Chapter 4 Thomas 9 Ed F 2008

CHRprsR

Integration

OvERWw This chapter examines two processes and their relation to one another.One is the process by which we determine functions from thefu dedvatives. Theother is the process by which we arrive at exact fomulas for such things as volumeand area through successive approximations. Both processes are called integration.

Integration and differentiation are intimately connected. The nature of the con-nection is one of the most important ideas in all mathematics, and its independentdiscovery by Leibniz and Newton still constitutes one of the greatest technicaladvances of modem times.

Indefinite IntegralsOne ofthe early accomplishments of calculus was predicting the future position of amoving body from one ofits known locations and a fomula for its velocity function.Today we view this as one of a number of occasions on which we determine afunction from one of its known values and a formula for its rate of change. It isa routine process today, thalks to calculus, to calculate how fast a space vehicleneeds to be going at a certain point to escape the earth's gravitational field or topredict the useful life of a sample of radioactive polonium-210 ftom its presentlevel of activity and its rate of decay.

The process of deterrnining a function from one of its known values and itsderivative /(;) has two steps. The first is to find a formula that gives us allthe functions that could possibly have / as a derivative. These functions are theso-called antiderivatives of /, and the formula that gives them all is called theindefinite integral of /. The second step is to use the known function value toselect the particular antiderivative we want from the indefinite integral. The firststep is the subject of the present section; the second is the subject of the next.

Finding a formula that gives all of a function's antiderivatives might seem likean impossible task, or at least to require a little magic. But this is not the case atall. If we can find even one of a function's antiderivatives we can find them all,because of the first two corollaties of the Mean Value Theorem of Section 3.2.

275

276 Ch,ap{er 4. t.teq(ation

Finding Antiderivatives-lndef ini te Integrals

DefinitionsA function F(r) is an antiderivative of a function /(r) if

F'(x) - f (x)

for all r in the domain of /. The set of all antiderivatives of / is theindefinite integral of / wiih respect to r, denoted by

t -

J I t r t ' t t ' '

The symbol J is an integral sign. The function l is the integrand of theintegral and x is the variable of integration.

According to Corollary 2 of the Mean Value Theorem (Section 3.2), once wehave found one antidedvative F of a function /, the other antiderivatives of l'differ from F by a constant. We indicate this in integral notation in the followingway:

I r a; a,: F(r) * c.

The constant C is the constant of integration or arbitrary constant. Equation (l)

is read, "The indelinite integral of f with respect to 'lr is F(,r) i C." When we find

F(t) + C, we say that we have integrated / and eYaluated the integral.

EXAMPLE 7 Evalurte / 2r c/r.

Solution

| ^"t

^n antidefivative of 2jf

| 2 x d x : x ' 1 ' C . .J

----\ the arbitrafy constant

The formula x2 + C generates all the antiderivatives of the function 2r. The func-

tions ,tr + l, x2 - t, and jr2 + vD are all antiderivatives of the function 2r, as you

can check by differentiation.

Many ofthe indefinite integrals needed in scientitic work are found by reversing

derivative formulas. You will see what we mean if you look at Table 4.1, whichlists a number of standard integral forms side by side with theit derivative-formulasources.

In case you are wondering why the integrals of the tangent, cotangent, secant,and cosecant do not appear in the table, the answer is that the usual formulas 1brthem require logarithms. In Section 4.7, we will see that these functions do haveantiderivatives, but we will have to wait until Chapte$ 6 and 7 to see what theyare.

( 1 )

Table 4.1 Integral formulas

EXAMPLE 2 Se/ected integrals from Table 4.1

a) [* 'a, : l+cJ O

al I Lnar:

l ' - ,r ,a,:2xu2+c:zJr+cO

,/ t" 2r7*: -"o\z' +c

ar /"o. io,: I "o"| 'a' :*y?tc:2sin

4.1 Indefinite Integrals 277

Fonnula I

fofinula I

with n = -1/2

Formula 2w i t h k : 2

Fomula 3with k: | /2

o

xt+c

Finding an integral formula can sometimes be difflcult, but checking it, oncefound, is relatively easy: differentiate the righrhand side. The derivative should bethe intesrand.

EXAMPLE 3f

Right: I x cos x dx : x sin r + cos x + CJ

Reason: The derivative of the righfhand side is the integand:

f iU " r nx +cos x IC I : xcos r+s in . r - s i n - r+0 : x cos x .

r. l ta,:#*", nl-r, n rutionor

lat= Itor=r+c (special case)

2. I

s i 'bdx=-"o io ' * "

t . lcoskxax:" 'Y ' * "

4. f

sec2 t d.x =tan,- 1-g

s . l c s c "

r a x = - c o t r + c

6, f

sec -tan x dx = secx q g

z. | """

r"ot ra, = *csc, + c

. | - l = x 'a x \ n + t /

d '. ( r ) = 1

d / coskx \_ _ l _ _ _ _ _ ; _ l : s m k x

d /sinkx\, | . l = c o s f . r

d

,; lanl : sec- t

d ^

,; (- cot ,) = csc' t

d

; ; s e c r = s e c r t a n t

d

;; t-csc l): csc 'rj cot .r

Z7A Chapter 4: lntegration

Wrong : J

xcosxdx :x s in ; -F C

Reason: The derivative of the right-hand side is not the integrand:,1

i 1x sin x + C) :.x cos x +sin x +0 + x cos x.dx ' ' - " " - - ' t r

Do not wony about how to derive the conect integral formula in Example 3.We will present a technique for doing so in Chapter 7.

Rules of Algebra for AntiderivativesAmong the things we know about antiderivatives are these:

1. A function is an antiderivative of a constant multiple ft/ of a function / if andonly if it is f times an antiderivative of l.

2. In particular, a function is an antiderivative of -/ if and only if it is thenegative of an antiderivative of /.

3. A function is an antiderivative of a sum or diffeience / a g if and only if itis the sum or difference of an antiderivative of / and an antiderivative of g.

When we express these observations in integral notation, we get the standard arith-metic rules for indefinite integration (Table 4.2).

Table 4.2 Rules for indefinite inteqration

Constant Muttipte Rute: I

olOlat -t I

tftla,

(Does not work if t varies with i.)

Rute for Negatives: l-frOor:- | fUlo"

( R u l e l w i t h k : l )

sum and Difference Rute: /t"ff' ltrt ' l lr '

- |

f<*la" + |

ef'la,

EXAMPLE 4 Rewriting the constant of integration

f l

. / 5 s e c x I a n x d x = 5 . / s e c x t a n r d r

' l \ l e l - . R 1 .

: 5 ( s e c ; * C )

: 5 s e c r * 5 C

: 5 s e c " r * C '

= 5 s e c , r * C

Trble 1.1. Fo nula 6

FiIst lb|ln

Shortef finm. where C is 5C

t lsur l tbnn no f r ime. Sincc 5 r jnrc\ rL i ritfbilfaf,"- co stJ l is rn arbilif] coDstrni.

What about all the different forms in Example 4? Each one gives all tilantiderivatives of /(r):5secrtanx, so each answer is corfect. But the least

4.1 Indefinite Integrals 279

complicated of the three, and the usual choice, is

t -

. / 5 s e c , v t a n x d x - 5 s e c . r + C .

Just as the Sum and Difference Rule for differentiation enables us to differenti-ate expressions teIm by tetm, the Sum and Difference Rule tbl integration enablesus to integrate expressions term by tem. When we do so, we combine the individualconstants of integmtion into a single arbitrary constant at the end.

EXAMPLE 5 7 . i ' , , , i . ' ' : ' ' , - i t t? ' - t ' . , : . j : l

EvaluatefIQ2 -2 : (+5 )dx .J

Solution If we recognize that (r.r/3) - x2 + 5x is an antiderivative of x2 - 2x +5, we can evaluate the integral as

antide vative /a$itrary

constant

/

Io t 2x+s)c ]x = l , '+sr+ c / .J3

If we do not recognize the antiderivative right away, we can generate it telmby term with the Sum and Difference Rule:

r - f f Il , r ' 2 r - 5 t d r -

| x - d ; - | 2 r d t - | 5 d tJ J J J

: {

*a , - x2 +c2*5 - r * c r .

This formula is more complicated than it needs to be. If we combine Cr, C2, andC3 into a single constant C : Ct I Czl Cz, the formula simpli l les to

, t ,- - r ! ) . { + (

and Jtlll gives all the antiderivatives there are. For this reason we recommend thatyou go right to the flnal form even if you elect to integrate term by term. Write

f - l f ll , r ' 2 x . 5 t d x - l r ' t l x - l 2 r , l r + l 5 J t

, I , J J J

n l=T- " ' * 5 r *C .

Find the simplest antiderivative you can for each part and add the constant at theend.

The Integrals of sin2 x and cos2 xWe can sometimes use trigonometric identities to transfbrm integrals we do notknow how to evaluate into integmls we do know how to evaluate. The integralformulas for sin2 x and cos2 r arise frequently in applications.

280 Chaoter 4: Inteqration

EXAMPLE 6

a) J

sin2 x,lxr n 1 - c o s 2 x:J , or

I l sin 2x ^ x sin 2.r ^- l (a . t ' , 1

/ . oJ *a " : I l cos 2x

o ,.l .l 2

l + ( ! \ l f

,r sin 2x: i , i C A s i n p l r l ( r ) . b u l

J r ' _ . ' r : r l . _ : r ) . ' r

=l l, cos2x)dx :l I r, -f,

| *"2,a*

b)

Exercises 4.1

Finding AntiderivativesIn Exercises 1-18, find an antiderivative for each function. Do asmany as you can mentally. Check your answers by differentiation.

c ) x 2 2 x + l

c ) - x 7 - 6 n + 8

c ) r - a + 2 r + 3

tt l] +; c) -'-3 +.r - 1

b)

-; c) 2- --x ' x '

l . l

2X' I

l - lz4x 4a

I " _ Is c ) 7 . r + - -)ax Jxt l

c t - - r ' -3 3

t ?c l - - x " / _

2 2

3 sin .Y c) sin zl - 3 sin 3rI 1tX , 'T,I

cos cl cos _ + 7r cor r2 2 2

sec- ct - sec_ -t r 2

- c s c c t t - 6 c s c " l r2 2

-

Evaluating IntegralsEvaluate the integrals in Exercises 19 58. Check your answers btdifferentiation.

D. IG + t)dx

n. | (2, '+'r) a'

zz. I e,' 5x : 7) dx

" IG * I),'zz. | ,- ', 'o'

zs. | {J, * +c) a,

-. l(,,- *z),'

b) "t

b) rt

b) .r- '

15, a) csc jr cot .{.. 1tx lfx

c, z csc' 2 2

16. a) sec.r tan.r, rx Tx

cl sec tan2 2

17, (sin r cos x)'?

b) - csc 5ir cot 5,r

b) 4 sec 3r tan 3x

18. (1 * 2cos.r)'z

l . a ) 2 x

2. a) 6x

3. a) -3r-a

4 , a ) 2 x 3

I5. a, -

2q . a l _ _ ,

t , a t t

\ / x

4 . _8. a)

1Vr

9 . a1 ? " - t t t

' 3

t -10 . a t

1x- " '

11. a) z sin zr

12, a) T cos 7rx

13. a) sec2 r

14. a) csc2;r

b)

b)

b)

b)

b)

zo. I rs e,ta,I t . l

zz. J \1++s)art ^

24. .l

tl - r' - J.t' rJx

f / 1 ) \

" . / l ; - -+z^)a 't -

28. J

x-'t. dx

30. [ ( t * 1 \ , 'J \ 2 Jx )

| / 1 I \3' . J \ ;_ n)d\

b)

b)

b)

b)

*. lz,Q-"'1 a'

s' l '$,!{ a'

n. |

{-z cos ) at

tn. | , "" laun. I ez """' ,) a,

*. I'"i#*s.

| $

"", " ta\\ x - 2 sec2 r) dx

ae. | )r*e

r - csc,r cot r)lx

el. |

<"in2x - cscz x)dx

at. |

{z "o" z" - 3 sin 3.r) d:r

o g . I o r i n ' r d ,J

sr. l\1!a,sz.

l : +tun' ae

(H in t : 1+ t ^n2 0 : sec2 0 )

55. .l

cot' a dx

(Hint: llcot2 r : csc2 r)

s z . / " o r e { t a n

0 + s e c d e

50. / "ott) dul ' 1

". l!-f9"st.

| {z + tun' el ae

sa. l0-cot")dx

sa. [ -:!J -asJ C S C d - S 1 n d

Checking Integration FormulasVerify fte integral formulas in Exercises 59-64 by differentiation. InSection 4.3, we will see where formulas like these come from.

I t ' l v - 1 \ 4

ss . J r t ' - z fa , - .=1 I . * ,

/ r l v - ! 5 l - lo o .

/ r : r - s r - ' a x : - l : 4 3 ' l a

f - I6 1 .

/ s e c ' { 5 r - t r d r : j r a n { 5 . r - l r C

f / v - l \ / r l \62.

J cscz ( - i ) r , = : cor ( f J+c

1 1 I6 3 . I ^ d t : - - + C

J \ 1 + \ ) ' r + r

Exercises 4.1 Za1

t r , x| - A X : i L

J ( . { + l ) r - r + I

Right, or wrong? Say which for each formula and give a briefreason lbr each answer.

t 1 2a)

./ :r sin.rd,r : I sin x + C

. tD t t , t s l n . t d i t : - . { c o s t r + c

c) f r s in , rd r : - r cos n +s in , r +CJ

Right, or wrong? Say which for each formula and give a briefrcason for each answer.

| " sec3 Ia )

J t a ] l 0 s e c ' 0 d 0 :

3 + C

tr f mB sec2 g dg : -l ,un' d + cJ 2

t " l "c )

. / tan 0 sec ' ede: sec '9 * C

Right, or wrong? Say which for each formula and give a bdefreason for each answer

r i t " r t r 3at

J tzx +1 t2dr : - t *a

f -b t

J 3 t 2 ^ + l t ' d r : ( 2 x + 1 ) ' + C

t -c l

J 6 t 2 t - l l t ' d x : ( 2 . r + l ) ' + C

Right, or wrong? Say which for each formula and give a briefreason for each answer.

t -

d J "tr, + t at : J x2 + x + C

r -b t

. l J 2 r + l d x : y ' r : + r + C

f _ I ? - _ \ re t

JJ2x - rdx :11 ' / z r t t ) t c

Theory and Examples69. Suppose that

d _ d1 6 ) = E ( l

- / r c ) a n d s G ) : * 0 , + 2 \ .

Find:

o lroa'

") Ir f r"la"

ey luGl + g1,11a,

u. l , ' { ,+t)a '36. l!4,1tra

/{-s sin r)ar

n . l 3cos50da* I(+),.

+e. l!

sec e taroao

64.

65.

67.

n) lg@ax

at fr sata,

rl /rf r'l - sl)tdx

282 Chapter 4: Integration

nr I k rx t - t )dxl k + f rx la '

70. Repeat Exercise 69, assuming that- d d

J ,r, : O,

c' and g(.r ') = - {.r sin ( L

This section shows how to use a known value of a function to select a particularantiderivative from the functions in an indef,nite integral. The ability to do this isimportant in mathematical modeling, the process by which we, as scientists, usemathematics to leam about reality.

Init ial Value ProblemsAn equation like

dy7 ,

= r t x t

that has a derivative in it is called a differential equation. The problem of findinga function y of .r when we klow its derivative and its value y0 at a particular pointxq is called an initial value problem. We solve such a problem in two steps, asdemonsffated in Example 1.

EXAMPLE 1 Finding a body's velocity from its acceleration and initi.vetoctty

The acceleration of gravity near the surface of the earth is 9.8 m/secz. This meansthat the velocity u of a body falling freely in a vacuum changes at the rate of

du- = 98m' /sec2 '

A I

If the body is dropped from rest, what will its velocity be I seconds after it isreleased?

Solution In mathematical terms, we want to solve the initial value problem tharconsists of

The differential equation: #:nt

The initial condition: u :0 when / :0 (abbreviated as u(0) : Q)

We first solve the differential equation by integrating both sides with respect to r:

!:n'd t

I hc di l lercnt i . r l cqul l io I

I r tcsf i l lc wi th re\ fcc l to r .

ln lesr r ls c\ r lu.Lrcd

Con\L: ! r I \ .ornbine. : l as or .

I :a t : I gs a tJ qt . l

D 1 - C l : 9 . 8 t + C 2

u : 9 .8 t + C.

Differential Equations, Initial Value Problems,and Mathematical Modeling

4.2 Differential Equations, Init ial Value problems. and l\4athematical Modelinq 2g3

This last equation tells us that the body's velocity r seconds into the fall is 9.gr +C m/sec for some value of C. What value? We find out from the initial condition:

u : 9 . 8 t I C

0 : 9 . 8 ( 0 ) + C , r 0 ) : 0

C : 0 .

Conclusion: The body's velocity I seconds into the fall is

u : q . 8 r + O : 9 . 8 r m / s e c . AThe indefinite integral F(,r) + C of the function /(,r) gives the general solu_

tion y : p1y) * C of the differenti al equation dy /dx : /(r). The general solutiongives all the solutions of the equation (there are infinitely many, one for each valueof C). We solye the differential equation by finding its general solution. We thensolve the initial value problem by finding the particurar sorution that satisfies theinitial condition )(r0) : .y0 (y has the value )0 when .x : x0).

EXAMPLE 2 Finding a curve from its slope function and a point

Find the curve whose slope at the point (;r, y) is 3.r2 if the curve is required topass though the point (1, -1).

sorufion In mathematical language, we are asked to solve the initial value problemthat consists of

The differential equation: *

: r* l l,c crn c s stopc lr 3 Ll

) ( l ) : - 1 .

To solve it we first solve the differential equation:dydx

The init ial condition:

[90,:J A A Iz"a*

x 3 + C .( lon\ t tu i \ o i inrer i i r t ro l lc(rnr t r incd. gr \ ing rhc !c | . r l t

,1.1 The curvesy =x3 + C fi l l the(oordinate plane without overlapping. rnExample 2 we identify the curvey : x' - 2 as the one that passes throughthe given point (1, 1).

This tells us that l equals x3 + C for some value of C. We find that value fromthe condition y(1) : -1:

y = - { - + L

1 : ( 1 ) 3 + C

c : -2 . t lThe curve we wanr is y: x3 -2 (Fig. . l).

In the next example, we have to integrate a second derivative twice to find thefunction we are looking for The first integration,

f ,J2s cl:I

-)at : ::! ' 6"

J q t . d l

gives the function's first derivative. The second integration gives the function.

2A4 Chapter 4: Integration

I I nojectileII

II

I

iI

EXAMPLE 3 Finding a projectite's height from its acceleration' initial

velociiy, and ittitial Position

A heavy projectile is fired straight up from a platform 3 m above the ground' with

an initial vefocity of 160 m,/sec. Assume that the only force affecting the projectile

i"tl"g i" flight is from gravity, which produces a downward acceleration of 9 8

m/sec"2. Findl equation lor the projectile's height above the ground as a function

of time t if t = 0 when the Fojectile is firecl. How high above the ground is the

projectile 3 sec after firing?

Solution To model the problem, we draw a figure (Fig' 4'2) and let s denote

the projectile's height above the ground at time t' We assume s to be a twlca-

OitrerentiaUte function of t and represent the projectile's velocity and acceleration

with the derivativesds

t) : = anod I

Since gravity acts in tlre direcrion of decreaslng s in our model' the initial value

problem to solve is the following:,12 s

The differential equation: - = -9 4

l 10 ; : 160d I

and s(0) = 3.

We integrate the differential equation with lespect to t to fir'd dsldtl

f d2s f

J *o '= / ( -e8)d '

| : -e .s r+cr .a t

We apply the first initial condition to find Cr:

1 6 0 : _ 9 . 8 ( 0 ) + c r

Cr : 160'

This completes the formula for ds/dr:ds : _9.8r + 160.a t

We integrate ds/dt with respect to t to find r:t ) . fI zat = , (-9.8r + 160) dr

J d I J

s : -4.9t? + 16ot -t cz.

We apply tie second initial condition to find C2:

3 : -4-9Q)2 + 160(0) + C, r(o) : l

C z : 3 '

This completes the formula for s as a function of t:

s -- -4.9t2 + 160/ + 3.

Ground level

a.z ln" ,r"t.t' for modeling theprojecti le motion in Example 3.

du d2sq : - = - - - .

d t d l '

The initial conditions:

*,0, = ,,tn

U

4.2 Differential Equations, Init ial Value Problems, and Mathematical Modeling 285

To find the projectile's height 3 sec into the flight, we set / :3 in the formulafor s. The height is

s : -4 .9(3) '?* 160(3) + 3 :438.9 m. A

When we find a function from its first derivative, we have one arbitrary constant,as in Examples I and 2. When we find a function from its second derivative, wehave to deal with two constants, one from each antidifferentiation, as in Example3. To find a function from its third derivative would require us to find the values ofthree constants, and so on. In eac,h case, the values of the constants are determinedby the problem's initial conditions. Each time we find ar antiderivative, we needan initial condition to tell us the value of C.

Sketching Solution CurvesThe graph of a solution of a differential equation is called a solution curv€ (integralcurve), The curves y : .d3 * C in Fig. 4.1 are solution curves of the differentialequaion dy/dx:3.r2. When we cannot find explicit formulas for the solutioncurves of an equation dy/dx: /(x) (that is, we camot find an antiderivative ofl), we may still be able to find their general shape by examining derivatives.

EXAMPLE 4 Sketch the solutions of the differential equation

, 1' - x 2 + 7 '

Step 2: Rise and fall. The domain of y' is (-co, oo). Thereare no critical points, so the solution curves have no cusps orextrema. The curves rise from left to right because y'> 0. At

"r :0, the curves have slope 1.

Solution

Step 1: y' aad y" . As in Section 3.4, the curve's general shapeis determined by y' and y". We already know y':

, 1t -

x 2 + 1 '

We find y" by differentiation, in the usual way:

, , d d l r \y : , t y ) : , | , . Iq x o x \ x ' 1 t /

- (x2 + I)2'

Step 3: Concqvity. The second derivative changes from (-|-)to (-) at r :0, so the curves all have an inflection point atx :0 .

I

t , - 2 , t ] l--___1+-I

0III

slope

conc up

)infl I

General shape :

0

point slope =0 conc

up oo!!n

infl point

J = --;---:i + +x ' + l l--T--'

Step 4: Summary:


4.3 The solution curve in Example 5.

The first derivative tells us still more:I

l i m Y ' : 1 1 6 * : 0 'J _ d x + l

so the curves level off as x -> f oo.

Step 5: Specific points and solutioncurves, We plot an assortment ofpoints on the l-axis where we knowlhe curves ' s lope ( i t is I a t x : '0), mark tangents with that slopefor guidance, and sketch "parallel"curves of the right general shape.

!

EXAMPLE 5 Sketch the solution of the initial value problem

1Differential equation: y' :

x, + I

/ : 0 w h e n ; : 0 .Initial condition:

sorutron We find the solution's general shape (Example 4) and sketch the solutioncurve that passes through the point (0, 0) (Fig. a.3). D

The technique we have learned for sketching solutions is particularly helpfulwhen we are faced with an equation dy/ax : f(x) that involves a function whoseantiderivatives have no elementary formula. The antiderivatives of the function

f(x):1/(x2 * l) in Example 4 do have an elementary formula, as we will see inChapter 6, but the antiderivatives of g(.r) : .rt a.r+ do not. To solve the equationdy /dx : ^,/1+ xa, we must proceed either graphically or numerically.

Mathematical ModelingThe development of a mathematical model usually takes four steps: First we obseruesomething in the real world (a ball bearing falling from rest or the trachea contractingduring a cough, for example) and construct a system of mathematical variables andrelationships that imitate some of its important features. We build a mathematicalmetaphor for what we see. Next we apply (usually) existing mathematics to thevariables and relationships in the model to draw conclusions about them. After thatwe ffanslate the mathematical conclusions into information about the system understudy. Finally we check the information against observation to see if the modelhas predictive value. We also investigate the possibility that the model applies toother systems. The really good models are the ones that lead to conclusions thatare consistent with observation, that have predictive value and broad application,and that are not too hard to use.

The natural cycle of mathematical imitation, deduction, interpretation, and con-flrmation is shown in the diagrams on the following page.

4.2 Differential Equations, Init ial Value Problems, and Mathematical Modeling 247

. Observedr Thingsfalling from rest

in a vacuum

L The velocity atlime f should be 32r.

2. Alt bodies fall withthe same constant

acceleration: 32 f t/sec2.

Assumptions abouthow measurable

quantities are related

Interpretation inreal world terms

Vadablesl

Initial values:s = 0 a n d v = 0w h e n t = 0

Assumed relation:s = l6t2

Applvcalculus

Mathematicalconclusions:l . . r = 32 t2 . a = 3 2

. ObseIved:Alight raybending as it passesfrom one medium to

another

Confirmationmeasur€ment

Sfiell's law: The sine ofthe angle of refraction

will always be c2lcttimes lhe sine of the

anBle of incidence.

Assumptions abouthow measulable

quantities are related

Interpretation inreal-world lerms

Variables:Angles, light speeds,distances, etc.

Assumed relations:Femaf s principleTime = distance/rate

Apply extremevalue calculation ofExample 5, Section 3.6

Mathematical conclusion:

sin 0. = ;! sin d,

4.4 (a) The modeling cycle for the shapesof coll iding galaxies. (b) The computer'simage of how galaxies are reshaped bythe coll ision.

Computer SimulationWhen a system we want to study is complicated, we can sometrmes expenment

first to see how the system behaves under different circumstances. But if this is

not possible (the experiments might be expensive, time-consuming, or dangerous),

we might run a series of simulated expedments on a computer-experiments that

behave like the real thing, without the disadvantages. Thus we might model the

effects of atomic war, the effect of waiting a year longer to ha est trees, the effect

of crossing particular breeds of cattle, or the effect of reducing atmospheric ozone

by 1.Vo, all without having to pay the consequences or wait to see how things work

out naturally.We also bring computers in when the model we want to use has too many

calculations to be practical any other way. NASA s space flight models are run on

computers-they have to be to generate course corections on time. If you want to

model the behavior of galaxies that contain biltions and billions of stars, a computer

offers the only possible way. One of the most spectacular computer simulations in

recent years, carried out by Alar Toomre at MI! explained a peculiar galactic shape

that was not consistent with our previous ideas about how galaxies are formed. The

galaxies had acquired their odd shapes, Toomre concluded, by passing through one

another (Fig. 4.4).

