61. Symmetry about a point A function / is symmetric about a point(c, d) ifwhenever (c - x,d - !) is on the graph, then so is(c + x, d + y). Functions that are symmetric about a point (c, d)are easily integrated on an interval with midpoint c.

a. Show that if / is symmetric about (c, d) and a > 0, then

Illirf-l dx : 2af(c) : 2ad.

b. Graph the function f(r) : sin2 x on the interval l},nl2landshow that the function is symmetric about the point @la,ll2).

c. Using only the graph ofl (and no integrarion), show that

rTl Z 1

Jo sln-xdx: T.

(See the Guided Projects for more on symmetry in integrals.)

Bounds on an integral Suppose / is continuous on [a, b] withf" (") > 0 on the interval. It can be shown that

5.5 Substitution Rule 357

a. Assuming / is nonnegative on [a, b), draw a figure to illustratethe geometric meaning of these inequalities. Discuss yourconclusions.

b. Divide these inequalities by (b - a) and interpret the resultinginequalities in terms of the average value of / on Ia, b].

QUICK CHECK ANSWERS

I. f(-x)Sex) : f(x)g(x);therefore, /g is even.2, The average value is the constant; the average value is 0.3. The average value is zero on the interval; by the MeanValue Theorem for Integrals, "f(")

: 0 at some point on theinterval.

62.

@ - 4f(+) =

5.5 Substitution Rule

l"' ,u, o*

We assume C is an arbitrary constant

without stating so each time it appears.

Given just about any differentiable function, with enough know-how and persistence, youcan compute its derivative. But the same cannot be said of antiderivatives. Many func-tions, even relatively simple ones, do not have antiderivatives that can be expressed interms of familiar functions. Examples are sin (x2), (sin x)f x, and x". At the moment, thenumber of functions for which we can find antiderivatives is extremely limited. The im-mediate goal of this section is to enlarge the family of functions for which we can findantiderivatives. This campaign resumes in Chapter 7, where additional integration meth-ods are developed.

Indefinite IntegralsOne way to find new antiderivative rules is to start with familiar derivative rules and workbackward. When applied to the Chain Rule, this strategy lea<ls to the Substitution Rule. Afew examples illustrate the technique.

EXAMPLE 1 Antideriyatives by trial and error Find !cos2x dx.

soLUTloN The closest familiar indefinite integral related to this oroblem is

I

/cosxdx:sinx*C.which is true because

d, (sinx + C): cos,r.

ax

.i358 Cnap'rrn 5 . INrecRATrol.r l

Therefore, we might incorrectly conclude that the indefinite integral of cos 2x issin 2x * C. However, by the Chain Rule,

L(rrn2x + C\: 2cos 2x * cos2x.dx'

Note that sin 2x fails to be an antiderivative of cos 2-r by a multiplicative factor of 2.

A small adjustment corrects this problem. Let's try )sin 2x:

dlr \ r

at*\Zsin2x ): r.2cos2x: cos2x.

It works! So we have

| "orz, d. ::sin2x -r c.

Related Exercises 9-12 L

The trial-and-error approach of Example I does not work for complicated integrals.To develop a systematic method, consider a composite function f(S(r)), where F is an

antiderivative of f ; that is, F' = /. Using the Chain Rule to differentiate the compositefunction F(S(")), we find that

d r ^r ' ,tt ft/ / \\ t/ \ ./ / \\ r/ \

*lrE{x))) : F' (sQDs' Q) : f?@))s'(').

-,-/'(8(.{))

This equation says that f(g(*)) is an antiderivative of /(g(x))S'("), which is written

fI f?QDs'(x)dx: r(g(x)) + c, (1)

is any antiderivati"ve of /.

the composite function /(g(t)) in equation (1), we identify the "inner function" as

u : gQ), which implies that du : g'(x) d;r. Making this identification, the integral in

I you can cail the new variable anything equation (1) is written

you want because it isjust another [ ", -, -rr.

f

variableorintegration.rypicary,aisa J t39rg: J f fu) du: F(u) + c'

standard choice for the new variable. J(u) du

we see that rhe inregral .[ f G@Dd(x) dx withrespecr to x is replaced by a new inregral

[71uSa" with respect to the new variable a. In other words, we have substituted the newvariable z for the old variable .r. Of course, if the new integral with respect to a is no eas-

ier to find than the original integral, then the change of variables has not helped. The Sub-stitution Rule requires some practice until certain patterns become familiar.

THEOREM 5.6 Substitution Rule for Indefinite IntegralsLet u : g(x), where g' is continuous on an interval, and let / be continuous on

the corresponding range of g. On that interval,

I rrrrav'rr) ctx : l t,a r,

>--.

5.5 Substitution Rule 359

PROCEDURE Substitution Rule (Change of Variables)

1.. Given an indefinite integral involving a composite function f (S}D,identify aninner function rz : g(,r) such that a constant multiple of g'(x) (equivalently,a'(x)) appears in the integrand.

2. Substitute u : gQ) anddu: g'(x) dxinthe inregral.

3. Evaluate the new indefinite integral with respect to u.

4. Write the result in terms of x using u : CQ).

Disclaimer: Not all integrals yield to the Substitution Rule.

EXAMPLE 2 Perfect substitutions Use the Substitution Rule to find the followinsindefinite integrals. Check your work by differentiating.

^. lt{r. + r)3 dx u. f rcn'o'a,

SOLUTION

a. Weidentify u:2x * l astheinnerfunctionof thecompositefunction (2x + l)3.Therefore, we choose the new variable u : 2x * 1, which implies that du : 2 dx.Notice that du : 2 dx appears as a factor in the integrand. The change of variableslooks like this:

tllQ* + t)3'zax: lu3du,-t

t,u au

,o_ , +c4

(2x + l)a: 4 +C Replaceuby2x+1.

Notice that the final step uses ,, : 2x * | to return to the original variable.

b. The composite function elo' has the inner function u : l}x, which implies thatdu : l0 dx.The change ofvariables appears as

fr| ,to* l0 d, : I e'du Subsrirure u : l}x.du : l0 dx.t;-;; r

:e'+C Antiderivative

ero, + C. Replace r.r by 10,r.

Inchecking, we see nut!("'u' + C) : ,'o*. l0 : l0er0".ax

Related Exercises I 3-16 I

iqtlcr cHEcK i Find a new variable u so rhar [ +x31xa + 5)to dx : .[ut\ du.<,

EXAMPLE 3 Introducing a constant Find the following indefinite integrals.

a. j *a1*t + 6)e dx D. ./ Cos- x sln -r d"x

Substitute u : 2x i 7,du : 2 dx.

Antiderivative

It is a good idea to check.the result.

By the Chain Rule, we have

a lQr+\)a 1

dxL 4 )I )t.

360 Csnrren 5 . ItrrecRA'IloN

T

,ii

11

ilil

i

,1'

SOLUTION

a. The inner function of the composite function (xs + O)e is x5 + 6 and its derivative

5xa also appears in the integrand (up to a multiplicative factor). Therefore, we use the

substitution , : x5 * 6, which implies that du : 5xa dx or xo dx : ll5 du. By the

Substitution Rule,

IGt*6)sxadx.:"t \_/ I.-ue ad"

[,, 1 Subsritute,:xs+6.I It -du I

J ) du:Sxadx=xadx: -au

::lro" l,161o":

:i #.t Antiderivative

rl

,lr{io*

I

= :(xs + 6)10 + C. Reptace uby xs + 6.50'

b. The integrand can be written as (cos l)3 sin r. The inner function in the composition is

cos x, which suggests the substitution u : cos x. Note thatdu : -sin x dx otsin x dx : -du. The change of variables appea$ as

/.or'*sinxdx:-[r'auJq J J

,3 du

Substitute r, : cos x,du : -sinxdx.

4u-, +C Antiderivative

+

cos4 x- o, + C' RePlace r'r bY cos x'

Related Exercises 17-28 I

Oijrir-tn-iCitT In Example 3a, explain why the same substitution would not work as wellforthe integral I r'Qt + 6)e dx."<

Sometimes the choice for a u-substitution is not so obvious or more than one

rz-substitution works. The following example illustrates both of these points.

fxEXAMPLE 4 Variations on the substitution method Find /

-

dx.J Yx+l

SOLUTION

Substitution 1 The composite function V" +-l suggests the new variable u : x -t l.You might doubt whether this choice will work because du : dx and the x in the numer-

ator ofthe integrand is unaccounted for. But let's proceed. Letting u : x I l, we have

x:u-l,du:dx,andf - f ,' - |

| -!:dr : | -jdu Substitute u: x + l,du : dx..l Vx+l .l Yu

f / r \: IIyA- -|au

Rewriteintegrand../ \ Vu/

f , r,. -1 bt ,

I lu"' - u ""1 du. Fractional powers \ _

. 5.5 Substitution Rule

We integrate each term individually and then return to the original variable x:

361

.*_,

It' ., l

I ,',' - u-u2) d, : ?u3t2 - zut/2 + c Antiderivatives

+C Replaceubyx*1.

Factor out (x + l)t/z un|simplify.

: iA + r)3t2 - 2e + r)uz

:lA+r)1t2e-z)+c

Substitution2 Anotherpossible substitution is u : fx + L Now u2 : x * 1.

x : u2 - 1, and dx : 2udu.Makinp.these substitutions leads to

In Substitution 2, you could also use the

fact that

I

2\G + t'which implies

Idtt::dt.

2\/x -t 1

Substitute u: t/x+ l,x = u2 - l.

fZ | (u'z - l) du Simplify the integrand.

J

/,,3 \'' :2(+-rl+c Antiderivatives\3 / .

