Substitution by Two M

ethods

Evaluate

SolutionW

e have two choices.

Method 1:

Transform

the integral and evaluate the transformed integral w

ith the trans-form

ed limits given in T

heorem 6.

Evaluate the new

definite integral.

Method 2:

Transform

the integral as an indefinite integral, integrate, change back to x, anduse the original x-lim

its.

=23

c23>2

-0

3>2d=

23 c22

2d=

422

3

=23

css1d 3+

1d 3>2-ss-

1d 3+

1d 3>2d L

1

-1 3x

22x

3+

1 dx=

23 sx

3+

1d 3>2d-1

1

=23

sx3+

1d 3>2+

C

=23

u3>2

+C

L3x

22x

3+

1 dx=L 1

u du

=23

c23>2

-0

3>2d=

23 c22

2d=

422

3

=23

u3>2d0 2

=L2

0 1u du

L1

-1 3x

22x

3+

1 dx

L1

-1 3x

22x

3+

1 dx.

x=

bx

=a

du=

g¿sxd dxu

=gsxd

=Lgsbd

gsadƒsud du

.

=Fsuddu=

gsad

u=gsbd

=Fsgsbdd

-Fsgsadd

Lb

aƒsgsxdd #g¿sxd dx

=Fsgsxdddx=

a

x=b

5.6Substitution and Area Betw

een Curves377

= ƒsgsxddg¿sxd

= F¿sgsxddg¿sxd

ddx Fsgsxdd

Fundam

entalT

heorem, Part 2

Let

When

When x

=1, u

=s1d 3

+1

=2

.x

=-

1, u=s-

1d 3+

1=

0.

u=

x3+

1, du=

3x2 dx

.

Let u

=x

3+

1, du=

3x2 dx

.


Replace u

by x3+

1.

Use the integral just found,

with lim

its of integration for x.

Which m

ethod is better—evaluating the transform

ed definite integral with trans-

formed lim

its using Theorem

6, or transforming the integral, integrating, and transform

ingback to use the original lim

its of integration? In Exam

ple 1, the first method seem

s easier,but that is not alw

ays the case. Generally, it is best to know

both methods and to use

whichever one seem

s better at the time.

EXAMPLE 2

Using the Substitution Formula

Definite Integrals of Symm

etric Functions

The S

ubstitution Formula in T

heorem 6 sim

plifies the calculation of definite integrals ofeven and odd functions (S

ection 1.4) over a symm

etric interval (Figure

5.26).[-

a, a]

=-c s0d 22

-s1d 2

2 d=

12

=-c u

22 d1 0

=-L

0

1u du

Lp>2p>4

cot u csc

2 u du

=L0

1u #s-

dud


x

y0a

–a

(b)

x

y0a

–a(a)

FIGURE 5.26

(a) ƒeven,

(b) ƒodd, 1

a-a ƒsxd dx

=0

1a-a ƒsxd dx

=21

a0 ƒsxd dx

Theorem 7

Let ƒ

be continuous on the symm

etric interval

(a)If ƒ

is even, then

(b)If ƒ

is odd, then La

-a ƒ(x) dx

=0.

La

-a ƒsxd dx

=2L

a

0ƒsxd dx.

[-a, a]. L

et

When

When u

=p>2, u

=cot (p>2)

=0.

u=p>4, u

=cot (p>4)

=1.

- du

=csc

2 u du.

u=

cot u, du=

-csc

2 u du,

Proof of Part (a)

The proof of part (b) is entirely sim

ilar and you are asked to give it in Exercise 86.

The assertions of

Theorem

7 remain true w

hen ƒis an integrable function (rather than

having the stronger property of being continuous), but the proof is somew

hat more diffi-

cult and best left to a more advanced course.

EXAMPLE 3

Integral of an Even Function

Evaluate

SolutionS

ince satisfies

it is even on the symm

et-ric interval

so

Areas Between Curves

Suppose w

e want to find the area of a region that is bounded above by the curve

below by the curve

and on the left and right by the lines and

(Figure5.27). T

he region might accidentally have a shape w

hose area we could find w

ithgeom

etry, but if ƒand g

are arbitrary continuous functions, we usually have to find the

area with an integral.

To see what the integral should be, w

e first approximate the region w

ith nvertical rec-

tangles based on a partition of [a,

b] (Figure5.28). T

he area of thekth rectangle (Figure

5.29) is

¢A

k=

height*

width

=[ƒsc

k d-

gsck d] ¢

xk .

P=5

x0 , x

1 ,Á, x

n 6

x=

bx

=a

y=

gsxd,y

=ƒsxd,

=2

a 325-

323+

12b=

23215

.

=2

c x55

-43

x3+

6xd0 2

L2

-2 sx

4-

4x2+

6d dx=

2L2

0sx

4-

4x2+

6d dx

[-2, 2],

ƒs-xd

=ƒsxd,

ƒsxd=

x4-

4x2+

6

L2

-2 sx

4-

4x2+

6d dx.

=2L

a

0ƒsxd dx

=La

0ƒsud du

+La

0ƒsxd dx

=La

0ƒs-

ud du+L

a

0ƒsxd dx

=-L

a

0ƒs-

uds-dud

+La

0ƒsxd dx

=-L

-a

0ƒsxd dx

+La

0ƒsxd dx

La

-a ƒsxd dx

=L0

-a ƒsxd dx

+La

0ƒsxd dx


een Curves379

Additivity R

ule forD

efinite Integrals

Order of Integration R

ule

Let

When

When x

=-

a, u=

a.

x=

0, u=

0.

u=

-x, du

=-

dx.

ƒis even, so

ƒs-ud

=ƒsud.

x

y

a

b

Low

er curvey �

g(x)

Upper curvey �

f(x)

FIGURE 5.27

The region betw

eenthe curves

and and the lines

and x=

b.

x=

ay

=gsxd

y=

ƒsxd

x

y

y � f(x)

y � g(x) b �

xn

xn�

1a

� x

0x

1

x2

FIGURE 5.28

We approxim

ate theregion w

ith rectangles perpendicularto the x-axis.

x

y

a

b

(ck , f(ck ))

f(ck ) � g(ck )

�A

kck

(ck , g(ck ))�

xk

FIGURE 5.29

The area

of the kthrectangle is the product of its height,

and its width, ¢

xk .

ƒsck d

-gsc

k d,

¢A

k


We then approxim

ate the area of the region by adding the areas of the nrectangles:

Riem

ann Sum

As

the sums on the right approach the lim

it because ƒ

andg

are continuous. We take the area of the region to be the value of this integral. T

hat is,

A=

limƒƒ P

ƒƒ :0 a n

k=1 [ƒsc

k d-

gsck d] ¢

xk=L

b

a[ƒsxd

-gsxd] dx.

1ba [ƒsxd

-gsxd] dx

7P 7:0

,

ALa n

k=1 ¢

Ak=a n

k=1 [ƒsc

k d-

gsck d] ¢

xk .