. Obs€Ned:Oddly shaped

galaxies

Comparison:The compufer pictures

look like the photographsof the obsefled galaxies.

Plausible conclusion: Theygot this way by passing

thrcugh one another,

Interpr€tatron: Picturcsof galaxies passing though

one anoth€f

Assumptions abouthow gravity would affectstars of galaxies passing

through one another

Computer-generatedvisual images

tal

Vadables andforce equations

Applv:Computer simulationof colliding galaxies

Data aboutchanging starposrt1ons

Shapes of galaxi€s


Exercises 4.2

ln i t ia l Value Problemsl. Which of the following graphs shows the solution of the inilial

value problem

dy

a t : 2 x . y = 4 w h e n - r : l ?

l ) '

(a)

Give reasons for your answer.

2, Which ol the following graphs shows the solution of the initialvalue problem

rO. f ,

: cosr +. in, , r (r) = l

l l . ! : -n sin n0, r(0):0

12. L - cosra. r(.0\ l

B . *

= : * " t t a n r , u ( o ) : I

r4 . { ! = 8 t + csc . t , , ( i ) : .

1l r't r . r , - 2 - 6 r : r r 0 ) = 4 . . ' r U ' I

rc. fi -- o: 1'(o) :2, .r,(o): o

n. - l : : ! , , ' l - r , , ( r , ld r . r ' t ! l , l

18. 1' : = r l . 1' r c1z,;:a/ r d d l

19. __:= :6: t,,,(0) : 8, r,(0) : 0.

d 'e r, r ,

U , , 0 4 f 0 ) - - ) . e t ( ) t =

2 .2 1 . 1 L ' J : - 5 i 6 1 4 c 6 5 I

dJ

" t : - ' l w h e n r v : - l ?

)(0) : s

eQ) : a

, " " ' (0 \ :7 . r " (0) =. r ' (0) = 1, y(0) : 0

2 2 . ) ( 1 1 : - c o s r + 8 s i n 2 x :),"'(0) = 0, _r"(0) - r'(0): l, ](0) : 3

Finding Position from VelocityExercises 23-26 give the velocity t:: ds/tlt and initial position ofa body moving along a coordinate line. Find the body's position attime /

2 3 . ! : 9 . 8 r + 5 , . r ( 0 ) : 1 0

24, r . t :32 t 2 , s ( l l2 ) :4

25. u : s in z r , s (0 ) :0) ':) |

2 6 , u : : c o s : : , s ( r 1 ) : I

Finding Position from AccelerationExercises 27 30 give the acceleration a: d2s/dtI, initial velocity,and initial position of a body moving on a coordinate line. Find thebody's posit ion at t ime l .

27. a :32: !(0) : 20, r(0) : 5

Give reasons for your answet

Solve the initial valoe problems in Exercises 3-22.

t . ' I : r , ?, ) (2) : o

+. ! - rc x , y(o) : - r

t + :++r . , ' l r>0 ; . y (2 ) : t

e' fr :e" - ' rr + 5, )( l ) : o

t . f i =3x ' / r . 1 r t r : 5

g. dr = =!. 1r+1 : o

g. * : ,+ cosr , . r (o ) :4

0

2 8 . a = 9 . 8 : o ( 0 ) : - 3 , s ( 0 ) : 0

29. cr: -4sin2t. u(0) = 2, s(0) : -3q l r

30 . a : , cos l ' : r , ' 10 ) - 0 . t { 0 ) - - lll

Finding Curves31. Find the curve ) : /(t) in the -r)-plane that passes through the

point (9, 4) and whose slope at each point is 3 .r'6.

32. a) Find a curve y: f(x) with the lollowing properties:

d,y

Its graph passes through the point (0, 1) and has a

hodzontal tangent there.

b) How many curves like this are there? How do you know?

Solution (lntegral) Curves

Exercises 4.2 zag

s. f : r - l , ' 40.

Use the technique described in Examples 4 and 5 to sketch the solu-rions ol the initial value problems in Exercises 41 44.

,1 r I4 1 , " ' : - - - = - , - l < x < l : y ( 0 ) = 0

dx J l - x 'n r

42. + -- J1+ ra, )(o) : I

, 1 \ t I '43 . + : , ' : - 1 . y l 0 r=1

4 X r ' - i I

. l \ '

44 . + : - - . ) t 0 t : od x . { ' - F I

Applications45. On the moon the acceleration of gravity is 1.6 m/sec2 If a rock

is dropped into a crevasse, how fast will it be going just belbre

it hits bottom 30 sec later?

46. A rocket lifts off the surlace of Earth with a constant acceleration

of 20 m/sec2. How fast will the rocket be going 1 min later?

47. With approximately what velocity do you enter the water if you

dive from a l0-m platform? (Use 8 - 9.8 r/sec'?.)

E C. cnlcurmoR The acceleration of gravity near the surface of

Mars is 3.72 m./sec2. If a rock is blasted straight up from the

surface with an initial velocity of 93 m/sec (about 208 mph)'

how high does it 8o'! (Hint: When is the velocity zero?)

49. Stopping a car in time. You are driving along a highway at a

steady 60 mph (88 ftlsec) when you see an accident ahead and

slam on the brakes. What constant deceletation is required to stop

your cat in 242 ft? To frnd out, cany out the following steps.

Step t; Solve the initial value problem

c l ' 'Dif lerential equalionl

,1,, : -k (( conslanl)

dsInitial conditions: Ii : 88 and s :0 when r = 0.

Nleasuring tinre .rnd distance lrom\ahcn the br, lkes afe aPPlicd

Sfep 2; Find the value of / that makes dj/dl : 0. (The answer

will involve t.)

step 3: Find the value of k that makes s = 242 for the value of

/ you found in step 2

50. Stopping a motorcycle. The State of Illinois Cycle Rider Salety

Program requires ride6 to be able to brake from 30 mph (44

fvsec) to 0 in 45 ft. What constant deceleration does it take to

do that?

Sl, Motion along a coordinate /ine. A particle moves on a co-

ordinate line with acceleration a = d2sldt2 - s.n - Gl\n),subject to the conditions thal ds ldt : 4 and s : 0 when I : l.

Find

a) the velocity ! : ds/dt in terms of l,

b) the position .t in terms of /

d1,

dx

D

ii)

Exercises 33 36 show solutioncach exercise, find an equationpoint.

33.

curves of differential equations. In

for the curve through the labeled

34,

; = stn . r - cos.r

)

Use the technique described in Example 4 to sketch some of the

solutions of the differential equations in Exercises 37 40 Then solve

tlr equations to check on how well yor-r did

o.H:" n. fr:-z'+z

dr

- + = - + , ' r s l n t r

(2)

I

290 Chapter 4: lntegration

Esz . . ' . - - - , . - r ' : : . . - . : , ' : : : whenApo /o / 5 as t ronau t Dav idScott dropped a hammer and a feather on the moon to demon_strate that in a vacuum all bodies fall with the same (coNtant)acceleration, he dropped them from about 4 ft above the ground.The television footage ofthe event shows the hammer and featherfalling more slowly than on Earth, where, in a vacuum, they wouldhave raken only half a second to fall the 4 ft. How tong did irtake the hanlrner and feather to fall 4 ft on the moon?

-To find

out, solve the following initial value problem for ., as a functionof t. Then lind the value of I that maKes s equal to 0.

)2 "DifferenLial equarion:

; : _5.2 f l /secr

I n i t i a l cond i r i ons : 1 :O unA " : 4

when t : 0dt

53.,, : . ' ; . , r i ' : . . : . - . r . :- . : ; . . : . ? ! . .r : . /o.. . The standard equation forthe position s of a body moving with a constant acceleration aalong a coordinate line is

a 1

" : t r + ' o r + ' u ' ( 1 )

where uo and ,ro are the body's v€locity and position at timet:0. Derive this equation by solving the initiai value problem

)2 "D i f f e r e n t i a l e q u a t i o n : 1 _ : a

als

i : u o a n d s : s o w h e n / : 0

where s is the body,s height above the surface. The equationhas a minus sign because the acceleration acts downwara, in thedirection of decreasing J. The velocity uo is positive if the objectrs rising at time / :0, and negative if the object is falling.

Instead of using the result of Exercise 53, you "ani".iu"

_Eq. (2) directly by solving an appropriate initial value problem.

What initial value problem? Solve it to be su.e you tau" tfr" righione, explaining the solution steps as you go along.

Theory and Examples55. Frn:r, ig t j istaJa(.: lrani f tai-,1 ei j at lr i (1efi /at ive cf vel()aity

a) Suppose thar the velociry of a body moving along rhes-axis is

ds- ; : u = 9 . 8 1 - 3 .A I

l) Find the body's displacement over the rime intervalfrom I = 1 to t : 3 given that r : 5 when I : 0.

2) Find rhe body's displacement ftom I : I to I : 3 giventlrat s: 2 when, = 0.

3) Now find rhe body,s displacement from, = 1 ro I :3grven that s =,ro when, : 0.

b) Suppose the position., ofa body moving along a coordinateline is a differcntiable ftrnction of time /. Is it true that onceyou kaow an antiderivative of the velocity ft1llctior dsldtyou can find the body's dispiacement froln t : a to t = beven if you do not know the body,s exact position at eitherof those times? Give reasons for you, answer

56, Lttlioueness rtf sciutions. If differentiable functions ), = F(.:r)and l, : G(r) borh solve rhe initial value problem

dr

d, = JA), l( .ro): yo,

on an interval 1, must F(,v): G(j) for evety j in 1? Givereasons for your answer

Initial conditions:

54. (Continuation of Exercise 53.). ."e i . i l ,1eet ihe,,Jrfa.c af. .';,-.:.. For free fall near the surface of a planet where theacceleration of gravity has a constant magnitude of g length_units/sec2, Eq. (1) takes the form

, : - l s rz+unr+s6 ,

A change of variable can often turn an unfamiliar integral into one we can evaluate.The method for doing this is called the substitution m;thod of iot.grution. ft i, on.of the principal methods for evaluating integrals. This section shoris how and whythe method works.

The Generalized power Rule in Integral FormWhen a is a differentiable function of ,r and ,? is a rational number ditl.erent from-1, the Chain Rule tells us

lntegration by Substitution-Running the

d ( u '+ t \ , , d "d r \ r+1 )= 'E

4.3 Integration by substitution-Running the Chain Rule Backward 291

This same equation, from another point of view, says that u"+' /(n * 1) is one ofthe antiderivatives of the function u'(duldx). Therefore,

f t - d u \ u u - lI l u ' " l d t : - + CJ \ d ^ / n + l

The integral on the left-hand side of this equation is usually written in the simpler"differential" form,

[ .u" au.,I

obtained by treating the d"r's as differentials that cancel. Combining the last twoequations gives the following rule.

Equation (1) actually holds for any realexponent'r f -1, as we will see inChapter 6.

If u is any differentiable function,

Iu"au="" ,c .. l n+ l

(n I -1. n rational) ( 1 )

In deriving Eq. (l) we assumed a to be a differentiable function of the variabler, but the name of the variable does not matter and does not appear in the finalformula. We could have represented the variable with 0, t, ), or any other letterEquation (l) says that whenever we can cast an integml in the form

I u"au, @+ -r),I

with z a differentiable function and da its differential, we can evaluate the integralas lu'+t 11n + t1) + c .

EXAMPLE 1 Evaluate [ (* +z)t a"..l

Sorution We can put the integral in the form

[ ," au

by substituting

dx. d x: x * 2 , d u : d ( x + 2 ) :

: 1 . d x : d x .

Thenf f _I \x +2) 'd t : I u 'du

, ] J

' , 6

o

(x i?)b , ^: _ _.t_ L_.

, ' : ' - 1 . J r ' = , 1 \

lnrcgfdc- using Eq.

R c p l r c c l l b y 1 I 2 .


EXAMPLE 2f - f

JJt+ f .2yr ty = . l u t t ' zau

u(t /2)+1= -- - *c

\ r / z ) _ f L

: u ' t ' I C

2: 3 ( l + ) ' ) ' i ' + C

EXAMPLE 3 Adjusting the integrand by a constant

| .n=0,: l, ' ,.!a,: I I u t lzdu

4 J

4 3/2 -

I " , ": - u t t ' + Co

I . . ^: - (4t 1)'' ' + Ct)

[ "o" u au: s in, , + c.

in tecr l ic . u\ in! . rE q . ( 1 ) \ t i r h

Sinrplef tiJl]r

I tc l l rcc r r b!I + ., ' .

With the I / ' .1 out l l |nrthc i r tc l ! rx l is no\r ' i r l

Integfate. s ing Lq. { I\ ! i t h r : l / 2 .

SiDrplr j r lo fnr

ReplLrce r b)' -1r -- l

Trigonometric FunctionsIf a is a differentiable function of .r, then sinll is a differentiable function of ,r.The Chain Rule gives the derivative of sin r as

d d u1- Sln Z : COS lt -;.ax dx

From another point of view, however, this same equation says that sin, is one ofthe antiderivatives of the product cos u , (duldx). Therefore,

| / d r \/ l . o t , r 4 x : s r n a - L .

J \ d x /

A formal cancellation of the dx's in the integral on the left leads to the followingrule.

If a is a differentiable function, then

-

4.3 lntegration by Substitution-Running the Chain Rule Backward 293

Equation (2) says that whenever we can cast an integral in the form

t. l

cos Lt du,

we can integrate with respect to ,l to evaluate the integral as sin a * C.

EXAMPLE 4f 1 ,

: . l

cosu . jdu

1 l: 1 J

cosu du

I: - S l n 4 + l,7

= I s i n r76 + 5 )+ C7

The companion formula for the integral of sina when a

function is

I sinu,lu: - cos r. ' + c. (3)

EXAMPLE 5

- / s i n 1 r ' 1 . x r d . r

L e t r : r l

: L 6 u . l d u J r r : l r r , r r. / 3 . , ' t r - \ ' t

r t:

J slru au

I ^ l r c . . . r v . , ' r . . t 1 r .: ( - C O S l . r l + l t L , , r

= - ] "o,

( r , ) + C Rcplacc rr b1 rr .- 3 " " " t " '

. - a

The Chain Rule lbrmulas for the derivatives of the tangent. cotangent, secant,

and cosecant of a differentiable function r,r lead to the following integrals.

I cos t\r) +std0

L e t ' l : 7 r + 5 .

wirh ( l /7) our 1ro|r1.the i i lcgral is no$in strndard tinm.

Integf t r tc $! lh

Rcplace u by'70+5. t r

is a differentiable

;/ r '?sin {rr),/r

l r " r t ua r : ta l ' t u+c

J" r " t , r l , : co tu+C

u tar\ u du: sec & + C

c o t L t d u : c s c r , + C

(4)

(s)

(6)

(7)

I

/ sec

I

.l '"" u


The Substitution Method ofIntegration

Take these steps to evaluate the integral

/ r t n , r t , n t . r t d . r .J "

when . f Jnd 8 r re cont inuou. luncr ion . l

5tep t. Substitute .l : g(.v) anddu : g'(r) dr to obtajn the integral

rI J Q ) d u '

Step 2. Integrate with respect to r/.step 3: Replace, by.g(-r) in the result.

[ \ae- [ , " , 'zeae. / cos' l t / . l

t ,1= .l

sec' u . - du

r t:1 .1 , r r_ ,au

= ] tan , , + C2

I- - an2e +c

In each fbrmula. a is a di1l-erentiable lunction of a real variable. Each formula canbe checked by difl-erentiating the right hand side with respect to that variable. Ineach case, the Chain Rule applies to produca the integrand on the left.

EXAMPLE 6

+0

d \^ (2e) |

I

Check:

d / 1 \ I ( 1I tun 20 C | = flan 20)

dA \2 ' - - - ' -

) 2 de ' - - '

l /= . l s e c " 2 0 .

2 \

1 ^= ' s e c - 2 8 . 2

2

The Substitution Method of IntegrationThe substitutions in the preceding examples are all instances of the lbllowing generalruie.

: r(s(r)) + cThese three steps arc the steps of the substitution method of integration. The methodworks because F(g(r)) is an antiderivative of /(g(,r)) .g'(,r) whenever F is anantiderivative of /:

d, F ( P ( x ) ) = F ( e ( r ) ) ' 8 ( r )

a \

- J r g { . \ , ) . g { r r

Implicit in the substitution method is the assumption that we are replacing r by afunction of r. Thus, the substitution :8(r) must be solvable for r to give x asa func t ionn:g r ( r ) ( "g inverse o f r i " ) . The domains o f a and. r mayneedtoberesfficted on occasion to make this possible. You need not be concemed with thisissue at the moment. We will discuss inverses in Section 6.1 and treat the theoryof substitutions in greater detail in SecLions 7.4 and 13.7.

I r rrr",, . s'(x)dx: l trro,: F (u) .t C

4.3 Integration by Substitution-Running the Chain Rule Backward 295

EXAMPLE 7f -

I rx ' + 2x - 3) ' ( . r + 1) drL c r r : \ r + l \ l .r i r : l t r i t l r l t

1 1 1 ' l ) r / r / = ( f l l l l

: l* )o':: I *0.:1.4*c: ! , ,+c

2 3 6t -: ur 'x ' +2x - 3) ' + c

lu lcgfnr . n i l r !c ' f . .1

,D

EXAMPLE 8f f t - e t r : s r n 1 -

/ s i n a t c o s 7 4 7 : I u a d u 1 r l = c o s / 1 / /J J

: l l * "

ntcsr.rc $jrh fc\fect r{i /,.- 5 ' "

sin5 /= _ r ( \ i t : ( ( r '

- l

The success of the substitution method depends on finding a substitution that

will change an integral we cannot evaluate direcdy into one that we can. If the first

substitution fails, we can tly to simplify the integrand further with an additional

substitution or two. (You will see what we mean if you do Exercises 47 and 48.)

Altematively, we can start afresh. There can be more than one good way to start,

as in the next example.

EXAMPLE I Evaluatef 2z.dz.

lwtSorution We can use the substitution method of integmtion as an exploratory tool:

substitute for the most troublesome part of the integrand and see how things work

out. For the integral here, we might try u : z2 + l or we might even press our luck

and take ,J to be the entire cube root. Here is what happens in each case.

solut ion 1 Subst i tu te u:22+1.

[ 2zdz I du

lw=: J *:

. l u ' / 'du

, ,2/3- L . t r -

2j

- - - 1 .

- ) l t z l ' - a R r ' 1 . . \2

L e L r r : : r l .

I n t h e l i n I r / r ' . / r

Intcgfulc \ \ i (h f t \pect


solut ion 2 Subst i tu te, , :7 : r + I instead.

[ 2: d:. | 3u1 du| -:

J J : r + 1 J u

=31, , t ,

u 2= . 1 . + c' 2

: 1 i ; r 1 I1r ' r -1- c:

l . r , ' / i . ' I1 1 :

l r i r L ' ! r i l i f $ i ( i ] r . ' f f r I

I t . l , l , ( | I ' r \

Exercises 4.3

Evaluat ing IntegralsEvaluate the irdefinite integrals in Exercises l-12 by using the gi lcnsubstitrltions to rcduce the integnls lo standard fbnn.

1 . / s i n 3 t 1 1 t , r r = 3 t

f ^2,

lxstn(2.r ') t l . t , u -). . t '

3 , I sec 2 t r t n2 t d t . u :2 r

l , r ' ! t4 . I l 1 c o s l s i n J r . r r I . o ,

J \ t ) 2 l

s . I z s 7 t z l t r ! . r . u : ' 7 t - 2

6. / ' r t( ,r ' - t) t r /r . u : t ] I

| 9 r : , l r

I J t - r r

s . / l : , ' - - + f I l r ( ' l l t J r ' . r , \ - l ' ' I. l

9 . / . r ' i n r r ' - l J , / r . I

ro. /1-" (1) , i , , , - !, t r \ r , /

11. ;/

csc'] 20 cot 2a d0

a ) U s i n g u : c o l 2 e

J J 5 . r + 8a ) L l s i n g a = 5 . r i l J

Elaluatc thc integlals in Exelcises

r3. /

r/3 2s r1s

r. I L:tJ J5.r i ,1

17. I t1 :/ | t1) Llq

t _19.

/ 3 r v '7 3 r r r1 r

^ . | |

. , ! / r l l + J r )

23. f cos (3: +'1) r/:

25 . /s .c r t i . r+2r . / . .

/ r \i / . I S l n C O S , / r

. / l l

' . / " ( " , - ' ) ' "

. l l , / rL r r s in ( . r r ' r+ I ) r / . , ;

D) U \ tng r / : v / ) 1 + I

I l '16.

14 . / t 2 r + l t r r l :

| 3 tlt16. l _

, l ( 2 : ) l

| ^ -r8 .

. / 80 Jdr r ta

2 0 . l :J u,,2t, + I

r , / r l + v / i r ' r .

24. / sin (8: - 5) r/;

I2 6 . I t . r n ' r s e e r J r

/ - \ . l:8. I un sel' J\

. l ) 2

' . / "

( ' - - ) ' , '

32. /-r 'r 'r sin (ra,r 8) z1-r.lb.i Using r : csc 2d

-

39.

41.

. / *" (, * i ) ,^(, +\) a"

1*"(T)*'("f)^f sin (2t * 1)

J *", fZ' + yo'f -

/ vcor l csc- )d)

/ l - ' / l - t ' l , 'J I ' \ I /

t l 1 1 . ^

. l o , "n e "o" eoo

f cos Jo| - /lH

J "te

sin'1 "G

-"

I r r t +zr ' -5.r+5)(3r '?+4s 5). /s

l@ ' - t t ' * t t - r )@3 e +2 )do

Simplifying Integrals Step by Steplfyou do not know what substitution to make, try reducing the integralstep by step, using a trial substitution to simplify the integral a bit andthen another to simplify it some more. You will see what we meanif you try the sequences of substitutions in Exercises 47 and 48.

f 18 tan2 x sec2 r47. I - dx

J V + t a n ' x r

a) a : tan r , fo l lowed by u : u3 , then by w:2+ 1)b) r = tan3 r , fo l lowed by u :2*uc) u =2 | tan i x

48 , /J1 + s in '? ( . r - 1 ) s in (x 1 )cos( r - l ) r / r

a ) a : * 1 , fo l lowed by u : s in r r , then by w: l *u2b) a :s in ( . r - 1 ) , fo l lowed by u= l *u2c \ u - l + s i n ' ? ( x - 1 )

Evaluate the integrals in Exercises 49 and 50.

t t2, ltcos,Et2, - l l \ 619. I

J J 3 t 2 r - l ) , + 6

f s\n J0l l t d H

J t/0 cos3 J0

Exercises 4.3 297

lnitial Value ProblemsSolve the initial value problems in Exercises 51 56.

st. f = nt { t t ' - l )3, s ( l ) :3

n t52.

; : 4x ( i ' + 8 ) ' " , y r01 = g

, 1 t, 3 . ; " : t " i # ( r r i r ) .

s ro r ' 8

) r5 4 . 1 : 3 c o s ' l 1 - 9 1 .

d e \ 4

,12,5 ! . - : - + S l n I Z r ' .

a t ' \ z '

, l -

56. ? - + sec2 2r tan 2x,

tu' Ie*]j]9"n,t,**. [Y!]a,

J Vsec I

n. l lcosr. f i+rrat7t

r ( 0 ) - -8

s ' (0) : 100, s(0) = 0

) ' (0) : a , ) (0) : -1

57. The velocity of a particle moving back and forth on a line is

u : ds ldt :6sin2t m/sec for all t. Il s = 0 when t - 0, find

the value of s when t : r /2 sec.

58. The acceleration of a particle moving back and fofth on a line

is a - d2s |dt2 : 12 costtt m/sec2 for al l r . I f s:0 and u:8

m./sec when l :0, f ind s when t: I sec.

Theory and Examples59, It looks as if we can integrate 2 sin.r cos x with respect to n in

three different ways:r l

a ) l 2 s i r x c o " x d x - l z u d u.t J

: a 2 + C r - s i n 2 x * C r

t rb t f 2 s i n r c o s r d x - l - ) u d u

.t J

- -u2 I Cz : cosz x * C:r f

" t / 2 t i n r c o s . r d r = / . i n 2 r / . r j ' , . ' '

.t J

cos2r , -)

' - "

Can all three integrations be correct? Give reasons lor your an-

60. The substitution !l : tan i gives

| , t u ' _ . a _ l a n : , r _ r a/ s e c ' r t a n x d r - l u l u = r ( -

) + ( .

J J

The substitution a : secr givesl l ' t ) , - \ e c i , -

/ s e c ' r t a n x d r . ' l u d u - I l C - 2 + C ..t J

Can both integrations be correct? Give reasons for your answer

| , ' r t * ,"1' o,

I


E Estimating with Finite SumsThis section shows how pmctical questions can lead in natural ways to upproximrtions by finite sums.

Area and Cardiac OutputThe number of l i ters of blood your heart pumps in a minute is called your cardizrcoLtlpLi. For a person at rest, the rate might be 5 or 6 liters per minute. Dudngstl€nuous exercise tlte rate might be as high as 30 liters per minute. It might alsobe altered signil icantly by disease.

Instead of mcasu ng a patient's caldiac output with exhaled carbon dioxicle, asin Exercise 25 in Section 2.7, a doctor nlay prefer to use the dye dilution techniquedescribed here. You inject 5 to 10 mg of dye in a main vein near the heart. The dyeis drawn inio the right side of the heart and pumped through the lungs and out theleft side of the heart into the aorta. where its concentmtion can be measured everyfew scconds as the blood flows past. The data in Table 4.3 and the plot in Fig.4.5show the response of a healthy, resting patient to an injection of 5.6 mg of dyc.

To calculate the patient's cardiac output, we divide the amount of dye by thearea under the dye concentration curve and multiply the result by 60:

Caldiac output :amount of dye

a.r The dye concentrations from Table4.3, plotted and fitted with a smoothcurve . T ime is measured w i th t :0 a t thetime of injection. The dye concentrationsare zero a t the beg inn ing , wh i le rhe dyepasses through the lungs. Tl.ey then riseto a maxtmum at about t :9 sec andtaper to zero by t :31 sec .

x 60. ( 1 )area under curve

You can see why the lbrmula works if you check the units in which the va ousquantit ies are measured. The amount of dye is in mill igrams and the area is in(mill igrans/l iter) x seconds. which gives cardiac output in l i ters/minute:

T

mg . sec

In the exanple that fbllows. we estimate the arca under the concentration curle inFig.4.5 and llnd the patient's cardiac output.

