a 6 integral solution yes. sincef(x) > oandg(x) < 0 ......692 chapter 6 i] applicationsoftheintegral...

APPLICATIONS OF THE6 INTEGRAL

6.1 Area Between Two Curves

Preliminary Questionsb1. Whatis the area interpretation of | (f(x) g(x)dx if fx) = g(x)?

aSOLUTION Because f(x) > g(x), a (f(x) g(x)) dx represents the area of the region bounded between the graphs ofy = f(x) and y = g(x), boundedontheleft by the vertical line x = a and on theright by the vertical line x = b.

b2. Is / (f@) - g(x)) dx still equal to the area between the graphs of f and g if f(x) > 0 but g(x) < 0?

a

SOLUTION Yes. Since f(x) > O and g(x) < 0,it follows that f(x) g(x) => 0.3. Suppose that f(x) > g(x) on [0,3] and g(x) > f(x) on [3,5]. Express the area between the graphs over [0,5] as a sum of

integrals.SOLUTION Rememberthat to calculate an area between two curves, one must subtract the equation for the lower curve fromthe equation for the upper curve. Over the interval [0, 3], y = f(x) is the upper curve. On the other hand, over the interval[3, 5],y = g(x)is the upper curve. The area between the graphs overthe interval[0, 5] is therefore given by

3 5/ (fe) g(x)dx + / (g(x) f@)) dex.0 3

4. Suppose that the graph of x = f(y) lies to theleft of the y-axis. Is [? FO) dy positive or negative?SOLUTION If the graph of x = f(,) lies to the left of the y-axis, then for each value ofy, the corresponding value of x is lessthan zero. Hence, the value of f£ FO») dyis negative. Exercises1. Find the area ofthe region between y = 3x? + 12 and y = 4x + 4 over [-3, 3] (Figure 1).

y = 3x2 + 12

y=4x+4

FIGURE1SOLUTION Asthe graph of y = 3x? + 12 lies above the graph of y = 4x + 4 overthe interval [-3, 3], the area between thegraphsis

3 3 3/ (6? + 12) (4x + 4)) dx = / (3x2 4x + 8) dx = (x? = 2x? + 8x)| ,=102.3 3 =

2. Find the area of the region between the graphs of f(x) = 3x + 8 and g(x) = x + 2x +2 over(0,2].SOLUTION From the diagram below, wesee that the graph of f(x) = 3x + 8 lies above the graph of g(x) = x? + 2x + 2 overthe interval [0, 2]. Thus, the area between the graphsis

Plex+o-(2+20+2)] av= [(24x46) dx = (++ +

y=x?+2x+2x

05 10 15 2.0

688 CHAPTER 6 || APPLICATIONS OF THE INTEGRAL

3. Find the area ofthe region enclosed by the graphs of f(x) = x? + 2 and g(x) = 2x + 5 (Figure 2).

FIGURE2

SOLUTION From thefigure, we see that the graph of g(x) = 2x +5 lies above the graph of f(x) = x? + 2 overthe interval[ 1, 3]. Thus, the area between the graphsis

F [Ox +5) (+? +2)] a= ff (-x? +2x +3) dx1 3

= (-3° +x + x)3 15 sy _ 32

3) 3°4. Findthe area of the region enclosed by the graphs of f(x) = x3 10x and g(x) = 6x (Figure 3).

y f= - 10x

g(x) = 6x

FIGURE3

SOLUTION From thefigure, we see that the graph of f(x) = x? 10x lies above the graph of g(x) = 6x overthe interval[ 4, 0], while the graph of g(x) = 6x lies above the graph of f(x) = x3 10x overthe interval [0, 4]. Thus, the area enclosed bythe two graphsis

1= [(- 10-6) a4 [(& @?-100) dx0 4=f (169 dx+ f (16x x3) dx4 01 5 1 4\\4= (+ _ sx?) + G -x44 4 4 0

= 64+ 64 = 128.In Exercises 5 and 6, sketch the region between y = sin x and y = cos x over the interval andfindits area.

[73]SOLUTION Overthe interval [7, 7], the graph of y = cos x lies below that of y = sin x (see the sketch below). Hence,the areabetween the two curvesis

1/2 7/2 2 2/ qe Y 2-1.bot (sinx cos x) dx = Goosx six), = (01)a

SECTION 6.1 | Area Between Two Curves 689

y=sinx

y=cosx

SOLUTION Overthe interval [0, Z], the graph of y = sinx lies below that of y = cos x, while over the interval [F, 11], theorientation of the graphs is reversed (see the sketch below). The area between the graphsover[0, x] is then

x/4 a[ (cos x sinx) dx +f (sinx cos x) dx0 x/4

o pe + | . TT= (sinx + cos x) o ( cos x sin x) mt

2 2 2 2Ae= 24/2.