2, . .\?/t -\r/l: a(" + 1)3/2 - 2Q + l)tl' + C Replace ubyt/x + t.

4

:'a(* + t)IlzQ - 2) + C. Factorout(x * 1)tl'und,simplify.

The same indefinite integral is found using either substitution. Related Exercises 29-341

Definite IntegralsThe Substitution Rule is also used for definite integrals; in fact, there are two ways toproceed.

' You may use the Substitution Rule to find an antiderivative F, and then use the Fun-damental Theorem to evaluate F (b) - F (a).

'Alternatively, once you have changed variables from x to lr, you may also change thelimits of integration and complete the integration with respect to u. Specihcally, ifu : g(x),the lowerlimitx : a is replacedby u : S@)andrhe upperlimitx : bis replaced by u : g(b).

The second option tends to be more efficient, and we use it whenever possible. A fewexamples illustrate this idea.

THEOREM 5.7 Substitution Rule for Definite IntegralsLetu : g(r), where g' is continuous on [a, bl, and let f be continuous on therange of g. Then

l"' reat)g' (x) itx = Irrl),'

ro 0,.

[-Lo*: ["-tz,,a,J Yx+l J u

EXAMPLE 5 Definite integrals Evaluare rhe

f'dxf4Ya. | - b. l-.La.,Jo(x+3)r Jox'+l

following integrals.

f o/2

", Jo sinaxcosxdx

\

'a;

362 CHep,ren5 . INrrcnerroN

SOLUTION

) When the integrand has the form a. Let the new variable be u : x * 3 and then du : dx.Because we have changedf(ar + b), the substitution u : ax t b the variable of integration from x to rz, the limits of integration must also be expressedis often effective. in terrns of a. In this case.

r : 0impliesu : 0 + 3:3 Lowerlimit

. x:2impliesu:2+3:5 Upperlimit

The entire integration is carried out as follows:

f2 dx fsI u-3 du Substitute u : x + 3,du -- dx.J0lxtJf J3

,') ta

-Tl^ Fundamental rheorem

| --, ^-1\ 8: -t (t - - 3') : n5 Simplify.

b. Notice that a multiple of the derivative of the denominator appears in the numerator;

therefore, we let r,r : x2 + 1. Then du : 2x dx, or x a, : [rar.Changing limits

r: 0impliesrr: 0 * 1 : 1 Lowerlimit

x: 4implies u : 42 + l : 17 Upperlimit

Changing variables, we have

fo x , r[t'J, j *1d,

: ;.1, u-t du substirute u : x2 + r,du : 2x dx.

1 111: ; (ln lul ) | Fundamental rheoremzlt

':l:;(ln 17 - ln l) Simplify.

z

I: tlnl7 = 1.417. ln I : 0

c. Let u : sin x, which implie s that du : cos -{ dx. Thelower limit of integrationbecomes a : 0 and the upper limit becomes a : l. Changing variables, we have

fr/2fl/ sinaxcosxdx : l radu u:sin x.du:cosxzlxJo Jo

/r.rs\ l' I: [ ; | | : ;. FundamentalTheorem\ 5 / lo 5

Related Exercises 3544 I

The Substitution Rule enables us to find two standard integrals that appear frequently inpractice, ,f sin2 r dx and ,[ cosz x d.r. These integrals are handled using the identities

.) 1-cos2x , 1 l1_cos2xsrn-x: 2 ano cos-,r: 2 .

EXAMPLE 6 Integralof cos20 Evaluare ff/2"o",ea0.SOLUTION Working with the indefinite integral first, we use the identity for cos2 g:

I ro,'oao : ft + ?os2e do : : I o, * ! [ rorzeae.J J z 2J 2.1

i

. 5.5 Substitution Rule

The change ofvariables u : 20 is now used for the second integral, and we have

363

| ,o,'o do : Til*il*

)|*"'o'rrft', I cosu du

I \ l'lu-sin20 ll4 /lor \ /^ r.no):2.4stn,r)-\u+4srr / 4

See Exercise 86 for a generalization ofExample 6. Trigonometric integrals

involving powers of sin x and cos,r are

explored in greater detail in Section 7.2.

u:20,du:2d0

0

2

Using the Fundamental rheorem of calculus, the value of the dehnite intesral isnn/2 / ^

I ,ortoae: (a- +Jo \2

Ir:t l\4

@@

f("):(x+l)'2f(,):t/r,+r

f(x) : ""*'f(*) : cos (2"r + 5)

t$-*-:tl-:4ry;gljFIGURE 5.57

1. On which derivative rule is the Substitution Rule based?

2. Explain why the Substitution Rule is referred to as a change ofvariables.

5. When using a change of variables u : SQ) to evaluate the defi-

nite integral J'! flSGDS'(llx) dr, how are the limits of integrationtransfbrmed?

_r\ .:;

SECTION 5.5 EXERCISESReview Questions

QUICK cHEcK 3l changes of variables occur frequently in mathematics. For example,suppose you want to solve the equation xa - l3x2 + 36 : 0. Ifyou use the substitutionL,t. : x2, what is the new equation that must be solved for a? what are the roots of theoriginal equation? ,,

6. Ifthe change ofvariables u : x2 - 4 is used to evaluate

the definite integral Ii tt-ldx, what are rhe new limits ofintegration?

7. Find /cos2,r z/x.

3. The composite function /(g(x)) consists of an inner function 6. _ g. What identity is needed to find I si* x dx?and an outer function /. When doing a change of variables, whichfunction is often a likely choice for a new variable a? Basic Skills

4. Find a suitable substitution for evaluating ./tan -r sec2 x dx, and, 9-12. Tiial and error Find an antiderivative ofthe followingexplain your choice. Junctions by trial and erron Check your answer by dffirentiation.

Related Exercises 45-50 I

Geometry of SubstitutionThe Substitution Rule may be interpreted graphicaily. To keepmatters simple, consider the integral JrZ(2x + l) dx. The graphof the integrandl : 2(2x + 1) on the inrerval [0,2] is shownin Figure -5.57, along with the region R whose area is given bythe integral. The change of variables u:2x i I,du:2dx,"(0)

: l,andu(2) : 5 leads to the new integral

f2 fsl2(2x*l)dx: ludu.JO JI

Figure 5.57 also shows the graph of the new integrand y : u onthe interval [1,5] and the region R' whose area is given by the newintegral. You can check that the areas of R and R' are equal. Ananalogous interpretation may be given to more complicated inte-grands and substitutions.

I+ ; sin 20 + C. Evaluate integrals; u : 20.

y=2(2x+l)

.:364 Crnprpn5 . INTEGRATToN

13-16. Substitution given Use the given substitution to find rhe

follon-ing indefinite integrals. Check your answer by dffirentiation.

(lY l2x(x2 + l\4dx. u: x2 + |.lf

U4 / 8*cos (4x2 + 31 dx, u:4x2 + 3 1:

,]

G) /sin3xcosxdx.u-_ sinx ' I:i.l

4\r-AA l6*+ t)\/3x2 * xdx, u:3i + x-.1

17-28. Indefinite integrals (Jse a change ofvariables to Jind the

following indefinite integrals. Check your work by dffirentiation.

/r,r* - t)es dx $. f xe" dx

r6/i + t\420. J- _--dx

z'' f*ho'u.

lsinloocos0d0

26. I fsin xro dx

@ Iu'+x)to(2x+r)dx

@ | ru' + t6)6 dx

@ Ii+- ""t'

27. l$u-3x2;11x5-x)dx.lTt

28. I ^- dx (Hinr: Let u : x 2S,l ^-L

29-34. Yariations on the substitution method Find the following

f2/) 3 r

43. | --:+: 44. l" -:1J!:0,J2l(s\/3) xY25x, - | Jo \/u3 + 3a + 4

45-50. Integrals with sin2x and cos2x Evaluate thefollowingintegrals.

+s. f_'

,o{ , a*

ez. | ,i^'(o . i) *

o. faoorin'zo

ao

statements are true and give an explanation or counterexample.Assume that J, f ' , and f " are continuous functions for all realnumbers.

fru.

.f r{it'{*)a. ::ue)), + cI'ru

.l tfl*)l't'e)dx : fi{t{*))*, + c, n * -rc. I'in2rrtx:2[rinrd*J,I

f (x2 + I)roa. I e'z + t)edx: to + c

fbe.

J" f'(x)f"(x)ax = f,(b) - f,(") . ;. .

52-64. Additional integrals [Jse a change ofvariables to evaluate thefollowing integrals.

sz. f ,rr4wran4wdw tr. lsec2loxdx

54. /

(sin5x * 3sinrx - sinx)cosxdx

f sin' * ax

fr''o .o""' o'

f * "or'(x2)

dx

35-44. Definite integrals (Jse a change ofvariables to evaluate thefol lowin g def ni te i n tegrals.

integrals.

zs. [ -.:: o,.l Yx-4

3r. [-:: o.J Vr + 4f_$.

Jx'Vzx+tax

?), 1)-

30. I L-a,.J (y + r)"

32. ["'.-'-'.Or.l e^-fe'f_

sa. .l

(x + l)V3x + 2dx

@ /" ":H,"40. f - -_z-0,Jo t/s -r p, '

"n/442. |

' "!or.lo cos- x

ss. ["""':o*./ cot, x

*. Isinxsecsrdx

f' .r/, - *, o.

l,'1;:,0,r2

ur. .ln

x3 \/t6 - xa dx

f56. | 1x3/2 + t)s tc dx

f )fI o'^58. | '*

dr.J ( TI

1,"'t#o-

f**.ffu-r)("'-2x)trctx

@ /'r'10 - x2) dx @ ['ffi,"@,f"t't2ocosodo

I xze"+r dxJl

rn/2/ cos.r ,l-axJrl4 Srn' x

61.