Tab le 4 .3 Dye-d i lu t ion da ta

sec- . .ntr n

mg

mg. sec

L

Secondsafter

injectiont

l 92 I232527293 l

Dyeconcentration(adjusted forrecirculation)

0 .910.570.360.230.140.090

5 7 9 1 1 t 5 1 9 2 3Timc (sec)

DyeSeconds concentrationafter (adjusted for

injection recirculation)t c

5 07 3 .89 8 .0

l t 6 . 113 3 .6r 5 2 . 3t7 | .45

U l ' 5 7 9 i l 1 5 1 9

The region under the concentrationcurve of Fig.4.5 is approximated withrectangles. We ignore the portion fromt :29 to t = 31 ; i t s concent ra t ion isneg l ig ib le .

4.4 Estimating with Finite Sums 299

EXAMPLE 1 Find the cardiac output of the patient whose data appear in Table

4.3 and Fig. 4.5.

solution We know the amount of dye to use in Eq. (1) (it is 5.6 mg), so all we

need is the area under the concentation curve. None of the area formulas we know

can be used for this in'egularly shaped regiorl. But we can get a good esti]nate ofthis

area by approximating the region between the cutve and the l-axis with rectangles

and adding the areas of the rectangles (Fig. 4.6). Each rectangle omits some of the

area under the curve but includes area from outside the cuNe, which compensates

In Fig.4.6 each rectangle has a base 2 units long and a height that is equal to the

height of the curve above the midpoint of the base The rectangle's height acts as a

sort of average value of the tunction over the time interyal on which the rectangle

stands. After reading rectangle heights from the curye, we multipty each rectangle's

height and base to find its area, and then get the following estimate:

Area under curve - sum of rectangle areas

x Jg ) .2+ / (8 ) .2+ f ( r0 ) .2 + - . . + f ( .28 ) .2

- (1.,1)(2) + (6.3)(2) + (7.s)(2) +...+ (0.1)(2)

I (28 .8) (2 ) :57 .6 mg .sec /L (2 )

Dividing this figure into the amount of dye and multiplying by 60 gives a coffe-

sponding estimate of the cardiac output:

amount of dyeCardiac output -

area estlmate

The patient's cardiac output is about 5.8

, 6 0 : : . 6 0 = 5 . 8 L / m i n .- ) / .o

L/min.

--- Technology llsing o Grapher to Cctlculate Finite Sums If youl graphing

utility has a method for evaluating sums' you might want to use it in this.. section. Later in the chapter, you will find it useful tbr approximating "delinite": integrals. There will be other uses still later in your study of calculus'

Distance TraveledSuppose we know the velocity function t) : ds /dt : /Q) m/sec of a car moving

rlown a highway and want to know how far the car will travel in the time interyal

oSt ab.If we know an antiderivative F of I we can find the car's position

function,r = F(t) + C and calculate the distance traveled as the difference between

the car's positions at t imes I :4 and t : b (as in Section 4.2, Exercise 55).

If we do not know an antiderivative of t] : l (/), we carl approximate the rnswer

with a sum in the following way. We partition [c, b] into short time intervals on each

ofwhich u is fairb constant. Since velocity is the rate at which the car is traveling,

we approximate the distanca traveled on each time interyal with tfte formula

Distance = rate x time: JQ) . Lt

and add the results across k, bl. To be specific, suppose the partitioned interyal

looks like this

+Arl+Ar--rF Al+l

| . . | . I ' t ( " . . rtu tt l: /l b


with the subintervals all of length Al. Let lr be a point in the fit.st subinterval. Ifthe interval is short enough so the rate is almost constant, the car will move about

/(Ir)AI m during that interval. lf L is a point in the second interyal, the car wil lmove an additional l (t2)At m during that interval, and so on. The sum of theseproducts approximates the total distance D traveled fron t - q to t : &. If we use/? subintervals, then

D x f ( t \ ) a I + /0 r ) a l + . . .+ l ( r , , ) A / . (3 )

Let's try this on the projectile in Example 3, Section 4.2. The projectile wasfired straight into the air. lts velocity r sec into the flight was u : "f(t)

: 160 9.8rand it rose 435.9 m fiom a height of 3 m to a height of 438.9 m during the first3 sec ol' I l ight.

EXAMPLE 2 The velocity function of a projectile frred straight into the air is

f(.t) - 160 9.8I. Use the summation technique just described to estimate howfar the projectile rises during the first 3 sec. How close do the sums come to theexact f igul€ of 435.9 m?

Solution We explore the results for different numbers of intervals and difl'erentchoices of evaluation points.

3 subinten'ols oJ lengtlt l, *-ith f evaluated at left-hand endpoints:

tL tZ l1

0 t 2 3jrAr>l

With / evaluated rt / = 0, I , and 2. we have

p x f (.t1) Lt + f (t2) Lt + /(h) a/

* [ 1 6 0 e . 8 ( 0 ) ] ( l ) + 1 6 0 - 9 . 8 ( 1 ) l ( r ) + 1 1 6 0 9 . 8 ( 2 ) l ( l )

- 450.6.

3 stLbintervals oJ length 1, rvith I evahuted at right-hand endpoints:

With / evaluated at t : 1,2, and 3, we have

D - f (.t\) a/ + /(r:) a/ + /(/3) a,- [ 6 0 - 9 . 8 ( 1 ) ] ( r ) + [ 1 6 0 9 . 8 ( 2 ) ] ( l ) + [ 1 6 0 - 9 . 8 ( 3 ) ] ( l )

x 421.2.

Wth 6 subintervtls of lertgth l/2, we get

l t

tr t: r-t /.t 1: /o lt t1 t\ t1 t5 t6

a--a--j--J--a- r

0 1 2 3

i..{=#10 1 2 3 0 1 2 3

Ar Al

4.4 Estimating with Finite Sums 301

Using left-hand endpoints: D r,! 443.25.

Using right-hand endpoints: D x 428.55.

These six-interval estimates are somewhat closer than the three-interyal estimates.The results improve as the subintervals get shorter.

Table 4.4 Travel-distance estimates

As we can see in Table 4.4, the left-endpoint sums approach the true value435.9 from above while the dght-endpoint sums approach it from below. The truevalue lies between these upper and lower sums. The magnitude of the error in theclosest entries is 0.23, a small percentage of the true value.

Enorpercentage - * j1 .oos+." 475.9

It would be safe to conclude from the table's last entries that the projectile roseabout 436 m during its first 3 sec of flight. LJ

Notice the mathematical similarity between Examples 1 and 2. In each case,we have a function / delined on a closed interyal and estimate what we want toknow with a sum of function values multiplied by interval lengths. We can usesirnilar sums to estimate volumes.

VolumeHere are two examples using finite sums to estimate volumes.

EXAMPLE 3 A solid lies between planes perpendicular to the r-axis at.{ : 2and x :2. The cross sections of the solid pelpendicular to the axis between theseplanes are vertical squares whose base edges run from the semicircle y : -./9 - ,ito the semicircle ! : t/9 - | (Frg. 4.7a, on the following page). The height ofthe squarc at x is 2J9 x2. Estimate the volume of the solid.

so/ufion We partition the interval | 2, 2l on the "r-axis into four subintervals oflength Ar : l. The solid's ffoss section at the left-hand endpoint ofeach subintervalis a square (Fig. 4.7b). On each ofthese squares we construct a right cylinder (square

slab) of height 1 extending to the dght (Fig. 4.7c). We add the cylinders' volumesto estimate rhe rolume of the solid.

Number ofsubintervals

36

t 2244896

192

Length of eaehsubint€rval

I0.50.250.t250.06250.031250.015625

Left-endpointsum

450.6443.25439.58437 .',l 4436.82436.36436.13

Right-endpointsum

421.2428.55432.23434.06434.98435.44435.6'r

Enor magnitude =

Itrue value - calculated value

302 ChaDter 4: lnteoration

... ' t r = . , 6 - r ' ?

(a) (c)

We calculate the volume of each cylinder with the formula y: Ah (basearea x height). The area of the solid's cross section at r is A (r) : (side)2 :QJO V 1z : 4 Q x2), so the sum of the volurnes of the cylinders is

S+ : A (cr) Ax * A (cz) A.t + A (cr) Ar * A (c+) A,r

:4 (9 - c r '? ) (1 ) + 4 (9 - c r '? ) ( l ) +4 (9 - ca , ) ( l ) + 4 (9 c4r ) (1 )

:4 [ (9 ( D2)+o ( t ) '? )+(9- (0 ) , ) + (9 - ( l ) r ) l1

: 4 lQ - 4 )+ (9 - 1 ) + (e -o ) + (9 - l ) ]

: 1(36 6) : 120.

This compares favorably with the solid's true volume V :368/3 x 122.67will see how to calculate V in Section 4.7). The difference between ,S and ysmall percentage of V:

tv ,s,Error percentage

; :2 .2Vc.

With a flner parlition (more subintervals) the approximation would be even better.

EXAMPLE 4 Estimate the volume ol a solid sphere of radius 4.

Solution We picture the sphere as if its surface were generated by revolving thcgraph of the function I (r) : !16 = about the .r axis tFig. 4.8a1. We partitionthe interval 4 S r S 4into8 subintervals oflength Ax: l. We then approxlmarethe solid with right circular cylinders based on cross sections of the solid by planesperpendicular to the.:r-axis at the subintervals' left-hand endpoints (Fig. 4.8b). (Thecylinder at x : 4 is degenerate because the cross section therc is just a point.)We add the cylinders' volumes to estimate the volume of a sphere.

(a) The solid in Example 3. (b) Squarecross sections of the solid at x = 2, 1,0, and 1. (c) Rectangular cylinders (slabs)based on the cross section5 toapproximate the solid.

(wei s a

(368/ l ) - 120(368/3)

8368

'/

4.4 Estimating with Finite Sums

v -

303

! s (a) The semicircle y = 14 6-7revolved about the x-axis to outl ine a$here. (b) The solid sphere approximatedwith cross-section-based cylinders.

We calculate the volume of each cylinder with the formula V : Tr2h. Thesum of the eight cylinders' volumes is

Ss = r r [ " f (cr ) ]2 Ar +xt l f (c)12 Lx +r l f (c)12 Lx I . . . * r f f (cs)12 Lx- 2

= t l J 1 6 c ' ? l A x l r l J 1 6 - c ] l t x + n l / 1 6 - c r 2 l a rL ' I L

- l L

- t

. -12- r ' . . * n l J l 6 - . * , 1 a ,

L - l

=z [ { t0 - ( -qr+(6- ( -3)1 +(16- ( -Dr+. .+(16- (3) '? ) ]

: n l 0 ' t 7 ' l 1 2 + 1 5 + 1 6 + 1 5 + 1 2 + 7 l

= 84n .

This compares favorably with the sphere's true volume,4 4 2562

V : - t r ' = - n @ ) , = .3 3 3

The difference between 58 and V is a small percentage of y:

Enor percentase - lv:stl :Qt1!:):= to"

v (25613)tr1 \a1 _ )< ) |

256 64

The Average Value of a Nonnegative FunctionTo find the average of a finite set of values, we add them and divide by the number ofvalues added. But what happens if we want to find the average of an infinite numberof values? For example, what is the average value of the function -f(x) :.r2 onthe interval [-1, 1]? To see what this kind of "continuous" average might mean,imagine that we are pollsters sampling the function. We pick random x's between-l and 1, square them, and average the squares. As we take larger samples, weexpect this average to approach some number, which seems reasonable to call theaverage off over l-1, 1f.


(D)

4.9 (a) The qaph of f(x) = x2,I : x : 1. (b) Values of I sampled at

regular intervals.

We will be able to show later that the avemge value is 1/3.Notice that

The graph in Fig. 4.9(a) suggests that the average square should be less than1/2, because numbers with squares less than l/2 make up more than 70Vo of theinteryal [ - 1 , 1 ] . If we had a computer to generate random numbers, we could carryout the sampling experiment described above, but it is much easier to estimate theaverage value with a finite sum.

EXAMPLE 5 Estimate the average value of the function /(,r) : x2 on thein te rva l [ - l . l l .

Solution We look at the graph of y : x2 and partition the interval [0, 1l into 6subintervals of length Ar : 1/3 (Fig. 4.9b).

It appears that a good estimate for the average square on each subinterval isthe square of the midpoint of the subinterval. Since the subinteryals have the samelength, we can average these six estimates to get a final estimate for the averagevalue over l-1. l l .

Average value I

I6

1' : - r . l

alength of [- 1, l]

I

( - ; ) ' . ( - : ) ' . ( - * ) ' . ( ; ) ' . ( : ) ' . ( : ) '

( - ; ) ' . ( - ; ) ' . ( - : ) ' . ( : ) ' . ( : ) ' . ( ; ) '

6

25+9+t+1+9+253 -o.troz l o

[ ' ( - : )(-:) i. .'(;)

f a sum of function values II multiplied by inte al lengths l'

jl

length of [-1, 1]

Once again our estimate has been achieved by multiplying function values byinterual lengths and summing the results for all the intervals. tl

ConclusionThe examples in this section describe instances in which sums of function valuesmultiplied by interval lengths provide approximations that are good enough toanswer pmctical questions. You will find additional examples in the exercises.

The distance approximations in Example 2 improved as the intervals involvedbecame shorter and more numerous. We knew this because we had already foundthe exact answer with antiderivatives in Section 4.2. If we had made our partitionsof the time interval still finer, would the sums have approached the exact answer asa limit? Is the connection between the sums and the antiderivative in this case justa coincidence? Could we have calculated the area in Examole l. the volumes in

(-:)' l* * (:)' l]r [ l s t ' ? 1: - l l - - l . - *

2L\ 6 / 3

Exercises 4.4 305

Examples 3 and 4, and the average value in Example 5 with antiderivatives as well?

As we will see, the answers are "Yes, they would have," "No, it is not a coincidence,"

and "Yes, we could have."

Exercises 4.4

Cardiac Output1. The table below gives dye concentrations for adye-dilution cardiac-

output determination like the one in Example I . The amount of dyeinjected in this case was 5 mg instead of 5.6 mg. Use rectanglesto estimate the area under the dye concentration curve and thengo on to estimate the patient's caJdiac output.

the data points with a smooth curve. Estimate the area under the

curve and calculate the cardiac output from this estimate.

Seconds Dyeconcentrationafter (adjusted for


Seconds Dyeconcentrationafter (adjusted for


Seconds afterinjection

t

24b

8101 2t41 61 82022

00.61.42.'13.'14 .13 .82.9t. '71 .00 .50

02468

1 0t2l4

000.10.62.0A )

6.3'7.5

l 6l 82D2224262830

7.86 . 14.7J . J

2 . 1o.'70

Dye concentration(adjusted for recirculation)

c

Distance3, The table below shows the velocity ol a model train engine mov-

ing along a track for l0 sec. Estimate the distance traveled by the

engine using 10 subinteNals of length I with (a) left-endpoint

values and (b) right-endpoint values.

Time(sec)

0I2345

Velocity(in./sec)

Time(sec,

Velocity(in-/sec)

0) 2221 05

13

61E9

1 0

l t6260

The accompalying table gives dye concentrations for a cardiac

output determination like the one in Example l. The amount ol

dye injected in this case was l0 mg. Plot the data and connect

You are sitting on the bank of a tidal dver watching the incoming

tide carry a bottle upstream. You record the velocity of the flow

every five minutes for an hour, with the results shown in the table

on the following page. About how far upstream did the bottle

travel during that hour? Find an estimate using 12 subinterlals

of length 5 with (a) Ieft-endpoint values and (b) righGendpointvalues.

2 4 b 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 aTime (sec)

__


Time(min)

05

1 0l 5202530

Velocity(m,/sec)

I1 . 2t. '72.01 . 81 .6l . i l

Time Velocity(min) (m/sec)

40

505560

t . 21 . 01 . 81 . 5t . 20

Velocity (conve{edTime to ft/sec)(sec) (30 mi,rh :44 fvsec)

VelocityTime (converted to frsec)(sec) (30 mi,/h : 4il fvsec)

5. You and a companion are about to drive a twisty stretch of dirtroad in a car whose speedometer works but whose odometer(mileage counter) is broken. To find out how long this pafiicularstretch of road is, you record the car's velocity at 10-sec intervals.with the results shown in the table below. Estimate the lengthof the road (a) using lefFendpoint values and (b) using rightendpoint values.

Volume7. (Corltinuation oJ Example 3.)

Suppose rve use only twosquare cylinders to estimatethe volume V of the solid inExample 3, as shown irlprofile in the figure here.

a) Find the sum ,St of thevolumes of the cyl inders.

b ) Exp ress Y -S r l asapercentage of y to thenearest percent.

8, (Contiuatiotr of Etample 3.)Suppose we use six squarecylinders to estimate the volumeY ol ihe solid in Example 3, asshown in the accompanyingprofile view

a) Find the sum 56 of thevolumes of the cylinders.

b) Express Y 56l asa percentage of Y to thenearest percent,

9, (Continuatiott of Erample 4.) Suppose we approximate the vol-ume V of the sphere in Example 4 by partitioning the interval-4: -r :4 into four subirtervals of length 2 and using cyl inders based on the qoss sections at the subintervals left-handendpoints. (As in Example ,1, the leltmost cylinder will have azero radlus.)

a) Find the sum S1 of the volumes of the cylinders.b) Express y - Srl as a percentage of y to the nearest percent.

10. To estimate the volume y of a solid sphere of radius 5 youpa ition its diamcter into nve subinteNals of length 2. You then

0l 02030405060

044I 53530

35

708090

1001 1 0)20

1 52235

3035

6. The table below gives data for the velocity of a vintage sportscar accelerating fiom 0 to 1.12 rni/h in 36 sec (10 thousandths ofan nour).

Time(h)

Velocity(mi/h)

Time(h)

Velocity(mi/h)

0 .00.0010.0020.0030.0040.005

040628296

108

0.0060.0070.0080.0090.010

l l 6t25

t31142

a )

b)

Use rectangles to estimate how far the car traveled duringthe 36 sec it took to reach 1,12 mi/h.Roughly how many seconds did ii take the car Io reach thehalfway point? About how fast was the car going then?

mi/h

slice the sphere with planes perpendicular to the diameter atthe subinteNals' left-hand endpoints and add the volumes ofcylinders of height 2 based on the cross sections of the spheredetemined by these planes.a) Find the sum 55 of the volumes of the cylinders.b) Express ly 55l as a percentage of y to the nearest percent.

ll. To estimate the volume V of a solid hemisphere of radius 4,imagine its axis of symmehy to be the inteNal [0, 4] on ther-axis. Partition [0, 4] into eight subintervais of equal length andapproximate the solid with cylinders based on the circular crosssections of the hemisphere pelpendicular to the r-a,\is at thesubintervals' left-hand endpoints. (See the accompanying profile

a) Find the sum 58 of the volumes of the cylinders. Do you

expect 58 to overestimate y, or to underestimate V? Giveteasons tor your answet

b) Express lV - .S8 as a percentage of y to the nearestpercent.

Repeat Exercise 1l using cylinders based on cross sections at theright-hand endpoinls of lhe .ubinlervals.

Estirnates with large errar. A solid lies between planes per-pendicular to the.y-axis at r - 0 and * - 4. The cross sectionsof the solid perpendicular to the axis between these planes arevertical squares whose base edges run ftom the parabolic curve

1, : -.rf to the parabolic curv€ y : .r'&.

a) Find the sum ,Sa of the volumes of the cylinders obtainedby partitioning 0 : jr : 4 into four subintervals of length I

Exercises 307

based on the cross sections at the subinterval's right handendpolnts.

b) The true volume is V : 32. Express I V - ,Sa as a percent-age of Y to the nearest percent.

c) Repeat pafts (a) and (b) for the sum 58.

14. Estirnates with large errar. A solid lies between planes perpen-dicular to the r-axis at r = 0 and;r :4. The cross sections ofthe solid perpendicular to the a,\is between these planes are verti-cal equilatepl triangles whose base edges run from the paraboliccurve y: -.r4 to the parabolic curve y: f.

a) Find the sum Sa of the volumes of the cylinders obtainedby partitioning 0 = r : 4 into four subintervals of length1 based on the crcss sections at the subinterval's lefFhandendpoints.

b) The true volume is y:8.n6. Express y - Sal as a per-centage of y to the nearest percent.

E ")

CALCULATOR Repeat parts (a) and (b) lor rhe sum J8.

15. A reservoir shaped like a hemispherical bowl of radius 8 m isfilled with water to a depth of 4 m. (a) Find an estimate Sof the water's volume by approximating the water with eightcircumsc bed solid cylinders. (b) As you will see in Section 4.7,Exercise 71, the water's volume is V - 320n/3 m3. Find theenor lV Sl as a percentage of y to the nealest percent.

16. A rectangular swimming pool is 30 ft wide and 50 ft long. Thetable below shows the depth ll(n) of the water at 5-ft intervalsfrom one end of the pool to the other. Estimate the volume ofwater in the pool using (a) left-endpoint values of rl G) rightendpoint values of ft.

12.

13.

Position.r ft

5l01 52025

Depthh(x) tt

Positionr f t

303540

50

Depthh(x) ft

6.08.29 . 19.9

10.51 1 . 0

1 1 . 5I 1 . 912.3t2.713.0

The nose "cone" of a rocket is a paraboloid obtained by re-volving the curve 1 : .rf, 0 5 n 5 5, about the r-axis, where.r is measured in feet. To estimate the volume y of the nose cone.

17.


we partition [0, 5] into five subintervals of equal lengrh, slice thecone with planes perpendicular to the r axis at the subintervals,left-hand endpoints, and construct cylinders of height I based oncross sections at these points. (See the accompanying figure.)

]

Velocity and Distance23. An object is dropped staight down from an airplane. The object

falls faster and faster but the acceleratior is decreasing over timebecause of air resistance. The acceleration is measured in ft/secrand recorded every second after the drop for 5 sec, as shown inthe following table.

a) Find the sum S5 of the volumes of the cylinders. Do youexpect S5 to overestimate y, or to undercstimate V? Givereasons for your answer.

b) As you will see in Section 4.7, Exercise 72, the volumeof the nose cone is V : 25n /2 ftj . Express y - 55 | as apercentage of V to the nearest percent.

18. Repeat Exercise l7 using cylinders based on cross sections at therighrhand endpoints of the subintervals.

Average Value of a FunctionIn Exercises 19 22, use a finite sum to estimate the average value of /on the given interval by pafiitioning the interval into four subintervalsof equal length and evaluating / ar rhe subinterval midpoints.

19. I (r) : ir3 on [0, 2] 20. f(r) = 1/:r on [1,9]21 . f ( t ) : ( l l2 )+s i ' t ' r t on 10 ,21

r n r t 422 . f ' r t = , { ' " *ZJ on t o .4 l

a) Find an upper estimate for the speed when t = 5.b) Find a lower estimate for the speed when l:5.c) Find an upper estimate for rhe distance fallen when r = 3.

24. An object is shot staight upward from sea level with an inirialvelocity of 400 ftsec. Assuming gravity is the only force actingon the object, give an upper estimate for its speed after 5 sechave elapsed. Use g:32 ft/sec2 lor the gravitational constant.Find a lower estimate for the height attained after 5 sec.

Pollution Control25. Oil is leaking out of a tanker damaged at sea. The damage to the

tanker is worsening as evidenced by the increased leakage eachhour, recorded in the followine table.

Time (hours)

Leakage (gaUhr)

Time (hours)

Leakage (gaUhr) 265

Give an upper and a lower estimate of the total quantity ofoil that has escaped after 5 hours.Repeat part (a) for the quantity of oil that has escaped aftet8 hours.

c) The tanker continues to leak 720 gal/h after the first 8 hours.If the tanker originally contained 25,000 gal of oil, approx,imately how many more hours will elapse in the worst casebefbre all of the oil has spilled? in the best case?

A power plant generates electricity by buming oil. pollutants pro-duced as a result ofthe buming process are removed by scrubbersin the smoke stacks. Over time the scrubbers become less effi-cient and eventually they must be replaced when the amount ofpollution released exceeds government standards. Measurementsare taken at the end of each month determining the rate at whichpollutants are teleased into the atmosphere, recorded as follows.

190136

a)

b)

t: )+ sin2 nt

-

4.5 Riemann sums and Definite lntegrals 309

0.81 0.85 0.89 0.95Pollutantrelease rate(tons/day)

Assuming a 30-day month and that new scrubbers allow

only 0.05 tons/day released, give an upper estimate of thetotal tonnage of pollutant released by th€ end ofJune. whatis a lower estimate?In the best case, approximately when will a total of 125tons of pollutant have been released into the atmosphere?

CAS Explorations and Proiectsln Exercises 27 30, use a CAS to perform the following steps:

a) Plot the functions over the given interval.

b) Partition the interval into /l : 100, 200, and 1000 subintervals of

MayApr

0.2 0.70

equal length, and evaluate the function at the midpoint of each

subinterval.Compute the average value of the f'unction values genemted in

part (b).

Solve ihe equation /(t) : (average value) for.t using the aver-

age value calculated in (c) for the r: 1000 paltitioning

-f (r) : sin r on LO, lr1 28. /(-r) : sin'? .r on [0, z]

a)

b)

c)

d)

27.

I29. /(r) :1 5i1

30. /(.r)-a 5i12I

* [ ; , " ]

f l t Ion Lt ' ' l

Riemann Sums and Definite IntegralsIn the preceding section, we estimated distances, areas, volumes, and average values

with linite sums. The terms in the sums were obtained by multiplying selected

function values by the lengths of interyals. In this section, we say what it means for

sums like these to approach a limit as the intervals involved become morc numerous

and shorter. We begin by introducing a compact notation for sums that contain large

numbers of terms.

Sigma Notation for Finite SumsWe use the capital Greek letter X ("sigma") to write an abbreviation for the sum

f (t\) Lt + f (t) Lt + " + f(.t^) Lt

as fi=l /1ri) lr, "the sum from k equals 1 to n of f of t1 times delta l-" whenwe write a sum this way, we say that we have written it in sigma notation.

DefinitionsSigma Notation for Finite SumsThe symbol li=r a1 denotes the sum ai + 42 + ' + ctk. The a's are the

terms of the sum: at is the first term, a2 is the second tem, ar is th€

kth term, and ao is the nth and last term. The variable I is the index of

summation. The values of k run through the iltegers from 1 to n. Thenumber 1 is the lower limit of summationl the number n is the upperlimit of summation.