In Exercises 7 and 8, let f(x) = 20+ x x? and g(x) = x? 5x.7. Sketch the region enclosed by the graphs of f(x) and g(x) and computeits area.SOLUTION Setting f(x) = g(x) gives 20 + x x? = x* 5x, whichsimplifies to

0 = 2x 6x 20 = 2(x 5)(x +2).Thus, the curves intersect atx = 2 andx = 5. With y = 204+ x x? being the upper curve (see the sketch below), the areabetween the two curvesis

24\[ 343/ (eo +x x?) (x? - 5x)) dx = / (20 + 6x 2x?) dx = (20x 432-134 =.2 _2 3 Jil. 3y=20+x-x?

y=x?-5x

8. Sketch the region between the graphs of f(x) and g(x) over [4, 8] and compute its area as a sum of two integrals.SOLUTION Setting f(x) = g(x) gives 20+ x x? = x? 5x, which simplifies to

0 = 2x? 6x 20 = 2(x 5)(x + 2).Thus,the curvesintersect at x = 2 and x = 5. Overthe interval [4, 5], y = 20 + x x? is the upper curve but overthe interval[5, 8], y = x? 5x is the upper curve(see the sketch below). The area between the two curvesover the interval [4, 8] is then

[ ((20 + x- x?) - (x? 5x)) a+ [. ((x? - 5x) - 20 + x -x)) dx4

-erro) a+ [(22-02) dx

(-3? 20x) & 20x)3 a M3 5 à 3


9. Find the area between y = e* and y = e?* over[O, 1].SOLUTION Asthe graph of y = e?* lies above the graph of y = e* overtheinterval[0, 1], the area between the graphsis

1 1 Li 1 1 1| (ear(qe = -e?-e-[=--1)=

SECTION 6.1 || Area Between Two Curves 691

In Exercises 13-16, find the area ofthe shaded region in Figures 4-7.13.

y= 3x? +4x 10

FIGURE4

SOLUTION Asthe graph of y = x3 2x? + 10 lies above the graph of y = 3x? + 4x 10, the area of the shaded regionis2Il2/ G 2x2 + 10) (3x2 + 4x 10)) dx / Gé 5x2 4x + 20) dx=2) 2

5 2=x3 2x? + 20%) = ID3 lao à

14.

FIGURE5

SOLUTION Setting 2x = xV1 x2 yields x = 0 or 4 = V1 x2, so that x = +2, Overthe interval pá 0], y = 4x isthe upper curve but over the interval [0, CE y = x~1 x?is the upper curve. The area of the shaded region is then

0 1X XxLa, (5

15. x E x6 3 2FIGURE 6

SOLUTION Theline on the top-left has equation y = BA, and the line on the bottom-right has equation y = x. Thus,thearea to the left of x = & is

6o 343. 3 BE 4 2 » IS 3/37 372 (243 1)xx - x] dx = x*- x ==0 JT 2x1 27 4x a 2x 36 41 36 48The area to the right of x = & is

x/3 3/ (cos. =) dx = (sins _ =)1/6 2x An

(23 1)x y $382 __12V3-12+(V3-2)x48 16 ~ 24

3 3./3-8-xx/6 TS Theentire area is then

692 CHAPTER 6 I] APPLICATIONS OF THE INTEGRAL

16. y=cos 2x

y=sinx

FIGURE 7

SOLUTION Overthe interval [0, 2/6], the graph of y = cos 2x lies above the graph of y = sinx. The orientation of the twographsreverses over [x/6, 51/6] and reverses again over [51:/6, 27r]. Thus, the area between the two graphsis given by

x/6 5x/6 27A = / (cos 2x sinx) dx + / (sin x cos 2x) dx + / (cos 2x sinx) dx.

0 x/6 5r/6

Carrying out the integration and evaluation, we find

1 x/6 1 5x/6 1 2x

A= (5 sin2x + 05x) + (-e05x-zsin2x) + G sin 2x 4006x)2 0 2 x/6 2 5x/6

V3 V3 V3 V3 V3 V3 V3 V3= + -1+ + -|-> >=> ]|+1-|[- -4 2 2 4 2 4 4 2= 343.

In Exercises 17 and 18, find the area between the graphs ofx = sin y and x = 1 cos y overthe given interval (Figure 8).