DEFINITION

Area Between Curves

If ƒand g

are continuous with

throughout [a,b], then the area ofthe region betw

een the curvesand

froma

tob

is the inte-gral of

from a

to b:A=L

b

a[ƒsxd

-gsxd] dx.

(f-

g)y

=gsxd

y=

fsxdƒsxd

Úgsxd

When applying this definition it is helpful to graph the curves. T

he graph reveals which

curve is the upper curve ƒand w

hich is the lower curve g. It also helps you find the lim

itsof integration if they are not already know

n. You m

ay need to find where the curves inter-

sect to determine the lim

its of integration, and this may involve solving the equation

for values of x. Then you can integrate the function

for the area be-tw

een the intersections.

EXAMPLE 4

Area Between Intersecting Curves

Find the area of the region enclosed by the parabola and the line

SolutionFirstw

e sketch the two curves (Figure

5.30). The lim

its of integration are foundby solving

and sim

ultaneously for x.

Equate ƒ(x) and g(x).

Rew

rite.

Factor.

Solve.

The region runs from

to

The lim

its of integration are T

he area between the curves is

=a4+

42-

83 b-a-

2+

12+

13 b=

92

=L2

-1 s2

+x

-x

2d dx=c2x

+x

22-

x33 d-

1

2

A=L

b

a[ƒsxd

-gsxd] dx

=L2

-1 [s2

-x

2d-s-

xd] dx

a=

-1, b

=2

.x

=2

.x

=-

1

x=

-1,

x=

2.

sx+

1dsx-

2d=

0

x2-

x-

2=

0

2-

x2=

-x

y=

-x

y=

2-

x2

y=

-x.

y=

2-

x2

ƒ-

gƒsxd

=gsxd

x

y0–1

12

(–1, 1)

(x, f(x))

y � 2 �

x2

(x, g(x))

�xy �

–x(2, –2)

FIGURE 5.30

The region in

Exam

ple 4 with a typical

approximating rectangle.

If the formula for a bounding curve changes at one or m

ore points, we subdivide the re-

gion into subregions that correspond to the formula changes and apply the form

ula for thearea betw

een curves to each subregion.

EXAMPLE 5

Changing the Integral to Match a Boundary Change

Find the area of the region in the first quadrant that is bounded above by and be-

low by the x-axis and the line

SolutionT

he sketch (Figure5.31) show

s that the region’s upper boundary is the graph ofT

he lower boundary changes from

for

to for

(there is agreement at

). We subdivide the region at

into subre-gions A

and B, show

n in Figure 5.31.T

he limits of integration for region A

are and

The left-hand lim

it for re-gion B

is To find the right-hand lim

it, we solve the equations

andsim

ultaneously for x:

Equate ƒ(x) and g(x).

Square both sides.

Rew

rite.

Factor.

Solve.

Only the value

satisfies the equation T

he value is an extrane-

ous root introduced by squaring. The right-hand lim

it is

We add the area of subregions A

and Bto find the total area:

Integration with Respect to y

If a region’s bounding curves are described by functions of y, the approximating rectangles

are horizontal instead of vertical and the basic formula has y

in place of x.

=23

s8d-

2=

103.

=23

s2d 3>2-

0+a 23

s4d 3>2-

8+

8b-a 23

s2d 3>2-

2+

4b =c 23

x3>2d0 2

+c 23 x

3>2-

x22

+2xd2 4

Total area

=

L2

0 1x dx

(')'*

area of A

+ L

4

2s1

x-

x+

2d dx('''')''''*

area of B

For 2

…x

…4:

ƒsxd-

gsxd=1

x-sx

-2d

=1x

-x

+2

For 0

…x

…2:

ƒsxd-

gsxd=1

x-

0=1

x

b=

4.

x=

11

x=

x-

2.

x=

4

x=

1, x

=4

.

sx-

1dsx-

4d=

0

x2-

5x+

4=

0

x=sx

-2d 2

=x

2-

4x+

4

1x

=x

-2

y=

x-

2y

=1x

a=

2.

b=

2.

a=

0

x=

2x

=2

2…

x…

4gsxd

=x

-2

0…

x…

2gsxd

=0

ƒsxd=1

x .

y=

x-

2.

y=1

x


een Curves381

HIS

TO

RIC

AL

BIO

GR

AP

HY

Richard D

edekind(1831–1916)

x

y0 1 2

42

y � �

x

y � 0

y � x

� 2

(x, f(x))

(x, f(x))

(x, g(x))

(x, g(x))

A

B(4, 2)

Area �

2

0 �x dx

Area �

4

2 (�x �

x � 2) dx

LL

FIGURE 5.31

When the form

ula for abounding curve changes, the area integralchanges to becom

e the sum of integrals to

match, one integral for each of the shaded

regions shown here for E

xample 5.

For regions like these

use the formula

In this equation ƒalw

ays denotes the right-hand curve and gthe left-hand curve, so

is nonnegative.

EXAMPLE 6

Find the area of the region in Exam

ple 5 by integrating with respect to y.

SolutionW

e first sketch the region and a typical horizontalrectangle based on a parti-

tion of an interval of y-values (Figure5.32). T

he region’s right-hand boundary is the lineso

The left-hand boundary is the curve

so T

he lower lim

it of integration is W

e find the upper limit by solving

andsim

ultaneously for y:

Rew

rite.

Factor.

Solve.

The upper lim

it of integration is (T

he value gives a point of intersection

belowthe x-axis.)

The area of the region is

This is the result of E

xample 5, found w

ith less work.

=4

+42

-83

=103

.

=c2y+

y22

-y

33 d0 2

=L2

0[2

+y

-y

2] dy

A=L

b

a[ƒsyd

-gsyd] dy

=L2

0[y

+2

-y

2] dy

y=

-1

b=

2.

y=

-1,

y=

2

sy+

1dsy-

2d=

0

y2-

y-

2=

0

y+

2=

y2

x=

y2

x=

y+

2y

=0

.gsyd

=y

2.x

=y

2,ƒsyd

=y

+2

.x

=y

+2

,

ƒsyd-

gsyd

A=L

d

c[ƒsyd

-gsyd] dy.

x�

f(y)

yy

x

x

x

y

x�

g(y)

0 c d

x�

g(y)

x�

f(y)

0 c d

0c dx

� f(y)

x�

g(y)


Equate

and gsyd=

y2.

ƒsyd=

y+

2

x

y

y � 0

24

0 1 2(g(y), y)

(f(y), y)f(y) �

g(y)

(4, 2)

x � y

� 2

x � y

2

�y

FIGURE 5.32

It takes two

integrations to find the area of thisregion if w

e integrate with respect to

x. It takes only one if we integrate

with respect to y

(Exam

ple 6).

Combining Integrals w

ith Formulas from

Geometry

The fastest w

ay to find an area may be to com

bine calculus and geometry.

EXAMPLE 7

The Area of the Region in Example 5 Found the Fastest W

ay

Find the area of the region in Exam

ple 5.

