SECTION 5.2 | The Definite Integral 571

Similarly, if f(x) is monotone decreasing,

b Ly = 7 2% Saw) > Mn = Y aD > kv= % fee)k=0 k=0 k=1

Thus, if f(x) is monotonic, then My always lies in between Ry and Ly.Now,as in Figure 6, consider the typical subinterval [x; , xj] and its midpoint x;*. We let A, B,C, D, E, and F bethe areas

as shownin Figure 6. Note that, by the fact that x is the midpointof the interval, A= D+ EandF = B+C.Let Er representthe right endpoint approximation error (= A+ B+ D), let Ey represent the left endpoint approximation error( = C + F + E)and let Ey represent the midpoint approximation error ( = |B E}).

o If B > E, then Ey = B E.In this case,Er-Em =A+B+D-(B-E)=A+D+E>0,

so Er > Ey, whileEL-=Em=C+F+E-(B-E)=C+(B+C)+E-(B-E)=2C+2E>0,

so Ey > Ey. Therefore, the midpoint approximation is more accurate than either the left or the right endpoint approximation.e If B < E, then Ey = E B. In this case,

Er-Em =A+B+D-(E-B)J=D+E+D-(E-B)=2D+B>0,so that ER > Ey while

EL-Em=C+F+E-(E-B)=C+F+B>0,so Ey > Ey. Therefore, the midpoint approximation is more accurate than either the right or the left endpoint approximation.

o If B = E,the midpoint approximation is exactly equal to the area.Hence, for B < E, B > E, or B = E,the midpoint approximation is more accurate than either the left endpoint or the rightendpoint approximation.

5.2 The Definite Integral

Preliminary Questions5

1. What is | dx [the function is f(x) = 1]?3

5. 9SOLUTION / dx =f 1-dx = 1(5 3) = 2.

3 37

2. Let] = / f(x) dx, where f(x) is continuous. State whethertrue orfalse:2(a) 7 is the area between the graph and the x-axis over[2, 7].(b) If f(x) > 0, then 7 is the area between the graph and the x-axis over [2, 7].(c) If f(x) < 0, then / is the area between the graph of f(x) and the x-axis over[2, 7].SOLUTION(a) False. /,E f(x) dx is the signed area between the graph andthe x-axis.(b) True.(c) True.

JT3. Explain graphically: / cosx dx =0.0SOLUTION Because cos(x x) = cos x, the negative area between the graph of y = cos x and the x-axis over [4, x]exactly cancels the positive area between the graph and the x-axis over [0, 4].

-5 -14. Which is negative, / 8dx or / 8 dx?-1 5

=5SOLUTION Because 5 ( 1) = 4, / 8 dx is negative.-1

572 CHAPTER 5 || THEINTEGRAL ExercisesIn Exercises 1-10, draw a graph ofthe signed area represented bythe integral and compute it using geometry.

31. | 2x dx3SOLUTION The region bounded by the graph of y = 2x and the x-axis overthe interval [ 3, 3] consists of two right triangles.One has area 4(3) (6) = 9 below the axis, and the other has area 36) (6) = 9 above the axis. Hence,

Î3

32; / (2x + 4) dx2SOLUTION The region bounded by the graph of y = 2x + 4 and the x-axis over the interval [ 2, 3] consists of a single righttriangle of area 5 (5)(10) = 25 abovethe axis. Hence,

3/ (2x + 4) dx = 25.2

13; / (3x + 4) dx2

SOLUTION The region boundedbythe graph ofy = 3x + 4 and the x-axis over the interval [ 2,1] consists oftworight triangles.Onehas area 120) = = below the axis, and the other has area 20 = 2 above the axis. Hence,

49 2 151 3x +4) dx = ===,[er +90 673 >

1

4. / 4dx2

SOLUTION Theregion boundedbythe graph of y = 4 and the x-axis over the interval [ 2, 1] is a rectangle of area (3)(4) = 12above the axis. Hence,

1/ 4 dx = 12.2

SECTION 5.2 | The Definite Integral 573

85. / (7 x) dx6SOLUTION The region bounded by the graph of y = 7 x and the x-axis over the interval [6, 8] consists of two right triangles.Onetriangle has area 400) = 4 abovethe axis, and the other has area 300) = 3 below the axis. Hence,

31/26. / sinx dx

I

3x1 consists of two parts of equal12

SOLUTION The region boundedbythe graph of y = sin x and the x-axis overthe interval [5area, one abovethe axis and the other below the axis. Hence,

|x/2

5

T, / V25 x?2 dx0

SOLUTION The region bounded by the graph of y V25 x2 and the x-axis overthe interval [0, 5] is one-quarter of a circle ofradius 5. Hence,