Y y= sin ya

x=1-cosy

FIGURE 8

17.02./ (sin y (1 cos y)) dy ( cos y y + sin »)| ( 3 + ) (-1) 5Thegraphs cross at y = 0. Since x = 1 cosylies to the right of x = sin y on the interval [ 4, 0] along the y-axis, the areabetweenthe graphs from y = 4 to y = 0 is

6 . . 0 TT T/ ((1 cos y) sin y) dy = (y sin y + cos y)| =1-(-5+1)=5.x/2 7/2Thetotal area between the graphs from y = % to y = + is the sum

0x/2/ (sin y (1 cos y)) dy + / ((1 cos y) sin y) dy =20 x/2 2

JT


19. Findthe area ofthe region lying to the right of x = y? + 4y 22 andto the left of x = 3y + 8.SOLUTION Setting y? + 4y 22 = 3y + 8 yields

0=y?+y-30= (y +60 5),so the two curves intersect at y = 6 and y = 5. The areain question is then given by

5 5 3 2 1331/ (Gy +8) - 0? + 4y -22)) dy = / (-y?-» +30) dy = |-2-F+30)]] = .=6 -6 3 2 620. Find the area of the region lying to the right of x = y? 5 andto theleft of x = 3 y?.SOLUTION Setting y? + 5= 3 y? yields 2y? = 8 or y = +2. Theareaofthe region enclosed by the two graphsis then

2. 2 22 64[,(e-»>-0*+5) o = [_ (&-2°) a = (w-3»°)| -&.2 $ 3 Jo 321. Figure 9 shows the region enclosed by x = y3 26y + 10 and x = 40 6y y3. Matchthe equations with the curves andcompute the area of the region.

FIGURE 9

SOLUTION Substituting y = 0 into the equations for both curvesindicates that the graph of x = y? 26y + 10 passes throughthe point (10,0) while the graph of x = 40 6y? y? passes throughthe point (40, 0). Therefore, over the y-interval [ 1, 3], thegraph of x = 40 6y? y? lies to the right of the graph of x = y? 26y + 10. Theorientation of the two graphsis reversedover the y-interval [ 5, 1]. Hence, the area of the shaded region is

3 ((40 6y? y?) (1? 26y + 10) dy1E (0? 26y + 10) (40 6y? »>)) dy +]-1 3/ (2° + 6y? 26y 30) dy +] (y? 6y? + 26y + 30) dy5 1

Il

1 = 1 3i + 29° 139? 30y + (>5 »* _ 2y3 + 1392 + 30y = 256.5 -1

22. Figure 10 showsthe region enclosed by y =x3 6x and y = 8 3x?. Match the equations with the curves and compute thearea ofthe region.

ll

FIGURE 10 Region between y = x? 6x and y = 8 3x2.SOLUTION Setting x? 6x = 8 3x? yields (x + 1)(x + 4)(x 2) = 0,so the two curvesintersect at x = 4, x = 1 andx = 2. Overtheinterval [-4, 1], y = x3 6x is the uppercurve, while y = 8 3x? is the upper curve overthe interval [ 1, 21.Thearea of the region enclosed by the two curves is thenE (6 6x) (8 3x?) dx + f le 3x1?) (x3 - 6x)) dx4 1 = 1 2? 81 81 81= (gets 8x+ + 8 cut 3x? = + =- .E 4 4 1 4 4 2

694 CHAPTER 6 | APPLICATIONS OF THE INTEGRAL

In Exercises 23 and 24, find the area enclosed by the graphs in two ways: by integrating along the x-axis and byintegrating alongthe y-axis.23. x=9-y?, x=5SOLUTION Along the y-axis, we havepoints of intersection at y = +2. Therefore, the area enclosed by the two curvesisLo-r-90-[ro(ol7Along the x-axis, we have integration limits of x = 5 and x = 9. Therefore, the area enclosed by the two curvesis

9 4 3/2| 32\ 32[ 2/9 xdx = ~(9 x) =0- (5 e5 3 5 3 324. The semicubicalparabola y? = x3 and the line x = 1.SOLUTION Since y? = x3, it follows that x > 0 since y? > 0. Therefore, y = +x3/2, and the area ofthe region enclosed bythe semicubical parabola and x = 1 is

1 1 4 14| (x8? (-x3?)) dx = | 2x3/2 dx = x5/?2} = =,0 0 5 o 5Along the y-axis, we have integration limits of y = +1. Therefore, the area enclosed by the two curvesis

11 UM dy 2 IO = 3 (144) 2 ftLG y ) (> 5 4 (: :) (a+; 5In Exercises 25 and 26, find the area ofthe region using the method (integration alongeither the x- or the y-axis) that requires youto evaluatejust one integral.25. Region between y? = x + 5 and y? = 3 xSOLUTION From the figure below, we see that integration along the x-axis would require two integrals, but integration along they-axis requires only oneintegral. Setting y? 5 = 3 y? yields points ofintersection at y = +2. Thus,the area is given by

[,e-»-0° +5) o= [E(8-27) dy = (5-5 3 Ay2 x=3-y?