SolutionT

he area we w

ant is the area between the curve

and thex-axis, m

inusthe area of a triangle w

ith base 2 and height 2 (Figure5.33):

Conclusion from

Exam

ples 5–7

It is sometim

es easier to find the area between

two

curves by integrating with respect to y

instead of x. Also, it m

ay help to combine

geometry and calculus. A

fter sketching the region, take a mom

ent to think about the bestw

ay to proceed.

=23

s8d-

0-

2=

103 .

=23

x3>2d0 4

-2

Area

=L4

0 1x dx

-12

s2ds2d y=1

x, 0…

x…

4,


een Curves383

02

2

4

1 2

Area �

2

(4, 2)

x

y

y � 0

y � x

� 2

y � �

x

2

FIGURE 5.33

The area of the blue region

is the area under the parabola m

inus the area of the triangle (Exam

ple 7).y

=1x

5.6 Substitution and Area Between Curves 383

EXERCISES 5.6

Evaluating Definite IntegralsUse the Substitution Formula in Theorem 6 to evaluate the integrals inExercises 1–24.

1. a. b.

2. a. b.

3. a. b.

4. a. b.

5. a. b.

6. a. b.

7. a. b.

8. a. b.

9. a. b. L23

-23

4x

2x2 + 1 dxL

23

0

4x

2x2 + 1 dx

L4

1

101ys1 + y3>2d2 dyL

1

0

101ys1 + y 3>2d2 dy

L1

0

5r

s4 + r2d2 drL

1

-1

5r

s4 + r2d2 dr

L0

-27 t st2 + 1d1>3 dtL

27

0t st2 + 1d1>3 dt

L1

-1t3s1 + t4d3 dtL

1

0t3s1 + t4d3 dt

L3p

2p3 cos2 x sin x dxL

p

03 cos2 x sin x dx

L0

-p>4tan x sec2 x dxLp>4

0 tan x sec2 x dx

L1

-1r21 - r2 drL

1

0r21 - r2 dr

L0

-12y + 1 dyL

3

02y + 1 dy

10. a. b.

11. a. b.

12. a. b.

13. a. b.

14. a. b.

15. 16.

17. 18.

19. 20.

21.

22. L1

0s y3 + 6y2 - 12y + 9d-1>2 s y2 + 4y - 4d dy

L1

0s4y - y2 + 4y3 + 1d-2>3 s12y2 - 2y + 4d dy

Lp>4

0s1 - sin 2td3>2 cos 2t dtL

p

05s5 - 4 cos td1>4 sin t dt

L3p>2p

cot5 au6b sec2 au

6b duL

p>60

cos-3 2u sin 2u du

L4

1

dy

21y s1 + 1yd2L1

02t5 + 2t s5t4 + 2d dt

Lp>2

0

sin w

s3 + 2 cos wd2 dwL

0

-p>2 sin w

s3 + 2 cos wd2 dw

Lp

-p

cos z

24 + 3 sin z dzL

2p

0

cos z

24 + 3 sin z dz

Lp>2

-p>2 a2 + tan t2b sec2

t2

dtL0

-p>2 a2 + tan t2b sec2

t2

dt

Lp>3p>6 s1 - cos 3td sin 3t dtL

p>60

s1 - cos 3td sin 3t dt

L0

-1

x3

2x4 + 9 dxL

1

0

x3

2x4 + 9 dx

23.24.

AreaFind the total areas of the shaded regions in E

xercises 25–40.

25.26.

27.28.

29.30.

t

y

y �sec

2 t12

�3�3

–0 1 2

–4

y � –

4sin

2 t

x

y

��2

y � cos

2 x

0 1y �

1

x

y0–1

–�

–1 1

�2–

y �(cos x)(sin(�

� �

sin x))�2

x

y0–1

–1–2–3

–2–

�

y � 3(sin x)�

1 � cos x

x

y0�

y�

(1 � cos x) sin x

02

–2x

y

y � x�

4 � x

2

L-

1>2-

1t -

2 sin2 a1

+1t b

dtL2

3p

2

02u

cos2 su

3>2d du

31.

32.

33.

34.35.

x

y01

2

1

y � xy �

1

y �x

24x

y

–10–2 1

1

y � x

2

y � –2x

4

x

y0 1

1

x � 12y

2 � 12y

3

x � 2y

2 � 2y

01

1

x

y

(1, 1)

x � y

2

x � y

3

x

y

–2–1

12

–1 8(–2, 8)

(2, 8)

y � 2x

2

y � x

4 � 2x

2

NO

T T

O SC

AL

E


36.

37.38.

39.40.

Find the areas of the regions enclosed by the lines and curves in Exer-

cises 41–50.

41.

42.

43.

44.

45.

46.

47.

48.y

=x2

a2-

x2,

a7

0, and

y=

0

y=

x4-

4x2+

4 and

y=

x2

y=

7-

2x2

and y

=x

2+

4

y=

x2

and y

=-

x2+

4x

y=

x2-

2x and

y=

x

y=

x4

and y

=8x

y=

2x-

x2

and y

=-

3

y=

x2-

2 and

y=

2

x

y

–11

23

–2

2–5 4

(3, –5)

(–2, 4)y �

4 � x

2y � –x �

2

x

y5

–4

(–3, 5)

(1, –3)(–3, –3)

10

–3

y � x

2 � 4

y � –x

2 � 2x

01

2

1

x

y

y � x

2x

� y �

2

49.(H

ow m

any intersection pointsare there?)

50.


cises 51–58.

51.

52.

53.

54.

55.

56.

57.

58.

Find the areas of the regions enclosed by the curves in Exercises 59–62.

59.

60.

61.

62.


cises 63–70.

63.

64.

65.

66.

67.

68.

69.

70.

71.Find the area of the propeller-shaped region enclosed by the curve

and the line

72.Find the area of the propeller-shaped region enclosed by thecurves

and

73.Find the area of the region in the first quadrant bounded by theline

the line the curve

and the x-axis.

74.Find the area of the “triangular” region in the first quadrantbounded on the left by the y-axis and on the right by the curves

and

75.T

he region bounded below by the parabola

and above bythe line

is to be partitioned into two subsections of equal

area by cutting across it with the horizontal line

a.S

ketch the region and draw a line

across it that looksabout right. In term

s of c, what are the coordinates of the

points where the line and parabola intersect? A

dd them to

your figure.

y=

c

y=

c.

y=

4y

=x

2

y=

cos x.

y=

sin x

y=

1>x2,

x=

2,

y=

x,

x-

y1>5

=0

.x

-y

1>3=

0

x-

y=

0.