65-68. Areas of regions Find the area of the following regions.

65. The region bounded by the graph of/(,r) : x sin (x2) and thex-axis between.r : 0and x : t/i

=v>i-'

:

:

fl'rtner Explorations

\$, Explain why or why not Determine whether the following

'l

41.

The region bounded by the graph of f(0) : cos 0 sin 0 and thed-axis between 0 : 0 and 0 -- rrl2

The region bounded by the graph of/(x) : (x - 4)a and ther-axisbetween x = 2andx = 6

The region bounded by the graph of /(x) : J+:and theYx"-9

"r-axisbetween x = 4andx:5Morphing parabolas The family of parabolas y = (11

") - ,'/ o3,

where a > 0, has the property that for "r > 0, the x-intercept is(a, 0) and the y-intercept is \0, I I a ). Let A(a) be the area of theregion in the first quadrant bounded by the parabola and thex-axis. Find A(a) and determine whether it is an increasing,decreasing, or constant function of a.

Applications*,.70. *."o*c motion An obiect moves in one dimension with a veloc-

ity in m/s given by o(r) : 8 cos (rrtl6).

a. Graph the velocity function.b. As will be discussed in Chapter 6, the position of the object is

given by s(r) : fio(y) dy for t > 0. Find the position func-tion for all I > 0.

c. What is the period of the motion-that is, starting at any point,how long does it take the object to return to that position?

Population models The population of a culture of bacteria has a

growth rate given by p'(t) : tl! Uu",.ria per hour, forIt + | I

I > 0, where r ) I is a real number. In Chapter 6 it will beshown that the increase in the population over the time interval

[0, r] is given by Iip'(t)ds. (Note that the growth rate decreasesin time. reflecting competition for space and fbod.)

a. Using the population model with r : 2,what is the increase inthe population over the time interval 0 < t < 4?

b. Using the population model with r : 3, what is the increase inthe population over the time interval 0 < t < 6?

c. Let AP be the increase in the population over a fixed timeinterval [0, Z]. For fixed Z, does AP increase or decreasewith the parameter r? Explain.

d. A lab technician measures an increase in the population of 350bacteria over the 10-hr period [0, l0]. Estimate the value ofrthat best fits this data point.

e. Looking ahead: Work with the population modet using r : 3(part (b)) and find the increase in population over the time in-terval [0, Z] for any T > 0.If the culture is allowed to growindefinitely (7 + m), does the bacteria population rncreasewithout bound? Or does it approach a finite limit?

Consider the right triangle with verrices (0, 0), (0, b), and (a, 0),where a > 0 and D > 0. Show that the average vertical distancefrom points on the :r-axis to the hypotenuse is b I 2 for all a ) 0.

Average value of sine functions Use a graphing utility to verifythat the functions f(r) : sin ft-r have a period of 2n f k, wherek : 1,2,3, . . . . Equivalently, the first "hump" of /(x) : sin ftxoccurs on the interval l0,nlk). Verify that the average value ofthe first hump of /(x) : sin ft.r is independent of /<. What is theaverage value? (See Section 5.4 for average value.)

76. Equal areas The area of the shaded region under the curve) : 2 sin 2r in (a) equals the area of the shaded region under thecurve ) : sin r in (b). Explain why this is true without computing

5.5 Substitution Ruld

Additional Exercises74. Looking ahead Integrals of tan x and cot x

a. Use a change of variables to show that

t'Itanxdx = -ln lcos xl + C: h lsec xl + C.

J

b. Show thatrlcotxdx = In lsin xl + C.I

.t

75. Looking ahead Integrals of sec,r and csc x

a. Multiply the numerator and denominator of sec .r bysec "r + tan -r; then use a change of variables to show that

fI sec x dx : ln lsec x * tanxl * C.

J

b. Show that

fI csc x dx : -ln lcsc x * cotxl * C.

,l

77. F,qaal areas The area of the shaded region under the curve(\,G - l\2

y : 2{x in (a) equals the area of the shaded region under

the curve ! : x2 in (b). Without computing areas, explain why.

78-82. General results Evaluate the following integrals in which the

function f is unspecffied. Note ftp) is the pth derivative of f and fp isthe pth power of f . Assume f and its derivatives are continuous for allreal numbers.

f78. I tsI'(*) + 7f2G) +,f(x))f'(x)dx

,l

r27e. | (sf3(,) + 7f2(x) + te\f'(x) dx.

.llwhere/(1) : 4, f(2) : s

365

66.

67.

68.

69.

71.

I

i

72.

Fr73.

>4/

366 CH,lPfin 5 . INTEGRATIoN

7l80. I f'Q)f"(r)dx.where/'(0) : 3andf'(1) : 2

.lo

rt. .l 17(d1r117b+r)1x1ax,where p is a positive integer, n * -1

,r. J

2u'G) + 2f(x))f(x)f'(x)dx

83-85. More than one way Occasionally, two different substitutionsdo the job. Use both of the given substitutions to evaluate the followingintegrals.

f1ta. l,

xVx i adx: a ) O

rl84. l*'(/*t-adx:a)0

.lo

I .^85. / sec'0 tan? d0 (a : cos 0 and a = sec 0)J:

86. sin2ax and cos2ar integrals Use the Substitution Rule toprove tnal

f a sin (2ax)

J sin2axor :i- -;"'

+ c and

f r sin (2ax)

lcos2ax/r:)+-/ +C

Integral of sin2 r cos2 r Consider the integral

I: J

sin2xcos2xdx.

a. Find l using the identity sin 2"r : 2 sin x cos x.b. Find 1 using the identity cos2 x : I - sin2 x.c. Confirm that the results in parts (a) and (b) are consistent and

compare the work involved in each method.

Substitution: shift Perhaps the simplest change ofvariables is theshift or translation given by u : x * c, where c is a real number.

REVIEW EXERCISES .

Explain why or why not Determine whether the followingstatements are true and give an explanation or counterexample.Assume / and /' are continuous functions for all real numbers.

a. If A("r) : I]fOdt andf(t):2r - 3,thenAisaquadraticfunction.

b. Given an area function A(x) : .[] 71t7 at and an antiderivativeF of f ,it follows that A'(x) : F(").

". I! r'r.)dx: f(b) - f(a)d. fi I:lIG)l dx : o.then /(x) : 0 on [a. bl.e. If the average value of / on la, bl is zero, then /(r) : 0

on la, b).

a. Prove that shifting a function does not change the net area

under the curve. in the sense that

rb rb'lcI r(, * c\dx: I rtu\ar.Jo"' Jo+r""

b. Draw a picture to illustrate this change ofvariables in the case

that/(x) : sin;r, a : 0,b : n,c : nl2.

89. Substitution: scaling Another change ofvariables that can beinterpreted geometrically is the scaling u -- cx, where c is areal number. Prove and intemret the fact that

l,' ,r,a o' : ! lu,'

r{,) au

Draw a picture to illustrate this change of variables in the case that

f(x) : sinx.a : 0.b : n.c = t.90.-93. Multiple substitutions Use two or more substitutions to findthe follow ing integrals.

l'90.

J x sin4 (-r2) cos (r2) d-r

tflint: Begin with a : x2, then use o : sin a.)

f )-9t. I -

-::: (Hinr:Besinwitha: Vl +,..)J r/t+r/t+x '

tJ

tanto (4x) sec2 (4x) dx (Hint: Beginwith z : 4x.)

["'' -2!::!: o, (Hint:Beginwithu : cosd.)Jo Vcos2o + 16

QUICK CHECK ANSWERS

l. u : xa + 5 2. with u : x5 * 6,wehavedu : 5xa,and ra does not appear in the integrand. 3. New equation:u' - l3u + 36 : 0; roots: x : 12, ]:3 q

r. fi1zy1x1 - 3s(")) ax : z fi flx) dx + 3 ff g(x) d.x

s. If'GeDs'(x)dx : f(g(x)) + c2. Velocity to displacement An object travels on the x-axis with a

velocity given by u(t) : 2, * 5, for 0 < t '< 4.

a. How far does the object travel for 0 < t = 4?b. What is the average value of 'r.r on the interval [0,4]?c. True or false: The object would travel as far as in part (a) ifit

traveled at its average velocity (a constant) for 0 < t < 4.

(": t/*+ianclr.r : x + a)

(r: l/rt oandu: x + a)

92.

93.

87.

88.

l.

Area by geometry Use geometry to evaluate In'f6) dx,wherethe graph of/ is given in the figure.

Displacement by geometry Use geometry to find the displace-ment of an object moving along a line for 0 < / < 8, where thegraph of its velocity o : C(t) is given in the figure.

5

4

3

2

I

12345678t

Area by geometry Use geometry to evaluate fift- - * a-(Hint: Complete the square of 8-r - ,r'frrst).

Bagel output The manager of a bagel bakery collects the follow-ing production rate data (in bagels per minute) at six differenttimes during the morning. Estimate the total number of bagelsproduced between 6:00 and 7:30 a.m.

Production rateTime of day (a.m.) (bagels/min)

6:006:15

6:306:45

7:007:15

Review Exercises

9. Sum to integral Evaluate the following limit by identifying theintegral that it represents:

n l/tr\8 1/t\

''.l--ZL(;') .'l(;)

10. Area function by geometry Use geometry to find the area A(x)that is bounded by the graph of/(r) : 2t - 4 and the r-axisbetween the point (2,0) and the variable point (x,0), wherex > 2.Yerify thatA'(.r) : f(t).