EN «

x=y2-5 | 2

26. Region between y = x and x + y = 8 over[2,3]

SOLUTION From the figure below, we see that integration along the y-axis would require three integrals, but integration alongthe x-axis requires only oneintegral. The area of the region is then

3 3/ ((B x) x) dx = (8x x?)| = (24 9) (16 4) =3.2 2

Asa check,the area of a trapezoid is given byh 1_ = (4 = de¿41 + ba) 54+2)=3y


In Exercises 27-44, sketch the region enclosed by the curves and compute its area as an integral along the x- or y-axis.27. y=4-x?, y=x?-4SOLUTION Setting 4 x? = x? 4 yields 2x? = 8orx? = 4. Thus, the curves y = 4 x? and y = x? 4 intersectatx = +2. From the figure below, we see that y = 4 x? lies above y = x? 4 overtheinterval [ 2, 2]; hence, the area of theregion enclosed by the curves is

Fla020) ax= fear - ar 5x3 +

28. y=x?-6, y=6-x°, y-axisSOLUTION Setting x? 6 = 6 x? yields

O= x+ x2 12 = (x 2)(x? + 3x +6),so the curves y = x* 6 and y = 6 x? intersect at x = 2. Using the graph shown below, we see that y = 6 x3 lies abovey = x? 6 overtheinterval[0, 2]; hence, the area of the region enclosed by these curves and the y-axis is

2 3 2 23 2 La 1 3 # m/ (-x3)-@ -9) ax= | (-x3 x? 4.12) dx = (--xt- 223 +12 =.o 4.3 > 3y

29. x+ty=4, x-y=0, y+3x=4SOLUTION From the graph below, wesee that the top of the region enclosed bythe three lines is always bounded by x + y = 4.Onthe other hand, the bottom of the region is bounded by y + 3x = 4 for 0 < x < 1 and by x y = 0 for 1 < x < 2. The totalarea ofthe region is then

1 2/ 2x dx + [ (4 2x) dx0 1x2

1 2[ ((4 x) (4 3x)) dx+ | ((4 x) x) dx0 1

Il 1 2+ (41x x2)|, =1+(8-4-(4-1)=2.

30. y =8-3x, yp=6-x, y=2SOLUTION From the figure below, we see that the graph of y = 6 x lies to the right of the graph of y = 8 3x,so integrationin y is most appropriate for this problem. Setting 8 3x = 6 x yields x = 1, so the y-coordinate of the point of intersectionbetween y = 8 3x and y = 6 x is 5. The area boundedby the three given curvesis thus

5 1a=| (6-»-36-») a5 10 2=| ( -

696 CHAPTER 6 || APPLICATIONS OF THE INTEGRAL10 1 AP(A)(1-96

= 3.

31 y =8 -/x, y=4/x, x =0SOLUTION Setting 8 /x = ./x yields ./x = 4 or x = 16. Usingthe graph shownbelow, we see that y = 8 ./xlies abovey = 4/x overthe interval [0, 16]. The area of the region enclosed by these two curves and the y-axis is then

[Pen dx= [ (86-248) dx = Ge016 16 128

0 3 y

2 4 6 8 10 12 14 16x x32. y= , ==2+1 =F

SOLUTION Setting

= u yields x = 2,0,2.5Fromthe figure below, we see that the graph of y = x/5 lies above the graph of y = x/(x? + 1) over [ 2, 0] andthat theorientation is reversed over [0, 2]. Thus,

0 (x x 2 x xA= ==> dx S -=)dLE ir) x+/ (= =) *0x2 | 2 1 2 x?= To 7 ne +1) > 3 in +D-7 ,

2 1 1 2,-i+- -1n5-2-0(0 s+ 5is) + (Fins E )4In5 -,5

Illl

33. x =|y|, x=1-|ylSOLUTION From the graph below, we see that the region enclosed by the curves x = |y| and x = 1 |y| is symmetric withrespect to the x-axis. We can therefore determine the total area by doubling the area in the first quadrant. For y > 0, settingy=1-yyields y = 4 as the pointof intersection. Moreover, x = 1 |y| = 1 y lies to the right of x = |y| = y, so the totalarea of the region is M 42 \ 1D) = 2 = - =2ar 20), =2(5-3)=3


xls y=x?-6

SOLUTION From the graph below, we see that the region enclosed by the curves y = |x| and y = x? 6 is symmetric withrespect to the y-axis. We can therefore determine the total area of the region by doubling the area of the portion of the regionto theright ofthe y-axis. For x > 0, setting x = x* 6 yields

0=x*-x-6= (x-3)(x +2),| = x lies above y = x? 6. Therefore, the area of the

so the curvesintersect at x = 3. Moreover, on theinterval [0, 3], y =region enclosed by the two curvesis

