c:swp2507calwaism2001cwa ch 7 pdf final€¦ · dy = 16y4 4 + 9y3 3 ¡ 6y2 2 +3y +c =4y4 +3y3 ......

Chapter 7

INTEGRATION

7.1 Antiderivatives

1. If F (x) and G(x) are both antiderivatives of f(x),then there is a constant C such that

F (x)¡G(x) = C:The two functions can di¤er only by a constant.

5.Z6dk = 6

Z1dk

= 6

Zk0 dy

= 6 ¢ 11k0+1 +C

= 6k +C

6.Z9 dy = 9

Z1 dy = 9

Zy0 dy

=9

1y0+1 + C

= 9y +C

7.Z(2z + 3)dz

= 2

Zz dz + 3

Zz0 dz

= 2 ¢ 1

1 + 1z1+1 + 3 ¢ 1

0 + 1z0+1 +C

= z2 + 3z +C

8.Z(3x¡ 5) dx

= 3

Zx dx¡ 5

Zx0 dx

= 3 ¢ 12x2 ¡ 5 ¢ 1

1x+C

=3x2

2¡ 5x+ C

9.Z(6t2 ¡ 8t+ 7)dt

= 6

Zt2 dt¡ 8

Zt dt+ 7

Zt0 dt

=6t3

3¡ 8t

2

2+ 7t+C

= 2t3 ¡ 4t2 + 7t+C

10.Z(5x2 ¡ 6x+ 3)dx

= 5

Zx2 dx¡ 6

Zx dx+ 3

Zx0 dx

=5x3

3¡ 6x

2

2+ 3x+C

=5x3

3¡ 3x2 + 3x+C

11.Z(4z3 + 3z2 + 2z ¡ 6)dz

= 4

Zz3 dz + 3

Zz2 dz + 2

Zz dz

¡ 6Zz0 dz

=4z4

4+3z3

3+2z2

2¡ 6z +C

= z4 + z3 + z2 ¡ 6z +C

12.Z(16y3 + 9y2 ¡ 6y + 3)dy

= 16

Zy3 dy + 9

Zy2 dy ¡ 6

Zy dy + 3

Zdy

=16y4

4+9y3

3¡ 6y

2

2+ 3y +C

= 4y4 + 3y3 ¡ 3y2 + 3y +C

13.Z(5pz +

p2)dz = 5

Zz1=2 dz +

p2

Zdz

=5z3=2

32

+p2z +C

= 5

μ2

3

¶z3=2 +

p2z +C

=10z3=2

3+p2z +C

14.Z(t1=4 + ¼1=4)dt =

t1=4+1

14 + 1

+ ¼1=4t+C

=t5=4

54

+ ¼1=4t+C

=4t5=4

5+ ¼1=4t+C

443

444 Chapter 7 INTEGRATION

15.Z5x(x2 ¡ 8)dx =

Z(5x3 ¡ 40x)dx

=5x4

4¡ 40x

2

2+C

=5x4

4¡ 20x2 +C

16.Zx2(x4 + 4x+ 3)dx =

Z(x6 + 4x3 + 3x2)dx

=x7

7+4x4

4+3x3

3+ C

=x7

7+ x4 + x3 +C

17.Z(4pv ¡ 3v3=2)dv

= 4

Zv1=2 dv ¡ 3

Zv3=2 dv

=4v3=2

32

¡ 3v5=2

52

+C

=8v3=2

3¡ 6v

5=2

5+C

18.Z(15x

px+ 2

px)dx

= 15

Zx(x1=2)dx+ 2

Zx1=2 dx

= 15

Zx3=2 dx+ 2

Zx1=2 dx

=15x5=2

52

+2x3=2

32

+C

= 15

μ2

5

¶x5=2 + 2

μ2

3

¶x3=2 +C

= 6x5=2 +4x3=2

3+C

19.Z(10u3=2 ¡ 14u5=2)du

= 10

Zu3=2 du¡ 14

Zu5=2 du

=10u5=2

52

¡ 14u7=2

72

+C

= 10

μ2

5

¶u5=2 ¡ 14

μ2

7

¶u7=2 +C

= 4u5=2 ¡ 4u7=2 +C

20.Z(56t5=2 + 18t7=2)dt

= 56

Zt5=2 dt+ 18

Zt7=2 dt

=56t7=2

72

+18t9=2

92

+C

= 16t7=2 + 4t9=2 +C

21.Z μ

7

z2

¶dz =

Z7z¡2 dz

= 7

Zz¡2dz

= 7

μz¡2+1

¡2 + 1¶+C

=7z¡1

¡1 +C

= ¡7z+C

22.Z μ

4

x3

¶dx =

Z4x¡3 dx

= 4

Zx¡3 dx

=4x¡2

¡2 +C

= ¡2x¡2 +C

=¡2x2+C

23.Z μ

¼3

y3¡p¼py

¶dy =

Z¼3y¡3 dy ¡

Z p¼y¡1=2 dy

= ¼3Zy¡3 dy ¡p¼

Zy¡1=2 dy

= ¼3μy¡2

¡2¶¡p¼

μy1=2

12

¶+C

= ¡ ¼3

2y2¡ 2p¼y +C

24.Z μp

u+1

u2

¶du =

Zu1=2 du+

Zu¡2 du

=u3=2

32

+u¡1

¡1 +C

=2u3=2

3¡ 1

u+C

Section 7.1 Antiderivatives 445

25.Z(¡9t¡2:5 ¡ 2t¡1)dt

= ¡9Zt¡2:5 dt¡ 2

Zt¡1 dt

=¡9t¡1:5¡1:5 ¡ 2

Zdt

t

= 6t¡1:5 ¡ 2 ln jtj+C

26.Z(10x¡3:5 + 4x¡1)dx = 10

Zx¡3:5 dx+ 4

Zx¡1 dx

=10x¡2:5

¡2:5 + 4 ln jxj+C