2574

> 1/ V25 x2 dx = n(5)? = .0 4y

3

8. / |x| dx2SOLUTION The region bounded by the graph of y |x| and the x-axis over the interval [ 2, 3] consists of tworight triangles,

4(2) (2) = 2, and the other has area (3) (3) = 3: Hence,both above the axis. One triangle has area$ 9 13Lu

574 CHAPTER 5 | THEINTEGRAL

29. Le- |x|) dx

SOLUTION Theregion boundedbythe graph of y = 2 |x| and the x-axis over the interval [ 2, 2] is a triangle above the axiswith base 4 and height 2. Consequently,

2 1[e-bbax =F=4.2y

510. / 3+x-2|x]|) dx2SOLUTION The region bounded bythe graph of y = 3 + x 2|x| and the x-axis over the interval [ 2, 5] consists of a trianglebelow the axis with base 1 and height 3, a triangle above the axis of base 4 and height 3 and a triangle below the axis of base 2 andheight 2. Consequently,

[ (3+x 2]x|) dx = loc + La) = 200) =?= P 2 2 2 2°

w+ T x1 N-2

1011. Calculate | (8 x) dx in two ways:0(a) As the limit lim RyN 00(b) By sketching the relevant signed area and using geometry

llSOLUTION Let f(x) = 8 over [0, 10]. Consider the integral So f(x) dx 0° x) dx.(a) Let N be a positive integer and seta = 0, b = 10, Ax = (b a)/N = 10/N. Also, let x, = a+kAx = 10k/N,k = 1,2,...,N bethe right endpoints of the N subintervals of [0, 10]. Then

N N N N10 10k 0 10Ry = Ax ==> 8- )= [8 1}- kw= dx) fo= 0 (b- 7) le 01)- y Ek=1 k=1 k=1 k=1_ 10 fem 10 N N _ 39 2N NN 2" 2)) > N°(50 7)N>o N>00 N

(b) The region bounded by the graph of y = 8 x and the x-axis over the interval [0, 10] consists of two right triangles. Onetriangle has area (8) (8) = 32 abovethe axis, and the other has area 30) (2) = 2 below the axis. Hence,

10 (8 x) dx = 32-2 = 30.0

SECTION 5.2 | The Definite Integral

575

412. Calculate / (4x 8) dx in two ways: As the limit ye Ry and using geometry.= >00

4 4SOLUTION Let f(x) = 4x 8 over [ 1, 4]. Consider the integral / f(x) dx = / (4x 8) dx.1 1

o Let N be positive integer and set a = 1, h = 4, Ax (b a)/N = S/N.Then xx = a + kAXx = 1 + 5Kk/N,k =1,2,...,N are the right endpoints of the N subintervals of [ 1, 4]. Then

N N N N5 20k 60 100k=1 k= k= k=1

_ O4 100 NNN N?\2 2= 60 + 50 + m 10+ =- N N°

H lim R li 10+ 7) 10ence lim = lim [- )=-10.N 00 a N oo Ne The region bounded by the graph ofy» = 4x 8 and the x-axis over the interval [ 1, 4] consists of a triangle below the axiswith base 3 and height 12 and triangle above the axis with base 2 and height 8. Hence,

4/ (4x 8)dx = 2(3)(12) + 2 (2)(8) = -10.1 3 a

In Exercises 13 and 14, refer to Figure 1.

FIGURE 1 The twoparts of the graph are semicircles.

2 613. Evaluate: (a) [ f(x) dx (b) / f(x) dx

0 0SOLUTION Let f(x) be given by Figure 1.(a) The definite integral E f(x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis. Therefore,

A wall(b) The definite integral E f(x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis and a semicircle ofradius 2 which lies above the x-axis. Therefore,

$ I > 1 2 _ 37[ F(x) dx = 5(2) 570) = 5

576 CHAPTER 5 | THEINTEGRAL4 6

14. Evaluate: @ | f(x) dx (b) i | f(x)| dx

SOLUTION Let f(x) be given by Figure1.(a) The definite integral f7 F(x) dx is the signed area of one-quarter of a circle of radius 1 which lies below the x-axis andone-quarter ofa circle of radius 2 which lies above the x-axis. Therefore,

4 1 1 3Proa ino? Fn? = x(b) The definite integral E |.f(x)| dx is the signed area of one-quarter ofa circle of radius 1 and a semicircle of radius 2, both ofwhichlie above the x-axis. Therefore,

on6 1 1[rot = 380: + ry?=In Exercises 15 and 16, refer to Figure 2.