3 8 ls ls 3 92] (x (x*-6)) dx =2 =x* =x° + 6x}| =2[= -9+18) = 27.0 2 3 o 2

35. x=y?-18y, y+2x =0SOLUTION Setting y? 18y = -3 yields

350 = y? y a=),Y 2? [y 235

470so the points of intersection occur at y = 0 and y = ++5~. Fromthe graph below, we see that both curves are symmetric withrespect to the origin. It follows that the portion of the region enclosed by the curves in the second quadrantis identical to the regionenclosed in the fourth quadrant. We can therefore determinethe total area enclosed by the two curves by doubling the area enclosedin the second quadrant. In the second quadrant, y + 2x = 0 liesto the right ofx = y? 18y,so the total area enclosed by the twocurves is

=2 ee gs SS4 4 0 8 16 8

70/2 35. 1 \|Y792 1225 1225\ 12252 | (-3 - 0° 18y)) dy =2{ y? -y?4 ==,0

x=y- 18y

36. y =xV4x-2, y=-xvVx-2 x=4SOLUTION Note that y = x4x 2 and y = x4x 2 are the upper and lower branches,respectively, of the curve y? =x2(x 2). The area enclosed by this curve and the verticalline x = is

4 4/ (x x=2-(-xvx 2)) dx= | 2xVx 2dx.2 2

Substitute uv = x 2. Then du = dx,x =u +2 and2 2(2u?/? + 4u1/2) du = 512 + 83/2 = sae5 3 o 15Para -/0À atu + 2adu = f0


o2)

2-4

37. x=2y, x+1=(y-1?SOLUTION Setting 2y = (y 1)? 1 yields

0= y? 4y = y(y 4),

so the two curvesintersect at y = 0 and at y = 4. From the graph below, wesee that x = 2ylies to the right of x + 1 = (y 1)?over theinterval [0, 4] along the y-axis. Thus,the area of the region enclosed by the two curvesis

[ @-@-»?-9) =f (x-») a= CADRES

x+1=(y-1)?

38. x + y = 1, 12 + yi/2 =1SOLUTION From the graph below, wesee that the two curves intersect at x = 0 and at xxU2 + y2 = 1. Thearea ofthe region enclosed by the two curvesis then

(a-9-0- 9%) ax= [ (ox+219 ax (2124 482] -j« x) (1 x) x= f ( 2x + 2/x) r= (= + 3% 73

= l and that x + y = 1 lies above

y

39. y=cosx, y=cos2x, x =0,

SOLUTION From the graph below, wesee that y = cos x lies above y = cos 2xovertheinterval [0, 22, The area ofthe regionenclosed by the two curvesis therefore

sin 2x1 2x/3 373

sin x = ,2x/3

/ (cos x cos 2x) dx =0y

y=cosx

40. y=tanx, y=- tanx, x= AIA


SOLUTION Because the graph of y = tanx lies above the graph of y = tanx overthe interval [0, 2/4], the area bounded bythe two curves is

1/4 1/4A= f (tan x ( tan x)) ax=2 | tanx dx

0 0CDS x 7/4

= 2 In| sec2| 0 A 1% Ds= An9 2e 2402. Ji SEC Lin SE CI

2. a41. y=sinx, y=csce*xX, x = 4], y = csc x lies above y = sin x. The area of the region enclosed by the two curvesis then

1/2 1/2/ (esc? x sin x) dx = (= cot x + cos x) = (0 0) 14.2 = A1/4 1/4 2 2

SOLUTION Overthe interval [F, 5

y=sinx

0.2 0.4 0.6 0.8 1.0 1.2 14a 22.x=siny x=- yT

SOLUTION Here, integration along the y-axis will require less work than integration along the x-axis. The curves intersect when2 = sin y or when y = 0, +4. From the graph below, wesee that both curves are symmetric with respectto the origin. It followsthat the portion of the region enclosed by the curves in the first quadrant is identical to the region enclosed inthe third quadrant.We can therefore determine the total area enclosed by the two curves by doubling the area enclosed in the first quadrant. Inthefirstquadrant, x = sin y lies to the right of x = 2, so the total area enclosed by the two curves is

41/2ja 2 12 | siny= y|] dy = 2 cos y y?0 x JE 0

43. y=e*, y=e*%, y=2SOLUTION From the figure below, we see that integration in y would be most appropriate - unfortunately, we have notyet learnedhow to integrate In y. Consequently, we will calculate the area using two integrals in x:

0 In 2A= | Cenas | (2 e*) dxIn2 0

0 In2= (2x +e7*) + (2x e*)

In2 01 ( 2In2 + 2) + (2In2 2) (-1) = 4In2-2.Il


02 04 06

44, y = ~, y=x x

SOLUTION SettingInx (Inx)?I = oa yields x=1,e.x 2

Fromthe figure below, we see that the graph of y = Inx/x lies above the graph of y = (In x)?/x overtheinterval[1, e]. Thus, thearea between the two curvesis

a= ff Inx _ (nx)*)1 x x

e1 2; 1 3(Fe ma) ) 1

45. LAS Plotx 2y= and p= (x1)Vx2 +1

on the sameset of axes. Use a computer algebra systemto find the points of intersection numerically and compute the area betweenthe curves.SOLUTION Using a computer algebra system, we find that the curves

+ 2y= = and y =(x-1)x2 +1intersect at x = 0.3943285581 and at x = 1.942944418. From the graph below,wesee that y = Ton lies above y = (x 1),

xso the area of the region enclosed by the two curvesis

/ (== 1)0.3943285581 \Vx? +1The value of the definite integral was also obtained using a computer algebra system.