1l-26. Evaluating integrals Evaluate the following integrals.

367

3.

14.

t6.

18.

20.

,)

13.

15.

17.

19.

2r.

23.

,\

27.

4.

6.

"2lr. Jr1xa-2x+t)dx D. f cos3x dx

Jo l' + \'a'

J{s*t - 7x6)dx

ft

J,*r* + r)dx

[' o*lo {4- rz

ft __y_o_.lo Y25 - x'z

[ ,in'so ao,I

f x2+2x-2I

-

dxJ x'l3x'-6x

fot ,or" - 2xt6 + t) dx

J-r"0"*r dt

l/o"lf{tf + t)a dy

I t sin x2 co{ *2 dt

}

I

I

I'1 |

I

I

I

I

l

l

I

I

zn. lrO-cos23o)do

7In2 ,r

'u' J, Tipo.Symmetry properties Suppose tt',at ff 71r1dx : l0 and

Ii S@ dx : 20. Furthermore, suppose that / is an even functionand g is an odd function. Evaluate the following integrals.

f4 14 f4a. lf1)a, b. ltg1,1ar c. lert*)-3g(x))dxJ-4 J+ .lt

Integration by Riemann sums Consider the integral

J'r'12* - 21a".

a. Give the right Riemann sum for the integral with n : 3.

b. Use summation notation to write the right Riemann sum for anarbitrary positive integer n.

c. Evaluate the definite integral by taking the limit as n --+ oo ofthe Riemann sum in part (b).

Evaluating Riemann sums Consider the function /( x) : 3x + 4on the interval [3, 7]. Show that the midpoint Riemann sum withn : 4 gives the exact area ofthe region bounded by the graph.

28. Properties ofintegrals The figure shows the areas ofregionsbounded by the graph of/ and the x-axis. Evaluate the followingintegrals.

f' fd fba. I f(x) dx b. I yg) ax c. 2 | 1(J') axJa Jb J.

a. + Lo

t{*)a* ". ,

f,u f{*)0, r. z

lro r{,)a*

60

75

605040

7.

8.

29-34. Properties of integrals Suppose tn"t [l fQ) dx - 6,

Ii s@ clx : 4, una fro f (x) rJx : 2. Evaluate the following integralsor state that there is not enough information.

2s. f,o

rr{i o, n. - fo'zr{.,,a*

n. f,o

t{")st,) a,t. f,o

{tt{i - 2g(x))dx

33. L"#0, ,0. I'rro) - g(x))dx

368 Cnaprpn 5 . INrscRATror\

-a. Evaluate f1(0).c. Evaluate 11'(2).

I JL 6\^,/I7 35. Displacement from velocity A particle moves along a line with a

velocity given by l(r) - 5 sin (z'l) starting with an initial positions(0) : 0. Find the displacement of the paticle between r : 0 and

t : 2, which is given by s(t) : Jou(t) dt.Find the distance trav-

eled by the particle during this interval, which is IilotOl r,.

36. Average height A baseball is launched into the outfield on a para-

bolic trajectory given by y : 0.01x(200 - "r). Find the averageheight of the baseball during its flight.

37. Average values Find the average value of the functions shown in(a) and (b). Integration is not needed.

5

I

3

2

I

t23456i8x(a)

v

5

4

3

2

I

'-'-----=--1

v:f(x) i

2345678r

An unknown function The function / satisfies the equation3xa Z : fft@ ttt.Find f andcheckyouranswerbysubstitution.

An unknown function Assume ;f is a continuous function,

tf 'Qi dx : lo,and f (2): 4. Evaluate /(4).Function defined by anintegral Let F/(x) : 1;'t/+ 4 at.

Function defined by an integral Make a graph of the functionf {.t,

f(x\ : I !fotx > l.Besuretoincludealloftheevidence.lt r

you used to arrive at the graph.

Identifying functions Match the graphs A, B, andC in the figurewith the functions /(x),/'(x), and Jl71t\at.

Geometry of integrals Without evaluating the integrals, explainwhy the following statement is true for positive integers n:

rl rlIlx"dx+ l{iax:r.Jo ,lo

Change of variablesUse the change of variables z3 : x2 - | toevaluate the integral frt t t/7 - t a*.

Inverse tangent integral Prove that, for nonzero constants a and b,

I dx I _/ax\I a\\ b'z: Ltan \. , /

46-51. Additional integrals Evaluate the following integrals.

f.^I Sln zxas' .l t * *r' ra' (sin 2x : 2 sin x cos x)

41.

42.

43.

44.

45.

Ii"^(+),'fao/ t,."-'-Xr - r',

li;5'.

/(tan-r x)s48. l

-,

dxJ l+x', qrn'Y

50. l--7==|dxJ Yl - x'-+-J

b. Evaluate 11'(1).d. Use geometry to evaluate 11(2).

47.

49.

51.

1738.

39.

40.

Area with a parameter Let a t 0 and consider the family offunctions f(r) - sin cx on the interval l},nla).a. Craphf fora:1.2.3.b. Let g(a) be the area ofthe region bounded by the graph

of / and the x-axis on the interval 10, n I al. Graph g forO I a I oo. Is g an increasing function, a decreasingfunction, or neither?

Equivalent equations Explain why a function that satisfies theequation u(x) + zl;u(t) dt : l0 also satisfies the equationu'(x) + 2u(x):0.

53. -..

-+*rI'" T*"

e. Find the value of s such that FI(x) : sH(-x).

F. 54. Area function properties Consider the function f(") :x2 - 5x * 4andtheareafunction A(r): fiy1t\at.a. Graph / on the interval [0,6].b. Compute and graph A on the interval [0,6 .

c. Show that the local extrema of A occur at the zeros of /.d. Give a geometrical and analytical explanation for the obser-

vation in part (c).

e. Find the approximate zeros of A, other than 0, and call themx1 and x2.

f. Find & such that the area bounded by the graph of / and thex-axis on the interyal [0, -r,] equals the area bounded by thegraph of/ and the x-axis on the interval [x1, b].

g. If/ is an integrable funcrion and A(x) : .ilffO dt,isitalways true that the local extrema of A occur at the zeros of /?Explain.

Guided Proiects 369

Function defined by an integral Letf(*) : Iif, - t)ts1t - z)o ctr.

a. Find the intervals on which / is increasing and the intervals onwhich / is decreasing.

b. Find the intervals on which / is concave up and the intervalson which / is concave down.

c. For what values of .x does .f have local minima? Localmaxima?

d. Where are the inflection points of /?Exponential inequalities Sketch a graph of/(l) = e' on an arbi-trary interval [a, b]. Use the graph and compare areas of regionsto prove that

eb-eo eolebe\qv''<-b_

o1 2 .

(Source; Mathematics Magazine 81, no. 5 @ecember 2008): 374)

56.

Chapter 5 Guided Projects

Applications of the material in this chapter and related topics can be found in the foltowing Guided Projects. For additionalinformation, see the Preface.

. Limits of sums

. Symmetry in integrals

. Distribution of wealth

6.2

6.3

6.4

6.5

6.6

6.7

Regions Between Curves

Volume by Slicing

Volume by Shells

Length of Curves

PhysicalApplications

Logarithmic and ExponentialFunctions Revisited

Exponential Models

Applications of Inte gration

Chaptef PfeVieW Now that we have some basic techniques for evaluatingintegrals, we turn our attention to the uses of integration, which are virtually endless.Some uses of integration are theoretical and some are practical. We first illustrate the gen-eral rule that if the rate of change of a quantity is known, then integration can be used todetermine the net change or future value of that quantity over a certain time interval. Next,we explore some rich geometric applications of integration: computing the area of regionsbounded by several curves, the volume of three-dimensional solids, and the length ofcurves. A variety of physical applications of integration include finding the work done bya variable force and computing the total force exerted by water behind a dam. All of theseapplications are unified by their use of the slice-and-sum strategy. We end this chapter byrevisiting the logarithmic function and exploring the many applications of the exponentialfunction.

6.8

6.1 Velocity and Net ChangeIn previous chapters we established the relationship between the position and velocity ofan object moving along a line. With integration, we can now say much more about thisrelationship. Once we relate velocity and position through integration, we can make analo-gous observations about a variety of other practical problems, which include fluid flow,population growth, manufacturing costs, and production and consumption of natural re-sources. The ideas in this section come directly from the Fundamental Theorem of Calculus,and they are among the most powerful applications of calculus.

Position atl: d Position att: b> a++Velocity, Position, and DisplacementSuppose you are driving along a straight highway and your position relativeto a reference point or origin is s(t) for times r > 0 (tii-rrue {;.1). Yourdisplacement over a time interval la, b) is the change in the position be-tween those times, or s(b) - s(a). If s(b) > r(o), then your displacement ispositive; when s(b) < s(a), your displacement is negative.

Now assume that o(l) is the velocity ol the object at a particular time /.Recall from Chapter 3 that u(/) : s'(/), which means that s is an antideriv-ative of o. From the Fundamental Theorem of Calculus, it follows that

fb fbI uQ) ar = / t'(t) at : s(b) - t(o) : displacement.

Ja Ja

s:0 s(a):-1ffi.

Displacement : s(b) - s(a) > 0

Position att: b> a Position att: a{v

s:0 s(b) s(a)

@Displacement : s(b) - s(a) < 0

FIGURL 6.1

r(b) s (line of motion)

s (line of motion)

370

14i:lfi { 'r 6.1 Velocity and Net Change 371

we see that the definite integral Ilr@ dr is the displacement (change in position) be-tween times t : a and r : b. Equivalently, the displacement over the time interval [a, b]is the net area under the velocity curve over [a, b] (Fi_eure 6.2a).