= ¡4x¡2:5 + 4 ln jxj+C

27.Z

1

3x2dx =

Z1

3x¡2dx

=1

3

Zx¡2dx

=1

3

μx¡1

¡1¶+C

= ¡13x¡1 +C

= ¡ 1

3x+C

28.Z

2

3x4dx =

Z2

3x¡4dx

=2

3

Zx¡4dx

=2

3

μx¡3

¡3¶+C

= ¡29x¡3 +C

= ¡ 2

9x3+C

29.Z3e¡0:2x dx = 3

Ze¡0:2x dx

= 3

μ1

¡0:2¶e¡0:2x +C

=3(e¡0:2x)¡0:2 +C

= ¡15e¡0:2x +C

30.Z¡4e0:2v dv = ¡4

Ze0:2v dv

= (¡4) 10:2e0:2v +C

= ¡20e0:2v +C

31.Z μ

¡3x+ 4e¡0:4x + e0:1

¶dx

= ¡3Zdx

x+ 4

Ze¡0:4x dx+ e0:1

Zdx

= ¡3 ln jxj+ 4e¡0:4x

¡0:4 + e0:1x+C

= ¡3 ln jxj ¡ 10e¡0:4x + e0:1x+C

32.Z μ

9

x¡ 3e¡0:4x

¶dx

=

Z9

xdx¡ 3

Ze¡0:4x dx

= 9 ln jxj ¡ 3μ¡ 1

0:4

¶e¡0:4x +C

= 9 ln jxj+ 15e¡0:4x

2+C

33.Z μ

1 + 2t3

4t

¶dt =

Z μ1

4t+t2

2

¶dt

=1

4

Z1

tdt+

1

2

Zt2 dt

=1

4ln jtj+ 1

2

μt3

3

¶+C

=1

4ln jtj+ t

3

6+C

34.Z μ

2y1=2 ¡ 3y26y

¶dy

=

Z2y1=2

6ydy ¡

Z3y2

6ydy

=1

3

Zy¡1=2 dy ¡ 1

2

Zy dy

=1

3

μy1=2

12

¶¡ y

2

4+C

=2y1=2

3¡ y

2

4+C

35.Z(e2u + 4u)du =

e2u

2+4u2

2+C

=e2u

2+ 2u2 +C

36.Z(v2 ¡ e3v)dv =

Zv2 dv ¡

Ze3v dv

=v3

3¡ e

3v

3+C

=v3 ¡ e3v

3+C


37.Z(x+ 1)2 dx =

Z(x2 + 2x+ 1)dx

=x3

3+2x2

2+ x+C

=x3

3+ x2 + x+C

38.Z(2y ¡ 1)2 dy =

Z(4y2 ¡ 4y + 1)dy

=4y3

3¡ 4y

2

2+ y +C

=4y3

3¡ 2y2 + y +C

39.Z p

x+ 13px

dx =

Z μpx

3px+

13px

¶dx

=

Z(x(1=2¡1=3) + x¡1=3)dx

=

Zx1=6 dx+

Zx¡1=3 dx

=x7=6

76

+x2=3

23

+C

=6x7=6

7+3x2=3

2+C

40.Z1¡ 2 3

pz

3pz

dz =

Z μ13pz+2 3pz

3pz

¶dz

=

Z(z¡1=3 ¡ 2)dz

=z2=3

23

¡ 2z +C

=3z2=3

2¡ 2z +C

41.Z10xdx =

10x

ln 10+C

42.Z32xdx =

32x

2(ln 3)+C

43. Find f(x) such that f 0(x) = x2=3; and¡1; 35

¢is on

the curve.

Zx2=3dx =

x5=3

53

+C

f(x) =3x5=3

5+C

Since¡1; 35

¢is on the curve,

f(1) =3

5:

f(1) =3(1)5=3

5+C =

3

53

5+C =

3

5

C = 0:

Thus,

f(x) =3x5=3

5:

44. Find f(x) such that f 0(x) = 6x2 ¡ 4x + 3; and(0; 1) is on the curve.

f(x) =

Z(6x2 ¡ 4x+ 3)dx

=6x3

3¡ 4x

2

2+ 3x+ C

= 2x3 ¡ 2x2 + 3x+CSince (0; 1) is on the curve, then f(0) = 1:

f(0) = 2(0)3 ¡ 2(0)2 + 3(0) + C = 1C = 1

Thus,f(x) = 2x3 ¡ 2x2 + 3x+ 1:

45. C 0(x) = 4x¡ 5; …xed cost is $8.

C(x) =

Z(4x¡ 5)dx

=4x2

2¡ 5x+ k

= 2x2 ¡ 5x+ kC(0) = 2(0)2 ¡ 5(0) + k = kSince C(0) = 8; k = 8:

Thus,C(x) = 2x2 ¡ 5x+ 8:

46. C0(x) = 0:2x2 + 5x; …xed cost is $10.

C(x) =

Z(0:2x2 + 5x)dx

=0:2x3

3+5x2

2+ k

C(0) =0:2(0)3

3+5(0)2

2+ k = k

Since C(0) = 10; k = 10:

Thus,

C(x) =0:2x3

3+5x2

2+ 10:


47. C0(x) = 0:03e0:01x; …xed cost is $8.

C(x) =

Z0:03e0:01x dx

= 0:03

Ze0:01x dx

= 0:03

μ1

0:01e0:01x

¶+ k

= 3e0:01x + k

C(0) = 3e0:01(0) + k = 3(1) + k

= 3 + k

Since C(0) = 8; 3 + k = 8; and k = 5:Thus,

C(x) = 3e0:01x + 5:

48. C0(x) = x1=2; 16 units cost $45.

C(x) =

Zx1=2 dx =

x3=2

32

+ k =2

3x3=2 + k

C(16) =2

3(16)3=2 + k =

2

3(64) + k =

128

3+ k

Since C(16) = 45;

128

3+ k = 45

k =7

3:

Thus,

C(x) =2

3x3=2 +

7

3:

49. C0(x) = x2=3 + 2; 8 units cost $58.

C(x) =

Z(x2=3 + 2)dx

=3x5=3

5+ 2x+ k

C(8) =3(8)5=3

5+ 2(8) + k

=3(32)

5+ 16 + k

Since C(8) = 58;

58¡ 16¡ 965= k

114

5= k:

Thus,

C(x) =3x5=3

5+ 2x+

114

5:

50. C0(x) = x+1

x2; 2 units cost $5.50, so

C(2) = 5:50:

C(x) =

Z μx+

1

x2

¶dx

=

Z(x+ x¡2)dx

=x2

2+x¡1

¡1 + k

C(x) =x2

2¡ 1

x+ k

C(2) =(2)2

2¡ 12+ k

= 2¡ 12+ k

Since C(2) = 5:50;

5:50¡ 1:5 = k4 = k:

Thus,

C(x) =x2

2¡ 1

x+ 4:

51. C0(x) = 5x¡ 1

x; 10 units cost $94.20, so

C(10) = 94:20:

C(x) =

Z μ5x¡ 1

x

¶dx =

5x2

2¡ ln jxj+ k

C(10) =5(10)2

2¡ ln (10) + k

= 250¡ 2:30 + k:Since C(10) = 94:20;

94:20 = 247:70 + k

¡153:50 = k:

Thus, C(x) =5x2

2¡ ln jxj ¡ 153:50:

52. C0(x) = 1:2x(ln 1:2); 2 units cost $9.44(Hint: Recall that ax = ex ln a:)

C(x) =

Z1:2x(ln 1:2)dx

= ln 1:2

Z1:2x dx

= ln 1:2

Zex ln 1:2 dx

= ln 1:2

μ1

ln 1:2ex ln 1:2

¶+ k

= ex ln 1:2 + k

C(2) = e2 ln 1:2 + k = 1:44 + k


Since C(2) = 9:44;

144 + k = 9:44

k = 8:

Thus,

C(x) = ex ln 1:2 + 8

= 1:2x + 8:

53. R0(x) = 175¡ 0:02x¡ 0:03x2

R =

Z(175¡ 0:02x¡ 0:03x2)dx

= 175x¡ 0:01x2 ¡ 0:01x3 +C:

If x = 0; then R = 0 (no items sold means norevenue), and

0 = 175(0)¡ 0:01(0)2 ¡ 0:01(0)3 +C0 = C:

Thus, R = 175x¡ 0:01x2 ¡ 0:01x3

gives the revenue function. Now, recall thatR = xp; where p is the demand function. Then

175x¡ 0:01x2 ¡ 0:01x3 = xp175x¡ 0:01x¡ 0:01x2 = p, the demand function.

54. R0(x) = 50¡ 5x2=3

R =

Z(50¡ 5x2=3)dx

= 50x¡ 3x5=3 +C

If x = 0; then R = 0 (no items sold means norevenue), and

0 = 50(0)¡ 3(0)5=3 +C0 = C

Thus, R = 50x¡3x5=3 gives the revenue function.Now, recall that R = xp; where p is the demandfunction. Then

50x¡ 3x5=3 = xp50¡ 3x2=3 = p;

which gives the demand function.

55. R0(x) = 500¡ 0:15px

R =

Z(500¡ 0:15px)dx

= 500x¡ 0:1x3=2 +C:If x = 0; R = 0 (no items sold means no revenue),and

0 = 500(0)¡ 0:1(0)3=2 +C0 = C:

Thus, R = 500x¡ 0:1x3=2 gives the revenue func-tion. Now, recall that R = xp; where p is thedemand function. Then

500x¡ 0:1x3=2 = xp500¡ 0:1px = p, the demand function.

56. R0(x) = 600¡ 5e0:0002x

R =

Z(600¡ 25,000e0:0002x)dx

= 600x¡ 25,000e0:0002x +C:If x = 0; then R = 0 (no items sold means norevenue), and

0 = 600(0)¡ 25,000e0:0002(0) +C0 = ¡25,000 + C0 = 25,000.

Thus, R = 600x¡ 25,000e0:0002x + 25,000= 600x+ 25,000(1¡ e0:0002x)

gives the revenue function. Now, recall thatR = xp; where p is the demand function. Then

600x+ 25,000(1¡ e0:0002x) = xp

600 +25,000x

(1¡ e0:0002x) = p;

which gives the demand function.

57. f 0(t) = 1:498t+ 1:626

(a) f(t) =Z(1:498t+ 1:626)dt

= 0:749t2 + 1:626t+C

In 1992 (t = 2); f(t) = 8:893; and

8:893 = 0:749(2)2 + 1:626(2) +C

8:893 = 6:248 +C

2:645 = C

Thus, f(t) = 0:749t2 + 1:626t+ 2:645:


(b) In 2006, t = 16; and

f(16) = 0:749(16)2 + 1:626(16) + 2:645

= 191:744 + 26:016 + 2:645

= 220:405

The function predicted approximately 220 millionsubscribers in 2006.

58. P 0(x) =px+ 1

2 ; pro…t is ¡1 when 0 hamburgersare sold.

P (x) =

Z μpx+

1

2

¶dx

=2x3=2

3+x

2+ k

P (0) =2(0)3=2

3+0

2+ k

Since P (0) = ¡1; k = ¡1:

P (x) =2

3x3=2 +

x

2¡ 1

59. (a) P 0(x) = 50x3 + 30x2; pro…t is ¡40 when nocheese is sold.