310 ChaDter 4: lnteoration

EXAMPLE 1

The sum insigma notation

The sum written out-{n€term for each value of /r

The value ofthe sum

I +2+3+4+5

(,1) ' ( l ) + (- i ) r (2) + (-1)3(3)

t 2_+_1+ 1 2+ |

The lower limit of summation does not have to be 1; it can be any integer.

EXAMPLE 2 Express the sum I * 3 + 5 +7 +9 in sigma notation.

Solution6

1+3+5+7+9:LQk-3)

i

5\- l.

3\ - r - t r * l -

2 1 .t ' '

- l + ) -

'1 a

Starting with k:2:

s ta r t i ng w i th k : -3 : 1+3 +5 +7 +9 : L Qk +7 )k : l

The formula generating the terms changes with the lower limit of summation, butthe terms generated remain the same. It is often simplest to start with t :0 ork : 1 .

Starting with k : 0: I + 3 + 5 + 7 + 9 : i 1rn * t1/r:0

Star t ingwi thk= l :5

l r 1 - ! q - ! 7 r O - \ - / ? r - l r

Algebra with Finite SumsWe can use the followins rules whenever we work with finite sums.

Algebra Rules for Finite Sums ,

l. Sum Rule:

2. Difference Rule:

3. Constant Multiple Rule:

L@r+b i :Lar *Lbr

l ( a r -b ) : l ao Lbot : r k-1 k=1

4. Constant Value Rule:

lcar = c ' 2oo {Any number c)t = l k = l

lc:n. c (c is any constant value.)

--

4.5 Riemann Sums and Definite Integrals 3'11

There ate no surprises in this list. The formal proofs can be done by mathe-matical induction (Appendix 1).

EXAMPLE 3

L ) LQk -k \=3Lk -Lk2f t : l k=r t : l

b) f , ( -or) : I ( -1) . a*: - t .k=l k=l

c) t (k+4) :Lk+L4k=1 k=l k: r

: (1 +2+3)+(3 .4 )

:6+12 :18

F - . - _ \ - . .

l ) i l l i i 'n .e I t f l :. l l r ( I ( ' ( r r l .L iur l

\ h l r i f l c R n l r

\ l r r l i l D r e R L r l .

Srni R'r lc

U

Sum Formulas for Positive IntegersOver the years people have discovered a variety of formulas for the values of finitesums. The most famous of these are the formula for the sum of the first r integem(Gauss discovered it at age 5) and the formulas for the sums of the squares andcubes of the nrst n intesers.

The first /, integers:

The first n squares:

The first n cubes:

\ - r . :l J '

\ - i . 2 :/ 2 -

\ - 1 3 :/ - -

n(n -l l)( 1 )

2

n(n -t l)(2n + 1) (2)6

/ n ( n * 1 , 1 \ 2\ 2 /

(3)

4

EXAMPLE 4 Evaluatel(t2-3le).

So/ution We can use the algebra rules and knolvn formulas to evaluate the sum

without writing out the terms.4 4 4

l<t ' -zt) : l t ' -z lr . .4 (4+ l ) ( 8+ l )

Dil i i ' f r f r . R Llct i rd a i i r . l rnt\ l l l r i r ) l r l t : r l c

- ( 4 ( 4 + 1 ) \ F i . r r . , l- t . | , \ L L r r : l\ , /

l

6

: 3 0 - 3 0 : 0

312 chapter 4: Integration

.r. tt The graph of a typical functiony = f(x) over a closed interval [a, b]. Therectangles approximate the regionbetween the graph of the function andthe x-axis.

(.,J(.,,))

Riemann SumsThe approximating sums in Section ,1.4 are examples of a more general kind of sumcalled a Riemann ("ree-mahn") sum. The functions in the examples had nonnegativevalues, but the more general notion has no such restriction. Given an arbitrarycontinuous function l: l(r) on an interval [a,b] (Fig.4.l0), we patit ion theinterval into r subinteryals by choosing n I points, say rl, x2, ..., x,, r, betweena and b subject only to the condition that

a < x t < X 2 < " ' < X , , t < b .

To make the notation consistent, we usually denote c by;e and & by r,. The set

p : { l e . , r 1 , . . . . ; r u j

is called a partition of La, bl.The partition P defines n closed subintervals

[ ,16 , r1 ] , [ r1 , r2 ] , . . . , [ , r , r , r , ] .

The typical closed subinterval f.r1 1, x1] is called the ftth subinteryal of P.

c r . J ( c r ) . 1

\

The length of the ith subinterval is Ax1 : ')e^ rr-1.

l<ar , f [ r ,+ r- A,'l +| <-A.r , ,+

In each subinterval [r1 1,;r1], we select a point c1 and construct a verticalrectangle liom the subinterval to the point (c*, J (.ci) on the curve J : I (r). Thechoice of c1 does not matter as long as it l ies in fxr r,xrl. See Fig.21.10 again.

If f(cr) is positive, the number /(c1) Axi = 6eig51 x base is the area of the

(c.. fk))

L i The curve o f F ig .4 .10 w i threctangles from finer partit ions of la, blFiner partit ions create more re€tangleswith shorter bases.

4 .5 R iemann Sums and Def in i te In tegra ls 313

rcctangle. If /(cr) is negative. then /(cr) A,ri is the negative of the area. In any

case. we add the r? products /(cr) A,r^ to folm the sum

s,, : f ;{c*)l;rr.

This sum, which depencls nn "

unj ,'n. choice ol the uumbers c1, is calied a

Riemann sum for / on the interval [a, ]1, atter German mathematician Georg

Friedrich Bernhard Riemann (1826 1866), who studied the l imits of such sums.

As the partitions of fn. bl become liner, the rectangles dcfined by the partition

approximate the region between the r axis and the graph of / with increasing

accuracy (Fig. 4.1 1). So v,'e expect the associated Riemann sums to have a limiting

value. To test this expectation, we need to develop a numedcal way to say that

partitions become finer and to determine whether the conesponding sums have a

limit. We accomplish this with the following definitions

The norm of a partition P is the partition's longest subirterval length. it is

denoted by

llP (fead "the norm of P").

The way to say that successive partitions of an interval become finer i5 to siry

that the norms of these partitions approach zero. As the norns go to zero, the

subintervals become shorter and their number approaches infinity.

EXAMPLE 5 The set P = [0. 0.2. 0.6, l. 1.5. 2] is a partition of [0, 2]. Thereare f ive subintervals of P: [0 ,0.2] , 10.2,0.61, f0 .6, l l , [ ] , 1 .5] , and L l 5 , 21.

l *A - r , *F l r . - . . . . *+a . r r++Ar r -+ l . ^ ' s '

0.6 1 .5

The lengths of the subintervals are Alr : 0.2. Arr - 0.4. A;r3 :0.4, Ar1 = 0.5.

and Ar5 :0.5. Tbe longest subinterval length is 0.5, so the norm of the pafi it ion

is Pll :0.5. In this example. there are two subintervals of this length

DefinitionThe Definite Integral as a Limit of Riemann SumsLet /(.r) be a function defined on a closed intenal [a, ]1. We say that the

limit of the Riemann sums Xi='/(c1) A,r1 on fa, bl as l lPll -+ 0 is the

number 1 if the following condition is satisfied:

Given any number e > 0, there exists a corresponding number 6 > 0

such that for every partition P of la, bl

l l P l=d + f . f t . u t a " , I < (

for any choice of the numbers ct in the subintervals ["r1 1,.t1].

If the limit exists, we wdte

l i m | / 1 r ' ^ 1 n r ^ - / .P - A -


We call .f the definite integral of / over [4. b], we say that / is integrable over

[a, b], and we say that the Riemann sums of f on [4. b] converge to the num-

bcr 1.We usually write 1 as .l! f f. lar, which is rearl "integral of / from.r to 0."

Thus. if the l imit exists,

The amazing fact is that despite the variety in the Riemann sums Il (cr) Arr

as the partitions change and the arbitrary choice of c('s in the intervals of each

new partition, the sums always have the same limit as I Pll -+ 0 as long as .f is

continuous. The need to establish the existence of this limit became clear as the

nineteenth century progressed, and it was finally established when Riemann proved

the following theorem in 1854. You can find a current version of Riemann's proof

in most advanced calculus books.

Theorem 1The Existence of Definite IntegralsAll continuous functions are integrable. That is, if a function / is continuous

on an interrral [a, b], then its definite integral over [4, b] exists.

Why should we expect such a theorem to hold? Imagine a typical partition

P of the interval la. bl.'lhe function /. being continuous, has a minimum value

minl ("min kay") and a maximum value maxl ("max kay") on each subinterval.

The products minlAxl associated with the minimum values (Fig.'1.12a) add up to

what we call the lower sum for / on P:

I : m inrA_r r * m in2A, r2 * . . . * m in , ,A- r , , .

The products maxÂ.rk obtained lrom the maximum values (Fig.4.12b) add up to

the upper sum tbr / on P:

U : maxlA.r1 + maxtA.rr2 + .+max,,Ar,,.

The difttlence U a between the uppel and lower sums is the sum of the areas ol

the shaded blocks in Fig. ,1.12(c). As I Pll + 0, the blocks in Fig. 4.12(c) become

more numerous, nanowel and shorter. As Fig. 4.12(d) suggests, we can make thc

nonnegative number U - t less than any prescribed positive e by taking Pll close

enough to zero. In other words,

l i m ( U - r ) : 0 .\ P + o

and, as shown in more advancad texts,

l i m L : l i m U .P - A l ' - 0

The fact that Eqs. (4) and (5) hold for any continuous function is a consequence

of a special property, called uni,lbrn contintrity, that continuous functions have on

closed interyals. This propefty guarantees that as llP I + 0 the blocks that make

up the difference between U and I in Fig.4.l2(c) become less tall as they become

less wide and that we can make them all as short as we please by making them

narrow enough. Passing over the € - d arguments assoaiated with uniform continujt]

u t b

l im ) - f t . ; I Ar i = / 11x) d , r .P - A - l "

(4)

(s)

Tq

I

4.5 R iemann Sums and Def in i te In tegra ls 315

1r.- h(d)

wc can n?&e U - L smxller than any grven posrtrve €by miking l lP small enough.

(c)

The diftercnce U I can be made vcry small:

less than € . (l, rr).

The dif ference between upper andlower sums.

keeps our derivation of Eq. (5) from being a proof. But the argument is right in

spirit and gives a faithful portrait of the proof

Assuming that Eq. (5) holds 1br any continuous function / on [a. Dl' suppose

we choose a point c1 from each subinten'al L,rl r, ril of P and form the Riemann

sum tri]=,.1 (c1) A:rr. Then minr < /(cr) < maxr lbr each k. so

/ - \ -1 _ l l c . ) L ^ t - U -l = !

The Rienann sum for f is sandwiched between I and U By a nodified version of

the Sandwich Theorem of Section 1 .2. the l imit of the Riemann sums as l l P -+ 0

exists and equals the common limit of U and l:

l i m L = l i m f f t , ^ t & r - l i r n U .I . n , . 0 = P - o

Pause for a moment to see how remarkable this conclusion really is lt says that

no matter how we choose the points cl. to form the Riemann sums as Pll -+ 0,

the l imit is always the same. We can take every /(cr) to be the minimum value of

I on [,r^ l,,rr]. The limit is the same. We can take every l (ci) to be the mrximum

value of / on [,r1 1. r1]. The limit is the same We can choose every cr at random

The limit is the same.Although we stated the integral existence theorem specifically for continuous

functions, many discontinuous f'unctions are integrablc as well. We ffeat the inte-

gration of bounded piecewise continuous firnctions in Additional Exercises I I l8

at the end of this chapter. We explore the integration of unbounded functions in

Section 7.6.

1

The lower sum L : ) minr A-r{ is lcss than . the upper sum U = i mrx^ tr-r,.

316 Chaoter 4: lnteoration

Functions with No Riemann IntegralWhile some discontinuous functions are integrable, others are not. The function

. lt when x is rational/ (n ' =

lo when x rs i r rar ional ,

for example, has no Riemann integral over [0, 1]. For any partition p of [0, 1], theuDDer and lowet sums are

Y : lmaxlAlp : f t . axr : f u* : t,

1 : f m iqA,4 : fo .a - ro :0 .

Erery subintcfval

Every subinteNalcontarns an i r fat i0nr l

For the integral of / to exist over [0, 1], U and Z would have to have the samelimit as ll P'l -- 0. But they do not:

l i m l : 0l tP t+0

Therefore. / has no integral on 10.eithel unless t is zero.

w h i l e l i m U : 1 .P t+0

ll. No constant multiple t/ has an integral

-1Lower linit of inregrarion

Integral offliom a to,Wlen you find the value of the integral,you have evaluated the integral.

The value of the definite integral of a function over any particular intervaldepends on the function and not on the letter we choose to represent its independentvariable. If we decide to use t or u instead of x, we simply write the integral as

J" fr'lat 1," ro o, instead or 1." f ,,,

o*.

No matter how we write the integral, it is still the same number, defined as alimit of Riemann sums. Since it dogs not matter what letter we use, the variable ofintegration is called a dummy variable,

EXAMPLE 6 Express the limit of Riemann sums

. lim-l13ca'? - 2ct l ' 5) A.ttI P n - o -

as an integral if P denotes a partition of the interval [-1, 3].

--!F

4.5 Riemann sums and Definite Integrals 317

Sotution The function being evaluated at cr in each term of the sum is ./(r) :

3x2 -2x +5. The interval being partitioned is [-1, 3]. The limit is therefore theintegral of / from 1 to 3:

, t ' ^l i m ) - r l c r ' 2 c ^ - 5 1 A l a - | r 3 x z - 2 x + 5 t d x

lp +oai J t LJ

Constant FunctionsTheorem 1 says nothing about how to calculate definite integrals. Except for a few

special cases, that takes another theorem (Section 4 7). Among the exceptions are

constant functions. Suppose that / has the constant value f(x): c over la, bl.

Then, no matter how the cr's are chosen,

-j- Af / t c r t l , , : \ c ' L x t

' , . 1

- c . ) A x / r " r ' r , r , r \ l ' r ' I l ( , r " s , l

: c (b - a ) . i r r r l : l c , r g th o l i n r c , r l i l a . / r l - l , r

Since the sums all have the value c(b - a), their limit, the integral, does too.

, : 160 9 .8 t

If /(,r) has the constant value c on [4, D], then

f b l b

J, fOla ' : J . 'o* : c(b - a) '

EXAMPLE 7

a ) I 3 d x : 3 { 4 l - l ) ) : { 3 r { 5 r : l 5

a,t 1",{

z,l a* : 3(4 - (-r)) : (-3)(5) : -15

The Area Under the Graph of a Nonnegative FunctionThe sums we used to estimate the height of the projectile in Section 4.4, Example

2, were Riemann sums for the projectile's velocity function

u : f ( t ) : 1 6 0 - 9 . 8 t

on the interval t0,31. We can see ftom Fig.4.13 how the associated rectangles

approximate the trapezoid between the t-axis and the curve u : 160 - 9.8r. As the

norm of the partition goes to zero, the rectangles fit the trapezoid with jncreasing

accuracy and the sum of the areas they enclose approaches the trapezoid's area,

which isb t I b t ^ 1 6 0 + 1 3 0 6

Trapezo id a rea=h ' j ' . - - - l j ) 9 .

D

l<- height -l

Region is a trapezoid with height = 3base (top) = 130.6

base (bottom) : 160.

4.13 Rectangles for a Riemann sum ofthe velocity fundion f(t) : 160 - 9.8tover the interval [0,3].

(3, 130.6)


The region in Example 8.

This confirms our suspicion that the sums we were constructing in Section 4.4.Example 2, approached a limit of 435.9. Since the limit of these sums is also theintegral of / from 0 to 3, we now know the value of the integral as well:

f lI (J60 - 9.8t) dt : trapezoid area : 435.9.

Jo

We can exploit the connection between integrals and area in two ways. Whenwe know a formula for the area of the region between the x-axis and the graph of acontinuous nonnegative function y : /(r), we can use it to evaluate the function'sintegral. When we do not know the region's area, we can use the function's integralto define and calculate the area.

DefinitionLet f (x) > 0 be continuous on [a, b]. The area of the region between thegraph of / and the ;-axis is

e: l,' f <oa,.

Whenever we make a new definition, as we have here, consistency becomes anissue. Does the definition that we have just developed for nonstandard shapes givecorect results for standard shapes? The answer is yes, but the proof is complicatedand we will not go into it.

EXAMPLE 8

Evaluate

x d x , 0 < a < b .

so/ufion We sketch the region under the cuwe :! : r, a < "r < b (Fig. 4.14), andsee that it is a trapezoid with height (b - a) and bases a and D. The value of theintegral is lhe area of this trapezoid:

l b , . 1 * b b z a 2I x d x = ( bI

o ' ' 2

: 1 z

Thus,

t r t , t J 5 f t l t 2I X q ) =

J , 2 2

and so on.Notice that x2l2 rs an antiderivative of -t, further evidence of a connecdon

between antiderivatives and summatlon.

EXAMPLE 9

Find the area of the region between the parabola y :,t2 and the r-axis on theinterval [0, b].

l<- Ar >k- A-r +<- Ar t l<-At-'l

r 75 The redangles of the Riemann sumsin Example 9,

Notice that r3/3 is an antiderivative of -r2.

4.5 Riemann sums and Definite Inteqrals 3'19

So/ution We evaluate the integral for the area as a limit of Riemann sums.We sketch the region (a nonstandard shape) (Fig.4.15) and partition [0, ,] into

n subintervals of length Ax : (b - 0) /n : b/n. ^fhe points of the partitton are

x o = 0 , r r : L x , x z : 2 L x , . . . , r n t : ( n - 1 ) A x , x r : n \ x : b .

We are free to choose the c/,'s any way we please. We choose each cr to be theright-hand endpoint of its subinterval, a choice that leads to manageable arithmetic.Thus, ct : ;rl, c2 : x2, and so on. The rectangles defined by these choices have

f (c) Lx : /(A.r) A,r : (Ar)2 A.{ : (1'?)(Ar)3

f(c) Lx : f (2Ax) Lx: (2\x)z Lx: (22)(Lx)3

f (c . ) tx : f (nLx) Lx: (nLx)2 Lx:1n2.11Ax;3.

The sum of these areas is

s, : t / (cr )^r

- \ - 1 2 t a x ) '- 1 ,

= r a ' r j I , t 2

b3 n(n t 1) (2n + 1)

n 3 6

b 3 @ t 1 ) ( 2 n + 1 )6 n 2

b 3 Z n z + 3 n + I6 n 2

b 3 / 3 l \-= . -2+:+ ,1 .o \ f i n /

We can now use the definition of definite integral

r b ns a "

| .[ (t) dx : l im ) / (c() A-rJ o ? l ' u - t _ t

to find the area under the parabola from ;r : 0 to .r : b as

r I , r ' i \ r L . , . i , : r

(6)

loo "a' :l i m ( t

h 3 / I t \l im ; ' 12+ :+ ,1 L .

n - F o \ t ? n " /

h t h J= ; . {2+0+0) : .

o . ,

r o = 0 r t r z x 3 x n 1 x n : b

320 chaoter 4: lnteqration

With different values of &, we get

f t . l r 1I x ' d x : - : - ,

J r J J

and so on.

r r s ^ { 1 . 5 ) r 3 . 3 7 5

J n x ' d x :

, : - - 1 . 1 2 5 .

J

Exercises 4.5

Sigma NotationWrite the sums in Exercises l-6 without sigma notation. Then evaluatethem.

Values of Finite Sums

17. Suppose that I ar : -5 and I br - 6. Find the values of

2 6 kt . t I

3. I cos frz

5 - J - l - l ) *+ r s i n -

a) t(t - 1)'?

l l . 1 + 2 + 3 + 4 + 5 + 6t t t t

13 ' t +4+ t+ t 6

rs . I 1+1-1* l2 1 4 5

r \- i____:

54. I sinftz

6 , l ( - l ) t cos tz

a) | 3a1

d) I(a* br)

a) | 8a1

c) t (4r + 1)

t 019. a) Ik

t 320. a) ! t

62 1 F / l - t 2 l

52s. Lk(3k + 5)

r , 1 / 5 \ -

2 7 . 5 - i + l l t l- . 115 l t r I

b r t aE , 6

e) iQtt - zat)

c\ t@r + bo)

18. Suppose ttrat ! a1 - 6 and I Dr : 1. Find the values oft:r &=l

7. Which ofthe following express 1 * 2 * 4 + 8 + 16 + 32 in sigmanotation?

8 . Wh icho f rhe fo l l ow inge \p ress | - 2 -4 -8 - l 6 . 32 ins igmanotation?

6

3^ ' \ - / _ t r t + l ) i + 2

Use the algebra rules on p. 310 and the formulas in Eqs. (l)-(3) toevaluate the sums in Exercises 19 28.

5b) I2o

b) i2sobk

d) i(rr 1)

t 0c) t i .

k: l

t 3c) tk,

5b) t ( -112f

t 0b) t t '

1 3b)

- : r

J '9. Which formula is not equivalent to the other two?

4 { - l \ ' - r ' t l r *a r I - b ) I , , c l

t _ 2 ^ - , t _ n ̂ |

10. Which formula is not equivalent to the other two?

I r t \ r

3b) t ( r+1) ,

- l

c) t r ,

u. L(k' - 5)

26. L k(2k + t)

^'(P,r)'-L+Express the sums in Exercises 1l-16 in sigma notation. The formof your answer will depend on your choice of the lower limit ofsummation. Rectangles for Riemann Sums

In Exercises 29-32, graph each function /(r) over the given inteNal.Partition the interval into four subinteNals of equal length. Thenadd to your sketch the rectangles associated with the Riemann sumttr/(cr) A,rr, given that ca is the (a) left-hand endpoinr, (b) right-hand endpoint, (c) midpoint of the ftth subinterval. (Mal(e a separatesketch for each set of rectaneles.)

1 2 . 1 + 4 + 9 + 1 6

1 4 . 2 + 4 + 6 + 8 + l 0

' ^ ! -2- - i ' 1- 1) ) ) t )

Exercises 4.5 3Z'l

29. f (x): xz - r, 10,21

30. /("{): --r '?, [0, U

31. / ( . r ) :5 tu 1 , 1 -7 ,71

3 2 . / ( r ) : 5 i 1 a 1 1 , I n , n )

33. Find the norm of the partit ion P = {0,1.2, 1.5,2.3,2.6,31.

3. Find the norm of the pa.rtition P : {-2, -1.6, -0.5, 0,0.8, 1}.

Expressing Limits as IntegralsExpress the limits in Exercises 35-42 as defrnite integrals.

35. l im I c, 'zax1. where P is a panitron ol [0. 2llPt-ok=- l

36. l im | 2c^r^x,. where P is a panition ol [-].01pI-o*=

37. l im f rcl r - Jc, t Ax, . where P is a panrtion ol l-7.5j

ss. l',<z - txl ax

Use areas to evaluate the inte$als in Exercises 57-60.

s7. to "

xdx , b>o

s9 . l "oz ra r ,

o .o . t

EvaluationsUse the resultsExercises 61 72.

1"56I. I xdx

F t i64. I rdr

l t / 2 ^67. I t ' d t

st. lou

*a", r,, o

o . l "o

z , a t , o .o . r '

of Examples 8 and 9 to evaluate the integrals in

63. J,

0d0

65.

I:.,I,*

)c dx

x2 dr

e2 deur' /"

,no366.

Jn s2 ds

69, t d x

x ' d r

38.

39.

40.

4t.

42.

*' l',s a'

*. fo' r-rcu at

n. l",o.s a,

Using Area toln Exercises 49-56,the integrals.

o. | " ( |+z)

a", 1

51. J,.t/9

x'z dx

sz. l_',1'l a'

l i . i f I

) Ax1. where P is a panition oi Ll.4lP -oEi \cr /

l im ; -larr. where P is a panidon of 12. JltP -oEi I - cr

tim f I - ri l lx^. where P is a panition oi [0. l]llP -o k=-t

-

,liE.r!C"" cr) atr, where P is a partition of L z/4, 0l

tirn ittan cr) A-rr, where P is a paruuon ot LO, trl4lPn-oEl

rJ5.,0.

J, rdx

| 3rE

71. Jn

x2 dx 'f',

I"Îo"

46. I :do

"Jn4s. / nEa,J!5

Evaluate Integralsgaph the integands ard use arcas to evaluate

Finding AreaIn Exercises 73-76, use a definite integral to find the arca of theregion between the given curve and the ir-axis on the inteNal 10, ,],

74. y : na2

76.v: i+r

Theory and Examples77, What values of a and , maximize the value of

t b| 1r - xzS dr't

J"

(l1inr. Where is the integland positive?)

What values of a ^nd b minimize the value oft b

I lrn - 2r') dr?J "

Upper and lower sums for increasing fun.tions

a) Suppose the graph of a continuous function / (r) rises stead-ily as r moves from left to right across an interval [a, b].Let P be a partition of [a, D] into z subintervals of lengthLx : (b - a)/n. Show by referring to the accompanyingfigure that the difference between the upper and lowet sumsfor / on this partition can be rcpresented graphically as thearea of a rectangle R whose dimensions are lf (U) f (")lby Lx. (Hint: The difference U , is the sum of areas

as in Example 9.

Constant FunctionsEvaluate the inteerals in Exercises 43-48.

aa. f,

r-zotdx

78.

79.

f3/2

J , ,, (-2x 'l 4) dx

f 0 . -

J ,nt/ 16 - x2 dx

f r

J_,(1 - lxl) dx


of rectangles whose diagonals QoQr, QtQz. . . . , Q, ,Q,lie along the cuwe. There is no overlapping when these

rectangles are shifted horizontally onto R.)

b) Suppose that instead of being equal, the lengths Axr of the

subintervals of the partition ot' [4, b] vary in size Show that

U L < f(b) l(a) a.rna.,

where A.r.o* is the norm of P, and hence that lim p1-n

(u - L ) : 0 .

lJpper and lower sums for decreasing fLln(t ions (Continua-

tion of Ercrcise 79)

a) Draw a figure like the one in Exercise 79 for a continuous

function /(r) whose values decrease steadily as ,rr moves

ftom left to right across the interval [a,r]. Let P be a

partition ol [d, ,] into subintervals of equal length. Find an

expression for U - L that is analogous to the one you found

for U - a in Exercise 79(a).

b) Suppose that instead of being equal, the lengths A,rr of the

subinteNals of P vary in size. Show that the inequality

u - L = l f ( b ) / ( a ) l a - r . *of Exercise 79(b) still holds and hence that lim p -6( u - L ) : 0 .

Evaluate f 12 dr, b > 0, bycarrying out the calculationsof Example 9 with inscribedrectangles, as shown here,instead of circumscribedrectangles.