Not to be confused with the displacement is the distance traveled over a time inter-val, which is the tcttal distance traveled by the object, independent of the direction of mo-tion. If the velocity is positive, the object moves in the positive direction and thedisplacement equals the distance traveled. However, if the velocity changes sign, then thedisplacement and the distance traveled are not generally equal.

WIWPIA policeman leaves his station at 9 a.m. on a north-south freeway, travel-ing north (the positive direction) for 40 mi between 9 a.m. and l0 a.m. From l0 a.m. toI 1 a.m., he travels south to a point 20 mi south of the station. What is the distance traveledand the displacement between 9:00 a.m. and 1l:00 a.m.? "q

To compute the distance traveled, we need the magnitude, but not the sign of the ve-locity. The magnitude of the velocity lu(t)l is called the speed. The disrance traveled overa small time interval dt is lu(t)l dt (speed multiplied by elapsed time). Summing these dis-tances, the distance traveled over the time interval la, b) is the integral of the speed; that is,

fbdistance rraveled : I lu\l at

Ja

As shown in Figure 6.2b, integrating the speed produces the area (not net area)bounded by the velocity curve and the r-axis, which corresponds to the distance traveled.The distance traveled is alwavs nonnesative.

L,.ot*"-"",:,^rU-r^^--

Ja

^)

(a)

ffi-*'tI Distance traveled : A1 + 42: J lXOl ar i

-.,.,____o _-___i(b)

FIGURE 6.2

I ^FFrrr.-.^r.-

ffifttfrffi Describe a possiblemotion of an object along a line for0 < / <

-5 for which the displace-ment and the distance traveled aredifferent. <

DEFINITIONS Position, Velocity, Displacement, and Distance

1. The position of an object at time /, denoted s(r), is the location of the objectrelative to the origin.

2. The velocity of an object at time t is u(t) : r'(r).3. Thedisplacementof theobjectbetweent: aandl : b t ais

fos(b)- s(a): I u(t)dt.

4. Thedistancetraveledbytheobjectb"M; t: aand,t: b ) ais

l,,o 1o1,110,,

L_Y: ]:(:l l " *"'T"o

:l :1",:_ol:"' u"'-"'

EXAMPLE 1 llisplacement from velocity A cyclist pedals along a straight road withvelocity a(t):2tz 8/ + 6mi/hrfor0 = / < 3,where/ismeasuredinhours.

a. Graph the velocity function over the interval [0, 3]. Determine when the cyclist movesin the positive direction and when she moves in the negative direction.

b. Find the displacement of the cyclist (in miles) on the time intervals [0, l], [], 3], and

[0, 3]. Interpret these results.

c. Find the distance traveled over the interval f0, 31.

5tz Crnr'ten 6 . ApprrcaroNs oF INTEcRATToN

SOLUTION

a. Bysolving a(t):2t2 - 8t + 6 : 2(t - l)(t - 3) :0,wefindthatthevelocityiszeroatr:1and/:3.Thevelocityispositiveontheinterval 0 < / < 1(Fi-sLrre6.3a),

which means the cyclist moves in the positive s direction. For 1 ( I < 3 the velocity isnegative and the cyclist moves in the negative s direction.

b. The displacement (in miles) over the interval [0, I ] is

frs(l)-s(0): I u(t)dt

JO

rl

I Qr' - 8t + 6) dt substitute for rl.Jo

/t \lt g

(rt' - +t + ot )1,

: ; Evaluate integral'

A similar calculation shows that the displacement over the interval [1, 3] is

f3.. 8s(3) - s(l) :

J, u?)dt : -;.

Over the interval [0, 3], the displacement ir 3 * (-3) : 0. This means that thecyclist returns to the starting point after three hours.

c. From paft (b), we can deduce the total distance traveled by the cyclist. On the interval

[0, 1] the distance traveled is I m; on the interval [1, 3], the distance traveled is also t mi.

Therefore, the distance traveled on [0, 3] is f mi. Alternatively (Figure 6.3b), we can

integrate the speed and get the same result:

13 71

llr@la,: llzt-ztJo Jo

/ota^: litt _ 4t2 +\J

L6

J

^7+ l eQ,'-Jr '

/1* | -;t' -t 4t'

\J

dt Definition of lo(r) |

Evaluate integrals.

Simplify.

(b)

Rilated Exercises 7-10 I

8/ + 6))

\t3- u')1,

+ 6)dt

\ tro')l

/ t0

v(t)=2P-8t+6

Distance traveled from t : 0 to r = 3

1

is llv(rrlar:i+i:f.

;i$;;;,1 ^:J xrr ar: !

FIGURE 5.3

s(0) =

Note that t is the independent variable ofthe position function. Therefore, another(dummy) variable, in this case x, must be

used as the variable of integration.

Theorem 6. I is a consequence (actually a

restatement) of the Fundamental

Theorem of Calculus.

6.1 Velocity and Net Changb 373

Future Value of the Position FunctionTo find the displacement of an object, we do not need to know its initial position. For ex-ample, whetheranobjectmoves from.r : -20 to s : -10 orfroms : 50 to s : 60, itsdisplacement is l0 units. What happens if we are interested in the actual position of theobject at some future time?

Suppose we know the velocity of an object and its initial position s(0). The goal is tofind the position s(r) at some future time t > 0. The Fundamental Theorem of Calculusgives us the answer directly. Because the velocity a is an antiderivative of the position s,we have

Io'rroa*: fo','{x)dx:,(",1r: s(r) - s(o).

Rearranging this expression, we have the following result.

Theorem 6.1 says that to find the position s(r), we add the displacement over the interval[0, r] to the initial position s(0).

aui&'eHl'tk s tIs the position s(r) a number or a funcrion? For fixed rimes t : a andt : b, is the displacement s(b) - s(a) a number or a function? q

There are two equivalent ways to determine the position function:

. Using antiderivatives (Section 4.8)

. Using Theorem 6. I

The latter method is usually more efficient, but either method produces the same result.The following example illustrates both approaches.

EXAMPLE 2 Position from velocity A block hangs at rest from a massless spring atthe origin

,(r : o). At t : 0, the block is pulled downward j m to its initial position

s(0) : -f and released (Figure 6.4). Its velocity is given by a(t) : j sin r (in m/s) for/ > 0. Assume that the upward direction is positive.

a. Find the position of the block for r > 0.

b. Graph the position function for 0 < t = 3rr.

c. When does the block move through the origin for the first time?

d. When does the block reach its highest point for the first time and what is its position atthat time? When does the block retuffl to its lowest point?

1.00

0.15

0.50

0.25

0

-0.25

FIGURE 5.4

"&d:'t

Jl** t -l;#_H

1.00=F0.?5€o.s0€Eorr@*)

o.2s --trl-4

1

j Position of 1

i the block I

i_"1" lil::!ry '.J

THEOREM 6.1 Position from VelocityGiven the velocity u(t) of an object moving along a line and its initial positions(0), the position function of the object for future times r > 0 is

st/r'\ : s/O\ +"\-/"\"/*position at initial. time / position

ftI u(x) dx.

JO;;;ilover [0, r]

374

jCnaptsn6 . ApprteelronsoFINTEGRATIoN

S'r:,r-t"::I'::]v(r):Isinr

SOLUTION

a. The velocity function (Figure 6.5a) is positive for 0 < t I 7T, which means the blockmoves in the positive (upward) direction. At t : z, the block comes to rest momentar-ily; for r < t < 2r, the block moves in the negative (downward) direction. We let

s(l) be the position at time r > 0 with the initial position s(0) : -i m.

' Method 1: Using antiderivatives Because the position is an antiderivative of thevelocity, we have

f f.r(r) : I u@dt : /fsinr a, : -Icost * c.

.tJTo determine the arbitrary constant C, we substitute the initial condition r(0) : -1into the expression for s(t):

-4:-4cos0*C

Solving for C, we find that C : 0. Therefore, the position for any time / > 0 is

Is(/): -4cosl

Method 2: Using Theorem 6.1 Alternatively, we may use the relationship

nt

s(r):s(0)+ luQ)dx.Jo

tlSubstituting u(x) : i sin x and s(0) : -i, the position function is

,(rr :-1 + ['lrin*a*4 Jo4s(o) ae)

| (r \l': - 4 - (a *t "/ l, Evaluate integral'

I T,: - 4- f(cost - 1) Simplify.

I: -- cos t.4

b. The graph of the position function is shown in Figure 6.5b. We see that s(0) : - j m,as prescribed.

c. The block initially moves in the positive s direction (upward), reaching the origin(s : 0) when s(r) : -| cos / : 0. So the block arrives at the origin for the

first time when / : Tl2.

d. The block moves in the positive direction and reache.s its high point for the first time

when / : rr;the position at that moment is s(z') : f m. The block then reverses

direction and moves in the negative (downward) direction, reaching its low point att : 2tr. This motion repeats every 2rr seconds. ReLated Exercises I l-l 8 I

i_Oulcr cHEcK 4 Without doing further calculations, what are the displacement and

distance traveled by the block in Example 2 over the interval 10,2r)? <

Simplify.'t

) It is worth repeating that to find the

displacement, we need to know only the

velocity. To find the position, we must

know both the velocity and the initialposition s(0).

.<'

FIGURE 6.5

The terminal velocity of an object

depends on its density, shape, size, and

the medium through which it falls.

Estimates for human beings in free fallvary from 120 miltu (54 m/s) to180 mi/hr (80 m/s).