x-

y3=

0

y=

sec2 sp

x>3d and

y=

x1>3,

-1

…x

…1

x=

3 sin

y 2cos y

and x

=0,

0…

y…p>2

x=

tan2 y

and x

=-

tan2 y,

-p>4

…y

…p>4

y=

sec2 x,

y=

tan2 x,

x=

-p>4,

and x

=p>4

y=

sin sp

x>2d and

y=

x

y=

cos spx>2d

and y

=1

-x

2

y=

8 cos x

and y

=sec

2 x, -p>3

…x

…p>3

y=

2 sin

x and

y=

sin 2x,

0…

x…p

x+

y2=

3 and

4x+

y2=

0

x+

4y2=

4 and

x+

y4=

1, for

xÚ

0

x3-

y=

0 and

3x2-

y=

4

4x2+

y=

4 and

x4-

y=

1

x=

y3-

y2

and x

=2y

x=

y2-

1 and

x=

ƒ yƒ 21

-y

2

x-

y2>3

=0

and x

+y

4=

2

x+

y2=

0 and

x+

3y2=

2

x-

y2=

0 and

x+

2y2=

3

y2-

4x=

4 and

4x-

y=

16

x=

y2

and x

=y

+2

x=

2y2,

x=

0, and

y=

3

y=

ƒ x2-

4ƒ

and y

=sx

2>2d+

4

y=2

ƒ xƒ and

5y=

x+

6


een Curves385

x

y

–10 2

1–1

–22

(–2, –10) y � 2x

3 � x

2 � 5x

y � –x

2 � 3x

(2, 2)

x

y

30 6

–2

y �3 x

y ��

x3 x3

(3, 6)

(3, 1)

⎛⎝⎛⎝

–2, –3 2

b.Find c

by integrating with respect to y. (T

his puts cin the

limits of integration.)

c.Find c

by integrating with respect to x. (T

his puts cinto the

integrand as well.)

76.Find the area of the region betw

een the curve and the

line by integrating w

ith respect to a.x,b.y.

77.Find the area of the region in the first quadrant bounded on theleft by the y-axis, below

by the line above left by the

curve and above right by the curve

78.Find the area of the region in the first quadrant bounded on theleft by the y-axis, below

by the curve above left by the

curve and above right by the line

79.T

he figure here shows triangle A

OC

inscribed in the region cutfrom

the parabola by the line

Find the limit of the

ratio of the area of the triangle to the area of the parabolic regionas a

approaches zero.

80.S

uppose the area of the region between the graph of a positive

continuous function ƒand the x-axis from

to

is 4square units. Find the area betw

een the curves and

from

to

81.W

hich of the following integrals, if either, calculates the area of

the shaded region shown here? G


a.b.L1

-1 s-

x-sxdd dx

=L1

-1 -

2x dx

L1

-1 sx

-s-

xdd dx=L

1

-1 2x dx

x=

b.

x=

ay

=2ƒsxd

y=

ƒsxdx

=b

x=

a

x

y

CA

O–a

a

y�

x2

y�

a2

(a, a2)

(–a, a2)

y=

a2.

y=

x2

x

y0 1 2

12

x � 2�

y

x�

3�

y

x�

(y�

1) 2

x=

3-

y.

x=sy

-1d 2,

x=

21y

,

y=

2>1x

.y

=1

+1x

,y

=x>4

,

y=

-1

y=

3-

x2

82.T

rue, sometim

es true, or never true? The area of the region be-

tween

the graphs

of the

continuous functions

andand the vertical lines

and is


er.

Theory and Examples

83.S

uppose that F(x) is an antiderivative of

Express

in terms of F

.

84.S

how that if ƒ

is continuous, then

85.S

uppose that

Find

if a.ƒis odd,

b.ƒis even.

86.a.

Show

that if ƒis odd on

then

b.Test the result in part (a) w

ith and

87.If ƒ

is a continuous function, find the value of the integral

by making the substitution

and adding the resultingintegral to I.

u=

a-

x

I=L

a

0

ƒsxd dxƒsxd

+ƒsa

-xd

a=p>2

.ƒsxd

=sin

x

La

-a ƒsxd dx

=0

.

[-a, a],

L0

-1 ƒsxd dx

L1

0ƒsxd dx

=3

.

L1

0ƒsxd dx

=L1

0ƒs1

-xd dx

.

L3

1 sin

2xx

dx

ƒsxd=ssin

xd>x, x7

0.

Lb

a[ƒsxd

-gsxd] dx

.

x=

b sa6

bdx

=a

y=

gsxdy

=ƒsxd

x

y

–1

–1 1

1

y � –x

y � x


88.B

y using a substitution, prove that for all positive numbers x

and y,

The Shift Property for Definite IntegralsA

basic property of definite integrals is their invariance under transla-tion, as expressed by the equation.

(1)

The equation holds w

henever ƒis integrable and defined for the nec-

essary values of x. For example in the accom

panying figure, show that

because the areas of the shaded regions are congruent.

x

y01

–1–2

y�

(x�

2) 3y

� x

3

L-

1

-2sx

+2d 3 dx

=L1

0x

3 dx

Lb

aƒsxd dx

=Lb-

c

a-c

ƒsx+

cd dx.

Lxy

x 1t dt

=Ly

1 1t dt.

89.U

se a substitution to verify Equation (1).

90.For each of the follow

ing functions, graph ƒ(x) over [a,b] and

over to convince yourself that E

quation(1) is reasonable.

a.b.c.

COMPU

TER EXPLORATIONS

In Exercises 91–94, you w

ill find the area between curves in the plane

when you cannot find their points of intersection using sim

ple alge-bra. U

se a CA

S to perform

the following steps:

a.P

lot the curves together to see what they look like and how

many points of intersection they have.

b.U

se the numerical equation solver in your C

AS

to find all thepoints of intersection.

c.Integrate

over consecutive pairs of intersectionvalues.

d.S

um together the integrals found in part (c).

91.

92.

93.

94.ƒsxd

=x

2 cos x, gsxd

=x

3-

x

ƒsxd=

x+

sin s2xd,

gsxd=

x3

ƒsxd=

x42

-3x

3+

10, gsxd

=8

-12x

ƒsxd=

x33

-x

22-

2x+

13,

gsxd=

x-

1

ƒ ƒsxd-

gsxdƒ

ƒsxd=2

x-

4 , a

=4,

b=

8, c

=5

ƒsxd=

sin x,

a=

0, b

=p

, c

=p>2

ƒsxd=

x2,

a=

0, b

=1,

c=

1

[a-

c, b-

c]ƒsx

+cd

3875.6

Substitution and Area Between Curves

Chapter 5Questions to Guide Your Review

387

Chapter 5Questions to Guide Your Review

1.H

ow can you som

etimes estim

ate quantities like distance traveled,area, and average value w

ith finite sums? W

hy might you w

ant todo so?

2.W

hat is sigma notation? W

hat advantage does it offer? Give ex-

amples.

3.W

hat is a Riem

ann sum? W

hy might you w

ant to consider such asum

?

4.W

hat is the norm of a partition of a closed interval?

5.W

hat is the definite integral of a function ƒover a closed interval

[a,b]? When can you be sure it exists?

6.W

hat is the relation between definite integrals and area? D

escribesom

e other interpretations of definite integrals.

7.W

hat is the average value of an integrable function over a closedinterval? M

ust the function assume its average value? E

xplain.

8.D

escribe the rules for working w

ith definite integrals (Table 5.3).G

ive examples.

9.W

hat is the Fundam

ental Theorem

of Calculus? W

hy is it so im-

portant? Illustrate each part of the theorem w

ith an example.

10.H

ow does the F

undamental T

heorem provide a solution to the ini-

tial value problem

when ƒ

is continu-ous?