FIGURE 23 5g(t) dt and | g(t) at.15. Evaluate /

30SOLUTION

e The region bounded by the curve y = g(1) and the r-axis overthe interval [0, 3] is comprised of tworighttriangles, one witharea 3 belowthe axis, and one with area 2 above the axis. The definite integral is therefore equal to 2 4 = 3.

e The region bounded by the curve » = g(f) and the 1-axis over the interval [3, 5] is comprised of another two righttriangles,one with area 1 above the axis and one with area 1 below the axis. The definite integral is therefore equalto 0.

ca16. Find a, b, and c suchthat / g(t) dt and / g(t) dt are as large as possible.

b0a

SOLUTION To make the value of | g(t) dt as large as possible, we wantto include as muchpositive area as possible. This0 chappens when we take a = 4. Now, to make the value of / g(t) dt as large as possible, we want to make sure to include all of

bthe positive area and only the positive area. This happens when we take b = 1 andc = 4,17. Describe the partition P and the set of sample points C for the Riemann sum shown in Figure 3. Compute the value of theRiemann sum.

sl ENI Nez0.5 1 225332 45 5

FIGURE3

SOLUTION Thepartition P is defined by

x=0 < x, =1 < x=25 < xg=32 < x4=5

The set of sample pointsis given by C = {c = 0.5,c2 = 2,c3 = 3,c4 = 4.5}. Finally, the value of the Riemann sum is34.25(1 0) + 20(2.5 1) + 88.2 2.5) + 15(5 3.2) = 96.85.

SECTION 5.2 || The Definite Integral 577

18. Compute R(f, P, C) for f(x) = x? + x forthe partition P andthe set of sample points C in Figure 3.SOLUTION

IlR(F, P,C)= f(0.5)(1 0) + f(2)(2.5 1) + f(3)(3.2 2.5) + f(4.5)(5 3.2)0.75(1) + 6(1.5) + 12(0.7) + 24.75(1.8) = 62.7Il

In Exercises 19-22, calculate the Riemann sum R(f, P,C) for the given function, partition, and choice ofsample points. Also,sketch the graph off and the rectangles corresponding to R(f, P, C).19. f(x) =x, P = {1,1.2,1.5,23, C = {1.1,1.4,1.9SOLUTION Let f(x) = x. With

P = {xo = 1,201 = 12,20, = 1.5,x3 = 2) and C = {cy = 1.1,c2 = 1.4, c3 = 1.9},weget

R(f, P,C) = Ax] f(c1) + Axo f(c2) + Axz f(c3)(1.2 1)(1.1) + (1.5 1.2)(1.4) + (2 1.5)(1.9) = 1.59,Il

Hereis a sketch of the graph of f and the rectangles.

| 05 1 15 2 2520. f(x) = 2x +3, P = { 4,-1,1,4,8}, C = { 3,0,2,5}SOLUTION Let f(x) = 2x + 3. With

P = {x = -4,x1 = -1,x2 = 1,x3 = 4,x4 = 8} and C = {cy = 3,co = 0,c3 = 2,c4 = 5},weget

RF, P,C) = Ax f(c ) + Ax2 f(c2) + Ax3 f(c3) + Axa f(ca)= (-1 (-4))(-3) + (1 (-1))G) + (4 1)(7) + (8 4)(13) = 70.

Hereis a sketch of the graph of f and the rectangles.

2. f(x) =x? +x, P = {2,3,4.5,5}, C = {2,3.5,5}SOLUTION Let f(x) = x? + x. With

P = {xo = 2,x1 = 3,x3 = 4.5, x4 = 5) and C = {cy = 2,c2 = 3.5,c3 = 5},

wegetR(F, P,C) Axi f(c1) + Ax2 f(c2) + Ax3 f(c3)

(3 2)(6) + (4.5 3) (15.75) + (5 4.5) (30) = 44.625.Il

Here is a sketch of the graph of f and the rectangles.y

302520 yA15 7

578 CHAPTER 5 | THEINTEGRAL

22. f(x) =sinx, PSOLUTION Let f(x)

I 2 LL ZY C=(04,0.7,1.2)xIl n me 5

P= (xo =0,x1 = 2x3 = 5x4 = 3} and C = {cy = 0.4, c2 = 0.7, c3 = 1.2),we get

R(f#,P,C) Il Ax1 f(c1) + Ax2 f(c2) + Ax3 f(c3)E 0) (sin 0.4) + E = =) (sin 0.7) + (5 = 5) (sin 1.2) = 1.029225.Il

Here is a sketch of the graph of f and the rectangles.y

10.840.6 + oe0.40.2

+ + ++ + + x0.2 0.4 0.6 0.8 1 1.214 1.6

In Exercises 23-28, sketch the signed area representedby the integral. Indicate the regions ofpositive and negative area.5

23. / (4x x?) dx0SOLUTION Here is a sketch ofthe signed area represented by the integral So (4x x?) dx.

/1/4

SOLUTION Hereis a sketch of the signed area represented by the integral /er5/4 tan x dx.