-.r

" " : , - 6.1 'Velocity and Net Change 375

EXAMPLE 3 skydiving suppose a skydiver leaps from a hovering helicopter andfalls in a straight line. He falls at a terminal velocity of 80 m/s for 19 seconds, at whichtime he opens his parachute. The velocity decreases linearly to 6 m/s over a two-secondperiod and then remains constant until he reaches the ground at t : 40 s. The motion isdescribed by the velocity function

a('|) :{:3' - ,,'if 0<t<19if 19 < t < 2Iif2l<t<40

Determine the altitude from which the skydiver jumped.

soLUTfoN we let the position of the skydiver increase downward with the origin(r : 0) corresponding to the position of the helicopter. The velocity (. ,gure 6.u) ispositive, so the distance traveled by the skydiver equals the displacement, which is

fo^o lrr,ll o, :

f'n ,o at + f' es3 - 3tt) o, * fo u o,

lle /:8orl + ( zs:rlo \

: 1120.

3Zr2\ 121 140_ . ,|l +6tlL / ltg 121

Fundameirtal Theorem

Evaluate and simplify.

The skydiver jumped from 1720 m above the ground. Notice that the displacement of theskydiver is the area under the velocity curve.

FIGURE 6.5

Related Exercises 19-20 I

reffiffe-ffiY Suppose (unrealistically) in Example 3 that the velocity of the skydiver is80m/sfor0 < t <20andthenitchangesinstantaneouslyto6m/sfor20 < t < 40.Sketch the velocity function and, without integrating, fincl the distance the skydiver fallsin40s.-

AccelerationBecause the acceleration ofan object moving along a line is given by a(t) : u'(/), the rela-tionship between velocity and acceleration is the same as the relationship between position

,21

l,(783 - 3tt) at

376

.:tCsapGn 6 . ApplicanoNs on lrvrrcnaroN

'a

Theorem 6.2 is a consequence of the

Fundamental Theorem of Calculus.

and velocity. Given the acceleration of an object, the change in velocity over an interval

la,blis fb fb

change in velocity : a(b) - u(a) : I u' (t) dt : I a(t) dt.Ja Jd

Furthermore, if we know the acceleration and initial velocity ?;(0), then the velocity at fu-ture times can also be found.

EXAMPLE 4 Motion in a gravitational field An artillery shell is fired directly up-ward with an initial velocity of 300 m/s from a point 30 m above the ground ( ;ure 6.7).

Assume that only the force of gravity acts on the shell and it produces an acceleration of9.8 m/s2. Find the velocity of the shell for r > 0.

SOLUTION We let the positive direction be upward with the origin (s : 0) conespondingto the ground. The initial velocity of the shell is 'u(0) : 300 m/s. The acceleration due togravity is downward; therefore, a(t) : -9.8 m/s2. The velocity for t > 0 is

ffi

I : 9.8 mls2

s(0) = 39.fl u(o; = 36s n/,

The velocity decreases from its initial value of300 m/s, reaching zero at the high point ofthe trajectory when o(t) : 300 - 9.8/ : 0, or at t x 30.6 s (Figure 6.8). At this point the

velocity becomes negative, and the shell begins its descent to Earth.Knowing the velocity function, you could now find the position function using the

methods of Example 3. Related E.rercise,s 21 27 <

Net Change and Future ValueEverything we have said about velocity, position, and displacement carries over to moregeneral situations. Suppose you are interested in some quantity Q that changes over time;

Q may represent the amount of water in a reservoir, the population of a cell culture, or theamount of a resource that is consumed or produced. If you are given the rate Q' atwhich Qchanges, then integration allows you to calculate either the net change in the quantity Q orthe future value of Q.

We argue just as we did for velocity and position: Because 0(r) is an antiderivative ofQ'ft), the Fundamental Theorem of Calculus tells us that

7b

I Q' G) dt : Q(b) - Q@) : net change in Q over la, b).Jo - ''

Geometrically, the net change in Q over the time interval la, bf is the net area under thegraph of Q' over la,b).

Altematively, suppose we are given both the rate of change Q' and the initial value Q(0).Integrating over the interval [0, r], where r > 0, we have

ftu(t) : o(0) + I o(*) dx : 3oo -t

+ ,lo300 m/s -9.8 m/s2

lo'en.rldx:3oo-e.8r.FIGURE 6.7

v

300

200

100

v(t):396-n.tt

Htshp.*;'."-**li-':r::""::::looj

> Note that the units in the iptegral are

consistent. For example, if Q' has units ofgallons/second, and t and "r have units ofseconds, then Q' (r) dx has units of(gallon/second) (seconds) : gal, which

are the units of O.

THEOREM 6.2 Velocity from AccelerationGiven the acceleration a(t) of an object moving along a line and its initial velocityo(0), the velocity of the object for future times r > 0 is

u(t):u(o) + fro|)ax.

FIGURE 6.8

fo' n't4 rtx : e(t) - o(o).

6.1 Velocity and Net Change

Rearranging this equation, we write the value of e atany future time / > 0 as

5t I

OU)

. flturevalue

THEOREM 6.3 Net Change and Future ValueSuppose a quantity Q changes over time at a known rate e, . Then thenetchangeinQbetween t: aandt: bis

fbQ@)-Q@): lQ'Q)a,.

Janet change in B

Given the initial value Q(o), the future value of e at future times r > 0 is7t

QQ):Q@)+ lq,1*7a,.Jo

rtI

=0(0)+ lg,1lax.+Joi n itialvalue net change

over [0, r]

I

I

t

I

I

I

I

I

1",*t;l"I

t

I

I

t

i

I

I

I

I

i

I

I

i

I

At the risk of being repetitious,

Theorem 6.3 is also a consequence ofthe Fundamental Theorem of CalculusWe assume that 0' is an integrablefunction.

) Although N is a positive integer (thenumber of cells), we treat it as a

continuous variable in this examole.

The correspondences between velocity-displacementlems are shown in Table 6.1.

problems and more general prob-

Table 5.1

Velocity-Displacement problems

Position s(ttVelocity: s'(t) : u1r]r

General Problems

Quantity Q(r) (such as volume or population size)

Displacement: s(b) - s(a) : f,ur1t1at

Rate of change: Q'(r)

Netchange: OQ) - e(o): l"un,Q)

Futurevalue otQ:Q(t): O@) * IFutureposition:s(r): s(Q+ f uQ)ax

EXAMPLE 5 ceil growth A culture of cells in a lab has a population of 100 cellswhen nutrients are added at time l : 0. Suppose the population l,r1r; increases at arategiven by

N'(t) : goe-oJt ceils/hr.

Find N(r) for r > 0.

soLUTloN As shown in Figure 6.9, the growth rate is large when / is small (plentyoffood and space) and decreases as / increases. Knowing that the initial population isN(0) : 100 cells, we can find the population N(/) at any future time / > 0 using

Theorem 6.3: rr

N(r):N(0)+ lw,1*1a*.lo

: loo + ['9or-0r,4*._l?J

N(0) N'(r)

: roo + f (:L)"-o' tt'L \ -o.r 7

- 'J l. Fundamental rheorem

: 1000 - 900e-o'r/ Simplify.

dt

Q'Q)dX

N'(r1 : 99"-o rt

se0o

IooE630o

FIGURE 6.9

Cnep,ren 6 . Appt-Icarroxs or IN'rscRA'[or'I

) Although I is a positive integer (the

number of books produced), we treat it as

a continuous variable in this example.

SECTION 6.1 EXERCISESReview Questions1. Explain the meaning of position, displacement, and distance

traveled as they apply to an object moving along a line.

2. Suppose the velocity ofan object moving along a line is positive.

Are position, displacement, and distance traveled equal? Explain.

The graph of the population function (Figure 6.10) shows that the population increases,

but at a decreasing rate. Note that the initial condition N(0) : 100 cells is satisfied and

that the population size approaches 1000 cells as / --+ oo.

1000 _ go0e-o.lt

I N(10) - N(5) - 215 ""ttt ,

I N(l5) N(10) = 130 cells I

I The population is growing i

i a1a <le919asing ,.1!9: _ __J

FIGURE 6.10

15 20

Time (hr)

Related Exercises 28-34 1,

EXAMPLE 6 Production costs A book publisher estimates that the marginal cost of a

particular title (in dollars/book) is given by

C'(x):12-0'0002x'

where 0 .: x 3 50,000 is the number of books printed. What is the cost of producing the

12,001st through the 15,000th book?

SOLUTION Recall from Section 3.5 that the cost function C(x) is the cost required to produce

x units of a product. The marginal cost C'(x) is the approximate cost of producing one addi-

tional unit after x units have already been produced. The cost of producing books x : 1 2,001

through x : 15,000 is the cost of producing 15,000 books minus the cost of producing the

hrst 12,000 books. Therefore, the cost ofproducing books 12,fi)1 through 15,000 is

r 15,000

c(rs,ooo) - c(l2,ooo) : I c'1r7a,J t2.ooo

r 15,000

: I OZ - O.}OO2x) dx Substitute for C'(x).J t 2.ooo

! aoo

A 600o

=Jl 400

I rs,00o

: (l2x - 0.0001x2) | fundamental Theorem| 12,000

: $27,900 Simplify.Related Exercises 35-38 {

'auiaKtHEiiiff Would the cost of increasing the production from 9000 books to 12,000

books be more or less than the cost of increasing the production from 12,000 books to

15.000 books? Explain.

3. Given the velocity function tr of an object moving along a line, ex-

plain how definite integrals can be used to find the displacement

of the object.

Explain how to use definite integrals to find the net change in a

quantity, given the rate of change of that quantity'4.

.;

5. Given the rare of change of a quantity e and its initial value e(0),explain how to find the value of e at aTuture rrme I > 0.