82. Let

l f r 2 3 n l l" - n l , n ' n ' ' t )

Calculate lin,,-- & by showing that S,, is an approximatingsum of the integral

whose value'we know from Example 8. (gir?L Partition [0, 1linto r? inteNals of equal length and write out the approximatingsum for inscribed rectangles.)

83. Let

1",

- ' ( ; ) '

80.

and interpret S, as an approximating sum of the integral

r l Î x 'dx ,

Jo

whose value we know from Exanple 9. (Hinlr Pafiition 10, ll

into , intervals of equal length and write out the approximating

sum for inscribed rectangles.)

84. Use the formula

sin h + sin 2h + stn 3h + . +sin , l i

_ cos (h /2) - cos ((m ] (t /2))h)2 sin (h 12)

to i ind the area underthe curve y - sin xfromx:0tox:r12.ln two stepsl

a) Partition the interval 10, 1t /2) into,? subintervals of equal

length and calculate the corresponding upper sum U; thenb) Find the limit of U as '? + co and Ar : (b - a)ln "' 0.

S cAs Explorations and ProjectsIfyour CAS can draw rectangles associated with Riemann sums, use it

to draw rectangles associated with Riemann sums that converge fo the

integrals in Exercises 85 90. Use z - 4, 10, 20, and 50 subintervalsof equal length in each case.

t t 18 s . / ( l r ) / r : ;

J o z

. l ' 2 : t n - l t jJ r | ,

- f -

n 1

To calculate lim,,-- t,, show that

r ^ 4( x ' + l ) d r : -

' -; i(;)' . .(+)']

86. I

r,. J-c o s r d x = 0

ss. /""

sec'r ar : t

89.

90.

91. a)

b)

t,t;I: d.r (The inteSral's value is In 2.)

Write the sum S, in Exercise 82 in sigma notation and useyour CAS to flnd lim,-- S,.Do the same for the sum & in Exercise 83.

4.6 Propert ies, Area, and the Mean Value Theorem 323

b) Use sigma notation to wdte the left-endpoint sum t fbl/l

subinteryals of length 4/r.

c) Find lim,-- S,. How does this limit appear to be related

to the volume of the solid?

94. (Continuation of Section 4.4 Example 4..) In sigma notation, the

lefFendpoint sum in Example 4, Section 4.4, is

8

5 8 : I r r f t o r + t r t - l r r ' ? l .. t=l

a) Use sigma notation to write the analogous left-endpoint

sums 516 for 16 subintervals of length 1/2 and Sso for 80

subintervals of length l/10.

b) Use sigma notation to write the left-endpoint sum S, fbr n

subintervals of length 8/n.c) Find hr!,-- S,. How does this limit appear to be related

to the volume of the sphere?

92. Wr i te the sum s in r+s in2 l r+ . . .+s in r r f t in Exerc ise 84 insigma notation and use your CAS to find lim,-- t,.

(Continuation of Section 4.4, Example 3.) It sigma notation, theleft-endpoint sum in Example 3, Section 4.4, is

4

s r : l + f o 1 l a , l - l r r ? l .

a) Use sigma notation to write the analogous left-endpointsums 53 for eight subinteNals of length 4/8 and S25 for25 subinteNals of length 4/25.

Properties, Area, and the Mean Value TheoremThis section describes working rules for integrals, examines the relationship between

the integral of an arbitrary continuous function and area, and takes a fresh look at

average value.

Properties of Def inite IntegralsWe often want to add and subtract definite integrals, multiply their integrands by

constants, and compare them with other definite integmls. We do this with the

rules in Table 4.5 (on the following page). All the rules except the first two follow

from the way integrals are defined with Riemann sums. You might think that this

would make them relatively easy to prove. After all, we might argue, sums have

these properties so their lirnits should have them, too. But when we get down to

the details we find that most of the proofs require complicated €-d arguments with

norms of subdivisions and are not easy at all. We omit all but two of the proofs.

The remaining proofs can be found in more advanced texts.

Notice that Rule I is a definition. We want every integral oYer an inteNal of

zero length to be zero. Rule 1 extends the definition of definite integral to allow for

the case 4 : b. Rule 2, also a definition, extends the definition of deflnite integral

to allow for the case b < a. Rules 3 and 4 are like the analogous rules 1br limits and

indefinite integrals. Once we know the integrals of two functions, we automatically

know the integrals of all constant multiples of these functions and their sums and

differences. We can also use Rules 3 and 4 repeatedly to evaluate integrals of

arbitrary finite linear combinations of integrable functions term by term. For any


l. Zem: 1." f Al o" : o (A definition)

2, order of Intesr.ttion: fo" frrro":- f.u fi'to, (Atso a definition)

3, constant Muttipler, t"'ty6a*=t l,' f<oa, (Any number t)

f.u ,r,ro,: - l.o ,oro, (k: -t)

4. sums and Dffirences: l"o {f li

* ,t.)) o, : f.u f a> a, +

l"u e@ a*

s. Additivit),: l"' ralo, * 1," tara* = 1"" f <oa,

6. Max-Min Inequality.' If max / and min / are the maximum and minimum valuesof / on [a, D], then

ntn f . (b -a)= f .o

f O lar smaxf . (b -a) .

7. Donination: f (x) z e@) orL la,bl + f.u f ,rro, z f"u s@a'

f@)zo oL la ,b j + l . ' f { ia , .o

(Special case)

Table 4.5 Rules for definite inteorals

constants cr,..., c,, regardless of sign, and functions fi(;),..., fi("r), integrableon la, bl,

1"" lnf,{,)+...+c,f,(x))dx:,, !"0 7,6) a' + ... + ", [" ' f,<') a*.

The proof, omitted, comes from mathematical induction.Figure 4.16 illustrates Rule 5 with a positive function, but the rule applies to

any integrable function.

Proof of Rule 3 Rule 3 says that the integml of t times a function is t times theintegral of the function. This is true because

r b

I kf(x\dx: l i t "L kf (c i \ ̂x iJ o l r l ' u

- '

: tim r f, f {,,)o,,lP l i+o ,_ ,

-r- fb: k l i m - ) . f t c i l \ x i = k I I t x t d x .tlPll-0 u' -

J" '

tr

4.td Additivity for definite integrals:t b t c g

I f1)dx+ I f|)dx = I fk)dxJa Jb Ja

. b

I ru)ax: I r(x)dx - I ruldx.J t Ja Ja

4.6 Properties, Area, and the Mean Value Theorem 325

Proof of Rule 6 Rule 6 says that the integral of f over la,bl is never smaller

than the minimum value of f times the length of the inte al and never larger than

the maximum value of / times the length of the inte al. The reason is that for

every partition of [a, D] and for every choice of the points ci,

m i n { . ( b , a ) = m r n | ) L t r

: ) m l n / . a r r

< ) l ( c f ) a x r

< ) - mar / . Ax*k=I

" \ - .: m a x I ' ) l ] x k

: m a x f . ( b - a ) .

In short, all Riemann sums for / on [4, D] satisfy the inequality

min f . (b - a) < L f @t) Lxr < max f ' (b - a).n:l

Hence their limit, the integral, does too

EXAMPLE I Suppose thatf t f 4

J , I t x )dx : s .

J t . f \ x \dx : -2 .

Then

r. fo' f r ia,: - l ,o f { ,)o*: -(-z):2

z. l_',rzrat i3h(x))dx:z l-',f ala, +z l,',nala,-- 2 (5) + 3 (7) :31 Rurcs I rnd -1

1 4 t t ft . I f l x )dx= | f l x l dx+ I f G )dx :5+ ( -z ) :3 RLr rcs

t rJ - t J - t J l

In Section 4.5 we leamed to evaluate thlee general integrals:

f"u ,a*:,{r, - o) rAny cons tan t c r (1 )

r b h 2 n 2

I x d x : ' - - " - l 0 < a < b )J" z z

f b h )

I x ' d x : ; { b > 0 ) .Jo r

The rules in Table 4.5 enable us to build on these results

n l n l : l l i r )

/ ( , r ) : n r a x /

Ru l , " :

o

l ' ,oat a' : '

(2)

(3)


1 r 2 1 2 r z

; J" uo'-7 Jo tdt * Jo

to '

I /q\ _, (4_9)*,,,_u,4 \ 3 / '

' \ 2 2 ) ' -

! -u+10: -+J J

Solutionf 2 t t 2 \

I l ; -7 t +s l a t :. / 0 \ i /

EoAMPLE 2 eua*t" /' (l -r, *t)

".

Rules i

E q s . ( l ) l

a

EXAMPLE 3 Evahnre J,

x2 dx.

Sorution We cannot apply Eq. (3) directly because tlle lower limit of inte-gmtion

is different from 0. We can, however, use the Additivity Rule to express f x2 dxas a difference of two integrals that can t)e evaluated with Eq. (3):

" 2 1 3 r l

I * 'd , + I , ' d r : | *2d ,Jo Jz Jo

r l t i f 2

I * ' d r= I * ' d * - | x2dxJ2 Jo Jo

(3)3 e)33 3

2 7 8 1 9-

3 3 - 3 '

In Section 4.7, we will see how to evaluate ll xz dx in a more direct way. al

The Max-Min Inequality for definite integrals (Rule 6) says that min f . (b - a) is

a lower bound for the value ot [ 161ax and that max f , (b - ")

is an upperbound.

Show that the value of

f t -

I Jl + cos xdxJo

Rule 5

Solvc for

Eq. (3) nowappl ies.

EXAMPLE 4

cannot possibly be 2.

Solutron The maximum value of .r4 +cos.r on t0, 1l is v1 + 1 : \,, so

f t -I ,{t +

"r" , d, < ma.r,( /JT;os r . (1 -0) rabre '1s'

J 0 - '

R r l c 6

=\4' t :Jr.The integral cannot exceed r/2, so it cannot possibly equal 2. l-)

EXAMPLE 5 Use the inequality cos .r > (1 - x2 /2), whtchholds for all .r, tofind a lower bound for the value of li cos x dx.

4.6 Properties, Area, and the Mean Value Theorem 327

Solutionf l f l / 1 2 \

; f cos xdx .

J " \ t - , )o -

f t 1 f 1

tJ, , to*-1J"" 'a '

>1. ( l o ) -1 . { l ) ' -1- 2 3 6

R r L l c I

R'r lcs I urd l

^r 0.83.

The value of the integral is at least 5/6.

Integrals and Total AreaIf an integrable function y: /(r) has both positive and negative values on aninterval [a, r], then the Riemann sums for f on la, bl add the areas of the rectanglesthat lie above the r-axis to the negatives of the areas of the rectangles that lie belowtt (Fig. 4.17). The resulting cancellation reduces the sums, so their limiting valueis a number whose magnitude is less than the total area between the cu e and the.r-axis. The value of the integral is the area above the axis rninus the area belowthe axis.

This means that we must take special care in finding areas by integration.

LI

4. ?7 (a) The Riemann sums are algebraicsums of areas and so is the integral towhich they converge. (b) The value of theintegral of f from a to b is

rb r t , f t z

I r\rax= I fk)dx+ I f?\dxJa Ja Jx '

tb+ I f (x)dx = A'' - AzI Az.

Jr)

lu y6v a,= t,

EXAMPLE 5 Find the area of the region between the curve ):4-xz,0 5 .r = 3, and the .t-axis.

so/ution The x-intercept of the curve partitions [0, 3] into subintervals on which

-f(;r) :4 -.r2 has the same sign (Fig.4.18). To find the area ofthe region betweenthe graph of / and the r-axis, we integrate / over each subinterval and add theabsolute values of the results.

Integral over 10, 2l:

f'fo) o' = o,

1 2 ? 2 1 2

I \+ x 'z )dx : I +ax- | r 'd *Jo Jo Jo

l ? \ l: 4 t 2 - 0 J - ' ;

:8-g:3

r'

1 6;

(b)

E ( s . 1 i r

4. tB Part of the region in Example 6 l iesbelow the x-axis.

Ifl(c*) > 0, /(c*)Ar. is an area...

...but ifl(c*) < 0, /(cr)A.xe isthe negative of an area.

328 Chapter 4: lntegratron

How to Find the Area of the

Region Between a Cu've Y - f(x)'

a = x < b, and the x-axis

1, Paftition [d, b] with the zeros oi Jf.2. lntegrafe / over each subinterval.3. Add the absolute values of the

integrals.

4.?9 A sample of values of a function onan interval [a, b].

Integrti over 12, 3l:

r.l -4dx -

J . x ' t lx

r t 3 r r t 2 t ' t=41J - 2 r - l - . I

\ J 3 |'7

3

1 6fhc region s tre,t: Area - -, t

The Average Value of an Arbitrary ContinuousFunctionIn Section 4.4, Example 5, we discussed the average value of a nonnegaflle con-

tinuous function. We are now ready to deline average value without requiring /

to be nonnegative, and to show that every continuous function assumes its averrge

value at least once.We start once again with the idea from arithmetic that the average of n numbers

is the sum of the numbers divided by n. For a continuous function / on a closed

interval [a, b] there may be inlinitely many values to consider' but we can sample

them in an orderly way. We partition [c, D] into lr subintervals of equal length (the

length is A.]r : (b - a) I n) and evaluate / at a point ct in each subinterval (Fig'

4.19). The average of the n sampled values is

/ (cr) * / (c:) + " ' + | (c, ' )J \ck)

l, '<+ "la": l, '

1 9

l | ) t

33

iAr

b - a

Ib - u

i r , , , 'L ) r ' - ^

1 - J - ' - '

.o"Jil*-"iliL.,. 'r

Thus, the average of the sampled values is always l/(b a) times a Riemann sum

for / on fa.bl. As we increase the size of the sample and let the norm of the

partit ion approach zero, the avemge must approach (l l@ -tD l, l l(;r)d'r ' We

are led by this remarkable fact to the following definition.

DefinitionIf I is integrable on [.], bl, its average (mean) value or [4, l'] is

1 tbav( f ) : uo J . f

6 )c tx '

E{AMPLE 7 Find the average value of l(r):4 12 on [0' 3] Does /

actually take on this value at some point in the given domain?

-

4.6 Properties, Area, and the Mean Value Theorem 32g

Solution

^ l l hr v ( l ) : - I f \ x ) d xD _ a J o

-* / ' ,0 - , , r - - - l r (

l l . - r3 r ' r I: _ (4 r3 _0 ,_ -s_ ) : ' ( r2_s ) = l

The average value of /(x):4-,r2 over the interval t0, 3l is l. The functionassumes this value when 4 - x2 : l or x :1n5. Since one of these polms, ,, :.,/i, lies in [0, 3], the function does assume rts average value in the given domainrF ig .4 .20) .

J

The Mean Value Theorem for Definite IntegralsThe statement that a continuous function on a closed interval assumes its averagevalue at least once in the interval is known as the Mean Value Theorem ior DefiniteIntegrals.

Theorem 2The Mean Value Theorem for Definite IntegralsIf / is continuous on [a, b], then at some point c in [a, bl,

Jrc\= 1

fu t '* 'a* '

(Frg. 4.21).

-. .In Example 7, we found a point where f. assumed its average value by setting

/("r) equal to the calculated average value and solving for x."But thls does notprove that such a point will always exist. It proves only that it existed rn Example7. To prove Theorem 2, we need a more general argument.

Proof of Theorem 2 If we divide both sides of the Max_Min Inequairty (Rule 6)by (b - a), we obrain

l l b^ i n f :

6 _ o J " l r x t d x l m a t f .

Since / is continuous, the Intermediate Value Theorem for Continuous t unctions(Section 1.5) says that / must assume every value between min / and max /. Itmust therefore assume the val:ue (l/(b _ a\ f i f(x)clx atso_. point cin[a,bl.

a.

The continuity of / is important here. A discontinuous function can srep overrts average value (Fig. 4.22).

What else can we learn from Theorem 2? Here is an example.

Average value ofyon [0, 3] is assumeda t x = J 1

4.20 The average value of f(x) = 4 x2on [0,3] occurs at x = ^r€ 1E16n,01" 7y.

4.21 Theorem 2 for a positive function:At some point c jn Ia, bl,

r bf(c). (b-a) = | f(x)dx.

J .

+a, - l,' -'a,)

Average value l/2not assumed

4.22 A discontinuous function need notassume tts average value.


EXAMPLE 8 Show that if / is continuous ot fa, bl, a t' b' and it

l bI fe )dx :0 ,J"

then /(t) :0 at least once in [4, b].

Solution TIlr- average value of / on [a, b] is

1 t b 1av ( / ) : - l - f r x t ax : , . 0 : 0

D A J " O - a

By Theorem 2, / assumes this value at some point c rn [a,b].

Exercises 4.6

Using Properties and Known Values to Find OtherIntegrals1. Suppose tlat / and I are continuous and that

t2 t5 15

J , f r , ta^ : - t .

J f r x tdx :6 . J , s * ta ' =a .

Use the rules in Table 4.5 to find

4. Suppose that /" 81)dt:Jr.Ftnd

d lo s(D at

o lo ?r{ta,

r) f" g@)au

at l'fra,

"\ l,' fft,) s{ita' ,1 l,' vy1"; - r1i1o'

2. Suppose that / and ft are continuous and that

l,' f ,,, o" : -r, l,n f at a, : s, l,' ,61a, : o.

Use the rules in Table 4.5 to find

f 2a)

J, s@) dx

f 2c)

J, 3 f(x) d.t

d f,n -zf G) a"

o l,n tzf l'> - 3h(t)ldx

"\ l,' f {i a,

3. Suppose that Ji /(x) dr : 5. Find

f 2a\

J, f (u) au

o l, 'r{,)a,

5. Suppose that / is continuous and ttrat f; fQ) d.z = 3 nd

I: f k) dz :'l Find

,'t 1," rr,) a, at fo' f at a,

6. suppose that l? is continuous and that /r, h(r) dr : o and

[3 ,h1r; dr : 6. Frrd

d f" n{ia' rt) - l,' n{,t a,

D l,' gG) a'

il l"' froa"

D Ln uG)*o(,)to,

fn' r {i a,

t l"' wra- fl)tdt

?2

J' J5fr'>a'f 2

J, tf r,)ta*

Evaluate the intelrals in Exercises 7-18.

t. l,' t a'

e. Jo

st dx

t. lo' {2, -!a,

f l ,13.

J , \ t+ ; ) dz

1 2 ^15.

J, 3u" du

n. lo' e,' +, - a,

?-2

s. / "4a*

t s -'o' J. io'

b)

d)

12. I ( t - r t \a tJ o '

u. f,o {2, - ! a,

rc. l',,2t u" au

n. l ro e"+ ' -$a"

AreaIn Exercises

19.

In Exercises 23-26, graph the function over the given interval Then

{a) integrate the function over the interval and (b) find the area of the

regjon belween the graph and $e.Y axi' '

23. y -- x2 6r + 8, [0, 3]

24, y : -x2 + 5x 4, 10,21

25. y - 2x - -,r'?, [0, 3]

26. ) : ).'?- 4,r, [0, 5]

Average ValueIn Exercises 27 34, graph the function and find its average value over

the given inteNal. At what point or points in the given inteNal does

the function assume its average value?

27. f rxt: r'? - 1 on 10. J3l

Exercises 4.6 331

ln Exercises 35-38, lind the avemge value of the function over the

given inteNal from the graph of / (without integrating)'

. r * 4 , 1 = x : - l- t : 2 , - l < x = 2 on [ 4 ,2 ]

36. f (t) -- - { - P o n [ - 1 , l ]

J

"f(r) - sin t

f ( 0 ) : t a n 0

Theory and ExamPles39. Use the Max-Min Inequality to find upper and lower bounds for

the value ofr t rI - , d x

J n l + " [ '

40. (Continuatibn of Erercise 39.) Use the Max-Min Inequality to

find upper and lower bounds for

r . 0 . 5 I f l l, _ ^ e , r a n d | . . , d t .

J o I - r ' J o . l - ' r '

Add these to arrive at an improved estimate of

r l lI - dx .

J n r + t 2 -

41. Show that the vatue of f] sin (x2) d, cannot possibly be 2'

42. Show that the value of I; uE + s,l, Iies between 2Jj = 18

and 3.

43. Suppose that / is continuous and ttrat f /(r) dr : 4 Shou

that /().) : 4 at least once on [], 2]

,f4. Suppose that / and g arc continuous on la, bl' a +D and ihal

.t Gtrl - eti lar:0. show that /( ') = s(i) at least once

in la, bl.

45. lnLegrals of nannegative functtons Use the Max-Min Inequal'

ity to show that if / is integrable then

/(r) - 0 on ta,bl + J"

J (.r) r1'r z 0'

19 22, find the toral shaded area.

20.

[0,2r)

| " " f| 4 ' 4 )38.

I '28. f ( t ) : - ; on

f (.x) - -3x2 - 1

f (x ) :3x2 - 3

f(t) : (t - r)2

t0. 3l

on [0, l]

on [0, 1]

on [0, 3]

to

30.

31.

32.33.31.

f ( t ) : t2 - t on [ -2 , l ]

s ( x ) : l x l - I o n ( a ) [ - 1 , 1 ] , ( b ) t l , 3 l , a n d ( c ) [ l 3 ]

l ( r ) - - l r l on (a ) [ -1 ,0 ] , (b ) t0 '11 ' and(c ) [ -1 , 1 ]


46. lntegrals of nonpositive functions. Show that il / is integrablethen

50. It would be nice if average values of integrable functions obeyedthe following rules on an interval [a, ,]:

a) av (f * g) : av(/)+av(s)b) av (t/) : lr av (/) (any number f)c) av (/) = av (g) if .f(,r) = e(") on fa, bl.

Do these rules ever hold? Give r€asons for your answels.

51, If you average 30 mi/h on a 150-mi tdp and then retum over thesame 150 mi at the rate of 50 mi/h, what is your average speedfor the trip? Give reasons for your answer. (Source: David H.Pleache\ The Mothenatics Tbachetr Vol. 85, No. 6, pp. 445+46,Seprember 1992. )

52. A dam released 1000 m3 of water at l0 m3/min and then releasedanother 1000 m3 at 20 m3/min. mat was the avenge mte atwhich the water was released? Give reasons for your answer.

47.

48.

49.

f b/ ( . t ) 5 0 on la .b l =+ I f 6 )dx =0 .

J,,

Use the inequality sin,r : r, which holds for "{ > 0, to find anupper bound for the value of /n sin .r dr.

The inequality sec a Z | + k212) holds on ( r 12, jt /2). tJseit to find a lower bound for the value of fi sec I dr.

If av (/) really is a typical value of the integrable function /(r)on [a, r], then the number av(/) should have the same integralover [a, r] that / does. Does it? That is, does

fb tbI a t r f t d : L -

| f r r t d x lJ O J A

Give reasons for your answer

The Fundamental TheoremThis section presents the Fundamental Theorem of Integral Calculus. The inde-pendent discovery by Leibniz and Newton of this astonishing connection betweenintegntion and differentiation started the mathematical developments that fueledthe scientific revolution for the next two hundred years and constitutes what is stillregarded as the most important computational discovery in the history of the world.

The Fundamental Theorem. Part 1If /(t) is an integrable function, the integral from any fixed number a to anothernumber x defines a function F whose value at .r is

r@: I f< t la t . ( . r )

For example, if / is nonnegative and .r lies to the right of a, F(x) is the areaunder the graph from a to ,r. The variable x is the upper limit of integration ofan integral, but F is just like any other real-valued function of a real variable. Foreach value of the input r there is a well-defined numerical output, in this case theintegral of / from a to x.

Equation (1) gives an important way to define new functions and to describesolutions of differential equations (more about this later). The reason for mentioningEq. (1) now, however, is the connection it makes between integrals and derivatives.For if / is any continuous function whatever, then F is a differentiable function of* whose derivative is / itself. At every value of .r,

d. F ( x ) :

ax

This idea is so important that it is the first part of the Fundamental Theorem ofCalculus.

fr l.' ,u,0,: t,,,

4.7 The Fundamental Theorem 333

Theorem 3The Fundamental Theorem of calculus, Part 1If / is continuous on {q,bl, then F(x): I; f (t)dt has a derivative atevery point of [a, D] and

< b.

This conclusion is beautiful, powerful, deep, and surprising, and Eq. (2) maywell be the most important equation in mathematics. It says that the differen-tial equation dF /dx : / has a solution for every continuous function /. It saysthat every continuous function / is the derivative of some other function, namely

I: fG) dt.It says that every continuous function has an antiderivative. And it saysthat the processes of integration and differentiation are inverses of one another.

Proof of Theorem 3 We prove Theorem 3 by applying the definition of derivativedirectly to the function F(-r). This means writing out the difference quotient

#:* l ' . r<oat=ô' a=x (2')

F (x i h) F(,r)

and showing that its limit as ft + 0 is the number /(.r).When we replace F(x * i) and F(,r) by their defining integrals, the numerator

in Eq. (3) becomes

f t + hF t x - h ) - F t x t : I [ r 0 d r - | f r r t d r .

J,, J, '

The Additivity Rule for integrals (Table 4.5 in Section 4.6) simplifies the right-handside to

I f ( t ) dt ,J t

so that Eq. (3) becomes

F ( x + h ) - F ( x ) : lrr,- + h) - F(x))

:i 1."' too'' (4)

(3)

According to the Mean Value Theorem for Definite Integrals (Theorem 2 in thepreceding section), the value of the last expression in Eq. (4) is one of the valuestaken on by / in the interval joining x and .r * li. That is, for some number c inthis interval,

I l. '. ' t<'to': rct' (5)

We can therefore find out what happens to (1/lz) times the integral as I -+ 0 bywatching what happens to f (c) as h --> 0.


4.23 The rate at which the carpet coversthe floor at the point x is the width ofthe carpet's leading edge as it rolls past x.ln symbols, dAldx: f(x).

What does happen to f (c) ash -+ 0? As h --+ 0, the endpointi * ft approaches

x. pushing c ahead of it like a bead on a wire:

-1----j--+--;1-t or ---r----$-------e-

So c approaches x, and, since / is continuous at x, f (c) approaches /(r):

um /(c): / ( r) . (6)h+A

Going back to the beginning, then, we have

F ( x * h ) - F ( x )Dcf in i t ion ol dcrNat ive

Eq. (+)

Eq. (5)

Eq. (6): f(x)'

This concludes the proof.

If the values of / are positive, the equation. t f t

+ I Iu td t : t tx )dx J"

has a nice geometric interpretation. For then the integral of / ftom a to -r is the

arca A(x) of the region between the graph of / and the .x-axis from a to .x' Imagine

covering this region from left to right by unrolling a carpet of variable width /(t)(F\g. 4.23). As the carpet rolls past r, the rate at which the floor is being covered

is /(r).