P (x) =

Z(50x3 + 30x2)dx

=25x4

2+ 10x3 + k

P (0) =25(0)4

2+ 10(0)3 + k

SinceP (0) = ¡40;¡40 = k:

Thus,

P (x) =25x4

2+ 10x3 ¡ 40:

(b) P (2) =25(2)4

2+ 10(2)3 ¡ 40 = 240

The pro…t from selling 200 lbs of Brie cheese is$240.

60. (a) f 0(t) = 0:01e¡0:01t

f(t) =

Z0:01e¡0:01t dt

= ¡0:01e¡0:01t

0:01+ k

= ¡e¡0:01t + k(b) f(0) = ¡e¡0:01(0) + k = ¡e0 + k = ¡1 + k

Since f(0) = 0;

0 = ¡1 + kk = 1:

f(t) = ¡e¡0:01t + 1

f(10) = ¡e¡0:01(10) + 1= ¡e¡0:1 + 1= ¡0:905 + 1= 0:095

0.095 unit is excreted in 10 min.

61.Zg(x)

xdx =

Za¡ bxx

dx

=

Z ³ax¡ b´dx

= a

Zdx

x¡ b

Zdx

= a ln jxj ¡ bx+CSince x represents a positive quantity, the absolutevalue sign can be dropped.Z

g(x)

xdx = a lnx¡ bx+C

62. (a) c(t) = (c0 ¡C)e¡kAt=V +M

c0(t) = (c0 ¡C)μ¡kAV

¶e¡kAt=V

=¡kAV

(c0 ¡C)e¡kAt=V

(b) Since equation (1) states

c0(t) =kA

V[C ¡ c(t)];

then from (a) and by substituting from equation(2), we obtain

(c0 ¡C)μ¡kAV

¶e¡kAt=V

=kA

VC ¡ kA

V[(c0 ¡C)e¡kAt=V +M ]

¡kAV

(c0 ¡C)e¡kAt=V +M

=¡kAV

(c0 ¡C)e¡kAt=V + kAVC ¡ kA

VM:

If t = 0; c(t) = c0; so

c0 = (c0 ¡C)e0 +M Equation (2 )

c0 = c0 ¡C +Mor C =M:


Thus,

¡kAV

(c0 ¡C)e¡kAt=V

=¡kAV

(c0 ¡C)e¡kAt=V + kAVM ¡ kA

VM

or¡kAV

(c0 ¡C)e¡kAt=V = ¡kAV

(c0 ¡C)e¡kAt=V :

63. N 0(t) = Aekt

(a) N(t) =A

kekt + C

A = 50; N(t) = 300 when t = 0:

N(0) =50

ke0 +C = 300

N 0(5) = 250

Therefore,N 0(5) = 50e5k = 250

e5k = 5

5k = ln 5

k =ln 5

5:

N(0) =50ln 55

+C = 300

250

ln 5+C = 300

C = 300¡ 250

ln 5¼ 144:67

N(t) =50ln 55

e(ln 5=5)t + 144:67

= 155:3337e0:321888t + 144:67

(b) N(12) = 155:3337e0:321888(12) + 144:67¼ 7537

There are 7537 cells present after 12 days.

64. V 0(t) = ¡kP (t)P (t) = P0e¡mt

V 0(t) = ¡kP0e¡mt

V (t) =k

mP0e

¡mt +C

V (0) =k

mP0e

0 +C

V0 ¡ k

mP0 = C

Therefore,

V (t) =k

mP0e

¡mt + V0 ¡ k

mP0

=kP0me¡mt + V0 ¡ kP0

m:

65. a(t) = 5t2 + 4

v(t) =

Z(5t2 + 4)dt

=5t3

3+ 4t+C

v(0) =5(0)3

3+ 4(0) +C

Since v(0) = 6; C = 6:

v(t) =5t3

3+ 4t+ 6

66. v(t) = 9t2 ¡ 3pt

s =

Zv(t)dt

=

Z(9t2 ¡ 3pt)dt

= 3t3 ¡ 2t3=2 +Cs = 3t3 ¡ 2t3=2 +C

Since s(1) = 8;

8 = 3(1)3 ¡ 2(1)3=2 + C8 = 1 +C

7 = C:

Thus,s(t) = 3t3 ¡ 2t3=2 + 7:

67. a(t) = ¡32

v(t) =

Z¡32 dt = ¡32t+C1

v(0) = ¡32(0) +C1Since v(0) = 0; C1 = 0:

v(t) = ¡32t

s(t) =

Z¡32t dt

=¡32t22

+C2

= ¡16t2 +C2At t = 0; the plane is at 6400 ft.That is, s(0) = 6400:

s(0) = ¡16(0)2 +C26400 = 0 +C2C2 = 6400

s(t) = ¡16t2 + 6400


When the object hits the ground, s(t) = 0:

¡16t2 + 6400 = 0¡16t2 = ¡6400

t2 = 400

t = §20

Discard ¡20 since time must be positive.The object hits the ground in 20 sec.

68. a(t) = 18t+ 8

v(t) =

Z(18t+ 8)dt

= 9t2 + 8t+C1v(1) = 9(1)2 + 8(1) +C1 = 17 +C1

Since v(1) = 15; C1 = ¡2:

v(t) = 9t2 + 8t¡ 2

s(t) =

Z(9t2 + 8t¡ 2)dt

= 3t3 + 4t2 ¡ 2t+C2s(1) = 3(1)3 + 4(1)2 ¡ 2(1) +C2

= 5 +C2

Since s(1) = 19; C2 = 14:Thus,

s(t) = 3t3 + 4t2 ¡ 2t+ 14:

69. a(t) =15

2

pt = 3e¡t

v(t) =

Z μ15

2

pt+ 3e¡t

¶dt

=

Z μ15

2t1=2 + 3e¡t

¶dt

=15

2

μt3=2

32

¶+ 3

μ1

¡1e¡t¶+C1

= 5t3=2 ¡ 3e¡t +C1v(0) = 5(0)3=2 ¡ 3e¡0 +C1 = ¡3 +C1Since v(0) = ¡3; C1 = 0:

v(t) = 5t3=2 ¡ 3e¡t

s(t) =

Z(5t3=2 ¡ 3e¡t)dt

= 5

μt5=2

52

¶¡ 3

μ¡11e¡t¶+C2

= 2t5=2 + 3e¡t +C2s(0) = 2(0)5=2 + 3e¡0 +C2 = 3 +C2

Since s(0) = 4; C2 = 1:Thus,

s(t) = 2t5=2 + 3e¡t + 1:

70. The acceleration of gravity is a constant with value¡32 ft/sec2; that is,

a(t) = ¡32:First …nd v(t) by integrating a(t):

v(t) =

Z(¡32)dt = ¡32t+ k:

When t = 0; v(t) = v0:

v0 = ¡32(0) + kv0 = k

andv(t) = ¡32t+ v0:

Now integrate v(t) to …nd h(t).

h(t) =

Z(¡32t+ v0)dt = ¡16t2 + v0t+C

Since h(t) = h0 when t = 0; we can substitutethese values into the equation for h(t) to get C =h0 and

h(t) = ¡16t2 + v0t+ h0:Recall that g is a constant with value ¡32 ft=sec2,so 1

2g has value ¡16 ft/sec2; and

h(t) =1

2gt2 + v0t+ h0:

71. First …nd v(t) by integrating a(t):

v(t) =

Z(¡32)dt = ¡32t+ k:

When t = 5; v(t) = 0:

0 = ¡32(5) + k160 = k

andv(t) = ¡32t+ 160:

Now integrate v(t) to …nd h(t):

h(t) =

Z(¡32t+ 160)dt = ¡16t2 + 160t+C

Since h(t) = 412 when t = 5; we can substitutethese values into the equation for h(t) to get C =12 and

h(t) = ¡16t2 + 160t+ 12:Therefore, from the equation given in Exercise 70,the initial velocity v0 is 160 ft/sec and the initialheight of the rocket h0 is 12 ft.


72. (a) First …nd v(t) by integrating a(t):

v(t) =

Z(¡32)dt = ¡32t+ k:

When t = 0; v(t) = v0:

v0 = ¡32(0) + kv0 = k

and v(t) = ¡32t+ v0:Now integrate v(t) to …nd s(t):

s(t) =

Z(¡32t+ v0)dt

= ¡16t2 + v0t+C

Since s(t) = 0 when t = 0, we can substitute thesevalues into the equation for s(t) to get C = 0 and

s(t) = ¡16t2 + v0t:

(b) When t = 14; s(t) = 0; so

0 = ¡16(14)2 + v0(14)v0 = 224

The velocity was 224 feet per second at time t = 0.

(c) v0t = 224(14) = 3136

The distance the rocket would travel horizontallywould be 3136 feet.

73. (a) First …nd B(t) by integrating B0(t):

B(t) =

Z9:2935e0:02955tdt

¼ 314:5e0:02955t + k

When t = 0; B(t) = 792:3:

792:3 = 314:5e0:02955(0) + k

477:8 = k

and

B(t) = 314:5e0:02955t + 477:8:

(b) In 2012, t = 42:

B(42) = 314:5e0:02955(42) + 477:8

¼ 1565:8

About 1,566,000 bachelor’s degrees will be con-ferred in 2012.

7.2 Substitution

2. (a)Z(3x2 ¡ 5)4 2xdx

Let u = 3x2 ¡ 5; then du = 6xdx:(b)

Z p1¡ xdx

Let u = 1¡ x; then du = ¡dx:

(c)Z

x2

2x3 + 1dx

Let u = 2x3 + 1; then du = 6x2 dx:

(d)Z4x3ex

4

dx

Let u = x4; then du = 4x3 dx:

3.Z4(2x+ 3)4 dx = 2

Z2(2x+ 3)4 dx

Let u = 2x+ 3; so that du = 2dx:

= 2

Zu4 du

=2 ¢ u55

+C

=2(2x+ 3)5

5+C

4.Z(¡4t+ 1)3 dt

= ¡14

Z¡4(¡4t+ 1)3 dt

Let u = ¡4t+ 1; so that du = ¡4 dt:

= ¡14

Zu3 du

= ¡14¢ u

4

4+C

=¡u416

+C

=¡(¡4t+ 1)4

16+C

5.Z

2dm

(2m+ 1)3=

Z2(2m+ 1)¡3 dm

Let u = 2m+ 1; so that du = 2dm:

=

Zu¡3 du

=u¡2

¡2 +C

=¡(2m+ 1)¡2

2+C

Section 7.2 Substitution 453

6.Z

3dup3u¡ 5 =

Z3(3u¡ 5)¡1=2 du

Let w = 3u¡ 5; so that dw = 3du:

=

Zw¡1=2 dw

=w1=2

12

+C

= 2w1=2 +C

= 2(3u¡ 5)1=2 +C

7.Z

2x+ 2

(x2 + 2x¡ 4)4 dx

=

Z(2x+ 2)(x2 + 2x¡ 4)¡4dx

Let w = x2 + 2x¡ 4, so that dw = (2x+ 2)dx:

=

Zw¡4dw

=w¡3

¡3 +C

= ¡(x2 + 2x¡ 4)¡3

3+C

= ¡ 1

3(x2 + 2x¡ 4)3 +C

8.Z

6x2 dx

(2x3 + 7)3=2

=

Z6x2(2x3 + 7)¡3=2 dx

Let u = 2x3 + 7; so that du = 6x2 dx:

=

Zu¡3=2 du

=u¡1=2

¡12

+C

= ¡2u¡1=2 +C

=¡2u1=2

+C

=¡2

(2x3 + 7)1=2+C

9.Zzp4z2 ¡ 5dz =

Zz(4z2 ¡ 5)1=2 dz

=1

8

Z8z(4z2 ¡ 5)1=2 dz

Let u = 4z2 ¡ 5; so that du = 8z dz:=1

8

Zu1=2 du

=1

8¢ u

3=2

32

+C

=1

8¢μ2

3

¶u3=2 +C

=(4z2 ¡ 5)3=2

12+C

10.Zrp5r2 + 2dr =

Zr(5r2 + 2)1=2 dr

=1

10

Z10r(5r2 + 2)1=2 dr

Let u = 5r2 + 2; so that du = 10r dr:

=1

10

Zu1=2du

=1

10¢ u

3=2

32

+C

=u3=2

15+C

=(5r2 + 2)3=2

15+C

11.Z3x2 e2x

3

dx =1

2

Z2 ¢ 3x2 e2x3 dx

Let u = 2x3; so that du = 6x2 dx:

=1

2

Zeu du

=1

2eu +C

=e2x

3

2+C

12.Zre¡r

2

dr

Let u = ¡r2; so that du = ¡2r dr:Zre¡r

2

dr = ¡12

Z¡2re¡r2 dr

= ¡12

Zeu du

=¡eu2+C

=¡e¡r22

+C


13.Z(1¡ t)e2t¡t2 dt

=1

2

Z2(1¡ t)e2t¡t2 dt

Let u = 2t¡ t2; so that du = (2¡ 2t)dt:

=1

2

Zeu du

=eu

2+C

=e2t¡t

2

2+C

14.Z(x2 ¡ 1)ex3¡3x dx

Let u = x3 ¡ 3x; so that

du = (3x2 ¡ 3)dx = 3(x2 ¡ 1)dx:Z(x2 ¡ 1)ex3¡3x dx

=1

3

Z3(x2 ¡ 1)ex3¡3x dx

=1

3

Zeu du =

eu

3+C

=ex

3¡3x

3+C

15.Ze1=z

z2dz = ¡

Ze1=z ¢ ¡1

z2dz

Let u = 1z ; so that du =

¡1z2 dx:

= ¡Zeu du

= ¡eu +C= ¡e1=z +C

16.Zepy

2pydy =

Zey

1=2

2y1=2dy

=

Z1

2y¡1=2ey

1=2

dy

Let u = y1=2; so that du = 12y¡1=2 dy:

=

Zeu du = eu +C

= ey1=2

+C = epy +C

17.Z(x3 + 2x)(x4 + 4x2 + 7)8dx

=1

4

Z(x4 + 4x2 + 7)8(4x3 + 8x)dx

Let u = x4 + 4x2 + 7, so that du = (4x3 + 8x)dx

=1

4

Zu8du =

1

4

Z μu9

9

¶+C

=u9

36+C =

(x4 + 4x2 + 7)9

36+C

18.Z

t2 + 2

t3 + 6t+ 3dx

Let u = t3 + 6t+ 3, so that du = (3t2 + 6)dt:Zt2 + 2

t3 + 6t+ 3dx =

1

3

Z(3t2 + 6)dt

t3 + 6t+ 3

=1

3

Zdu

u

=1

3ln juj+C

=1

3ln¯t3 + 6t+ 3

¯+C

19.Z

2x+ 1

(x2 + x)3dx

=

Z(2x+ 1)(x2 + x)¡3 dx

Let u = x2 + x; so that du = (2x+ 1)dx:

=

Zu¡3 du =

u¡2

¡2 +C

=¡12u2

+C =¡1

2(x2 + x)2+C

20.Z

y2 + y

(2y3 + 3y2 + 1)2=3dy

=

Z(y2 + y)(2y3 + 3y2 + 1)¡2=3dy

Let u = 2y3+3y2+1; so that du = (6y2+6y)dy:

=1

6

Z6(y2 + y)(2y3 + 3y2 + 1)¡2=3dy

=1

6

Zu¡2=3du

=1

6¢ u

1=3

13

+C

=u1=3

2+C

=(2y3 + 3y2 + 1)1=3

2+ C


21.Zp(p+ 1)5 dp

Let u = p+ 1; so that du = dp; also, p = u¡ 1:

=

Z(u¡ 1)u5 du

=

Z(u6 ¡ u5)du

=u7

7¡ u

6

6+C

=(p+ 1)7

7¡ (p+ 1)

6

6+C

22.Z4rp8¡ r dr

=

Z4r(8¡ r)1=2 dr

Let u = 8¡ r; so that

du = ¡dr; also, r = 8¡ u:

= ¡4Z¡r(8¡ r)1=2 dr

= ¡4Z(8¡ u)u1=2 du

= ¡4Z(8u1=2 ¡ u3=2)du

= ¡4μ8u3=2

32

¡ u5=2

52

¶+C

=8(8¡ r)5=2

5¡ 64(8¡ r)

3=2

3+C

23.Z

upu¡ 1 du

=

Zu(u¡ 1)¡1=2 du

Let w = u¡ 1; so that dw = du and u = w + 1:

=

Z(w + 1)w¡1=2 dw

=

Z(w1=2 +w¡1=2)dw

=w3=2

32

+w1=2

12

+C

=2(u¡ 1)3=2

3+ 2(u¡ 1)1=2 +C

24.Z

2x

(x+ 5)6dx

=

Z2x(x+ 5)¡6 dx = 2

Zx(x+ 5)¡6 dx

Let u = x+ 5; so that

du = dx; also, u¡ 5 = x:

= 2

Z(u¡ 5)u¡6 du = 2

Z(u¡5 ¡ 5u¡6)du

= 2

μu¡4

¡4¶¡ 10

μu¡5

¡5¶+C

= ¡u¡4

2+ 2u¡5 +C

=¡1

2(x+ 5)4+

2

(x+ 5)5+C

25.Z(px2 + 12x)(x+ 6)dx

=

Z(x2 + 12x)1=2(x+ 6)dx

Let x2 + 12x = u; so that

(2x+ 12)dx = du

2(x+ 6)dx = du:

=1

2

Zu1=2 du =

1

2

μ2

3

¶u3=2 +C

=(x2 + 12x)3=2

3+C

26.Z(px2 ¡ 6x)(x¡ 3)dx

=

Z(x2 ¡ 6x)1=2(x¡ 3)dx


du = (2x¡ 6)dx = 2(x¡ 3)dx:

=1

2

Z(x2 ¡ 6x)1=22(x¡ 3)dx

=1

2

Zu1=2 du =

1

2

μu3=2

32

¶+C

=u3=2

3+C =

(x2 ¡ 6x)3=23

+C


27.Z

t

t2 + 2dt

Let t2 + 2 = u; so that 2t dt = du:

=1

2

Zdu

u

=1

2ln juj+C

=ln (t2 + 2)

2+C

28.Z ¡4xx2 + 3

dx

Let u = x2 + 3; so that du = 2x dx:Z ¡4xx2 + 3

dx = ¡2Z2x dx

x2 + 3= ¡2

Zdu

u

= ¡2 ln juj+C= ¡2 ln (x2 + 3) +C

29.Z(1 + 3 lnx)2

xdx

Let u = 1 + 3 lnx; so that du = 3x dx:

=1

3

Z3(1 + 3 lnx)2

xdx

=1

3

Zu2 du

=1

3¢ u

3

3+C

=(1 + 3 lnx)3

9+C

30.Z p

2 + ln x

xdx

Let u = 2 + ln x; so that

du =1

xdx:

Z p2 + ln x

xdx =

Z pu du

=

Zu1=2 du

=u3=2

3=2+C

=2

3u3=2 +C

=2

3(2 + ln x)3=2 +C

31.Z

e2x

e2x + 5dx

Let u = e2x + 5; so that du = 2e2x dx:

=1

2

Zdu

u

=1

2ln juj+C

=1

2ln¯e2x + 5

¯+C

=1

2ln (e2x + 5) +C

32.Z

1

x(ln x)dx

Let u = ln x; so that

du =1

xdx:Z

1

x(ln x)dx =

Z1

udu

= ln juj+C= ln jln xj+C

33.Zlog x

xdx

Let u = log x; so that du = 1(ln 10)x dx:

= (ln 10)

Zlog x

(ln 10)xdx

= (ln 10)

Zudu

= (ln 10)

μu2

2

¶+C

=(ln 10)(logx)2

2+C

34.Z[log2(5x+ 1)]

2

5x+ 1dx

Let u = log2(5x+ 1); so that

du = 5(ln 2)(5x+1) dx:

=ln 2

5

Z5[log2(5x+ 1)]

2

(ln 2)(5x+ 1)dx

=ln 2

5

Zu2 du

=ln 2

5

μu3

3

¶+C

=(ln 2)[log2(5x+ 1)]

3

15+C


35.Zx83x

2+1dx

Let u = 3x2 + 1; so that du = 6xdx:

=1

6

Z6x ¢ 83x2+1dx

=1

6

Z8u du

=1

6

μ8u

ln 8

¶+C

=83x

2+1

6 ln 8+C

36.Z105

px+2

px

dx

Let u = 5px+ 2; so that du = 5

2pxdx:

=2

5

Z5 ¢ 105

px+2

2px

dx

=2

5

Z10udu

=2

5¢ 10

u

ln 10+C

=2 ¢ 105px+25 ln 10

+C

39. (a) R0(x) = 4x(x2 + 27,000)¡2=3

R(x) =

Z4x(x2 + 27,000)¡2=3 dx

= 2

Z2x(x2 + 27,000)¡2=3 dx

Let u = x2 + 27,000, so that du = 2xdx:

R = 2

Zu¡2=3 du

= 2 ¢ 3u1=3 +C= 6(x2 + 27,000)1=3 +C

R(125) = 6(1252 + 27,000)1=3 +C

Since R(125) = 29:591;

6(1252 + 27,000)1=3 +C = 29:591C = ¡180

Thus,

R(x) = 6(x2 + 27,000)1=3 ¡ 180:

(b)

R(x) = 6(x2 + 27; 000)1=3 ¡ 180 ¸ 406(x2 + 27,000)1=3 ¸ 220(x2 + 27,000)1=3 ¸ 36:6667

x2 + 27,000 ¸ 49,296.43x2 ¸ 22,296.43x ¸ 149:4

For a revenue of at least $40,000, 150 players mustbe sold.

40. (a) D0(t) = 90(t+ 6)pt2 + 12t

= 90(t+ 6)(t2 + 12t)1=2

D(t) =

Z90(t+ 6)(t2 + 12t)1=2dt

= 45

Z(2t+ 12)(t2 + 12t)1=2dt

Let u = t2 + 12t; so that du = (2t+ 12)dt:

D = 45

Zu1=2du

= 45 ¢ 2u3=2

3+C

= 30u3=2 +C

D(t) = 30(t2 + 12t)3=2 +C

Since D(4) = 16,260,

30(42 + 12 ¢ 4)3=2 +C = 16,26015,360 +C = 16,260

C = 900

Thus,

D(t) = 30(t2 + 12t)3=2 + 900:

(b) D(t) = 30(t2 + 12t)3=2 + 900 = 40,00030(t2 + 12t)3=2 = 39,100(t2 + 12t)3=2 = 1303:333

t2 + 12t = 119:317

t2 + 12t¡ 119:317 = 0

x =¡12§p122 ¡ 4(1)(¡119:317)

2(1)

x ¼ ¡12§ 24:932(1)

x ¼ 6:465 or x ¼ ¡18:464

7 years must pass.