6. What is the result of integrating a population growth rate betweentwotimes t: aand,t = b?

Basic Skillsm 7-10. Displacement from velocity Assume t is time measured in sec_ontls and velocities ltave units of mf s.

a' Graph the verocity function over the given intervar. Thendetermine when the motion is in the jositive direction andwhen it is in the negative direction.

b. Find the displacement over the given interval.c. Find the distance travelecl overlhe given interval.

W aQ):6 - 2t:0 < r < 6where / is measured in seconds and o has units of m/s.

"": {i|2, - ^,

ifO</<20if20<t<45if t > 45

8.

@10.

r 1-1wirh

u(t):10sin2r;0<t=21ru(t):72-5P+6t;0<t<5u(t):59"-zt'0=t<4

i. f.oyjtlon from velocity Consider an object m,oving along a linethe folkning velocities and initial positions.

'.. " "o

a.

b.

c.

d.

c.

Graph the velocitr Junction on the given intervar and deter-mine when the object is moving in the positive direction andwhen it is moving in the negattve clirection.

letlrmlne the position function for t > 0 using both the anti_derivative method and tlt" Fundamentat fn"orZ* o1 Cotrutw(lheo,rem 6.1). Checkfor agreement between the two methods.Graph the position function on the given interval.

16.

26. Deceleration A car slows down with an acceleration ofa(t) : -15 ftls2. Assume that o(0) : 60 ft/s and s(0) = 6.a. Determine and graph the position function fbr r > 0.b. How far does the car travel in the time it takes to come to rest?

@ tpprouching a station Afu : 0, a train approaching a station be_gins decelerating from a speed of g0 mi/hr according to the accel_eration function a(t) : -1280(1 + Bt)'-3 mif hr2, *fr"; ; =;:"'How far does the train travel between r : 0 and t : 0.2? Be_tween/:0.2andt = 0.4?

@ ,(r) :6 - 2ton[0,5]; s(0): e

12. u(t) : 3 sin z-r on [0, a]; s(6; : 1

@ ,(r) : 9 - t2 onlO,4l; .r(0) : -214.

s,i@

u\t) = 111, + t)onl0,8l; r(0) = _4

Oscillating motion A mass hanging from a spring is set in motionand its ensuing velocity is given by a(r ) = 2; "oi

7'l to1. r = O.Assume that the positive directionis upward ""a <Ol : O.

a. Determine the position function for I > 0.b. Graph rhe position function on the interval i0, 3].c. At what times does the mass reach its tor""rt point the frstthree times?

d. At what times does the mass reach its highest point the firstthree times?

Cycling distance A cyclist rides down a long straight roadat a velociry tin m/min.r given by o(r) : +0d _ 26r, for0=r<l0min.

l. Io* far does the cyclist travel in the first 5 min?b. How tar does the cyclist travel in the first 10 min?c. How far has the cyclist traveled when her u"fo.lty i, 250 m/min?Flying into a headwind The velocity of an airplane flying into aheadwind is giv-en by u(t) : 30(ro _ l) _t# f";; < I < 3 hr.Assume that s(0) = 0.

a. Determine and graph the position function for 0 s I < 3.b. How far does the airplane travel in the first 2 hr?c. How far has the airplane traveled at the instant its velocityreaches 400 milfu?

Graph the velocity function for 0 = r < 100. When is thevelocity a maximum? When is the velocity zero?What is the distance traveled by the automobil. in tt

" first 30 s?What is the distance traveled by the automobil" inii. fir.t OO rfWhat is the position of the automobile when t : 75?

20. Probe speed A data c.ollection probe is dropped from a stationaryballoon and it falls ,

negrecting .o ,.,,.,l,ll1 ",ii,'.",'i.J J:: Yilili"?I ;j?,n; i.rj;immediately slows t

tains unt' ir enrers ,i"""".lllt"tt speed of l0 m/s, which it main-

a. Graph the velocity function.b. How far does the probe fall in the first 30 s after it is released?c. If the probe was released from an altitude of 3 km, when doesIt enter the ocean?

2l-24. Position and velocity from accelera tion Find the positionand v-elocit,v- of an object ntoving yilong o ,rroigi, tin, ,rith the givenacceleration, inititil velocity, and initiil positiin. A.ssume units oJ.meters and seconds.AC\QJi o(r) : -e.8. u(0) = 20,s(0) : 0

22. a(t) : e-,,u(O) = 60,s(0) : 46/6-:

Y) oVl: -0.01r.r.r(0) : 16.s(0) = s24. a(t):2olQ + 2)2,u(0): 20,s(0) : 10

Acceleration A drag racer accelerates at a(t) : gg ft/sr. Assumethat a(0) : 0 and s(0) : 0.

a. Determine and graph the position function for I > 0.b. How far does th" .0... travel in 4 s? .l

c. At this rate, how lone will it take the racer to travel ] mi?d. How Iong does it take tne racer to travel 300 ft?e. How far has the racer traveled when it..u.h., u speed of178 ft/ s?

6.1 Velocity and Net Change 379

18. Day hike The u"lo.l,y (in mi/hr) of a hiker walking along astraight trail is given by u(t) : 3sin2 (ntl2),for0 = t = 4hr.Assume that s(Q) : 6.

a. Determine and graph the position function for O < t < 4.b. What is the distance traveled by thehiker in the first 15minutes of the hike? lHinr: sin2t : 1r I _ .os 2t).)c. Whar is the hiker.s position u, , : il'

-'f.\(!ri Piecewise velocity The velocity of a (fast) automobile on astraighr highway is given by the function

28.

380 Cneyrnn 6 . ApprrcenoNs op Ivrncnaton

Peak oil extraction The owners of an oil reserve begin extractingoil at r : 0. Based on estimates ofthe reserves, suppose theprojected extraction rate is given by Q'(t) : 3t2 (40 - r)2, where0 < r < 40, Q is measured in millions of barrels, and t ismeasured in years.

a. When does the peak extraction rate occur?b. How much oil is extracted in the first 10, 20, and 30 years?c. What is the total amount of oil extracted in 40 years?d. Is one-fourth of the total oil extracted in the first one-fourth of

the extraction period? Explain.

Oil production An oil refinery produces oil at a variable rategiven by

35-38. Marginal cost Consider the foltowing marginal cost functions.a. Find the additional cost incurred in dollars when production is

increasedfrom 100 units to 150 units.b. Find the additional cost incuned in dollars when production is

increasedfrom 500 units to 550 units.

Q7 c'(*) = 2000 - 0.5x 36. C'(x) : 200 - 0.05.r

@ ,'e): 3oo + lox - o.o1x2

38. C'(x) = 3000 - .r - 0.001x2

Further Explorations39. Explain why or why not Determine whether the following state_

ments are true and give an explanation or counterexample.

a. The distance traveled by an object moving along a line is thesame as the displacement of the object.

b. When the velocity is positive on an interval, the displacementand the distance traveled on that interval are equal.

c. Consider a tank that is filled and drained at a flow rate ofR(r) :1 - f1100(gat/min),forr > 0. Itfoltowsthatthevolume of water in the tank increases for 10 min and thendecreases until the tank is empty.

d. A particular marginal cost function has the property that it ispositive and decreasing. The cost of increasing productionfrom A units to 24 units is greater than the cost of increasinqproduction from2A units to 34 units.

40-41. Vefocity graphs The Jigures show velocity functions for motionalong a straight line. Assume the motion begins with an initial positiono/s(O) : 0. Determine the following:

a The displacement befii)een t : 0 and t : 5b. The distance traveled between t : 0 and t : 5c. The position at t : 5

d, A piecewise function for s(t)

4245. Equivalent constant velocity Consider the following velocityfunctions. In each case, complete the sentence: The same distancecould have been traveled over the given time period at d constantvelocity of _.

u(t):2r*U for0=1<go(t)=1-Pll6 for0<r<4D(t):zsht for}<t<no(t): t(25 - f)t/2 for0 < r < 5

Where do they meet? Kelly started at noon (r : 0) riding a bikefrom Niwot to Berthoud, a distance of 20 km, with velocityu(t) : 15lQ * 1)2 (decreasing because offatigue). Sandy startedat noon (r : 0) riding a bike in the opposite direction from

lit

'1

11

,lifO<r<30if30<t<40if t > 40

where / is measured in days and e is measured in barrels.

a. How many barrels are produced in the first 35 days?b. How many barrels are produced in the first 50 days?c. Without using calculus, determine the number of barrels

produced over the inrerval [60, 80].

30-33. Population growth

30. Starting with an initial value of P(0) : 55, the popularion of aprairie dog community grows at a rate of p,(r) : Z0 - tl5 (inunits of prairie dogs/month), for 0 < t < ZOQ.

a. What is the population 6 months later?b. Find the population P(r) for O < t < 200.

Q! When records were frst kept (r : 0), the population of a ruraltown was 250 people. During the following years, the populationgrew at a rate of P'(l) : 30(1 + \4).a. What is the population after 20 years?b. Find the population P(t) at any time / > 0.

32. The population of a community of foxes is observed to fluctuateon a 10-year cycle due to variations in the availability ofprey.When population measurements began (r : 0 years), the popula_tion was 35 foxes. The growth rate in units of foxes/vr wasobserved to be

P'(t) : 5 "''

a. What is the population 15 years later? 35 years later?b. Find the population P(t) at any rime / > 0.

A culture of bacteria in a petri dish has an initial population of1500 cells and grows at a rate of N'(r) : l00e 025'cells/day.

a. What is the population after 20 days? after 40 days?b. Find the population N(/) ar any rime r > 0.