EXAMPLE 1

+ /' "", tdt : cos x

d Fdx h+o

: lim

: lim

h

= I f l t ) dt

f (c)

tr

*1,'#,,:1 l x 2

E q . ( 2 ) \ ! r r h l ( / ) : c o s I

IE q . ( l ) $ i r h / ( l ) : -

fl

EXAMPLE 2 Find dy /dx if

y : ["' "o" t dt.

J j

Solution Notice that the upper limit of integration is not x but xz ' To frrtd dy I dx

ws must therefore tleat y as the composite of

y : I c o s t d t a n d u : x 2J 1

and apply the Chain Rule:

Differential equation:

Initial condition:

so/ut on The function


Chrii Rlllc

du Srbstiture the lbrmula' E l i ) r 1 . .

dr :dx

dy du

a

du dx

l- I costdtdu J t

duc o s t . - -

ax

"o" ,'

'.

2,

2x cos xz ,

Eq. ( l ) wi lh . l t ) : cos t

Usual fbrn

EXAMPLE 3integral.

Express the solution of the following initial value problem as an

dy-- : tan .da)c

y(1) : s

F (x) -- Jt

ta\tdt

is an antiderivative of tan x. Hence the general solutiirn of the equation is

Y : J l

t an td t +C '

As always, the initial condition determines the value of C:

5 : I t ? [ J ' t d t + C r ' ( 1 ) : s

5:0+c (7 )

The solution of the initial value problem is

y : I t z n t d t + 5 .l t

How did we know where to start integrating when we constructed F(.rX Wecould have started anywhere, but the best value to start with is the initial value ofr (in this case x : 1). Then the integal will be zero when we apply the initialcondition (as it was in Eq. 7) and C will automatically be the initial value of y.

tr

The Evaluation of Definite IntegralsWe now come to the second part of the Fundamental Theorem of Calculus, the part

that describes how to evaluate definite integals.

336 Chaoter 4: lnteoration

Theorem 4The Fundamental Theorem of Calculus, Part 2If / is continuous at every point of la, rl and F is any antiderivative of /on [a, D], then

f .o f , , ,o , -F(b) -F(0) .

f bHow to Evaluate I fq)dx

J ,

l . F i nd an an r i de r i va r i l e f o f / . Anyantidedvative will do, so pick thesimplest one you can.

2. Calculate lhe number ,/^(bl Fla).

f bThis number will be

J" ft)a*.

Theorem 4 says that to evaluate the definite integral of a continuous function

f from a to D, all we need do is find an antiderivative F of / and calculate thenumber F(D) - F(a). The existence of the antiderivative is assured by the first pafiof the Fundamental Theorem.

Proof of Theorem 4 To prove Theorem 4, we use the fact that functions withidentical derivatives differ only by a constant. We aheady know one function whosederivative equals /, namely,

cr.4 = [ 114at.

Therefore, if F is any other such function, then

F ( x ) : A ( 1 ; 1 6 ( 9 )

throughout fa, Dl for some constant C. When we use Eq. (9) to calculate F(D) -

F(a), we lind that

F(.b) - F (a) - LG(.b) + Cl - lc(a) + Cl

: G(b) - G(.a)r b r "

: l f 1d t - l f1d tJ . J .

: [ ' f , ro , -o- [ ' f rDa, .J,, J.

This establishes Eq. (8) and concludes the proof. tr

EXAMPLE 4

Notation

The usual notation for the numberF(h) F (tr) is F(.r)l: when F(r) hasa single te.m, or tF(r.)l: for F(r) F(a)when F(i) has more than one term.

u ) / " " o . , r d r : s i n t ] . ,

= t - z - s i n o : 0 - 0 : 0

, , l - ' , ,rsec

r tan rdr = r. . ' ] :" , , : sec 0- *. (- i)

o l,' ()"a - i) ,, -["''.1]l:

[,r,','* 11] [rrr"'* f]: 18+ l l _ t 5 l : 4 .

. , ^

**-l-tl83


Theorem 4 explains the formulas we derived for the integrals of x and x2 in

Section 4.5. We can now see that without any restriction on the signs of a and b,

fb x21u b2 a2I xdx : r l : , -T

' J a

fb ^ x3lu b3 alI x ' dx : ; l : ; - ;

J o J . ) " J J

51 a

T1 l I=0 l ;+ ;_ l l :

t ' t . " - ) - r l -

Jo u '_" ) _2t t r t t :

L ; ; . . lo

: [o -9- IL 3 o l -o :

lJcclL l \c I r ' r 1\ : | ] r

rnt iL lcr i \ r t i \ . o l r l

4.24 fhe rcgion between the curvey : x3 x2 - 2x and the x-axis(Example 5).

4.25 The graph of the household voltagev = vnaxsin 12or t over a full cycle. ltsaverage value over a half-cycle is 2V^.,1r.Its average value over a full cycle is zero.

EXAMPLE 5 Find the area of the region between the x-axis and the graph of

f ( x ) : x 3 x 2 - 2 x . , - 1 a x a 2 -

sorution First find the zeros of /. Since

f ( x ) : x 3 - x 2 - z x : x ( x ' - x 2 ) : x ( x + 1 ) ( x - 2 ) ,

the zeros are .r :0, -1, and 2 (Fig. 4.24). The zeros partition [-1,2] into twosubintervals: [-1,0], on which / > 0 and [0,2], on which / < 0. We integrate /over each subinterval and add the absolute values of the calculated values.

f " a " l

. 10

l n t e g r a l o v ? r l - 1 . 0 1 ' . I t r t - * ' - 2 x t d x = 1 " , - - - " ' IJ - r L + J I r

Integral over [0, 2]'.

8- 3

_ul:lEnclosed area: Total enclosed ar., : f

* U

EXAMPLE 6 Household electricity

We model the voltage in our home wiring with the sine function

V : V^* sin lZjtrt,

which expresses the voltage V in volts as a function of time t in seconds. Thefunction runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz).The positive constant y.* ("vee max") is the Peak voltage.

The average value of V over a half-cycle (duration 1/120 sec; see Fig. 4.25)

IS1 ft / t2ov - ________________ l V^^, sin l2jntdr' " ' -

t l 1 1 2 0 ) - v t o

t 1 11/120: l20Y'- | ,.,^- cos l20zt I

L I ZV7{ -lo

v_-..: __U:t [_ ces 7r . cos 0]

If

: 2 ' ^ '1T


The average value of the voltage over a full cycle, as we can see from Fig. 4.25, iszero. (Also see Exercise 64.) If we measured the voltage with a standard moving-coilgalvanometer, the meter would read zero.

To measure the voltage effectively, we use an instrument that measures thesquare root of the average value of the square of the voltage, namely

v^": t\ir*.The subscript "rms" (read the letten separately) stands for "root mean square."5 fnce t he ave rage va lue o l y ' = ( ym. i ) ' S tn l 2OT l Ove raCyc le l s

t f ' u o

. . { Y , * }{ v - } . " -

( t ^ f ) _ 0 , / 0 ( y . d . ) ' s i n ' t 2 o n t d t -

2 - ( 1 0 )

(Exercise 64c), the rms voltage is

( 1 1 )

The values given for household curents and voltages are always rms values. Thus,"115 volts ac" means that the rms voltage is 115. The peak voltage,

v^ ̂ : Jzv^" - J i . t tS .9 163 vol ts ,

obtained from Eq. (11), is considerably higher tr

Exercises 4.7

Evaluating IntegralsEvaluale rhe integrals in Exercises I 26.

17 . I (8y ' + s in ) ) . / )

f - ltn. J,

(r I r)'dr

r l t , , 1 I \u' l , \z -;)o'

f ! . c z L / c,,,J, T,"2s.

J nlx dx

n. fo' 0 - zi' a"

f t _29,

Jr t\/t2 + 1 (lt

B. [ " ' ' ( + r . . ' r + { )a rJ , t t ' t ' l

,0. fn f, * r)(tz + 4) tlt

f ' / | I \

J, , , \ i - x )o '

f r ,a ,

f " l

.il , (cos .t * | cos rl) a,r

f 2 -28, | 43t t t l t lx

30. / aJ t J2t r + 8

r . lo {2,+i la"

,. l"'(u-f;)0"f 1 ^

5. Jo

tx' + \/x)dx

t327,

J, 16/5 dtt

s. lo" "in ' a,

tt. lo"/'

z. """' , d,

13. f,t,'o/n

u" e *t e de

u- 1,",,!-f!*

z. l ' .(s-i)a,n. l ' ,6 ' -z '+!a'6. [ ' , ' ' ' d,

t l , )8. / la '

rc. lo" {t + "o" i a'

lz. [t" ./u """' , d ,

u. fo"

'' I

""" u tan u du

rc. l-",'"!-;2a,

Evaluating Integrals Using SubstitutionsIn Exercises 21-34, we a substitution to find an antiderivative andthen apply the Fundamental Theorem to evaluate the integral.

do 32. [" ''

"""' 1o - 2q deJ i f t

dt

!a "4

Exercises 4.7 339

Derivatives of IntegralsFind the derivatives in Exercises 45-48 (a) by evaluating the integral

and differentiating the result and (b) by differentiating the integral

3r. ;f

sin'

33. /

sin'?

34. IJ2r /3

Area

| 3 t 'd t

I sec' r, dy

/ e \l l+ ; l

cos -4 4

4tan3

directly.

* . ; J "

cos r / t

4i. + I J;duar Jo

Find dy ldt in Exercises 49-54.

4 9 . y - I \ / l + t 2 d t

pE5r. y- , s in( t ' )dr

, r . r=[" '^ 'L.J o \ / l 1 '

s4. ) ' : I nJ n r i t -

brief reasons for your answers.

a ) r : J t : d t - 3

" ) , : J , s e c t d t + 4

-- dv I" " . l 1 I : ; ,

} r f t r : - r

5 / , ] : s e c , t , ) ( U , : 4

d t , x > O,

cos "rt

dt

b) t : I " sec td t+4

d) v:1" ' !a ' z

56 . y ' : 5gq , r , f ( - 1 ) = 4

s8. y ' - 1, )(1) : -3

dou,E

or.hIn Exercises 35-40, find the total area between the region and ther-axts,

3 5 . Y - - 1 2 - 2 r , 3 = r = 2

3 6 , y : 3 x 2 - 3 , - 2 = x = 2

3 7 . y - a 3 - 3 x 2 + 2 x , 0 = t = 2

3 8 . y : a : - 4 r , 2 = , < 2

3 9 . y : a t / t , 1 = r = 8

4 0 . Y - a 1 / t - r , - l : - ' : 8

Find the areas of the shaded regions in Exercises 41-44.

41.

, ,_ I' - t .

,: J,n2

lnitial Value ProblemsEach of the following functions solves one of the initial value prob-lems in Exercises 55-58. Which function solves which problem? Give

Express the solutions of the initial value problems in Exercises 59-62in terms of integrals.

59. l l : sec x, yl2) - 3lJx

/ l \6 0 .

' ; : \ / l + ) , ' z , ) ( 1 ) : - 2

61 . ; : / ( r ) . r f f o ) : so

62. *

= g(t), uG6): un

Applications63. Archimedes' area formula for paftbola' Archimedes (287-

212 B.c.), inventor, military engineer, physicist, and the greatest

mathematician ofclassical times in the westem world, discovered


that the area under a parabolic arch is two-thirds the base timesthe height.

a) Use an integral to find the area under the arch

J : 6 - x - t 2 , - 3 = x 5 2 .

b) Find the height of the arch.c) Show that the area is two-thirds the base D times the

height 1,.d) Sketch the parabotic arch y : h _ (4h/brx2, _b12 = )c 3

,/2, assuming that lr and D are positive. Then use calculusto find the area of the region enclosed between the arch andthe r-axis.

64. (Continuation of Exampk 6.)

a) Show by evaluating the integral in the exprcssion

1 | t/6J

** | V^^sin t20nr dt\ r / u v , - u J 0

that the ave&ge value of y = y.* sin 1202t over a fullcycle is zero.The circuit that runs your elechic stove is rated 240 voltsrms. What is the peak value of the allowable voltage?Show that

t1/60

J" ,.r- f sin2 lzoitt d.t :

meters. Use the graph to answer the following questions.reasons for your answers.

Give

(v,*)'?120

Cost from marginal co5t. The marginal cost of printing a posterwhen r poste$ have been pdnted is

d c ldx 2rE

dollars. Find (a) c(100) c(l), the cost of printing posters 2_100;(b) c(400) - c(100), the cost of printing posre$ 101_400.Revenue from marginal revenue. Suppose that a company,smarginal revenue from the manufacture and sale of egg beatersis

drE =2 -2 / t x+ r l .

where r is measured in thousands of dollars and -r in thousandsof units. How much money should the company expect ftoma production mn of * : 3 thousand egg beate$? To find out,lntegrate the marginal rcvenue from.r = 0 to.x :3.

Drawing Conclusions about Motion from Graphs67. Suppose that / is the differentiable function shown in the accom-

panying gaph and that the position ar time I (sec) of a particlemoving along a coordinate aKis is

a) What is the particle,s velocity at time t :5?b) is the acceleration of the particle at time t : 5 positive, or

negative?c) What is the particle's position at time I :3?d) At what time during the fust 9 sec does ., have its largest

value?e) Approximately when is the acceleration zero?f) When is the particle rnoving toward the odgin? away from

the origin?g) On which side of the origin does the particle lie at time

t = 9'!

68. Suppose that g is the differentiable function $aphed here andthat the posirion at time I (sec) of a particle moving along acoordinate axis is

s= I ge)d)cJ O

meters. Use the gmph to atswer the following questions. Givereasons for your arnswers.

What is the particle's velocity at r : 3?Is the acceleration at time t : 3 positive, or negative?What is the particle's position at rime r = 3?When does the particle pass through the origin?When is the acceleration zero?When is the particle moving away from the origin? towardthe origin?On which side of the origin does the particle lie z,t t : 92

b)

c)

66.

a)b)c)d)e)f)

1 2 3 4 5

, : lo ' f oa,

69.

70.

7t.

1','

Volumes f rom Section 4.4(Continuation of Section 1.4, Example j.) The approximatingsum for the volume of the solid in Example 3, Section 4.4, wasa Riemann sum for an integral. Wlat integral? Evaluate it to findthe volume.

(Continuation of Section 4.1, Exanple 1.) The approximatingsum for the volume of the sphere in Example 4, Section 4.4, was

a Riemann sum for an integral. What integral? Evaluate it to findthe volume.

(Continuation of Section 1.4, Exercise 15.) The approximatingsums for the volume of water in Exercise 15, Section 4.4, areRiemann sums for an integral. What integral? Evaluate it to findthe volume.

(Corltinuation rt Section 4.4, Ercrcise 17.) The approximatingsums for the volume of the rocket nose cone in Exercise 17,

Section 4.4, is a Riemann sum for an integral. What integral?

Evaluate it to find the volume.

Theory and Examples73, Show that if t is a positive constant, then the area between the

-t-axis and one arch of the curve ) : sin t.r is 2/t.

7.1. Find

tin '

I -!- ar., . n r r J u r a _ l

75 . suppose I i f $ )d t :x2 -2x+ l .F ind / (x ) .

76. Find f(4) 1f I; f$)dt: r cos r.r.

77. Find the linearization of

Exercises 4.7 341

80. Suppose that / has a negative derivative for all values of r andthat /(1) :0. Which of the following statements must be trueof the function

r'h (x ) : I fG ) d t2

Jo

Give reasons for your answers.

a) , is a twice-differentiable function ol r.b) h and clh/tlx are both continuous.c) The giaph of lx has a horizontal tangent at r - 1.d) I has a local maximum at ,r : l.e) l? has a local minimum at x : l.f) The graph oflr has an inflection point at.r : 1.g) The graph ol dft/dir crosses the x-axis at ir : l.

i3 Grapher Explorations81. The Fundamental Theorem. If jf is continuous, we expect

to equal /(r), as in the prool of Part 1 ol the FundamentalTheorem. For instance, if /(t) : sq5 1, 15en

ty I L'.' t,o,

(12)

The right-hand side of Eq. (12) is the difference quotient for thederivative of the sine, and we expect its limit as l? -+ 0 to becos r.

Graph cos -r for I : .n : 2z. Then, in a different color ifpossible, graph the dght-hand side of Eq. (12) as a function of .xfor h:2, l ,0.5, and 0.1. Watch how the latter curves convergeto the graph of the cosine as l, + 0.

82. Repeat Exercise 81 for /(r) :3r'?. What is

I J '+n sin (.r * l l ) - srn r

h J, tosrar =

h

I t r , t t : ) \ U] lJt

9- d tt + t

7& Find the linea zation of

sec (r - l ) dl

7f. Suppose that ./ has a positive derivative for all values of ,r andthat /(l) : 0. Which of the following statements must be tlueof the function

t 'g(x) : I f ( t) dt ' !

.,r0


a )b )c )d )e )f )

t r^ t =z- l , ' t '

8 ( . r ) : 3 + /

I r+hl i m - ,

8 is a differentiable function of ,x.g is a continuous function of ,r.The graph of g has a ho zontal tangent at "y : l.g has a local maximum at -r : 1.g has a local minimum at jr : LThe graph of I has an inflection point at r : 1.The graph of dg ldx crosses the i -axis at i - 1

Graph /(r) = 3.r2 for -1 5,x : 1. Then graph the quotient((x + h)1 - rt)/h as a function of i for ft : 1, 0.5, 0.2, and0.1- Watch how the latter curves converge to the graph of 3r2 ash --+ 0.

S CAS Explorations and ProjectsIn Exercises 83-86, let l,(r) : Ii f f,) a, for the specified function

/ and interval [a, ]1. Use a CAS to perform the following steps andanswer the questions posed.

a) Plot the functions / and F together over la,bl.b) Solve the equation F'(r.) : 0. What can you see to be true about

the graphs of / and F at points where F'(r|:) - 0? Is your obser-vation borne out by Part I of the Fundamental Theorem coupledwith information provided by the first derivative? Explain youranswer,

c) Over what intervals (approximately) is the function F increasingand decreasing? What is fiue about / over those inteNals?


d) Calculate the derivative // and plot it together with F. What canyou see to be tlue about the graph of F at points where //(r) :0? Is your observation bome out by Part I of the FundamentalTheorem? ExDlain vour answer.

83. /(;r) : r i 4x2 a3x, [0,4]

84. f(t):2xa 17 x3 + 4612 - 43x + 12,

,f (-r) : sin 2r "or l, to, zol

"f(x) :,r cos itx, 10,27'l

ln Exercises 87-90, let F(.r): I'\') f (rdt for the specified a,l,and /. Use a CAS to perfom the following steps and answer thequestions posed.

a) Find the domain of F.

b) Calculate F'(x) and detemine its zeros. For what points in irsdomain is F increasing? decreasing?

c) Calculate F" (.x) and determine irs zero. Identify the local ex-trema and the points of inflection of F.

d) Using the information from parts (a)-(c), draw a rough hald_sketch of ), : F (-r) over its domain. Then graph F (j) on yourCAS to suppoft your sketch.

87 . a : l . a i x ) - . r2 . f (x t - - , r I - x ,

8 8 . a : 0 . d { x f : . r 2 . f t ^ t = , f i - ^ ,

8 9 . a : 0 , u ( x ) : l - x , f ( t ) = x 2 - 2 x - 3

9 0 . a : 0 , u ( x ) : t - x 2 , f G ) : x 2 2 x - 3

91. calcutate * l"''"

, U, o, ̂ "ocheck your answer using a cAS.

92. calculate # 1."" ,uro, and check your answer using

a CAS.

t' ;l85.

86.

This formula first appeared in a book writtenby Isaac Barrow (1630-1677), Newton'steacher and predecessor at CambridgeUnive$ity.

Substitution in Definite Integrals

f l -Evaluate

;l_, 3xzJ# + | dx.

There are two methods for evaluating a definite integral by substitution, and theyboth work well. One is to find the corresponding indefinite integral by substitutionand use one of the resulting antiderivatives to evaluate the definite inteeral bv theFundamental Theorem. The other is to use the followins formula.

To use the formula, make the same u-substitution you would use to evaluate thecorresponding indefinite integral. Then integrate with respect to u from the valueu has at;r : a to the value I has at:r : D.

EXAMPLE 1

Substitution in Defnite Integrals

TIfi FoRMTILA

f G6)) .e f (u) du

How To USE IT

Substitute u : I Qc), du : gt (x) dx, and integrate from g(a) to g(b).

(1)l.' r s@)' (,tr) dx : I

Solution We have two choices.

4.8 Substitution in Definite Integrals 343

Method 1: Transform the integral as an indefinite integml, integrate, change backto r, and use the original ;rJimits.

f _

J 3x2Jx3- r td .x : l . c t r ! = \ ' + l . r i r - l 1 l i .

l r i r , r : r f rne r \ i lh ic \Lrccr ro ! / .

I l . r r l . rcc r r br r l

t . r ' rhr i r l . ! i r l iL is i l .Lrni .r i r l i l i r r r l \ o l ; u r . s f x l r o L r l ( r ' ,t,3r'Jr416":

- ( ( -1)3 + l )3 / , ]

I Judu

?,,n * ,3

? {,' + r1'r' *"J

, t 1 1

: (x3 + D3/2 |J l ,

? l ( ( t ) '+ t ) 'P- ) -

A ; -

: ! 12, z _ or ) l : l l zJz l = ? l '' l L . l 3

Method 2: Transform the integral and evaluate the transformed integral with thetransformed limits given by Eq. (1)

f l

I 3x2JF +rdxJ - l

1 2 - L f l r : "

1 . d r - l r r l L .:

| , J u d u \ \ ' h c n , = t , , = 1 i i + t ( l

J O \ \ ' h . r f - l . r ! - 1 L r ' l - l

: ? u 'n] ' F. \ r lLrxt . Ih. r r r \ , ! l . i i f i ( . i r r rc! ' r l .3 ln- t ) r - t t ^ 5

_ '_ l)3 2 _ n3/21

'_ 1.>-/) l _ "

-- 3 r - " r - 31 - " " J - 3 J

Which method is better hansforming the integral, integrating, and transform-ing back to use the odginal limits of integration, or evaluating the transformedintegral with transformed limits? In Example 1, the second method seems easier,but that is not always the case. As a rule, it is best to know both methods and touse whichever one seems better at the time.

Here is another example of evaluating a transformed integral with transformedlimits.

EXAMPLE 2fn/2

l''o "o' t csc' o do :

f o

l, u . (-d.u)

7 O- J,

udu

[ , ' lo_LT),

| (o) '? ( l) ' l- l ,

- , I

I : t l = c ( , 1 r r . d r l r r f r . / i /r / i , = c s r r r / l i r .

\ \ h c r r / r r . 1 . r / = c o l ( r l t - |L h c f , r - r I r i : . o t r r l : 0

I

2 l


Technofogy Visualizing Integrals with Elusive Antiderivatives Many inte-grable functions, such as the important

f Q ) : e "from probability theory, do zot have antiderivatives that can be expressed interms of elementary functions. Nevertheless, we know the antiderivative of /exists by Part 1 of the Fundamental Theorem of Calculus. Use your graphingutility to visualize the integral function

161 : [ ' " ' ' a t

What can you say about F (-r)? Where is it increasing and decreasing? Whereare its extreme values, if any? What can you say about the concavity of itsgraph?

Exercises 4.8

Evaluating Definite IntegralsEvaluate the integrals in Exercises l-24.

r . a) Jn

' / t +1dt

t t _2. a)

Jn r,/ | - 12 dr

3, ^l lo

tan x sec2 x dx

4. ") I"'

3 cos2 .r sin ; dr

5. Jo

t3 (t + t4)3 dt

f o -b)

J_,Jt + rdt

b) J_,

rr/ 1 12 dr

61 fo 6n " ".."

t d,,

O) Jr.

, cos2 .t sin x d-r

f lb )

J , t 3 ( t + t 4 ) 3 d t

b) fo t (t2 + D1n at

bt J" ,a l , ' , 'd '14 1n [ ;' ' l, ffir^t r r L -

b) I -L rlxJ-r t Jx ' )+ |

nt fo 4:a,J t . / xa 19

b) / (1 - cos 3/) sin 3/df

r 0 /

" , J " , , \z+w r)nc ,arrE /2 /

o' " / , , ,

( ' * * i ) su'

,ar

t2"a) I _S=:clz bl

Jo J4 ] 3s inz .

. I s l n L )at | ;;--:--- - clw

J -r /2 l ) + Z COS Ul '

ln l2 <in r rb ) , # d *

Jo (J +z cos rur '

Jo Jrs +2|5ta +2tdr tu.

J,

/"'" "o,

t 29 sin 2o do

f i r / 2 / A \ / A \

J" *t' (i,) *.' (i) aa

/ 5{5 - 4 cos r)t la srn t dr

t" cos aI - " '

J , V 4 + J S r n z13.

14.

6.

,|

8.

7rtil

Jo tft2 + r)tt3 dt

o|"ffi*o l"'ffi^. lJt 4x

a ) I : d xJ o J x 2 + l

ut [' -jl:a,Jo ' / xa -19

o Io"'' lt-

cos 3r) sin 3rdr

zo. lo"'"

<t - sin 2t)3/2 cos 2t dt

2r. Ia'

gt t' + 4yt + 11 2/3 (r2y2 - 2y + 4) dy

15.

t7.

18.

19.

2!5(r + rry)'

9,

10.

l l .

zz. lo' {r'+srt rzy +g)-t/2 (y2 +

t f - -,r.

J, J0 cos2 G1t2\ d0 24.

AreaFind tle total areas of the shaded regions in Exercises 25-28.2s. 26.

Exercises 4.8 345

32. a) Show that

f0 i f i i s oddr a l

I nr* ta, : I| ) I h t x t d y i i f i i r e v e n .t J u

b) Test the result in part (a) with ,(r) : sin .t and with l(.r) :cos jr, taking a : z/2 in each case.

33. If / is a continuous function, find the value of rhe integral

, t ' f (x \d tJ o f k ) + f ( a x )

by making the substitution u : a x and adding the resultingintegral to 1.

34. By using a substitution, prove that for all positive numbers rand ),

r ' I t ' II - d r : 1 - o '

J , r J t I

The Shift Property for Definite IntegralsA basic property of delinite integrals is their invariance under trans-lation, as expressed by the equation.

fb rb -c

I f@'t : I f Grc)t t t (2)J O J A T

The equation holds whenever / is integrable and defined for thenecessary values of r. For example (Fig. 4.26),

l , ' {, *D'a, : fo' " a".