11.H

ow is integration by substitution related to the C

hain Rule?

12.H

ow can you som

etimes evaluate indefinite integrals by substitu-

tion? Give exam

ples.

13.H

ow does the m

ethod of substitution work for definite integrals?

Give exam

ples.

14.H

ow do you define and calculate the area of the region betw

eenthe graphs of tw

o continuous functions? Give an exam

ple.

dy>dx=

ƒsxd, ysx0 d

=y

0 ,


Chapter 5Practice Exercises

Finite Sums and Estim

ates1.

The accom

panying figure shows the graph of the velocity (ft

sec)of a m

odel rocket for the first 8 sec after launch. The rocket accel-

erated straight up for the first 2 sec and then coasted to reach itsm

aximum

height at

a.A

ssuming that the rocket w

as launched from ground level,

about how high did it go? (T

his is the rocket in Section 3.3,

Exercise 17, but you do not need to do E

xercise 17 to do theexercise here.)

b.S

ketch a graph of the rocket’s height aboveground as afunction of tim

e for 2.

a.T

he accompanying figure show

s the velocity (msec) of a

body moving along the s-axis during the tim

e interval fromto

About how

far did the body travel duringthose 10 sec?

b.S

ketch a graph of sas a function of tfor

assuming

3.S

uppose that and

Find the value of

a.b.

c.d.a 10

k=1 a 52

-b

k ba 10

k=1 sa

k+

bk-

1d

a 10

k=1 sb

k-

3ak d

a 10

k=1 a

k4

a 10

k=1 b

k=

25.

a 10

k=1 a

k=

-2

0 1

24

68

10

2 3 4 5

Tim

e (sec)

Velocity (m/sec)

ss0d=

0.

0…

t…

10

t=

10 sec.t=

0

>0

…t…

8.

24

68

0 50

100

150

200

Tim

e after launch (sec)

Velocity (ft/sec) t=

8 sec.

>4.

Suppose that

and Find the values of

a.b.

c.d.

Definite IntegralsIn E

xercises 5–8, express each limit as a definite integral. T

hen evalu-ate the integral to find the value of the lim

it. In each case, Pis a parti-

tion of the given interval and the numbers

are chosen from the

subintervals of P.

5.w

here Pis a partition of [1, 5]

6.w

here Pis a partition of [1, 3]

7.w

here Pis a partition of

8.w

here Pis a partition of

9.If

and find the values of the follow

ing.

a.b.

c.d.

e.

10.If

and find

the values of the following.

a.b.

c.d.

e.

AreaIn E

xercise 11–14, find the total area of the region between the graph

of ƒand the x-axis.

11.

12.ƒsxd

=1

-sx

2>4d, -

2…

x…

3

ƒsxd=

x2-

4x+

3, 0

…x

…3

L2

0sgsxd

-3ƒsxdd dx

L2

0 22

ƒsxd dxL

0

2ƒsxd dx

L2

1gsxd dx

L2

0gsxd dx

110 gsxd dx

=2

,1

20ƒsxd dx

=p

, 120 7gsxd dx

=7

,

L5

-2 a ƒsxd

+gsxd

5b

dx

L5

-2 s-p

gsxdd dxL

-2

5gsxd dx

L5

2ƒsxd dx

L2

-2 ƒsxd dx

15-2 gsxd dx

=2

,1

2-2 3ƒsxd dx

=12, 1

5-2 ƒsxd dx

=6

,

[0, p>2]lim

ƒƒ Pƒƒ :

0 a n

k=1 ssin

ck dscos c

k d ¢x

k ,

[-p

, 0]lim

ƒƒ Pƒƒ :

0 a n

k=1 acosa c

k2 bb ¢

xk ,

limƒƒ P

ƒƒ :0 a n

k=1 c

k sck

2-

1d 1>3 ¢x

k ,

limƒƒ P

ƒƒ :0 a n

k=1 s2c

k-

1d -1>2 ¢

xk ,

ck

a 20

k=1 sa

k-

2da 20

k=1 a 12

-2b

k

7 ba 20

k=1 sa

k+

bk d

a 20

k=1 3a

k

a 20

k=1 b

k=

7.

a 20

k=1 a

k=

0

13.

14.

Find the areas of the regions enclosed by the curves and lines in Exer-

cises 15–26.

15.

16.

17.

18.

19.20.

21.

22.

23.

24.

25.

26.

27.Find the area of the “triangular” region bounded on the left by

on the right by and above by

28.Find the area of the “triangular” region bounded on the left by

on the right by and below

by

29.Find the extrem

e values of and find the area of

the region enclosed by the graph of ƒand the x-axis.

30.Find the area of the region cut from

the first quadrant by the curve

31.Find the total area of the region enclosed by the curve and the lines

and

32.Find the total area of the region betw

een the curves and

for 0…

x…

3p>2

.y

=cos x

y=

sin x

y=

-1

.x

=y

x=

y2>3

x1>2

+y

1>2=

a1>2.

ƒsxd=

x3-

3x2

y=

1.

y=

6-

x,

y=1

x,

y=

2.

y=

x2,

x+

y=

2,

y=

8 cos x,

y=

sec2 x,

-p>3

…x

…p>3

y=

2 sin

x, y

=sin

2x, 0

…x

…p

y=

ƒ sin xƒ ,

y=

1, -p>2

…x

…p>2

y=

sin x,

y=

x, 0

…x

…p>4

y2=

4x+

4, y

=4x

-16

y2=

4x, y

=4x

-2

x=

4-

y2,

x=

0x

=2y

2, x

=0,

y=

3

x

y

01

1x

3�

�y

� 1, 0 �

x�

1

x3+1

y=

1, x

=0,

y=

0, for

0…

x…

1

x

y101

�x

� �

y�

1

1x

+1y

=1,

x=

0, y

=0

y=

x, y

=1>1

x, x

=2

y=

x, y

=1>x

2, x

=2

ƒsxd=

1-1

x , 0

…x

…4

ƒsxd=

5-

5x2>3,

-1

…x

…8


33.S

how that

solves the initial value problem

34.S

how

that solves

the initial

valueproblem

Express the solutions of the initial value problem

s in Exercises 35 and

36 in terms of integrals.

35.

36.

Evaluating Indefinite IntegralsE


37.38.

39.

40.

41.42.

43.44.

Evaluating Definite IntegralsE


45.46.

47.48.

49.50.

51.52.

53.54.