Endangered species The population of an endangered specieschanges at a rate given by p'(t) : 30 - Z\t (individuals/year).Assume the initial population of the species is 300 individuals.

a. What is the population after 5 years?b. When will the species become extinct?c. How does the extinction time change if the initial population is

100 individuals? 400 individuals?

I soo

o'G):i2600-60r( 200

41.40.

ro,i"(9).\)./

42.

43.

44.

45.

46.

34.

;l

47.

J

Berthoud to Niwot with velocity u(t) : 2OlQ + t)2 (alsodecreasing because of fatigue). Assume distance is measured inkilometers and time is measured in hours.

a. Make a graph of Kelly's distance from Niwot as a functionof time.

b. Make a graph of Sandy's distance from Berthoud as a functionof time.

c. How far has each person traveled when they meet? When dothey meet?

d. Ifthe riders' speeds areo(t) : AIQ + l)z andu(t) : BIQ + l)2 and the distance between the towns is D,what conditions on A, B, and D must be met to ensure that theriders will pass each other?

e. Looking ahead: With the velocity functions given in part (d),make a conjecture about the maximum distance each Dersoncan ride (given unlimited time).

Bike race Theo and Sasha start at the same place on a straightroad riding bikes with the following velociries (measure in mi/hr):

Theo: o.(r) : l0 for r > 0Sasha:trr(t) : 15t for0 < r < I and'ur(r): 15 for/ ) I

a. Graph the velocity functions for both riders.b. Ifthe riders ride for t hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a).c. Ifthe riders ride for 2 hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a).d. Which rider arrives first at the 10, 15, and 20 mile markers of

the race? lnterpret your answer geometrically using the graphsof part (a).

e. Suppose Sasha gives Theo a head start of0.2 mi and the ridersride for 20 mi. Who wins the race?

f. Suppose Sasha gives Theo a head start of0.2 hr and the ridersride for 20 mi. Who wins the race?

Two runners At noon (t : 0), Alicia starts running along a longstraight road at 4 mi/hr. Her velocity decreases according to thefunction u(l -- alQ + 1) for / > 0. Ar noon, Boris also startsrunning along the same road with a 2-mi head start on Alicia; hisvelocity is given by u(t) : 2lU + 1) for / > 0.

a. Find the position functions for Alicia and Boris, where .r : 0corresponds to Alice's starting point.

b. When, if ever, does Alicia overtake Boris?

Running in a wind A strong west wind blows across a circularrunning track. Abe and Bess start at the south end of the track andat the same time, Abe starts running clockwise and Bess startsrunning counterclockwise. Abe runs with a speed (in units ofmi/hr) given by u(E) : 3 - 2 cos g and Bess runs with a speedgiven by D(0) : 3 * 2 cos 0, where g and 0 are the cenrral anglesof the runners.

wind

6.1 'Velocity and Net Change 381

a. Graph the speed functions u and a, and explain why theydescribe the runners' speeds (in light of the wind).

b. Which runner has the greater average speed for one tap?

c. Challenge: If the track has a radius of ,I mi, how long does it takeeach runner to complete one lap and who wins the race?

Applications50. Filling a tank A 200-L cistem is empty when water begins

flowing into it (au : 0) at a rate in L/min given by e,Q) : 3\/i.a. How much water flows into the cistern in I hr?b. Find and graph the function that gives the amount of water in

the tank at any time t > 0.c. When will the tank be full?

51. Depletion ofnatural resources Suppose that r(r) : r0e-i1 is therate at which a nation extracts oil, where ro : l0Tbarrels/yris the current rate of extraction. Suppose also that the estimate ofthe total oil reserve is 2 a l0e barrels.

a. Find the minimum decay constant ft for which the total oilreserves will last forever.

b. Suppose ro : 2 x 107 banels/yr and the decay constant ft isthe minimum value found in part (a). How long will the total oilreserves last?

52. Snow plow problem With snow on the ground and falling at a con-stant rate, a snow plow began plowing down a long straight road atnoon. The plow traveled twice as far in the fust hour as it did in thesecond hour. At what time did the snow start falling? Assume the

' plowing rate is inversely proportional to the depth ofthe snow.

53. Filling a reservoir A reservoir with a capacity of 2500 m3 is filledwith a single inflow pipe. The reservoir is empty and the inflowpipe is opened at t : 0. Lenlng eG) be the amount of water inthe reservoir at time t, the flow rate of water into the reservoir1in mr/hr) oscillates on a 24-hr cycle (see figure) and is given by

48. | /-'\lQ'U) : 20

Ll + .", ( ,.;J ]

49.

I1I

a. How much water flows into the reservoir in the first 2 hr?b. Find and graph the function that gives the amount ofwaterin

the reservoir over the interval [0, r] where r > 0.c. When is the reservoir full?

Blood flow A typical human heart pumps 20 mL of bloodwith each stroke (stroke volume). Assuming a heart rate of60 beats/min, a reasonable model for the outflow rate ofthe heartis V'(t) : 2O(l + sin (2nr)), where V(r) is the total volume ofblood pumped attime t measured in milliliters. Assume thatv(0) : 0.

54.

Csesrnn 6 ApplrcerloNs op INrncnerroN

Gmph the outflow rate function.

Verify that the amount of blood pumped over a one-second

interval is 20 mL.Find the function that gives the total blood pumped between

I : 0 andafuturetime t ) 0.

d. What is the cardiac output over a period of I min? (Use calcu-

lus, then check your answer with algebra.)

55. Air flow in the lungs A reasonable model (with different param-

eters for different people) for the flow of air in and out of the

lungs is

v'(t\ : -lYo rtn f +),lu \ ) /

where V(r) is the volume of air in the lungs at time t 2 0, meas-

ured in liters, t is measured in seconds, and Vs is the capacity ofthe lungs. The time t : 0 corresponds to a time at which the

lungs are full and exhalation begins.

a. Graph the flow rate function with Ve : l0 L.

b. Find and graph the function V, assuming that

v(0):vo:10L.c. What is the breathing rate in breaths/min?

56. Oscillating growth rates Some species have growth rates that

oscillate with an (approximately) constant period P. Consider

the srowth rate function

where A and r are constants with units of individuals/yr. Aspecies becomes extinct if its population ever reaches 0 after

t :0.a. Suppose P : 10, A : 20, and r : 0. Ifthe initial population

is N(0) : 10, does the population ever become extinct?

Explain.b. Suppose P : 10, A : 20, and r : 0. Ifthe initial population is

N(0) : 100, does the population ever become extinct? Explain.

c. Suppose P : 10, A : 50, and r : 5. If the initial population

is N(0) : 10, does the population ever become extinct?

Explain.d. Suppose P : 10, A : 50, and r : -5. Find the initial popu-

lation N(0) needed to ensure that the population never

becomes extinct.

57. Power and energy Power and energy are often used interchange-

ably, but they are quite different. Energy is what makes matter

move or heat up and is measured in units of joules (J) or Calories(Cal), where I Cal : 4184 J. One hour of walking consumes

roughly I 06 J, or 250 Cal. On the other hand, power is the rate at

which energy is used and is measured in watts (W; I W : 1 J/s).

' Other useful units of power are kilowatts (l kW : 103 W; and

megawatts (1 MW : 106 w). If energy is used at a rate of I kwfor I hr, the total amount of energy used is I kilowatt-hour(kWh), which is 3.6 x 106J.

Suppose the power function of a large city over a24-hrperiod is given by

/ z-l\P(t): E'(t):300 - 200sin\12l

where P is measured in MW and I : 0 corresponds to 6:00 p.m.

(see frgure).

a. How much energy is consumed by this city in a typical 24-hr

period? Express the answer in MWh and in J.

b. Burning I kg of coal produces about 450 kWh of energy. How

many kg of coal are required to meet the energy needs of the

city for I day? For 1 yr?

c. Fission of I gram of uranium-235 (U-235) produces about

16,000 kwh of energy. How many grams of uranium are

needed to meet the energy needs ofthe city for I day? For 1 yr?

d. A typical wind turbine can generate electricity at a rate ofabout 200 kW. Approximately how many wind turbines are

needed to meet the average energy needs of the city?

Variable gravity At Earth's surface the acceleration due to

gravity is approximately I : 9.8 m/s2 twith local variations).

However, the acceleration decreases with distance from the

surface according to Newton's law of gravitation. At a distance

of y meters from Earth's surface, the acceleration is given by

a(v) :0 + vlR)z

where R : 6.4 X 106 m is the radius of Earth.

a. Suppose a projectile is launched upward with an initial veloc-

ity of z16 m/s. Let o(l) be its velocity and y(t) its height (in

meters) above the sutface / seconds after the launch. Neglecting

forces such as air resistance, explain *hy X: afu) and

dtt, : a(r).dt da ld ,.

b. Use the Chain Rule to show that i :

Z *lu').c. Show that the equation of motion for the projectile is

1d,,., or(u')

: a(y), where a(y) is given previously'

d. Integrate both sides ofthe equation in part (c) with respect to y

using the fact that when y : 0, u : o0. Show that

t,, .. ^/ t .\i@" - oo') : sn\, * 17*

- 1/.

e, When the projectile reaches its maximum height, o : 0.

Use this fact to determine that the maximum height is

R,U,?/md 2gR - u]Graph y.u* as a function of oe. What is the maximum height

when oo : 500 m/s, 1500 m/s, and 5 km/s?Show that the value of'uo neededao Put the projectile into orbit(called the escape velocity) is V2gR.

a.

b.

58./ c*t\

N',(1): Asin({' } + r,' \r/

g.

P(t) : E'(t)

Time (hr)