0.2 0.4 0.6

2725. Î sin x dx

JT

SOLUTION Hereis a sketch of the signed area represented by the integral ñsin x dx.

3x26. Î sin x dx0

SECTION 5.2 || The Definite Integral 579

SOLUTION Here is a sketch of the signed area represented by the integral So* sin x dx.

Î1/2SOLUTION Hereis a sketch of the signed area represented by the integral / In x dx.

0.60.40.2

-0.2-0.4-0.6

/1

SOLUTION Here is a sketch of the signed area represented by the integral A, tan ! x dx.

4 eo ty = u m

In Exercises 29-32, determine the sign ofthe integral without calculating it. Draw a graph ifnecessary.1

29. | xt dx2SOLUTION Theintegrandis alwayspositive. The integral must therefore be positive, since the signed area has only positive part.

130. / x dx2SOLUTION By symmetry, the positive area from theinterval [0, 1] is cancelled by the negative area from [ 1, 0]. With the interval[ 2, 1] contributing more negative area, the definite integral must be negative.

27/ xsinx dx0

SOLUTION Asyou can see from the graph below,the area below the axis is greater than the area above the axis. Thus, the definiteintegral is negative.

580 CHAPTER 5 | THE INTEGRAL

SOLUTION From the plot below, you can see that the area above the axis is bigger than the area below the axis, hence the integralis positive.

In Exercises 33-42, use properties ofthe integral and theformulas in the summary to calculate the integrals.4

33. Î (61 3)dt04 4 4 1SOLUTION / (61 3) dí = 6] tdt =: | 1dt 6: 3 4° -34 0) = 36.

0 0 0234. / (4x + 7) dx3

SOLUTION2 2 2

Î Gx+ Dax 4] xax +7 dx3 3 30 2

-4([ xax+ [ x dx + 7(2 ( 3))3 02 3

= 4 / xdx- | x dx +350 0=4(12 IL 3)? +35 = 25_ A2 2 2

935. I x? dx0

? 1SOLUTION By formula 6, / x* dx = 30) = 243.05

36. / x? dx25 5 2 1 1SOLUTION / x? dx = / x? dx - [ x2dx = (5)? = (2)? = 39.

2 0 0 3 31

37. Î (u? 2u) du0

SOLUTION

Pu 2u) du [ 24 2f d Lay of! (1)? ry 22u = u du udu == =21 = =--1=--.0 0 ! 0 3 2 3 3

1/238. / (12y? + 6y) dy0

SOLUTION

| an ©o

De a D=D & + On

S, = D

Y ES $

1/2Î (12y? + 6y) dy =0

Il mi wDl]

CA:

NIA

w+ a

Diem

PS

Nile

NSD

SECTION 5.2 | The Definite Integral 581

so. |3

SOLUTION First, write

Il1 0 1/ (1? +1 +1)d / ++ Dart | (1? +1 +1)dt3 3 0-3 1=-f OP +r+nare PoP ere nat

0 0Then,

1 1 1 19 7 1 196e E selle(+3-3)+(@+3+1)-3

Il

1/ (1? +1 +1)dLa

ll

340. / (9x 4x ) dx3

SOLUTION First write3 0 3/ (9x 4x?) dx = / (9x 4x?) dx + Î (9x 4x?) dx3 3 0

3 3= -/ (9x 4x?) dx + / (9x 4x?) dx.0 0

Then,

Il3Les 4x?) dx G¿y 4. 5) + 6 50? 4. 50)

Ms + sl 36) = -722 2 >Il

141. (x? + x) dx

a

SOLUTION First, IMez + x) dx = p x? dx + E xdx= 4b? + 37. Therefore1(x? + x) dx

=a

Il 0 1 1 arar| (? +x)dx= | (a)dx | (x? + x) dxa 0 0 0

1 1 1 1 1 1 5= (;+ 3: 2) (5-0 + sa) = ra 20? +2.2a

a. | x? dxa

SOLUTION2 2a 4 2 1 3 1 1 1[ Pax = f sax | x? dx = = (a?) -3 (a)? = 20° - za .

In Exercises 43-47, calculate the integral, assuming that5 5

/ FG) dx=5, / g(x) dx = 120 0

543. / (16) +8 (1)ax5 5 5

SOLUTION Î (f(x) + g(x)) dx = / Fo) dx + / g(x)dx =5412=17.0 0 044. [ (216) _ 320) dx

582 CHAPTER 5 | THEINTEGRAL

15 5 5SOLUTION / (re) 5500) dx = 2] f(x) dx af g(x) dx = 2(5) 302 = 6.