55.56.L

p>40 cos

2 a4t-p4 b

dtLp

0 sin

2 5r dr

L1>20

x3s1

+9x

4d -3>2 dx

L1

1/8 x-

1>3s1-

x2>3d 3>2 dx

L1

0

dr

23

(7-

5r) 2L

1

0

36 dxs2x

+1d 3

L4

1 A 1

+1uB 1>2

1u

duL

4

1

dtt1

t

L27

1x-

4/3 dxL

2

1 4y

2 dy

L1

0s8s

3-

12s2+

5d dsL

1

-1 s3x

2-

4x+

7d dx

L sec u

tan u 2

1+

sec u du

L 1

t sin s2t 3>2d dt

L st

+1d 2

-1

t 4 dt

L at-

2t bat+

2t b dt

L a1

22u

-p

+2 sec

2 s2u

-pdb

du

Ls2u

+1

+2

cos s2u

+1dd d

u Lstan

xd -3>2 sec

2 x dxL

2scos xd -1>2 sin

x dx

dydx=2

2-

sin2 x ,

ys-1d

=2

dydx=

sin x

x,

ys5d=

-3

d2y

dx2

=2sec x tan

x;

y¿s0d=

3, ys0d

=0

.

y=1

x0 A 1+

22sec tB dt

d2 y

dx2

=2

-1x2 ;

y¿s1d=

3, ys1d

=1

.

y=

x2+L

x

1 1t dt

Chapter 5Practice Exercises

389

57.58.

59.60.

61.62.

63.64.

65.66.

67.68.

69.70.

Average Values71.

Find the average value of

a.over

b.over

72.Find the average value of

a.over [0, 3]

b.over [0, a]

73.L

et ƒbe a function that is differentiable on [a,b]. In C

hapter 2w

edefined the average rate of change of ƒ

over [a,b] to be

and the instantaneous rate of change of ƒat x

to be In this

chapter we defined the average value of a function. For the new

def-inition of average to be consistent w

ith the old one, we should have

Is this the case? Give reasons for your answ

er.

74.Is it true that the average value of an integrable function over aninterval of length 2 is half the function’s integral over the interval?G


75.C

ompute the average value of the tem

perature function

for a 365-day year. This is one w

ay to estimate the annual m

eanair tem

perature in Fairbanks, Alaska. T

he National W

eather Ser-

vice’s official figure, a numerical average of the daily norm

alm

ean air temperatures for the year, is 25.7ºF, w

hich is slightlyhigher than the average value of ƒ(x). Figure 3.33

shows w

hy.

76.Specific heat of a gas

Specific heat

is the amount of heat re-

quired to raise the temperature of a given m

ass of gas with con-

Cy

ƒsxd=

37 sin

a2p

365 sx

-101db

+25

ƒsbd-

ƒsadb

-a

=average value of ƒ¿ on [a, b]. ƒ¿sxd.

ƒsbd-

ƒsadb

-a

y=2

ax

y=2

3x

[-k, k]

[-1, 1]

ƒsxd=

mx

+b

Lp

2>4p

2>36

cos1t

2t sin

1t dt

Lp>30

tan

u

22 sec u

du

Lp>40

sec

2 x

s1+

7 tan

xd 2>3 dxLp>20

3

sin x cos x

21

+3

sin2 x

dx

L2p>3

0 cos -

4 ax2 b

sin a

x2 b dx

Lp>2-p>2 15

sin4 3x cos 3x dx

L1

-1 2x sin

s1-

x2d dx

Lp>20

5ssin xd 3>2 cos x dx

L3p>4

p>4 csc z cot z dz

L0

-p>3 sec x tan

x dx

Lp

0 tan

2 u3 du

L3p

p

cot 2 x6 dx

L3p>4

p>4 csc

2 x dxLp>30

sec2 u du

stant volume by 1ºC

, measured in units of cal

deg-mole (calories

per degree gram m

olecule). The specific heat of oxygen depends

on its temperature T

and satisfies the formula

Find the average value of for

and the tem-

perature at which it is attained.

Differentiating IntegralsIn E

xercises 77–80, find .

77.78.

79.80.

Theory and Examples

81.Is it true that every function

that is differentiable on [a,b] is itself the derivative of som

e function on [a,b]? Give rea-

sons for your answer.

82.S

uppose that F(x) is an antiderivative of

Ex-

press in term

s of Fand give a reason for your

answer.

83.Find

if E

xplain the main steps in

your calculation.

84.Find

if E

xplain the main steps

in your calculation.

85.A

new parking lot

To meet the dem

and for parking, your town

has allocated the area shown here. A

s the town engineer, you have

been asked by the town council to find out if the lot can be built

for $10,000. The cost to clear the land w

ill be $0.10 a square foot,and the lot w

ill cost $2.00 a square foot to pave. Can the job be

done for $10,000? Use a low

er sum estim

ate to see. (Answ

ersm

ay vary slightly, depending on the estimate used.)

0 ft

36 ft

54 ft

51 ft

49.5 ft

54 ft

64.4 ft

67.5 ft

42 ft

Ignored

Vertical spacing �

15 ft

y=1

0 cos x s1>s1-

t 2dd dt.dy>dx

y=1

1x 2

1+

t 2 dt.dy>dx 1

10 2

1+

x4 dx

ƒsxd=2

1+

x4.

y=

ƒsxd

y=L

2

sec x 1

t 2+

1 dt

y=L

1

x

63

+t 4 dt

y=L

7x2

2 2

2+

cos3 t dt

y=L

x

2 22

+cos

3 t dt

dy>dx

20°…

T…

675°CCy

Cy

=8.27

+10

-5 s26T

-1.87T

2d.

>390

Chapter 5: Integration

T T

86.S

kydivers A and B

are in a helicopter hovering at 6400 ft. Sky-

diver A jum

ps and descends for 4 sec before opening her para-chute. T

he helicopter then climbs to 7000 ft and hovers there.

Forty-five seconds after A leaves the aircraft, B

jumps and de-

scends for 13 sec before opening his parachute. Both skydivers

descend at 16 ftsec w

ith parachutes open. Assum

e that the sky-divers fall freely (no effective air resistance) before their para-chutes open.

a.A

t what altitude does A

’s parachute open?

b.A

t what altitude does B

’s parachute open?

c.W

hich skydiver lands first?

Average Daily InventoryA

verage value is used in economics to study such things as average

daily inventory. If I(t) is the number of radios, tires, shoes, or w

hateverproduct a firm

has on hand on day t(we call I

an inventory function),

the average value of Iover a tim

e period [0, T] is called the firm

’s av-erage daily inventory for the period.

If his the dollar cost of holding one item

per day, the product is the average daily holding cost

for the period.avsId #h

Average daily inventory

=avsId

=1T

LT

0Istd dt.

>

87.A

s a wholesaler, T

racey Burr D

istributors receives a shipment of

1200 cases of chocolate bars every 30 days. TB

D sells the choco-

late to retailers at a steady rate, and tdays after a shipm

ent ar-rives,

its inventory

of cases

on hand

is W

hat is TB

D’s average daily inventory for the 30-

day period? What is its average daily holding cost if the cost of

holding one case is 3¢ a day?

88.R

ich Wholesale Foods, a m

anufacturer of cookies, stores its casesof cookies in an air-conditioned w

arehouse for shipment every 14

days. Rich tries to keep 600 cases on reserve to m

eet occasionalpeaks

in dem

and, so

a typical

14-day inventory

function is

The daily holding cost for each

case is 4¢ per day. Find Rich’s average daily inventory and aver-

age daily holding cost.