4.26 The integrations in Eq. (3). The shaded regions,being congruent, have equal areas.

Use a substitution to verify Eq. (2).

For each of the following functions, graph /(jr) over [d, r] andf(r + c) over [a - c, b - cl to convince yourself that Eq. (2) isrcasonable.

a \ f ( x ) : a 2 , a : 0 , b = 1 , c = lb ) / ( r ) = s i n r , a = 0 , b = n , c : i t / 2c ) / ( r ) : 1E=4, a :4 , b :8 , c :5

4y - 4) tly

l-,t ' ' ,-' "in' (t- i) "

28.

) : t(cos,r)(sin(rr + ?rsin r))

(3)

Theory and Examples29. Suppose that F(-x) is an antideivative of /(r) - (sin r)/r,

.r > 0. Express

l' srt zx

drJt x

in terms of F.

30. Show that if / is continuous, thent l r l

I t t x t d x : | 1 t t - x 1 d ' .Jr Ja

31. Suppose that

t lI fG)a" : t'

J O

Find

loI f G)dtt

J _ 1

if (a) / is odd, (b) / is even.

J5.

36.

y : (l cos.r) sin r

2

r = 3(sinr)rl i+ cosr


The length h = (b - a)/n is called the stepsize. It is conventional to use i in thiscontext instead of A -I.

Numerical IntegrationAs we have seen, the ideal way to evaluate a definite integrd f y{r; d:r is to finda formula F(;) for one of the antiderivatives of /(;) and calculate the numberF (b) - F(a). But some antiderivatives are hard to find, and still others, like theantiderivatives of (sin.r)/x ^d Jl + f , have no elementary formulas. We do notmean merely that no one has yet succeeded in finding elementary formulas for theantiderivatives of (sin .r)/r and

"/l aF. We mean it has been proved that no such

formulas exist.Whatever the reason, when we cannot evaluate a definite integral with an an-

tiderivative, we turn to numerical methods such as the bapezoidal rule and Simpson'srule, described in this section.

The Trapezoidal RuleWhen we cannot find a workable antiderivative for a function / that we have to

integmte, we partition the interval of integration, replace / by a closely fittingpolynomial on each subinterval, integrate the polynomials, and add the results to

approximate the integral of /. The higher the degrees of the polynomials for a given

partition, the better the results. For a given degree, the finer the paftition, the better

the results, until we reach limits imposed by round-off and truncation errors.The polynomials do not need to be of high degree to be effective. Even line

segments (graphs of polynomials of degree 1) give good approximations if we use

enough of them. To see why, suppose we partition the domain [a, b] of / into z

subintervals of length A,r : ft: (b - a)/n and join the corresponding points on

the curve with line segments (Fig. 4.27). The vertical lines from the ends of the

segments to the partition points create a collection of trapezoids that approximate

the region between the curve and the .x-axis. We add the areas of the hapezoids,

counting area above the x-axis as positive and area below the axis as negative:

. f _: ( y o + ) r ) , + ^ { y r + ) 2 ) f t - L r ;I , r ,_ , * , , , )n + ! ry ,_ t j - y hI

2 2

: : ( ) o | 2v t+2yz+ +2y , t +y , ))

where

yo: f@) , y r : fG) , yn- l : f l xn t ) . y^ : f (b ) .

The trapezoidal rule says: Use 7 to estimate the integral of f from a to b.

:, ()* * r, + v2 + " + r,- ' + )t,)

The Tiapezoidal Rule

To approximate Il f @) ar, """

hT:

l (Yo + 2tr I2Yz + " '12Y" t ' lY")

(for n subintewals of length ft : (b - a) /n and y1 : /(r1)).

( 1 )

4.27 The lrapezoida I rule approximatesshort stretches of the curve y = f(x) withline segments. To estimate the integral off from a to b, we add the "signed" areasof the trapezoids made by joining theends of the segments to the x-axis.

Table 4.5

P4

B

., -2

Pl

PJ

4

P

I1 6

36P.

I

A

b16

0 l r ! q 2

4.28 The trapezoidal approximation ofthe area under the graph of y = x2 fromx : 1 to x = 2 is a slioht overestimate.

4.9 Numerical Integration 347

EXAMPLE 1 Use the traoezoidal rule with r : 4 to estimate

r2| ,2 d* .

J t

Compare the estimate with the exact value of the integal.

Solution To find the trapezoidal approximation, we divide the interval of inte-gration into four subinteryals of equal length and list the values of y = .r2 at theendpoints and partition points (see Table 4.6). We then evaluate Eq. (1) with n :4

ard h : l /4 :

_ o f ,4 1_7 <

The exact value of the integral is1 2 - 3 1 2 e r 1I . 2 s - - ^ l - : - l - 1 - r r

J r J - l r J J J

The approximation is a slight overestimate. Each trapezoid contains slightly morethan the corresponding strip under the curve (Fig. 4.28). tr

Controlling the Error in the Trapezoidal ApproximationPictures suggest that the magnitude of the error

E7 : I f(x) d.x -r (2)

in the trapezoidal approximation will decrease as the step size ft decreases, becausethe trapezoids fit the curve better as their number increases. A theorem from ad-vanced calculus assures us that this will be the case if / has a continuous secondderivative.

r : f, {yo + ztt + 2y2 + 2y3 + ya:)

: * (' *' (*-) *' (*t; *' (#). ̂) :#

! = x 2

I25-t o

36t 649==l o

4

I54

o4'1

nz


4-29 Graph of the integrand inExample 3.

Although theory tells us there will always be a smallest safe value of M, in

practice we can hardly ever find it. Instead, we find the best value we can and go

on from there to estimate lEzl. This may seem sloppy, but it works' To make |trlsmall for a given M, we make h small.

EXAMPLE 2 Find an upper bound for error in the approximation found in

Example 1 for the value ofr 2

J, " a''

Sorution We first find an upper bound M for the magnitude of the second derivative

of f (x) -- xz on the interual \ < x <2' Since /"("r) :2for allx' we may safely

take M : Z.With b - a : I ar'd h : l/4, Eq. (3) gives

lnr l :bln'zu: , ! r( I) 'o,This is precisely what we find when we subtract T : 7 5 132 from li xz dx : 7 13'since li/3 -75/321:l-11961. Here our estimate gave the enor's magnitude

exdcfl), tut this is excePtional. tl

Find an upper bound for the error incuned in estimating

[ " t " i n r d ,Jo

The Error Estimate for the Trapezoidal Rule

If /" is continuous and M is any upper bound for the values of l/"1

la, bl, then

Erl < b-!

1,2 Y.1 2

(3)

EXAMPLE 3

with the bapezoidal rule with n : 10 steps (Fig. a.29).

S o r u f i o n W i t h a : 0 , b : n , w f i h : ( b - 4 ) / n : z / 1 0 , E q . ( 3 ) g i v e s

tr,t. b-1 h'M = l- ( L\ ' m : J- u.r ! . r r _ - n , " . , 1 2 \ t 0 / . - 1 2 0 0

The number M can be any upper bound for the magnitude of the second derivative

of f (x): a sin x on [0, z]. A routine calculation gives

f " ( x ) : 2 c o s r - x s i n * '

so

lf"(x)l -- l2 cos.r - -r sin.rl

S 2lcos xl * I .x l ls in x l

=2 . ! +T ' L : 2+1 r .

196

Tiansle i l rcqual i ty :

l t + h l ' - a \ + h

cos.r l rnd s in . r nevele x c c e d l . a D d 0 : | : 7

4.9 Numerical lntegration 349

We can safely take M : 2 + z. Therefore,

tt3 trtl2 + tt\ Roundcd uP tol E , l < ' " - M : ' "

- ' - ' < 0 . 1 3 3 . L ' e \ r r ;' - " - 1200' 1200

The absolute error is no greater than 0.133.For greater accuracy, we would not try to improve M but would take more

steps. With n : 100 steps, for example, h: r ll00 and

Mult ip ly both s ic les by l0rrr .

Sc]uilr€ roots ol bolh sides

/l is positive.

Rounded up. to be srfe

_ l r . r | - ,

F"r < l l ( - \ u' ' - 12 \ loo/ " ' '12o,ooo tr

EXAMPLE 4 As we will see in Chapter 6, the value of ln 2 can be calculatedf. - rl'- i-r--..]

nz: [ ' !4 ' 'J t x

How many subintervals (steps) should be used in the trapezoidal rule to approximatethe integral with an enor of magnitude less than l0 4?

solution 'fo determine n, the number of subintervals, we use Eq. (3) with

b - a : 2 - 1 : 1 , h : b - o : ! ,n n

I u t : ! - ( x - t t=2x - r :+ .a)('

Then

l - l b -a . , | " , , l / l \ ' l 2 llE r l 5 - , 'max l I ( x ) = ; l - lmax l ; 1 .| | t 2 | t z \ n t ' x - l

where max refers to the interval [, 2].This is one of the rare cases where we cal find the exact value of max]/"1.

Ot [], 2], y :2/x3 decreases steadily from a maximum of y = / 1s a minimumof y: | /4. Therefore,

I / t \ 2 1l E r l < - l : l . 2 : - .t t \ n / o r '

The enor's absolute value will therefore be less than l0 4 if,|

-: < lo-4,on'

104 )--- < n''o

100o < W l '

v o

100/ ; <n '

V O

40.83 < n.


Simpson's one-third ruleThe idea of using the formula

hA : - ( ) o + 4 y l + y 2 )

J

to estimate the area under a curve is knownas Simpson's one-third rule. But the rule wasin use long before Thomas Simpson(1720-1761) was bom. It is another ofhistory's beautiful quirks that one of theablest mathematicians of eighteenth-centuryEngland is remembered not for his successfultexts and his contributions to mathematicalanalysis but for a rule that was never his, thathe never laid claim to, and lhat bears hisname only because he happened to mentionit in a book he wrote.

4.3f, Simpson's rule approximates shortstretches of curve with Darabolic arcs,

The first integer beyond 40.83 is n : 41. With n : 41 subintervals we can guaranteecalculating ln 2 with an error of magnitude less than 10-4. Any larger n willwork, too.

Simpson's RuleSimpson's rule for approximaiing [: f\)dx is based onquadratic polynomials instead of linear poll,nomials. Wewith parabolic arcs instead of line segments (Fig. 4.30).

l

approximating / withapproximate the graph

! = A x 2 + B r c + C

Parabolic arc Parabolic arc

The integral of the quadratic polynomial y : Axz + B"r -l- C in Fig. 4.31 fromx : - h t o x : h r s

r h hI (e*t + n, i C) dx : - lyo + 4y, + yr) (4)

J-n J

(Appendix 4). Simpson's rule follows ftom partitionhg [a, b] into an even numberof subintervals of equal length lr, applying Eq. (4) to successive interval pairs, andadding the results.

Sirnpson's Ruler b " , ,ro approxrmare J, J \x I ax. rse

hS : ; ( y o { 4 y 1 - 1 2 y 2 * 4 } : - | . . . + 2 y " z + 4 y " 1 + y " ) . ( 5 )

-1

The y's are the values of / at the partition points

x o : a , x l : q J h , x . 2 : a * 2 h , , . . , r n t : c I t ( n - 1 ) h , x " : b ,

The number z is even, and fu: (S - a)/n.

4.3f By integrating from -h to h, wefind the shaded area to be

Error Control for Simpson's RuleThe magnitude of the Simpson's rule error,

" : l " ' f (x)dx- s '

Parabolic arc

l tro*or',nr,1

-

4.9 Numerical Integration 35'l

decreases with the step size, as we would expect from our experience with the

ffapezoidal rule. The inequality for controlling the Simpson's rule error, however,assumes Jf to have a continuous fourth dedvative instead of merely a continuoussecond derivative. The formula, once again from advanced calculus, is this:

As with thr trapezoidal rule, we can almost never find the smallest possiblevalue of M. We just find the best value we can and go on from there to estimateE s l .

EXAMPLE 5 Use Simpson's rule with n : 4 to approximatef 1

| 5xo dx.J O

What estimate does Eq. (7) give for the error in the approximation?

so/ution Again we have chosen an integral whose exact value we can calculatedirectly:

lo' ,ro o*: rt], : t

To find the Simpson approximation, we partition the interyal of integration

into four subintervals and evaluate f (x) :5xa at the partition points (Table 4.7).We then evaluate Eq. (5) with n :4 afi, h : 1/4:

hS : , (yo *4r r + 2y2 + 4h + j4 )

| / / 5 \ / R O \ / 4 0 5 \ \: : lo 4 l : - l -2 l= l - 4 l_ l+s l^ r .oozoo.

12 \ \ 256 / \ 2s6 l \ 256 l /

To estimate the error, we first find an upper bound M for the magnitude ofthe fourth derivative of /(,x):5.x4 on the interval 0 <x: 1. Since the fourth

dedvative has the constant value /(a) (x) = 120, we may safely take M :120.

With & - c : I and h: 1/4, Eq. (7) gives

h n I r l \ a IlE5 l < " - - - - ! ! 7 'a 74 : . : . | . I r rz0r : - - - < 0 .0026t . D' " - r 80 r80 \4 / 384

Which Rule Gives Better Results?The answer lies in the error-control fomulas

h n h - nlETl < "-----1 pt v. lE, < "1;- h1 M.

The Error Estimate for Simpson's Rule

If /(a) is continuous and M is any upper bound for the values of l/(a)l[.r, b], then

It'1 < b---! 7'o 1'4 '

180(7)

v =s l

05

_2)h

80z)o405

5

0I4

24

34

I


Trapezoidal vs. Simpson

If Simpson's rule is more accurate. whybother with the trapezoidal rule? There aretwo reasons. First, the tlapezoidal rule lsuseful in a number of specific applicationsbecJU.e ir lead. ro much :. i rnplcr erp|e.. ionr.Second, the trapezoidal rule is the basis forRhomberg integration, one oi the mostsatisfactory machine methods when highprecision is required.

rgnored

Honzontal spacing : 20 fl

: . : : The swamp in Example 6 .

The M's of course mean different things, the first being an upper bound on lf, ' land the second an upper bound on ./(a) . But there is more. The factor (b - q)/lS0in the Simpson formula is one-fifteenth of the factor. (b, a)/12 in the trapezoidalfbrmula. More important still, the Simpson formula has an la while the trapezoidalformula has only an /r2. If h is one-tenth, then ,2 is one-hundreclth but /ri is onlyone ten-thousandth. If both M's are l, for example, and b - a = l, then, withh = t / 1 0 .

lE, /-L\ ' . , : '\ r0 l -

1200 '

: 1 l1,800,000 1500 1200

l c . I <

For roughly the same amount of computational effort, we get better accuracy withSimpson's rule-at least in this case.

The /r2 versus /za is the key. If lr is less than I, rhen ha can be significantlysmaller than /r2. On the other hand, if /r equals l, there is no difference between iland /ra. If i? is grcater than l, the value of hr may be significantly larger than thevalue of /r:. In the latter two cases, the error-control formulas offer. Iittle help. Wehave to go back to the geometry of the curve l : /(,r) to see whether trapezoidsor parabolas, if either, are going to give the results we want.

Working with Numerical DataThe next example shows how we can use Simpson's rule to estimate the integralof a tulction from values measured in the ]aboratory or in the field even when wehave no formula fof the lunction. We can use the trapezoidal rule the same way.

EXAMPLE 6 A town wants to drain and fill a small polluted swamp (Fig.4.32).The swamp averages 5 ft deep. About how many cubic yards of dirt will ir take tofill the area after the swamp is drained'l

Solution To calculate the volume of the swamp. we estimate the surf'ace area andmultiply by 5. To estimate the area, we use Simpson's rule with r:20 ft and thel's equal to the distances measured across the swamp, as shown in Fig.4.32.

h5 - . { ) b . 4 r , r 2 l : . 4 r ' . 2 r ; * 4 1 . r 1 " r3

20= - ( 146+ ,188+ 152+216+80+ 120+ 13 ) : 3196 .

The volume is about (8100)(5) : 40,500 ftr or 1500 yd3.

Round-off ErrorsAllhough decreasing the step size /r reducas the enor in the Simpson and trapezoidalapproximations in theory, it may fail to do so in practice. When /r is very small.say /r : 10 5, the round off errors in the arithmetic requireil to evaluate S ancl ?may accumulate to such an extent that the error tbrmulas no longcr describe what isgoing on. Shrinking /l below a certain size can actually make things worse. Whilethis will not be an issue in the present book, yon should consult a text on numencalanalysis for alternative methods if you are having problens with round-off.

while

I

t 2

I / l \ r_ t tr 8 0 \ 1 0 /

Exercises 4.9 353

Exercises 4.9

Estimating lntegralsThe instructions for the integnls in Exercises I 10 have two parts'

one for the trapezoidal rule and one ior Simpson s rule.

I. Using the trapezoidal rule

Estimate the integral with n :4 steps and use Eq (3) tofind an upper bound for lEil.Evaluate the integral directly, and use Eq. (2) to find Er .CALCULATOR Use the formula (lErl/true value) x 100to express lErl as a percentage of the integral's true value.

l l . U t ing Sintpson's rule

Estimate the integral with n:4 steps and use Eq. (7) tofind an upper bound for E5l.Evaluate the integral directly, and use Eq. (6) to flnd Erl.CALCULAToR Use the formula (l-Es /true value) x 100to express Its as a percentage ofthe integral's fiue value.

a)

m b )f f i c )

a)

b )f f i c )

0.3750.751 . 1 2 5I .5r .8752.252.6253.0

p. [' =!-aeJ o J 1 6 I 0 :

l ' i2 3 cos t

J - , , 1zy i , n ;o '

l 2rs.

J, x ttx

17 . L ( x '

+ r ) dx

etJte + et

0.00.093340.184290.2'70'7 50 . 3 5 1 1 20.12.1430.490260.584660.6

(3 cos lY(2 + sin t)']

l .

7.

sin t dt

sin rt dt

t ll l .

Jn x t / l -x2dx

z. l,' {2, - D a,

+. f 'e l ,a,

s. l ' , t ' + ua,

s' / 'fna'

[' r,t"

J tG"+ r )dxr 2

J, tt '+la,

1.570801 17810

-0.785400.3927000.392'700.785401 . 1 7 8 1 01.57080

0.00 .99 r38r.269061.059610.150.488210.289460.134290

l r"t;I,'1",

(csc'? y)y'cot y-

9.

10.

0.785400.883570.981751.0'7992I . 1 7 8 1 01.2162'71.3'7 445L412621.57080

2.0l .5 r606L r82370.939980.7 54020.601450..1636.10 .316880

ln Exercises 1l-14. r-rse the tabulated values of the integrand to es-

timate the integral with (a) the fapezoidal rule and (b) Simpson's

rule with r : 8 steps. Round your answers to 5 decimal places. Then

(c) f,nd the integral's exact value and the approximatjon eror Er or

Er, as appropriate, lrom Eqs. (2) and (6)

00.1250.250.3750.50.6250.750.8751 . 0

*rfl17

0.00.124020.242060.34'7630.433010.487890.,196080.423610

The Minimum Number of SubintervalsIn Exercises 15-26, use Eqs. (3) and (7), as appropriafe, to estimate the

minimum number of subintervals needed to approximate the integrals

with an eror of magnitude less than 10-r by (a) the trapezoidal rule

and (b) Simpson's rule. (The integrals in Exercises 15 22 are the

intesrals from Exercises 1-8.)

rc. l,' lz, - D a,

ra. ;fn {''

- t) a'

354 chapter 4: lntegration

o. 1,, ' l t '+\at

\o '

J"ia"

sin (.v * l) dx

n. I",r.r

J 2

,n. I"'A, 1,,

t;1,,t;

21.

.t1

) q

(t1 + l) (tt

I-a ls(s 1 ) '

I ,

v j + l

cos ( ir + r) dt

Applications27. As the fish-and-game warden of your township, you are respon-

sible for stocking fte town pond with fish before fishing season

The average depth of the pond is 20 ft. You plan to start the

season with one f,sh per 1000 fC You intend to have at leasi

25Ea of rhe opening day's lish population left at the end of the

season. What is the maximum number of licenses the town can

sell if the average seasonal catch is 20 fish per license?

" 52O fL \

800ft \

l000ft \

1140 fr \

1160 fr

\ 1110 f r

\ aoo tt--t\

Oft -,,"

Venical sPacing : 200 ft

ffi x. clrcurnroR The design of a new airplane requires a gasoline

tank of constant cross-section area in each wing. A scale drawing

of a cross section is shown here. The tank must hold 5000 lb of

gasoline, which has a density ol42 1b/ft3. Estimate the length of

the tank.

!2 )3 )4 r5)6 )1

] , 0 = 1 .5 f t , ) l : l 6 f t , ' ' 2= l - 8 f l , i 3=1 .9 f t '

la: 2.0 ft ' 1r:1=z.1ft Horizontal spacing = 1lf

E zg. calcuraroR A vehicle's aerodynamic drag is determined in

part by its cross-section area and, all other things being equal'

engineers trJ to make this area as small as possible Use Simp-

son's rule to estimate the cross-section area of James Worden's

solar-powered Solectda car at MIT (Fig. 4.33)

4.33 Solectria cars are produced by Selectron Corp,Arlington, MA (Exercise 29).

The accompanying table shows time-to-speed data for a 1994

Ford Mustang Cobra accelerating ftom rest to 130 mph. How far

had the Mustang traveled by the time it reached this speed'l

Speed change Seconds

Zero to 30 mph40 mph50 mph60 mph70 mph80 mph90 mph

100 mphl l0 mph120 mph130 mph

2.23.2

5.97 .8

10.212.116.020.626.2

Sonrce'. Car and Driyer, Aprll

Theory and Examples

31. Polynomials of low degree. The magnitude of the eror in thetrapezoidal approximation of f f(x)dx is

b alErl -

" :: h2lf ' (c) '

where c is some point (usually unidentified) in (a, D). If / is alinear function ofr, then /" (c) : 0, 5q Er : 0 and T gives theexact value of the integral for any value of ft. This is no surprise,really, for if / is linear, the line segments approximating the graphof / fit the graph exactly. The sulprise comes with Simpson'srule. The magnitude of the error in Simpson's rule is

t E , t - b . ^ :

h ' . I ' ' ' t c t .180

where once again c lies in (a, b). If / is a polynomial of degreeIess than 4, then /(o) :0 no matter what c is, so Es:0 and Sgives the integral's exact value-even if we use only two steps.As a case in point, use Simpson's rule with n :2 to estimate

t2I x j dx .

Jo

Comparc your answer with the integral's exact value.

E 32. Usabte values of the sine-integral function. The sine-integralfunction,

si (,v) : / si l /

dt, sinc intcgral oi r"Jo t

is one of the many functions in engineering whose formulascannot be simplified. There is no elementary formula for theantiderivative of (sin t)/t. The values of Si (.r), however, arereadily estimated by numerical integration.

Although the notation does not show it explicitly, the func-tion being integrated is

I sin r| - , t + 0

f t t t : l tI I . r = 0 .

the continuous extension of (sin t)/t to the interval [0, r]. Thefunction has dedvatives of all orders at every point of its domain.Its graph is smooth (Fig. 4.34) and you can expect good resultsfrom Simpson's rule.

4.34 The continuous extension of y = Gin t)/t. Thesine-integral function Si(x) is the subject of Exercise 32.

a) Use the fact that l/(4) | : 1 on 10, r, /21to give an upper

Exercises 4.9 355

bound for the error that will occur if

^ . l f t \ [ "12 srn ts i ( - , : I d '

. J A

is estimated by Simpson's rule with n : 4.Estimate Si (n /2) by Simpson's rule with n : 4.Express the erlor bound you found in (a) as a percentage of thevalue you found in (b).

(Continuaion of Example J.) The efior bounds in Eqs. (3) and(7) are "worst case" estimates, and the trapezoidal and Simpsonrules arc often more accurate than the bounds suggest. The trape-zoidal rule estimate of

r s l n r

b)c)

E r:.

Jn r s , 'xdx

in Example 3 is a case inpoint.

a) Use the trapezoidalrule with ', : l0 toapproximate the valueof the integral. Thetable to the right givesthe necessary )-values.

b) Find the magnitude of the difference between z, the inte-gral's value, and your approximation in (a). You will findthe difference to be considerably less than the upper boundof 0.133 calculated with n : 10 in Example 3.

ll cl GRAPHER The upper bound of0.133 for Er in Example3 could have been improved somewhat by having a betterbound for

lf" (x)l : lZ cos .r -:r sin .vl

on [0, /]. The upper bound we used was 2 + E. Gta,ph/" over [0, z] and use TRACE or ZOOM to improve thisupper bound.

Use the improved upper bound as M in Eq. (3) tomake an improved estimate of lErl. Notice that the trape-zoidal rule approximation in (a) is also better than thisimproved estimate would suggest.

E gl. CnlCUUfOn (Continuation of Exercise 33)

tZ a) GRAPHER Show that the fourth derivative of /(.r):' t s l n t l s

"fL*) (.r) : 4 cos r +.r sin r.

Use TRACE or ZOOM to find an upper bound M for thevalues of /(u) on [0,2].

b) Use the value of M from (a) together with Eq. (7) to obtainan upper bound for the magnitude of the error in estimatingthe value of

0(0.1)/(0.2)tt(0.3)n(0.4)n(0.s)z(0.6)n(0.'7)n(0.8)r(0.9)z

1f

00.097080.369320.'762481.195 r31.57080|.'792',70t. '7'79t21.4',7'12'10.8'73',720

/ r s l n - rd t

: i ( r )=f+d,

with Simpson's rule with n : 10 steps.


c) Use the data in the table in Exercise 33 to estimate

l{ x sln x dx with Simpson's rule with z : l0 steps.d) To 6 decimal places, find the magnitude of the difference

between your estimate in (c) and the integral's true value,z. You will find the error estimate obtained in (b) to bequite good.

You are planning to use Simpson's iule to estimate the values of theintegrals in Exercises 35 and 36. Before proceeding, you tum to Eq.(7) to detemine the step size l? needed to assurc the accuracy youwant. What happens? Can this be avoided by using the tapezoidalrule and Eq. (3) instead? Give reasons for your answers.

35. I x3/2 dt

36. I x5/2 tlx

l. Can a function have morc than one antidedvative? If so, how arelhe antiderival ives related? Explain.

2. What is an indefinite inte$al? How do you evaluate one? Whatgeneral formulas do you know for evaluating indefinite integrals?