0ss. | g(x) dx5 0 5SOLUTION Î g(x) dx = - | g(x) dx = 12.5 0

546. [ (f(x) x) dx

5 5SOLUTION Poo x)dx = / f(x) dx - [ xdx=5- 56 =a0 0 0

5/0

SOLUTION It is not possible to calculate 5 g(x)f(x) dx from the information given.48. Prove by computing the limit of right-endpoint approximations:

b btfa [9]0 4

SOLUTION Let f(x) = x3,a = 0 and Ax = (b a)/N =b/N.ThenN N Nb b3 b* b4 [N4 N? N? b* pt b*Ry = Ax == kB.

|

= Pio

1

ep S| ee en ee,w= Ax) /00=3 2 ye)

=

ya

|

UF wela tata 41 oN + aN?k=1 k=1 k=1

b bt b4 b* b4H 3dx = lim Ry= li | e,ence |2INCFenIn Exercises 49-54, evaluate the integral using theformulas in the summary and Eq. (9).

3a. [ a? de0

3 34 8SOLUTION By Eq.(9), [ idx ==.o 4. 4350. Î x3 dx1 3 3 1 1 1SOLUTION / x3 dx -/ x? dx| x3 dx = -(3)* - -(1)? = 2%.1 0 0 4 43

51. / (x x3) dx03 3 3

SOLUTION Î (x x°)dx = / xdx - [ x? dx =0 0 0Le = 144 = ae2 4 4

152. Î (2x3 x + 4) dx

0SOLUTION Applyingthe linearity of the definite integral, Eq. (9), the formula from Example 4 and the formula forthe definiteintegral of a constant:

1 1 1 1 1 1| er-x+0a=2| «dx[ xdx+ [ 4dx =2--(1)4-=(1)? +4=4.0 0 0 0 4 21

53. Î (12x? + 24x? 8x) dx0SOLUTION

1 1 12| dx +24 f 3/ xdx0 0 0

1 1 112:14 494: 29 -3+=123 T3 2=3+8-4=7

1/ (12x3 + 24x? 8x) dx0

SECTION 5.2 | The Definite Integral2st. | (2x3 3x?) dx2

SOLUTION

2 0 2/ (2x3 3x7) dx = / (2x9 3x?) dx+ / (2x3 3x?) dx2 2 0

2 2= | (2x3 3x2) dx / (2x3 3x2) dx0 0

2 2 2 2=2f dxf dx 2f xdx+o / x? dx0 0 0 0

2.4 (43.1 3 2. en 3.4 23=2:30 30 -2 48:22=8-8-8-8=-16.

In Exercises 55-58, calculate the integral, assuming that1 2 4

/ f(x) dx = 1, / f(x) dx = 4, / f(x) dx =70 0 14

55. / f(x) dx04 1 4

SOLUTION Î f(x) dx = / f(x) dx +] f(x)dx =14+7=8.0 0 12

56. / F(x) dx12 2 1

SOLUTION / f(x) dx = / T(x) dx - [ fa)dx=4-1=3.1 0 0

157. / F(x) dx

41 4

SOLUTION / f(x) dx = -/ f(x) dx = 7.4 14

ss. | f(x) dx2SOLUTION From Exercise 55, E F(x) dx = 8. Accordingly,

4 4 2/ F(x) dx = | fxyax | f(x)dx =8-4=4.2 0 0

In Exercises 59-62, express each integralas a single integral.3 7

59, [ fax + | f(x) dx0 3sovunion [soya + [ra [roO O = .o x) dx , x) dx . x) dx

9 960. / Wax| f(x) dx3. 4

9 9 4 9 9SOLUTION / FO) ax- | f(x) dx = / f(x) dx +/ f(x) dx -| FO) dx = f(x) dx.2 4 2 4 49 5

61. / Seras FO) dx2 2

9 5 5 9 5 9SOLUTION L'roax- f(x) dx = [ sears f FO) dx -/ f(x) dx = F(x) dx.5

583

584 CHAPTER 5 | THEINTEGRAL3 9

oa. | rar + | f(x) dxÀ 3

3 9 7 7 9 9SOLUTION / FO) dx +] f(x) dx = -{ F(x) dx + / f(x) dx +f f(x) dx = / f(x) dx.

7 3 3 3 q 7

bIn Exercises 63-66, calculate the integral, assuming that f is integrable and / fa)dx=1- b7* for all b > 0.

15

63. Î f(x) dx15 4SOLUTION / f@)dx=1-5"1 = =1

564. Î f(x) dx

3 5 5 3SOLUTION / f(x) dx = / FO) dx -| f(x) dx = (: - 5) - (: _ 5) = =3 1 1

665. / (f(x) 4) dx1

6 6 6SOLUTION / (3/(%) 4) dx = | f(x) dx -4f 14x = 3(1 671) 4(6 1) =1

166. F(x) dx

1 1/2 ySOLUTION f(x) dx = - [ fa)dx=-I1- (5) =];

1/2 1 2b b

67. Es Explain the difference in graphical interpretation between / f(x) dx and / | f(x)| dx.a a

SOLUTION When f(x) takes on both positive and negative valueson [a, b], /a f(x) dx represents the signed area between f(x)and the x-axis, whereas /,E |f(x)| dx represents the total (unsigned) area between f(x) and the x-axis. Any negatively signedareas that were part of JL2M (x) dx are regarded as positive areas in JLD |f(x)| dx. Here is a graphical example of this phenomenon.