89.S

olon Container receives 450 drum

s of plastic pellets every 30days. T

he inventory function (drums on hand as a function of

days) is Find the average daily inventory. If

the holding cost for one drum is 2¢ per day, find the average daily

holding cost.

90.M

itchell Mailorder receives a shipm

ent of 600 cases of athleticsocks every 60 days. T

he number of cases on hand tdays after the

shipment

arrives is

Find the

averagedaily inventory. If the holding cost for one case is

¢ per day,find the average daily holding cost.

1>2Istd

=600

-202

15t.

Istd=

450-

t 2>2.

Istd=

600+

600t, 0…

t…

14.

0…

t…

30.

Istd=

1200-

40t,

391Chapter 5

Practice Exercises

Chapter 5 Additional and Advanced Exercises 391

Chapter 5 Additional and Advanced Exercises

Theory and Examples

1. a. If

b. If

Give reasons for your answers.

2. Suppose

Which, if any, of the following statements are true?

a. b.

c. on the interval

3. Initial value problem Show that

y = 1aL

x

0ƒstd sin asx - td dt

-2 … x … 5ƒsxd … g sxdL

5

-2sƒsxd + g sxdd = 9L

2

5ƒsxd dx = -3

L2

-2ƒsxd dx = 4, L

5

2ƒsxd dx = 3, L

5

-2g sxd dx = 2.

L1

02ƒsxd dx = 24 = 2?

L1

0ƒsxd dx = 4 and ƒsxd Ú 0, does

L1

07ƒsxd dx = 7, does L

1

0ƒsxd dx = 1?

solves the initial value problem

(Hint: )

4. Proportionality Suppose that x and y are related by the equation

Show that is proportional to y and find the constant ofproportionality.

5. Find ƒ(4) if

a. b.

6. Find from the following information.

i. ƒ is positive and continuous.

ii. The area under the curve from to is

a2

2+ a

2 sin a + p

2 cos a .

x = ax = 0y = ƒsxd

ƒsp/2dL

ƒsxd

0t2 dt = x cos px .L

x2

0ƒstd dt = x cos px

d2y/dx2

x = Ly

0

1

21 + 4t2 dt .

sin sax - atd = sin ax cos at - cos ax sin at .

d2y

dx2 + a2y = ƒsxd, dy

dx= 0 and y = 0 when x = 0.

7.T

he area of the region in the xy-plane enclosed by the x-axis, thecurve

and the lines and

is

equal to for all

Find ƒ(x).

8.P

rove that

(Hint:

Express the integral on the right-hand side as the difference

of two integrals. T

hen show that both sides of the equation have

the same derivative w

ith respect to x.)

9.F

inding a curveFind the equation for the curve in the xy-plane

that passes through the point if its slope at x

is always

10.Shoveling dirt

You sling a shovelful of dirt up from

the bottomof a hole w

ith an initial velocity of 32 ft/sec. The dirt m

ust rise 17ft above the release point to clear the edge of the hole. Is thatenough speed to get the dirt out, or had you better duck?

Piecewise Continuous Functions

Although w

e are mainly interested in continuous functions, m

anyfunctions in applications are piecew

ise continuous. A function ƒ(x) is

piecewise continuous on a closed interval

Iif ƒ

has only finitelym

any discontinuities in I, the limits

exist and are finite at every interior point of I, and the appropriate one-sided lim

its exist and are finite at the endpoints of I. All piecew

isecontinuous functions are integrable. T

he points of discontinuity subdi-vide I

into open and half-open subintervals on which ƒ

is continuous,and the lim

it criteria above guarantee that ƒhas a continuous exten-

sion to the closure of each subinterval. To integrate a piecewise con-

tinuous function, we integrate the individual extensions and add the

results. The integral of

(Figure5.34) over

is

The F

undamental T

heorem applies to piecew

ise continuous func-tions w

ith the restriction that is expected to equal

ƒ(x) only at values of xat w

hich ƒis continuous. T

here is a similar re-

striction on Leibniz’s R

ule below.

Graph the functions in E

xercises 11–16 and integrate them over

their domains.

sd>dxd1xa ƒstd dt

=32

+83

-1

=196

.

=cx-

x22 d-

1

0

+c x33 d0 2

+c-xd2 3

L3

-1 ƒsxd dx

=L0

-1 s1

-xd dx

+L2

0x

2 dx+L

3

2s-

1d dx

[-1, 3]

ƒsxd=•

1-

x,-

1…

x6

0

x2,

0

…x

62

-1,

2

…x

…3

limx:

c- ƒsxd and

limx:

c+ ƒsxd

3x2+

2.

s1, -1d

Lx

0 aLu

0ƒstd dtb

du=L

x

0ƒsudsx

-ud du

.

b7

1.

2b

2+

1-2

2

x=

bx

=1

y=

ƒsxd, ƒsxdÚ

0,

11.

12.

13.

14.

15.

16.

17.Find the average value of the function graphed in the accom

pany-ing figure.

18.Find the average value of the function graphed in the accom

pany-ing figure.

x

y1

12

30

x

y01

2

1

hsrd=•

r,-

1…

r6

0

1-

r2,

0…

r6

1

1,1

…r

…2

ƒsxd=•

1,-

2…

x6

-1

1-

x2,

-1

…x

61

2,1

…x

…2

hszd=e 2

1-

z,0

…z

61

s7z-

6d -1>3,

1…

z…

2

gstd=e

t,0

…t6

1

sin p

t,1

…t…

2

ƒsxd=e 2

-x ,

-4

…x

60

x2-

4,0

…x

…3

ƒsxd=e

x2>3,

-8

…x

60

-4,

0…

x…

3


x

y2

20

31

–1

1 3 4–1

y � x

2

y � 1 �

x

y � –1

FIGURE 5.34

Piecew

ise continuous functionslike this are integrated piece by piece.

Leibniz’s RuleIn applications, w

e sometim

es encounter functions like

defined by integrals that have variable upper limits of integration and

variable lower lim

its of integration at the same tim

e. The first integral

can be evaluated directly, but the second cannot. We m

ay find the de-rivative of either integral, how

ever, by a formula called L

eibniz’s Rule.

ƒsxd=L

x2

sin x s1

+td dt

and gsxd

=L21

x

1x

sin t 2 dt,

at which the carpet is being unrolled. T

hat is, A(x) is being in-

creased at the rate

At the sam

e time, A

is being decreased at the rate

the width at the end that is being rolled up tim

es the rate . T

henet rate of change in A

is

which is precisely L

eibniz’s Rule.

To prove the rule, let Fbe an antiderivative of ƒ

on [a,b]. Then

Differentiating both sides of this equation w

ith respect to xgives the

equation we w

ant:

Chain R

ule

Use L

eibniz’s Rule to find the derivatives of the functions in E

x-ercises 19–21.

19.20.

21.

22.U

se Leibniz’s R

ule to find the value of xthat m

aximizes the value

of the integral

Problem

s like this arise in the mathem

atical theory of politicalelections. S

ee “The E

ntry Problem

in a Political Race,” by S

tevenJ. B

rams and P

hilip D. S

traffin, Jr., in Political Equilibrium

, PeterO

rdeshook and

Kenneth

Shepfle,

Editors,

Kluw

er-Nijhoff,

Boston, 1982, pp. 181–195.