3. How can you sometimes use a tdgonometric identity to transforman unfamiliar intregal into one you know how to evaluate?

4. How can you sometimes solve a differential equation of the folm

dy ldx = f (x)'l

5. What is an idtial value problem? How do you solve one? Givean example.

6. If you know the acceleration of a body moving along a coordinaieline as a function of time, what more do you need to know tofind the body's position function? Give an example.

7. How do you sketch the solutions of a differential equation d)/dt- / (.,r) when you do not know an antiderivative of /? Howwould you sketch the solution of an initial vaheproblemdy /dx :

"f(x), y(ro) = )0 under these circumstances?

E. How can you sometimes evaluate indefinite integrals by substi-tution? Give examples.

9. How can you sometimes estimate quantities like distance traveled,

arca, volume, and average value with finite sums? Why might you

want to do so?

10. What is sigma notation? what advantage does it offer? Giveexamples.

H Numerical IntegratorAs we mentioned at the beginning of the section, the definite in-tegrals of many continuous functions cannot be evaluated with theFundamental Theorem of Calculus because their antiderivatives lackelementary formulas. Numerical integration offers a practical way toestimate the values ofthese so-call€d 'lorele mentary integrals,If yourcalculator or computer has a numerical integration routine, tly it onthe integrals in Exercises 37-40.

l t - . A n o n e l e m e n t a D i n t e s f r l t h a t37.

I \ / l + x4 dx c , ' 'e up ,n Nc\ron. \ r .scrrch

The integr.rl fron L).efcisc ll. Io

r . r 2 < i n , . , . , 1 . i \ r , r r ' t . t , ' , , " :3 8 . ,

- : d x l . , , . : , . , ' , , : ,lo x

illi:lJ:;[- nu,rbe,,ike ,0 "

ra/2tn.

Jn sin (r') dr An intesnl rs\oci l lcd $i lh lhe

di t l iact io. of l lsht

?a/2

n. | +ov4 - ctst "o*

ta, l.n' length orthe ell ip' ie

J o ' f r / 2 5 ) + ( " r 9 ) : I

11. What rules are available for calculating with sigma notation?

12. What is a fuemann sum? Why might you want to consider sucha sum?

13. What is the norm of a paflition of a closed interval?

14. what is the deflnite integral of a function / over a closed intervalfa, &l? When can you be sure it exists?

15. What is the relation between delinite integrals and area? Describesome other interpretations of dennite integrals.

16. Describe the rules for working with definite integrals (Table 4.5).Give examples.

17. What is the average value of an integrable function over a closedintefr'al? Must the function assume its average value? Explain.

18. Wtat does a function's average value have to do with samplinga function's values?

19. What is the Fundamental Theorem of Calculus? Why is it soimportant? Illusfiate each part of the theorem with an example.

20. How does the Fundamental Theorem provide a solution to theinitial value problem dy /dx : f(r), y(xfi = yo, when / is con-tinuous?

21. How does the method of substitution work for definite integrals?Give examples.

22. How is integation by substitution related to the Chain Rule?

23. You are collaborating to prcduce a short "how-to" manual for

CrnprpR Quesrrors ro GUIDE Youn Rsvtew

-

numerical integration, and you are writing about the trapezoidalrule. (a) What would you say about the rule itself and how touse it? how to achieve accuncy? (b) What would you say if youwere writing about Simpson's rule instead?

Practice Exercises 357

24. How would you compare the relative merits of Simpson's ruleand the trapezoidal rule?

Finite Sums and Estimates1. The accompanying figure shows the graph of the velocity (frsec)

of a model rocket for the first 8 sec after launch. The rocketaccelerated straight up for the first 2 sec and then coasted toreach its maximum height at I : 8 scc.

3. Suppose that L a*: -2 and t ,r :25. Find rhe value off t= l &=t

20 204, Suppose that t 4r :0 and L bk:'t. Find rhe values of

t= l t= l

20b ) L @ * + b i

2nd ) L @ * - 2 )

Definite IntegralsIn Exercises 5 8, express each limit as a definite integral. Then eval-uate the integral to find the value of the limit. In each case, p is apartition of the given interval and the numbers c1 are chosen fromthe subintervals of P.

5 . l i . ^ i {2c , - l ) 2A. r1 . uhere P rs a pan i r ion o t l t .5 lP - o r l l

o l iE. -E

cr(cr2 l)r/3 Arr, where P is a panition of [1,3]

:- / /cr \\t . , l l o |

( - " ( z l , ax , . where P is c pan i t ion o f | - r .0 l

t. ,}its|r -4

f.t" cr)(cos cr) Arr, where P is a panition of [0, r/2]

s. rf l:,3 f (x) dx : rz. I:2 f(x) dx: 6, and fz sG) dx :2,find the values of the following.

t a n , l oL+ b t f r h , - 3aa r* : t r k : lt o l o / 5 \

F,ro ' *u , l ) d , F \ ; -^ )

a)

150

100

50

6

b

I

5

d

;6 '

II

a)

c)

c)

c)

2I)

), 3ak

2 A / l t A , \

0 2 4 6 8

Time after launch (sec)

a) Assumitrg that the rocket was launched from ground level,about how high did it go? (This is the rocket in Section 2.3,Exercise 19, but you do not need to do Exercise 19 to dothe exercise here.)

b) Sketch a graph of the rocket's height aboveground as afunction of time for 0 : I : 8.

2. a) The accompanying figure shows the velocity (m/sec) of abody moving along the J-axis during the time interval fromt:0 to l : l0 sec. About how far did the body rravelduring those l0 sec?

b) Sketch a graph of .r as a function of / for0=r = l0 as-suming ,t (0) : 0.

a)f 2

J , f{r) a' b)

l 2

J, ek)dt d)

/ ' / "f(") + 8(') \l . \ s )" '

1,,f,

J Q ) d x

( lt g@)) dx


r0. tt I: f (x) dr : ft , I: '7

cG) tLx =r' and fr sQ)dx:2,flndthe values of the following

f2a)

Jn gG) dx

l 2b)

J, eQ) dr

c) [ f{i a"

. l

dt Jn

rt lota*

e\ Jn GG)-3f (t))dx

AreaIn Exercises l1-14, find the total area of the region between the graph

of / and the t-axis.

l l . f ( x ) : 1 2 - 4 x + 3 , 0 : r = 3

12. fG) : | - (x2 /4) , -2 - r 33

13. f ( r ) :5 - 5x2 /3 , - l : . j r = 8

1 4 . f ( x ) = t - J i , 0 < x = 4

lnitial Value ProblemsSolve the initial value problems in Exercises 15-18

/ ! | t 2 t Ir s . ; : ; . ) ( 1 ) : - l

. 2

r c .? : ( ,+1) . y r r r -1d \ \ j r . /

17.

18,

19.

i - = t S . , . 4 + i , . r ' r l t : 8 . / { l r = 0dt' \ / t

1 ' : c o s l : / ' { 0 1 : r ' t o r : 0 . r t 0 l : - I

r . lSho$ lhaf y : t2 - I - ,1t solves the init ial value problem

' J t r

, . t2 . I: 1 . : 2 ; : y ' r 1 t : 1 . y r l ) = 1 .t1x . I '

Show that J: I;0+2Jsect)dr solves the init ial value

problem

) 2 , ,

; i .,Gc x tan x: v tot : 3. Y(o) : o

Express the solutions of the initial value problems in Exercises 21

and 22 in terms of integrals.dv s in x

2 1 . * : ; . ) ( r ) : - r

,2. *:Jt ; , ] ; , yeD:2

Evaluating lndefinite IntegralsEvaluate the integrals in Exercises 23-44.

3s. / "* Jle "otnT.eae

:e. /."c i,^, 'r*

37. I sin" ; dx zt. | "os'!ra"

s, I

zl"o" x) t/2 sln x dx a0. /

ttan t) 3/2 sec2 x dt

n. I fze + | + 2 cos(20 + rD de

42. [ ( -L 1z 'ec ' )1ze - r r )deI \ J 2 0 - n /

*. I(, ?)(,.?)" 44' Iej!-]d,

Evaluating Definite IntegralsEvaluate the integrals in Exercises 45-70.

s. l_'-r{tf

-a"+Ta' ne. ;['

ttr' - 72s2 + s) ds

zt. [ { , ' +s, -Ta,

*' l ('a. !) "" Ii+* 'zs. I zeJz - * ae

st. | "g

+ ro;'ro 4'

r. / '*fta'

. 2 ^q. I l , tu

49. I _Jt t../t

t1 36dxsl ' / (zr+rt

53. / ' . , , , r11 - r r / r ) r t , d ,Jrft

55. Jn

sin' 5r dr

tr. J"

sec" e de

to . | ( r , ' - ' r * , )o '

" I(+-i)"t 6r2drzt ' J u'-{u

zo. lffiaen. I Q-,)'/'a"

34. I

csc2 ns ds

r2148.

J, x-at3 d:t

so. l,'Lffyd"

54.

56.

58.

. t /2

Jo ,3ll + 9"4) '/' d,

tn /4

J" cos')\4t -

;) dt

| .. csc2 x dt

sz. [' -=!!:la l/ ti - 5r))

Average Values71. Find the average value of f(x) : mt + b

a) over [ -1 ,1 ]b) over [-t, t]

72. Find the average value of

a) , - vjx over LU. Jlb) y - .,/6a ovet 10, al

73. Let f be a function that is differentiable on [a, ,]. h Chapter Iwe defined the average rate of change of / over la, bl to be

f(b) - f(a)b - u

and the instantaneous rate of change of / at r to be /'(,t). Inthis chapter we defined the average value of a function. For thenew definition of average to be consistent with the old one, weshould have

f t b ) f l a lt :

a \e rage va lue o f / ' on ld . b l

Is this the case? Give reasons for your answer.

74. Is it tlue that the average value of an integrable function overan interval of length 2 is half the function's integral over theinteNal? Give reasons for your answer

Numerical IntegrationE zs. cnlcuufoR According to the enor-bound formula for Simp-

son's rule, how many subintervals should you use to be sure ofestimating the value of

[3 1b t3 : I _ dx

J t x

by Simpson's rule with an error of no more than 10-a in abso-lute value? (Remember that for Simpson's rule, the number ofsubintervals has to be even.)

Practice Exercises 359

A brief calculation shows that if 0 = r : 1, then the secondderivative of f (t): .,/T +7 fies between 0 and 8. Based onthis, about how many subdivisions would you need to estimatethe integral of / from 0 to I with an error no greater than 10 3

in absolute value using the trapezoidal rule?

A direct calculation shows that

I 2 s i n ' . t d t - n .Ja

How close do you "oln"

to this value by using the trapezoidalrule with n - 6? Simpson's rule with n - 6? Try them and findout.

You are planning to use Simpson's rule to estimate the value ofthe integral

f 2I f(x)dx

J r

with an error magnitude less than 10-5. You have determinedthat l/(al(,t) :3 throughout the interval of integration. Howmany subintervals should you use to assure the required accuracy?(Remember that for Simpson's rule the number has to be even.)

ffi zg. calcurnroR Compure rhe average value of rhe temperaturefunction

for a 365-day year. This is one way to estimate the annual meanair temperature in Fairbanks, Alaska. The National Weather Ser-vice's official figure, a numerical average of the daily normalmean air temperatures for the yeat is 25.7'F, which is slightlyhigher than the average value of /(:r). Figure 2.42 shows why.

B 80. Specltc heat of a gas. Specific heat C. is the amount of heatrequired fo raise the temperature of a given mass of gas with constant volume by l'C, measured in units of cavdeg-mole (caloriesper degree gram molecule). The specilic heat of oxygen dependson its temperature f and satisfies the formula

c, :8.2 '1 + lO 5 (26T l .81Ta.

Find the average value of C. for 20' : f : 675'C and the tem-peraftre at which it is attained.

Theory and Examples81. Is it tme that every function ): /(r) that is differentiable on

[l', r] is itself the derivative of some function on [a, r]? Givereasons for your answer.

82. Suppose lhat Ft-r r is an anl iderivl ive of J lr r . / f x ' . Erpre.. f . , , / t la* in lerms o[ F and gire a reason for )ouranswet

83. Finddy/dx if y : 11 .s1i7 at.explain the main steps in yourcalculation.

84. Finddy/dt i t t :" [ t" . , ( l /( l t2))dt. Explain rhe main srepsin your calculation.

se. J,

cotz idx

6t. lo ,,. """

,t^ rd"

63. lo'/'

51rin ,1'r' "o"

* 4, 64.

*. 1",''ru

sina 3.r cos 3r d.r

ce. lo"'' "o" ' (i) '" (i) ,'

67. ["/'2!:2:d, 68./o ./1 + 3 sin' ,

es. ["t' Y:ao 70.Ja ./2 sec 0

eo. J,

tanz \ae

62.t"::,t,

cos "f

1""îl:

'- asa zaot z rJz

2x sin(l x2) dr77.

78.sec2 l

- r t t( 1+ 7 t an x ) l / 3

- - "

dt

ftxt : 37,r" (f i , ,. - to',) * tt


85. Anewparkinglof. To meet the demand ior parking, your town

has allocated the area shown here. As the town engineer, you

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

be built for $11,000. The cost to clear the land will be $0.10 a

square foot, and the lot will cost $2.00 a square foot to pave

Can the job be done for $11,000?

0 f t

Vertical sPacing : 15 ft

86, Skydivers A and B are in a helicopter hovering at 6400 ft.

Skydiver A jumps and descends for 4 sec before opening her

parachute. The helicopter then climbs to 7000 ft and hovers

there. Forty-five seconds after A leaves the aircraft, B jumps

and descends for 13 sec betbre opening her parachute. Both sky

divers descend at 16 ft/sec with palachute open Assume that

the skydivers fall freely (no effective air resistance) before their

parachutes open.

a) At what altitude does A s parachute open?

b) At what altitude does B's parachute open?

c) Which sk)diver lands first?

Average Daily InventoryAvemge value is used in economics to study such things as average

daily inventory. If 1(t) is the number of radios, tires, shoes, or what-

ever product a finn has on hand on day t (we call ,f an inventory

function), the average value ol 1 over a time period [0' ?] is called

the film's average daily inventory for the period.

1 l lAverage daily inventory : av (1) :

1 Jo I tt) dt.

If ft is the dollar cost ofholding one item per day, the product av(1) ' h

is the average daily holding cost for the period.

87. As a wholesaler, Tracey Burr Disffibutors receives a shipment

of 1200 cases of chocolate bars every 30 days. TBD sells the

chocolate to retailers at a steady rate, and / days after a shipment

arrives, its inventory of cases on hand is 1(t) : 1200 - 401' 0 :

t < 30. What is TBD's average daily inventory for the 30-day

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

holding one case is 3t a dayl

88. Rich Wholesale Foods, a manufacturer of cookies, stores its cases

of cookies in an air-conditioned warehouse for shipment every 14

days. Rich t es to keep 600 cases on reserve to meet occasional

peaks in demand, so a typical l4-day inventory function is 1(1) :

600 + 600r, 0 : t : 14. The daily holding cost for each case is

4e per day. Find Rich's average daily inventory and average daily

holding cost.

89, Solon Container receives 450 drums of plastic pellets every 30

days. The inventory function (drums on hand as a function of

days) is I(t) : 450 - t2 /2. Find the average daily inventory If

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

daily holding cost.

90. Mitchell Mailorder receives a shipment of 600 cases of athletic

socks every 60 days. The number of cases on hand t days after

the shipment arives is 1(t) :600 2o'l4i Find the average

daily inventory. lf the holding cost for one case is 1120 pet day'

find the average daily holding cost.

AppIttoNeI- ExBncrsBs-TrmoRv, ExAMeLES, AppLIcRrroNs

Theory and Examplesr 1 f

1 . a) I f l ,

1 f (x)dx:1, does /

/ (x)d ' : 1?

r r f t -b) If

/ f( jr)d.r :4 and /(;r) : 0, does

;1, ^/ f(r)dx:

/; ..


f 22. Suppose

/ ,g(x) dx = 2.f{ t) dx = +.

1,5 f(,) . tr :3, ls

t r / ' r / r , . r +s ( - r ) ) : 9

a) J, f \ )dx: 3

Which, if any, of the following statements are truel

c) /(,r) < s(t) on the interval 2 = -r :5

Show thatI t '

L f ( ) s i t a ( r t ) d ta Jn

solves the initial value problen

d _ \ , ^ a \ J- ; - - - a " J - l ( t \ . , = 0 r n d ) = 0 \ r h e n r = 0 .tlx' talx

(Hint: sin (ax at): slr ax cos.r1 cos d-r sin 4t.)

4. Suppose r and v are relared by the equarion

f ' I

Jo ./1 + 4f

Show that d'?y/dx'?is proportional to J and lind the constant ofproporrionality.

5. Find /(4) if

a ) I f G) . l t : . r : cosr i r .

f t \ \ ) ^b ) I t ' d t : x c o s i r x .

6, f ind fIn /2t from rhe lollouing inlonnarion.

i) / is positive and continuous.ii) The area under the curve r/: f(.r) from.r :0 to r : a is

a 2 a rt + t s t n a + t c o s a .

7. The area of the region in the ,ry-plane enclosed by the r-axis,the curye ) : /(;),,f (.r) > 0, and the lines .x : I and x - b isequal to Jb\ 1- J- for all b > LFind/(r).

8. Prove that

I" ' (1, ' ru,")au: ln r{u){t ,4a,.(llirt. Express the integral on the righFhand side as the differenceof two integrals. Then show that both sides of the equation havethe same derivative with respect to n.)

9. Find the equation for the curve in the ir]l-plane that passes throughthe point (1, - 1) if its slope at r is always 3*2 1 2.

10. You sling a shovelful of dift up from the boftom of a hole withan initial velocity of 32 ft/sec. The din must rise 17 ft above rherelease point to clear the edge of the hole. Is that enough speedto get the dirt out, or had you better duck?

Bounded Piecewise Continuous FunctionsAlthough we are mainly inte.ested in continous functions, many func-tions in applications are piecewise continuous. All bounded piecewisecontinuous functions are integrable (as are many unbounded iunctions,as we will see in Chapter 7). Bounded on an interval 1 means thatfor some finite constant M, .f 6) = M for all rv in 1. Piecewisecontinuous on 1 means that 1 can be partitioned into open or halfopen subintervals on which / is continuous. To integrate a boundedpiecewise continuous function that has a continuous extension to each

Addit ional Exercises-Theory, Examples, Applications 361

closed subinterval of the pa ition, we integrate the individual extensions and add the results. The integral of the function

f 1 - , r . l < x < 0/ L , t = l x ' . 0 1 x . 2'

I r . r - ^ . i ,(Fig. 4.35) over [-1,3] is

rJ t0 r2 11I y r . t ' t t : - | t t - t t r t . , . | , : d x + | 1 . t 1 t t x

J J J" Jt

| . t l n f r ' l t I l 'f

- r l ,*131.*L '1,,3 8 , 1 92 3 6

4.35 Piecewise continuous functions l ike this areintegrated piece by piece.

The Fundamentai Theorem applies to bounded piecewise contin-uous functions with the restricrion rhar (d/dr) J: fO dr is expectedto equal /(,{) only at values of )l: at which / is continuous. There isa similar restriction on Leibniz's rule below

Graph the functions in Exercises 1l-16 and integrate them overtheir domains.

x 2 1 3 , - 8 : i < o- 4 , 0 S - r : 3 .

J=, -4 :.r < 0r ) 4 , 0 : , r : 3

1 , 0 : r < 1s i n r r . l = t < 2

^ / T - , o : r < lQ z - 6 ) 1 / 3 , l 3 z s 2

11. / ( . r ) :

r2. f (x) :

13. ,(r) : {

14. h(z) :

362 ChaPter 4: Integration

l r - 2 < . { < - l

rs . F( r ) : I i , , . - r : " . r[ 2 . l = t 5 2

{ , - l < r < 0

16 . r ( r ) : l f - r : . 0 ! r . ll r l . : r < 2

17. Find the average value of the function graphed in Fig 4 36(a)'

18, Find the average vaiue of the function graphed in Fig 4 36(b)'

0 1 2 3(b)

4.36 The graphs for Exercises'17 and 18'

Leibniz's RuleIn applications, we sometimes encounter functions like

I t 2u '. f @ : I ( r + t ) d t a n d g ( f l - l - s i n t ' d t '

J"'" ' J tt

defined by integrals that have variable upper limits of integration and

variable lower limits of integration at the same time The first integral

can be evaluated directly but the second cannot We may find the

derivative oi either integral, however, by a fomula called Leibniz's

rule:

Figure 4.37 gives a geometric interpretation of Leibniz's rule lt

shows a carpet of variable width /(/) that is being rolled up at the

left at the s;me time t as it is being unrolled at the right (In this

interpretation time is 'I, not r.) At time,r' the floor js covered from

,r(j) to i](-jr). The (Ite dul.lx at which the carpet is being rotled up

need not be the same as the rate do/d.r at which the carpet is being

"f(vG))

4.37 Roll ing and unroll ing a carpet: a geometric

interpretation of Leibniz's rule:

dA .' . "dv A"

; = f (v(xl)"d'x - f(u(x))

;;

Iaid down. At any given time r, the area covered by carpet is

l ' ( r )A ( r ) : / f ( t ) d t .

At what rate is the covered area changing? At the instant 'r' A(t) is

increasjng by the width /(u(-{)) of the unrolling carpet times the rate

dl)ldx at \ahich the carpet is being unrolled. That is' A(rc) is being

increased at the mte

du/ (u r r ) i - '

At the same time, A is being decreased at the rate

duf (u(x) ) - '

ax

the width at the end that is being rolled up times the rate du/ln The

net rate of change in A is

dA ^ du ' l u- , ; - / ' ' ' ' " o ^ -

t ( u l x \ ) *

which is precisely Leibniz's rule.

To piove the rule, let F be an antiderivative of / on [d' D] Then

l ! ( r )

I f f , la , - F(r (x) ) F(a(x)) . (1)J , ( J )

Differentiating both sides of this equation with respect to i gives the

equation we want:

) r ' " n f I" I 1 s 1 4 1 - l . l F 1 u t . r 1 1 l ( r r { r ) ) ld ^ J , . , , ' ' - ' / , 1 I

th du: r '@QD

fi F'(u(r))

dr ch'rin R'Lr'

du du= / r u t r t t . f f u l t \ ) ,- a:l x ax

You will see another way to derive the rule in Chapter 12' Additional

Exercise 3.

Leibniz's Rule

If / is continuous on La, bl' and u(:r) and u('d) are differen

tiable functions of r whose values lie in [r7, ']' then

dn duf I t d r = f t n ( Y ) ) i J \ u \ x t ) i- t

Use Leibniz's rule toExercises 19 21.

t r l19. f (x) : I .d t

J t / r I

lind the derivatives of the functions in

Addit ional Exercises-Theory, Examples, Applications 363

Therefore, when r is large, S, will be close to 2/3 and we will have

R o o t s u m : J t + J z + + " G : 5 , . r 2 / z ; . ? ; / 2 .- 3

The lbllowing fable shows how good the approximation can be.

Root sum \213)nr' ' Relative error,,.

r2Jtg ( y ) : I s i n t ' d t

JJr

Use Leibniz's rule to find the value ofl that maximizes the valueof the integral

[ ^ ' " , t , , t 0 , .J.

- 11 r l i tJ , : , / + , / ' - + +\ n n \ n n

J I + , / 2 + . . . + J nn3l2

4.38 The relation of the circumscribed rectangles to theintegral /.,4dx leads to an estimate of the sum,4+^, t r+" t r+. . .+ ,6 .

1 050

1001000

22.468239.046',7 | .46

2t,09'7

21.082 1.386/22.468x6Ea235.10 1.4%666.6'1 0.1Vo

21,082 0.0'71a

Problems like this arise in the mathematical theory of politi-cal elections. See "The Entry Problem in a Political Race," bySteven J. Brams and Philip D. Straflln, Jr, in Political Equilib-

um, Peter Ordeshook and Kenneth Shepfle, Editors, Kluwer-N i j ho f f . Bo l ron . 1q82 . pp . l 8 l - 195 .

Approximating Finite Sums with IntegralsIn many applications of calculus, integrals are used to approximatefinite sums-the rcverse of the usual procedure of using finite sumsto approximate integrals. Here is an example.

EXAMPLE 7 Estimate the sum of the souare roots of the firstr r po . i l i ve in reger . . J t I r t . . .+ Jn .

Soruflon See Fig.4.38. The integral

is the limit of the sums

23. Evaluate

, . l ' 2 ' ' . ] 5 - . . . * n ,

by showing that the limit is

t t -I x 'dx

Jo

and evaluating the integral.

24. See Exercise 23. EvaluateI

l i m , r l r - L 2 ' + l ' - . . . | , 1 r ) .

25. Let f(x) be a continuous function. Express

t T / t \ " r r r lt im - l l l l t l l : l + . . . - r r r r, 6 n L \ n , / \ n , /

' ' n ' l

as a delinite integral.

26. Use the result of Exercise 25 to evaluate

a) Iim

b) l im

\e *o*a* . . . * rn ) .

* 1 1 l ' 5 + 2 ' 5 + 3 , t + . . . + 2 , 5 . y ,

tr

[ ' Jro^ = ]"" ' l ' : lJ o . l o r

,tr, :

d)

t / n ) n r -c , l i m - | s i n " | . i n 1 + " i n 1 , . . - s i n a I .

n ' N r \ n n n n /

What can be said about the following limits?

b)

l i m ; 1 l l t * 2 r 5 - 1 - 3 r : + . . . + n t t )

l

I i m + ( 1 1 5 + 2 r 5 + 3 t 5 + . . . + n t t )

Show that the area A, of an n-sided regular polygon in acircle of radius r is

, nr- ,fA r :

n s rn _ .

Find the limit of A, as z + oo. Is this answer consistentwith what you know about the arca of a circle?

e )

27. ̂ )


28, The effor function. The error function'

2 f 'ertQ): I I e-t' tu,

4n Jo

important in probability and in the theories of heat flow and signal

transmission, must be evaluated numerically because there is no

elementary expression for the antiderivative of e-r .

Use Simpson's rule with rr : l0 to estimate erf(l),In [0, U,

t ) 4 I

l " ( " " \ < rz .l d t . \ ' l

Give an upper bound for the magnitude ol the error of theestimate in (a).

a)b)

-

Date post:	19-Jul-2015
Category:	Education
Upload:	angelovedamla
View:	210 times
Download:	1 times

Chapter 4 Thomas 9 Ed F 2008

Education