Graph off(x) Graph of|/()|

68. Es Use the graphical interpretation of the definite integral to explain the inequality

/" Fen as| < ["alawhere f(x) is continuous. Explain also why equality holds if and only if either f(x) > 0 for all x or f(x) < 0 forall x.SOLUTION Let A+ denote the unsigned area under the graph of y = f(x) overthe interval [a,b] where f(x) > 0. Similarly,letA- denote the unsigned area when f(x) < 0. Then

b/ f(x) dx = Az A.

Moreover,b b[ teas} saat a= [seal ae.

Equality holds if and only if one of the unsigned areas is equal to zero; in other words, if and only if either f(x) > 0 for all x orf(x) < 0 for all x.

SECTION 5.2 | The Definite Integral 585

69. ÉS Let f(x) = x. Find aninterval [a, b] such thatb b/ fajdx[=2 and [ (FOIX = 5

SOLUTION Ifa > 0, then f(x) > 0 for all x [a,b], so

/"fa /1100|dxby the previous exercise. We find a similarresult if b < 0. Thus, we must have a < 0 and h > 0. Now,

[F resiar = Le + Lei x)| dx = za 5bBecause

then

b 1/ f(x) dx} = -|b? -a?|.a 2Ifb? > a , then

ln lo 3 Ion |=a? 4+ b% == and (b2 = -+7 md =>yield a = 1 and b = /2. Ontheother hand, if b? < a2, then

1, 15 3 Ly 1= b* = d =-(a° b*)=-=20 +59 =; and a =;yield a = 4/2 and b = 1.

27 2770. ES Evaluate J = [ sin? x dx and J = | cos? x dx as follows. First show with a graph that 7 = J. Thenprove0 0that J + J = 2x.

SOLUTION The graphs of f(x) = sin? x and g(x) = cos x are shown below atthe left and right, respectively. It is clear thatthe shadedareas in the two graphs are equal, thus

20 20i= | six dx = | cos? x dx = J.0 0

Now, using the fundamental trigonometric identity, we find27 2x

1+1= | (sin? x + cos? x) dx = | 1-dx = 2n.0 0Combining this last result with J = J yields J = J =x.

y y

In Exercises 71-74, calculate the integral.

71. Po-xa0

586 CHAPTER 5 | THE INTEGRAL

SOLUTION Overthe interval, the region between the curve and the interval [0, 6] consists of two triangles above the x axis, eachofwhich has height 3 and width 3, and so area 3. Thetotal area, hence the definite integral, is 9.

Alternately,

[p-sax = Poor[aaa

3 3 6 3 6=3/ ax- | x dx + Î xdx- | x dx -3[ dx

0 0 0 0 31 1 1=9--2 4-8 --37-9=9.> +7 2

3n. | 12x 4| dx1

SOLUTION The area between |2x 4] and the x axis consists of two triangles above the x-axis, each with width 1 and height2,and hence with area 1. The total area, and hencethe definite integral, is 2.

Alternately,3 2 3

/ 24 dx = f (4 2x) dx + | (2x 4) dx1 1 22 2 1 3 2 3

-4[ dx 2 / xax- | xdx|+2 / xdx- | x dx -4f dx1 0 0 0 0 2

=4-2|-22--12) +22 2 2 2173. / |x3| dx1

SOLUTION

3 x>03, jxI=) 3x|x x <0.

Therefore,1 0 1 1 1 1 1 1

/ Sax = | +| sax= | dx| x3 dx = (-1)* + (1)* = =.1 -1 0 0 0 4 4 2

274, / lx? 1| dx0

SOLUTION

1-1 1<x<22x*=1|=| | (121) O<x<l.Therefore,

2 1 2/ 2 11dx= | Ajax | (2 1) dx0 0 1

SECTION 5.2 || The Definite Integral1 1 2 1 2

-/ ax- | x? dx + | sax| x? dx -[ 1dx0 0 0 0 11 1 1

75. Use the Comparison Theorem to show that1 1 2 2

/ x dx </ x4 dx, | xt dx < / x dx0 0 1 1SOLUTION Ontheinterval [0, 1], x? < x*, so, by Theorem 5,