46. Sketch a region whoseareais represented by and evaluate using geometry.


SOLUTION Matching the integrand V1 x? |x| with the yrop ypor template for calculating area, we see that the region inquestion is bounded alongthe top by the curve y = V1 x? (the upper half of the unit circle) and is bounded along the bottom bythe curve y = |x|. Hence,the regionis + of the unit circle (see the figure below). The area of the region must then be

l370) =.4

47. ÉS Athletes 1 and 2 run alonga straight track with velocities v1 (1) and v2(t) (in m/s) as shown in Figure 11.(a) Whichof the following is represented by the area of the shaded regionover[0, 10]?

i. The distance between athletes | and 2 at time = 10s.ii. The difference in the distance traveled by the athletes over the time interval[0, 10].

(b) Does Figure 11 give us enough information to determine whois aheadat time = 10 s?(c) Ifthe athletes begin at the same time and place, whois ahead att = 10s? Att = 25 s?

(m/s) vi/

NwFAA

NA

t(s) 5 10 15 20 25 30FIGURE11

SOLUTION(a) The area of the shaded region over [0, 10] represents (ii): the difference in the distance traveled by the athletes over the timeinterval [0, 10].(b) No, Figure 11 does not give us enough information to determine who is ahead at time ¢ = 10 s. We would additionally need toknow the relative position of the runners atí =0s.(c) Ifthe athletes begin at the same time and place, then athlete 1 is ahead at 1 = 10 s because the velocity graph for athlete 1 liesabove the velocity graph for athlete 2 over the interval [0, 10]. Over the interval [10, 25], the velocity graph for athlete 2 lies abovethe velocity graph for athlete 1 and appears to have a larger area than the area between the graphs over [0, 10]. Thus, it appears thatathlete 2 is ahead at f = 25 s.48. Express the area (not signed) of the shaded region in Figure 12 as a sumofthree integrals involving f(x) and g(x).

FIGURE 12SOLUTION Becauseeither the curve boundingthetop ofthe region orthe curve bounding the bottom of the region or both changeat x = 3 and at x = 5,theareais calculated using three integrals. Specifically, the area is3 5 9/ (160 8(0) dx + / (f(x) 0) dx + [ (0 f(x) dx0 3 53 5 9= (f(x) g(x)) ax+ [ 1x| F(x) dx.


249. Find the area enclosed by the curves y = c x? and y = x c as a function ofc. Find the value of c for whichthis area isequal to 1.SOLUTION The curvesintersect at x = +./c, with y = c x* above y = x? c overtheinterval [ ./c, ./c]. The areaof theregion enclosed by the two curvesis then

/ 6 x?) (1? c)) = | (2c 2x?) (20x =Ye Ye 3 yo 3In orderfor the area to equal 1, we must have 8c3/2 = 1, which gives

91/3c= #& 0.520021.4

50. Set up (but do notevaluate) an integral that expresses the area betweenthe circles x? + y? = 2 and x? + (y 1)? = 1.SOLUTION Setting 2 y? = 1 (y 1)? yields y = 1. The twocircles therefore intersect at the points (1,1) and ( 1, 1).From the graph below, we see that over the interval [ 1, 1], the upper half ofthe circle x? + y? = 2 lies above the lower half ofthe circle x? + (y 1)? = 1. The area enclosed by the two circles is therefore given by the integral

f, (V2-2? -a- vi-2)) dx.y

51. Set up (but do not evaluate) an integral that expresses the area between the graphs of y = (1 + x )~! and y = x?.SOLUTION Setting (1 + x*)~! = x? yields x4 + x? 1 = 0. This is a quadratic equation in the variable x2. By the quadraticformula,

x2_ lt J1-4F1) _ -14 V5=a |2

1+V5As x? must be nonnegative, we discard pig. Finally, we find the two curvesintersect at x = +y >*~. From the graphbelow, we see that y = (1 + x2)71 lies above y = x . The area enclosed by the two curvesis then

JE[35(a +)= 12) dx.==1{[»=( +x?)