Approximating Finite Sum

s with Integrals

In many applications of calculus, integrals are used to approxim

ate fi-nite sum

s—the reverse of the usual procedure of using finite sum

s toapproxim

ate integrals.

Lx+

3

x ts5

-td dt.

gsyd=L

21y

1y

sin t 2 dt

ƒsxd=L

sin x

cos x

11

-t 2 dt

ƒsxd=L

x

1/x 1t dt =ƒsysxdd d

ydx-

ƒsusxdd dudx.

=F¿sysxdd d

ydx-

F¿susxdd dudx

ddxLysxd

usxdƒstd dt

=ddx

cFsysxdd-

Fsusxddd

Lysxd

usxdƒstd dt

=Fsysxdd

-Fsusxdd.

dAdx=

ƒsysxdd dydx

-ƒsusxdd dudx

,

du>dxƒsusxdd dudx

,

ƒsysxdd dydx

.

dy>dx

Chapter 5Additional and Advanced Exercises

393

t

y

0

Uncovering

Covering

f(u(x))y�

f(t)f(y(x))

u(x)

y(x)A(x)

�f(t)dty(x)

u(x)L

FIGURE 5.35

Rolling and unrolling a carpet: a geom

etricinterpretation of L

eibniz’s Rule:

dAdx=

ƒsysxdd dydx

-ƒsusxdd dudx

.

Leibniz’s RuleIf ƒ

is continuous on [a, b] and if u(x) and y(x) are dif-ferentiable functions of x

whose values lie in [a,

b],then

.ddx

Lysxd

usxdƒstd dt

=ƒsysxdd d

ydx-

ƒsusxdd dudx

Figure5.35

gives a geometric interpretation of L

eibniz’s Rule. It

shows a carpet of variable w

idth ƒ(t) that is being rolled up at the leftat the sam

e time x

as it is being unrolled at the right. (In this interpre-tation, tim

e is x, not t.) At tim

e x, the floor is covered from u(x) to y(x).

The rate

at which the carpet is being rolled up need not be the

same as the rate

at which the carpet is being laid dow

n. At any

given time x, the area covered by carpet is

Asxd=L

ysxd

usxdƒstd dt.

dy>dx

du>dx

At w

hat rate is the covered area changing? At the instant x,A

(x) is in-creasing by the w

idth ƒ(y(x)) of the unrolling carpet times the rate

For example, let’s estim

ate the sum of the square roots of the first

npositive integers,

The integral

is the limit of the upper sum

s

Therefore, w

hen nis large,

will be close to

and we w

ill have

The follow

ing table shows how

good the approximation can be.

nR

oot sumR

elative error

1022.468

21.08250

239.04235.70

1.4%100

671.46666.67

0.7%1000

21,09721,082

0.07%

23.E

valuate

by showing that the lim

it is

and evaluating the integral.

24.S

ee Exercise 23. E

valuate

limn:

q 1n

4 s13+

23+

33+

Á+

n3d.

L1

0x

5 dx

limn:

q 1

5+

25+

35+

Á+

n5

n6

1.386>22.468L

6%

s2>3dn3>2

Root sum

=21

+22

+Á

+2n

=S

n #n3>2

L23

n3/2.

2>3S

n

x

y0

y � �

x

11

n2n

n � 1

n

= 21

+22

+Á

+2n

n3>2

.

Sn=A

1n # 1n

+A2n

# 1n+

Á+A

nn # 1n

L1

0 1x dx

=23

x3>2d0 1

=23

21

+22

+Á

+2n

.25.

Let ƒ(x) be a continuous function. E

xpress

as a definite integral.

26.U

se the result of Exercise 25 to evaluate

a.b.c.

What can be said about the follow

ing limits?

d.e.

27.a.

Show

that the area of an n-sided regular polygon in a circle

of radius ris

b.Find the lim

it of as

Is this answer consistent w

ithw

hat you know about the area of a circle?

28.A

differential equationS

how that

satisfies both of the following conditions:

i.ii.and

when

29.A

function defined by an integralT

he graph of a function ƒconsists of a sem

icircle and two line segm

ents as shown. L

et

a.Find g

(1).b.

Find g(3).

c.Find

d.Find all values of x

on the open interval at w

hich ghas a relative m

aximum

.

e.W

rite an equation for the line tangent to the graph of gat

f.Find the x-coordinate of each point of inflection of the graphof g

on the open interval

g.Find the range of g.

s-3, 4d.

x=

-1

.

s-3, 4d

gs-1d.

y

13

–3

y � f(x)

–1–1 1 3

x

gsxd=1

x1 ƒstd dt.

x=p

.y¿

=-

2y

=1

y–=

-sin

x+

2 sin

2x1px

cos 2t dt+

1y

=sin

x+

n:q

.A

n An=

nr2

2 sin

2pn

.

An

limn:

q

1n15 s1

15+

215

+3

15+

Á+

n15d

limn:

q

1n17 s1

15+

215

+3

15+

Á+

n15d

limn:

q 1n

asin pn

+sin

2pn

+sin

3pn

+Á

+sin

npn b

.

limn:

q

1n16 s1

15+

215

+3

15+

Á+

n15d,

limn:

q 1n

2 s2+

4+

6+

Á+

2nd,

limn:

q 1n

cƒ a 1n b

+ƒ

a 2n b+

Á+

ƒ a nn bd


Chapter 5Technology Application Projects

395

Chapter 5Technology Application Projects

Mathem

atica/Maple M

oduleU

sing R

ieman

n S

um

s to Estim

ate Areas,Volu

mes,an

d Len

gths of C

urves

Visualize and approxim

ate areas and volumes in Part I.

Mathem

atica/Maple M

oduleR

ieman

n S

um

s,Defin

ite Integrals,an

d the F

un

damen

tal Th

eorem of C

alculu

sParts I, II, and III develop R

iemann sum

s and definite integrals. Part IV continues the developm

ent of the Riem

ann sum and definite integral

using the Fundam

ental Theorem

to solve problems previously investigated.

Mathem

atica/Maple M

oduleR

ain C

atchers,E

levators,and R

ocketsPart I illustrates that the area under a curve is the sam

e as the area of an appropriate rectangle for examples taken from

the chapter. You w

illcom

pute the amount of w

ater accumulating in basins of different shapes as the basin is filled and drained.

Mathem

atica/Maple M

oduleM

otion A

long a S

traight L

ine,P

art IIY

ou will observe the shape of a graph through dram

atic animated visualizations of the derivative relations am

ong the position, velocity, andacceleration. Figures in the text can be anim

ated using this software.

Mathem

atica/Maple M

oduleB

endin

g of Beam

sS

tudy bent shapes of beams, determ

ine their maxim

um deflections, concavity and inflection points, and interpret the results in term

s of a beam’s

compression and tension.

Documents

5.3 The Definite Integral · 2017-01-22 · 5.3 The Definite Integral 343 The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over