1 1/ x? dx < / x4 dx.0 0

Onthe other hand, x* < x* for x [1,2], so, by the same Theorem,2 2

/ x4 dx < / x? dx.1 11 6] 176. Prove that < dx < =.3 4 X 2

SOLUTION Ontheinterval [4, 6], < 4, so, by Theorem5,1 61 6 |5=/ pax sf 4x.3 4 6 4 X

Onthe other hand, 1 < + on the interval [4, 6], so

61 61 1 1/ ir | dx=-(6-4=-.4 X 4 4 4 2

Therefore 3 = £ 4 dx < À, as desired.77. Prove that 0.0198 < [> sin x dx < 0.0296. Hint: Show that 0.198 < sinx < 0.296 for x in [0.2, 0.3].SOLUTION For0<x< % * 0.52, we have e (sin x) = cos x > 0. Hence sin x is increasing on[0.2, 0.3]. Accordingly, for0.2 < x < 0.3, we have

m = 0.198 < 0.19867 = sin 0.2 < sinx < sin0.3 & 0.29552 < 0.296 = MTherefore, by the Comparison Theorem, we have

0.3 0.3 0.30.0198 = m(0.3 0.2) = / m dx < / sinx dx <0 M dx = M(0.3 0.2) = 0.0296.

0.2 2 0.2x/4

78. Prove that 0.277 < / cosx dx < 0.363.n/8

SOLUTION cos x is decreasing on the interval [7/8, m/4]. Hence, for 7/8 < x < 1/4,cos(x/4) < cos x < cos(x/8).

Since cos(x/4) = /2/2,

Since cos(m/8) < 0.924,

| cos x dx < / 0.924 dx = (0.924) < 0.363.1/8 1/8 8Therefore 0.277 < ER cosx < 0.363.

587

588 CHAPTER 5 |] THE INTEGRAL

n/2 any 779. Prove that 0 < / DE ex V2x/4 x 2

SOLUTION Letsin xIQ) = =

As wecansee in the sketch below, f(x) is decreasing on the interval [7/4, 2/2]. Therefore f(x) < f(x/4) forall x in [/4, 2/2].1/4) = 48,so:

dx =n/a 7

ro sin x ro 2/2 x2V2 V2dx < 4

n/a x

1 dx80. Find upper and lower bounds for / a0 w5x3 +4SOLUTION Let

1V5x3 +4

f(x)is decreasing for x onthe interval [0, 1], so #(1) < f(x) < f(0) for all x in [0,1]. f(0) = 3 and f(1) = 4, so11 1 13sas | f(x) dx [oe[ sesf roel 5

1rs [ f@)dx <5

f(x) =

IA

7°IA

A81. ÉS Suppose that f(x) < g(x) on [a,b]. By the Comparison Theorem, In f(x) dx < £ g(x) dx. Is it also true thatF'Ge) < g'(x) for x [a, b]? If not, give a counterexample.

llSOLUTION The assertion f (x) < g (x) is false. Consider a = 0, b = 1, f(x) = x, g(x) 2. f(x) S g(x)for all x in theinterval [0, 1], but f (x) = 1 while g (x) = 0 forall x.82. ES State whether true or false. If false, sketch the graph of a counterexample.

b(a) If f(x) > 0, then | f(x) dx > 0.

b(b) it | f(x) dx > 0, then f(x) > 0.

SOLUTION

(a) It is true that if f(x) > 0 for x [a, b], then fe f(x) dx > 0.(b) It is false that iff? f(x) dx > 0, then f(x) > 0 for x [a, b]. Indeed, in Exercise 3, we saw that fx +4) dx =7.5 >0,yet f( 2) = 2 < 0. Hereis the graph from thatexercise.

SECTION 5.2 || TheDefinite Integral 589 Further Insights and Challenges83. Explain graphically: If f(x) is an odd function, then

af(x) dx = 0.

a

SOLUTION If f is an odd function, then f(-x) = f(x) for all x. Accordingly, for every positively signed area in the righthalf-plane where f is above the x-axis, there is a corresponding negatively signed area in the left half-plane where f is belowthe x-axis. Similarly, for every negatively signed area in the right half-plane where f is below the x-axis, there is a correspondingpositively signed areain theleft half-plane where .f is above the x-axis. We conclude that the net area between the graph of f andthe x-axis over [ a, a] is 0, since the positively signed areas and negatively signed areas cancel each other out exactly.