52. CAS Find a numerical approximation to the area above y = 1 (x/x) and below y = sin x (find the points ofintersectionnumerically).SOLUTION Theregion in question is shown in the figure below. Using a computer algebra system, we find that y = 1 x/m andy = sin x intersect on the left at x = 0.8278585215. Analytically, we determine the two curves intersect on the right at x = x.The area above y = 1 x/z and below y = sinx is then

Nena: (sin o (1 ~ +) dx = 0.8244398727,where the definite integral was evaluated using a computer algebra system.


y=sinx

0 1 2 3

53. LAS Find a numerical approximationto the area above y = |x| and below y = cos x.SOLUTION Theregion in question is shown in the figure below. Wesee that the region is symmetric with respect to the y-axis,so we can determinethe total area of the region by doubling the area of the portion in the first quadrant. Using a computeralgebrasystem, wefind that y = cosx and y = |x| intersect at x = 0.7390851332. The area of the region between the two curvesis then

0.73908513322/ (cos x x) dx = 0.8009772242,0

where the definite integral was evaluated using a computer algebra system.y

y=cosx

54. £AS Use a computeralgebra systemto find a numerical approximationto the numberc (besides zero) in [0, 3), where thecurves y = sin x and y = tan? x intersect. Then find the area enclosed bythe graphs over[0, cl.SOLUTION Theregion in question is shown in the figure below. Using a computer algebra system, we find that y = sinx andy = tan? x intersect atx = 0.6662394325. The area of the region enclosed by the two curvesis then

0.6662394325| (sinx tan? x) dx = 0.09393667698,0

where the definite integral was evaluated using a computer algebra system.y

y=sinxy=tan?x

0 02 04 0.6

55. The back of Jon s guitar (Figure 13) is 19 inches long. Jon measured the width at 1-in. intervals, beginning and ending 3 in.from the ends, obtaining the results

6, 9, 10.25, 10.75, 10.75, 10.25, 9.75, 9.5, 10, 11.25,12.75, 13.75, 14.25, 14.5, 14.5, 14, 13.25, 11.25, 9

Use the midpoint rule to estimate the area of the back. FIGURE13 Back ofguitar.


SOLUTION Note that the measurements were takenat the midpoint of each one-inch section of the guitar. For example, in the 0to 1 inch section, the midpoint would be at 4 inch, and thus the approximate area ofthe first rectangle would be 1 - 6 inches . Anapproximationfor the entire area is then

A Il 1(6 + 9 + 10.25 + 10.75 + 10.75 + 10.25 + 9.75 + 9.5 + 10 + 11.25+ 12.75 + 13.75 + 14.25 + 14.5 + 14.5 + 14 + 13.25 + 11.25 + 9)214.75 in?.Il

56. Referring to Figure 1 at the beginning ofthis section, estimate the projected number of additional joules produced in the years2009-2030 asa result of government stimulus spending in 2009-2010. Note: One watt is equal to one joule per second, and onegigawatt is 10? watts.SOLUTION We make some rough estimates of the areas depicted in Figure 1. From 2009 through 2012, the area between thecurves is roughly a right triangle with a base of 3 and a height of 40; from 2012 through 2020, the area is roughly an 8 by 40rectangle. Finally, from 2020 through 2030, the area is roughly a trapezoid with height 10 and bases 40 and 27. Thus, additionalenergy produced is approximately

1 136) (40) + 8(40) + 300) (40 + 27) = 715 gigawatt-years.

Because 1 gigawatt is equal to 10? joules per second and 1 year (assuming 365 days) is equal to 31536000 seconds, the additionaljoules producedin the years 2009-2030 asa result of governmentstimulus spending in 2009-2010 is approximately 2.25 x 1017.Exercises 57 and 58 use the notation and results ofExercises 49-51 ofSection 3.4. Fora given country, F(r) is thefraction oftotalincome that goes to the bottom rth fraction ofhouseholds. The graph ofy = F(r)is called the Lorenz curve.

57. ÉS Let A be the area between y = r and y = F(r) over the interval [0, 1] (Figure 14). The Gini indexis the ratioG = A/B, where B is the area under y = rover[0,1].

1(a) Show that G = 2f (r F(r)) ar.0(b) Calculate G if

r forO


(b) Withthe given F(r),

a Il 1/2 1 ! § 22 | r cr dr+2] r=(5r==)) dr0 3 1/2 3 34 1/2 4 1

=; | r dr (r=-1) dr03 3 1/2

2, PP 4/1 1= =r? ==73 lo 3 \2 1/2_1_4/(_1),4(_3)_126 31 2) 31 8) 3

(c) If F(r) =r,then1

6=2/ (r r)dr=0.0(d) If F(r) = 0 for0


60. £AS Let c be the number such that the area under y = sin x over [0, x] is divided in half by the line y = cx (Figure 15).Find an equation for c and solve this equation numerically using a computer algebra system.