184. Compute | sin(sin(x))(sin? (x) + 1) dx.

1SOLUTION Let f(x) = sin(sin(x)) (sin?(x) + 1)). sin x is an odd function, while sin? x is an even function, so:

f(x) = sin(sin( x)) (sin? ( x) + 1) = sin(- sin(x)) (sin? (x) + 1)sin(sin(x)) (sin(x) + 1) = f(x).ll

Therefore, f(x) is an odd function. The function is odd and the interval is symmetric aroundtheorigin so, by the previous exercise,the integral mustbe zero.85. Let k and b be positive. Show, by comparing the right-endpoint approximations,that

b 1I xk dx = beth / xk dx0 0

SOLUTION Let k and b be any positive numbers. Let f(x) = x* on [0, 5]. Since f is continuous, both ye f(x) dx andf(x) dx exist. Let N be a positive integer and set Ax = (b 0) /N = b/N.Let xj =a+ jAx =bj/N, j =1,2,...,N

bethe right endpoints of the N subintervals of [0, b]. Then the right-endpoint approximationto fe f(x) dx = Ke xk dx isN

In particular, if b = 1 above, then the right-endpoint approximation to £ f(x) dx = Jo x* dx is

IA y 1N N= Ar N Iy__1 kSu 4x LG)x FN) = 5.1 2,0 = TANJ= = J=

In other words, Ry = pktlgN Therefore,b 1Î xkdx = lim Ry = lim Det! gy = p+! Jim Sn = Bt | xk dx.

0 0No No No86. Verify for 0 < b < 1 by interpreting in termsof area:

b 1/ V1 x2 dx = VIA + 5 sind0 2

SOLUTION The function f(x) = V1 x2 is the quarter circle of radius 1 in the first quadrant. For 0 < b < 1, the arearepresented by the integral I V1 x? dx can be divided into two parts. The area of the triangular part is 3 (b)V 1 b? using thePythagorean Theorem. The area of the sector with angle O where sin 9 = b, is given by (110). Thus

b 1 1 1 1/ V1l x2dx = -bV1 b2 4+ -6 = bV1 b2 + =sin7! b.À 2 2 2 2

590 CHAPTER 5 || THE INTEGRAL

| b 187. Ss Supposethat f and g are continuousfunctions suchthat, forall a,

a afejdx =faxa a

Give an intuitive argument showingthat f(0) = g(0). Explain your idea with a graph.SOLUTION Let Sea f (x)dx = Sea g (x) dx. Consider what happens as a decreases in size, becoming very close to zero.Intuitively, the areas of the functions become (a ( a))(f(0)) = 2a(f(0)) and (a (-a))(g(0)) = 2a(g(0)). Because we knowthese areas must be the same, we have 2a(f(0)) = 2a(g(0)) and therefore f(0) = g(0).88. Theorem 4 remains true without the assumption a < b < c. Verify this for the cases b < a < c and c < a < b.SOLUTION The additivity property of definite integrals states for a < b < c, we have [7 f(x) dx = e fx) dx + [y f(x) dx.

e Suppose that we have b < a < c. By the additivity property, we have tb f(x) dx = tp FO) dx + Se F(x) dx. Therefore,Me fc)dx = ff FO dx JE FO) dx = JE SO) dx + JE Fax.

e Now suppose that we have c < a < b. By the additivity property, we have e Fo) dx = fé f(x) dx + I FO) dx.Therefore, /* f(x) dx =-f% fo) dx = SP Sa)dx- [> Io)dx = fe f(x) dx + fé f(x) dx.

e Hencethe additivity property holds for all real numbersa, b, and c, regardless oftheir relationship amongst eachother.

5.3 The Fundamental Theorem of Calculus,Part|

Preliminary Questions1. Suppose that F (x) = f(x) and F(0) = 3, F(2) =7.

(a) Whatis the area under y = f(x) over[0, 2] if f(x) = 0?(b) Whatis the graphicalinterpretation of F(2) F(0) if f(x) takes onboth positive and negative values?SOLUTION(a) If f(x) > 0 over(0, 2], then the area under y = f(x) is F(2) F(0) =7-3 =4.(b) If f(x) takes on both positive and negative values, then F(2) F(0) givesthe signed area between y = f(x) and the x-axis.2. Suppose that f(x) is a negative function with antiderivative F such that F(1) = 7 and F(3) = 4. Whatis the area (a positive

number) between the x-axis and the graph of f(x) over[1, 3]?3

SOLUTION Î f(x) dx represents the signed area bounded by the curve and the interval [1, 3]. Since f(x) is negative on [1,3],13

| f(x) dx is the negative of the area. Therefore, if A is the area between the x-axis and the graph of f(x), we have:1

3Asn / FD dx = UFO) FU) = 4-7) =-() =3.3. Are the following statements true or false? Explain.

(a) FTC Lis valid only for positive functions.(b) To use FTC I, you haveto choosethe right antiderivative.(c) If you cannotfind an antiderivative of f(x), then the definite integral does not exist.SOLUTION(a) False. The FTC is valid for continuous functions.(b) False. The FTC I worksfor any antiderivative of the integrand.(c) False. If you cannotfind an antiderivative of the integrand, you cannot use the FTC I to evaluate the definite integral, but thedefinite integral maystill exist.