FIGURE 15

SOLUTION First note thatJT TmÎ sin x dx = cos x| =2,

0 0sina _ _ sina7 and y = cx = "7x. Then Now, let y = cx and y = sin x intersect at x = a. Then ca = sina, which gives c =

af. sina sina \|° asinasinx x}) dx = cos x x = 1 cos a .0 a 2a 0 2

We needi 11 cosa TT = (2) = 1,2 2

which gives a = 2.458714176 and finallysin ac = = 0.2566498570.a

61. ÉS Explain geometrically (without calculation):1 1/ x" dx+ | xn dy =1 (for n > 0)0 0

SOLUTION Let A, denote the area of region 1 in the figure below. Define A2 and A3 similarly. It is clear from the figure thatA, + A2+ A3 = 1.

Now, note that x and x!/ are inverses of each other. Therefore, the graphs of y = x and y = x1/n are symmetric about theline y = x, so regions 1 and 3 are also symmetric about y = x. This guarantees that 41 = A3. Finally,

1 1/ x" dx + | xl/" dx = Az + (Aa + A3) = Ai + A2 + 43 = 1.0 0

y

62. ES Let f(x) be an increasing function with inverse g(x). Explain geometrically:

/ /0 F(0)a

SOLUTION The region whose areais represented by / F(x) dx is shown as the shaded portion of the graph below on theleft,0Fa)and the region whose area is represented by / g(x) dx is shown as the shaded portion of the graph below on the right. Because

F(0)f and g are inverse functions, the graph of y = f(x)is obtained byreflecting the graph of y = g(x) throughthe line y = x. It

SECTION 6.2 | Setting Up Integrals: Volume, Density, Average Value 707

then follows that if we were to reflect the shaded region in the graph below on the right through the line y = x, the reflected regionwould coincide exactly with the region R in the graph below on the left. Thus

[ /0 F(0)Fa)

oka

a o Na)

6.2 Setting Up Integrals: Volume, Density, Average Value

Preliminary Questions1. Whatis the average value of f(x) on [0, 4] if the area between the graph of f(x) and the x-axis is equal to 12?SOLUTION Assuming that f(x) > 0 over the interval[0, 4], the fact that the area between the graph offand the x-axis is equal to12 indicates that to f(x) dx = 12. The average value off'over the interval [0, 4] is then

Jo fx) dx 124-0 4

2. Find the volumeofa solid extending from y = 2 to y = 5 if every cross section has area A(y) = 5.3;

SOLUTION Because the cross-sectional area of the solid is constant, the volume is simply the cross-sectional area times thelength, or 5 x 3 = 15.3. What is the definition offlow rate?SOLUTION The flow rate of a fluid is the volumeoffluid that passes through a cross-sectional area at a given point perunit time.4. Which assumption about fluid velocity did we use to compute the flow rate as an integral?SOLUTION To express flow rate as an integral, we assumedthat the fluid velocity depended only on theradial distance from thecenter of the tube.

45. The average value of f(x) on [1, 4] is 5. Find Î f(x) dx.1SOLUTION

4/ F(x) dx = average value on [1, 4] x length of[1, 4]1

=5x3=15. Exercises1. Let V be the volume of a pyramid of height 20 whose baseis a square ofside 8.

(a) Use similartriangles as in Example | to find the area ofthe horizontal cross section at a height y.(b) Calculate V by integrating the cross-sectional area.SOLUTION(a) Wecan use similar triangles to determine the side length, s, of the square cross section at height y. Using the diagram below,wefind

8 Ss 5Bo = £(20- y).2 2-7 * *=300-yThearea ofthe cross section at height y is then given by 35 (20 y)?.


|/\ A

20 4 2 4 3| z@-» = -5@0-»(b) The volumeof the pyramid is

20 12800 3

2. Let V be the volumeofa right circular cone of height 10 whose baseis a circle of radius 4 [Figure 1(A)].(a) Use similar triangles to find the area of a horizontal cross sectionat a height y.(b) Calculate V by integrating the cross-sectional area.

(A) (B)

FIGURE 1 Rightcircular cones.

SOLUTION(a) Ifris the radius at height y (see Figure 1), then

1010 y4 r

from similar triangles, which implies that r = 4 3 y. The area of the cross-section at height y is then

(b) The volumeofthe cone is

v-[" a AVo li? _ 1607FO TEAA 5 om3. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose baseis a circle of

radius R [Figure 1(B)].SOLUTION(a) From similar triangles (see Figure 1),

h Rh-y rn

wherero is the radius of the cone at a height of y. Thus, rg = R 77.(b) The volume ofthe cone is

a 6 integral solution yes. sincef(x) > oandg(x) < 0 ......692 chapter 6 i] applicationsoftheintegral...

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