41. C0(x) =60x

5x2 + e

(a) Let u = 5x2 + e; so that du = 10xdx:

C(x) =

ZC0(x)dx =

Z60x

5x2 + edx = 6

Zdu

u= 6 ln juj+C = 6 ln ¯5x2 + e¯+C

Since C(0) = 10; C = 4:

Therefore,

C(x) = 6 ln¯5x2 + e

¯+ 4 = 6 ln(5x2 + e) + 4:

(b) C(5) = 6 ln(5 ¢ 52 + e) + 4 ¼ 33:099

Since this represents $33,099 dollars which is greater than $20,000, a new source of investment income shouldbe sought.

42. (a) P 0(x) = xe¡x2

Let ¡x2 = u; so that ¡2xdx = du, or xdx = ¡12 du:

P (x) =

Zxe¡x

2

dx

= ¡12

Zeu du = ¡e

u

2+C

= ¡e¡x2

2+C

P (3) = ¡e¡9

2+C

Since 10,000 = 0.01 million and P (3) = 0:01;

¡e¡9

2+C = 0:01

C = 0:01 +e¡9

2= 0:01006 ¼ 0:01:

P (x) =¡e¡x22

+ 0:01

(b) limx!1 (x) = lim

x!1

Ã¡ex22

+ 0:01

!= limx!1

μ¡ 1

2ex2+ 0:01

¶= 0:01

Since pro…t is expressed in millions of dollars, the pro…t approaches 0.01(1,000,000) = $10,000.


43. f 0(t) = 4:0674 ¢ 10¡4t(t¡ 1970)0:4(a) Let u = t ¡ 1970: To get the t outside the parentheses in terms of u; solve u = t ¡ 1970 for t to gett = u+ 1970: Then dt = du and we can substitute as follows.

f(t) =

Zf 0(t)dt =

Z4:0674 ¢ 10¡4t(t¡ 1970)0:4dt

=

Z4:0674 ¢ 10¡4(u+ 1970)(u)0:4du

= 4:0674 ¢ 10¡4Z(u+ 1970)(u)0:4du

= 4:0674 ¢ 10¡4Z(u1:4 + 1970u0:4)du

= 4:0674 ¢ 10¡4μu2:4

2:4+1970u1:4

1:4

¶+C

= 4:0674 ¢ 10¡4·(t¡ 1970)2:4

2:4+1970(t¡ 1970)1:4

1:4

¸+C

Since f(1970) = 61:298; C = 61:298:

Therefore,

f(t) = 4:0674 ¢ 10¡4·(t¡ 1970)2:4

2:4+1970(t¡ 1970)1:4

1:4

¸+ 61:298:

(b) f(2015) = 4:0674 ¢ 10¡4·(2015¡ 1970)2:4

2:4+1970(2015¡ 1970)1:4

1:4

¸+ 61:298 ¼ 180:9:

In the year 2015, there will be about 181,000 local transit vehicles.

44. f 0(t) = 0:001483t(t¡ 1980)0:75(a) Let u = t ¡ 1980: To get the t outside the parentheses in terms of u; solve u = t ¡ 1980 for t to gett = u+ 1980: Then dt = du and we can substitute as follows.

f(t) =

Zf 0(t)dt =

Z0:001483t(t¡ 1980)0:75dt

=

Z0:001483(u+ 1980)(u)0:75du

= 0:001483

Z(u+ 1980)(u)0:75du

= 0:001483

Z(u1:75 + 1980u0:75)du

= 0:001483

μu2:75

2:75+1980u1:75

1:75

¶+C

= 0:001483

·(t¡ 1980)2:75

2:75+1980(t¡ 1980)1:75

1:75

¸+C

Since f(1980) = 262:951; C = 262:951:Therefore,

f(t) = 0:001483

·(t¡ 1980)2:75

2:75+1980(t¡ 1980)1:75

1:75

¸+ 262:951:

(b) f(2012) = 0:001483·(2012¡ 1980)2:75

2:75+1980(2012¡ 1980)1:75

1:75

¸+ 262:951 ¼ 992:8

In the year 2012, there will be about 993,000,000 outpatient visits.


7.3 Area and the De…nite Integral

2.Z 4

0

(x2 + 3)dx = limn!1

nPi=1

(xi2 + 3)¢x; where ¢x = 4¡0

n = 4n and xi is any value of x in the ith interval.

3. f(x) = 2x+ 5; x1 = 0; x2 = 2; x3 = 4; x4 = 6; and ¢x = 2

(a)4Xi=1

f(xi)¢x

= f(x1)¢x+ f(x2)¢x+ f(x3)¢x+ f(x4)¢x

= f(0)(2) + f(2)(2) + f(4)(2) + f(6)(2)

= [2(0) + 5](2) + [2(2) + 5](2) + [2(4) + 5](2) + [2(6) + 5](2)

= 10 + 9(2) + 13(2) + 17(2)

= 88

(b)

The sum of these rectangles approximatesZ 8

0

(2x+ 5)dx:

4. f(x) = 1x and x1 =

12 ; x2 = 1; x3 =

32 ; x4 = 2; and ¢x =

12

(a)4Xi=1

f(xi)¢x = f(x1)¢x+ f(x2)¢x+ f(x3)¢x+ f(x4)¢x

f(x1) = f

μ1

2

¶=112

= 2

f(x2) = f(1) =1

1= 1

f(x3) = f

μ3

2

¶=132

=2

3

f(x4) = f(2) =1

2

Thus,

4Xi=1

f(xi)¢x

= (2)

μ1

2

¶+ (1)

μ1

2

¶+

μ2

3

¶μ1

2

¶+

μ1

2

¶μ1

2

¶

= 1 +1

2+1

3+1

4=12 + 6 + 4 + 3

12=25

12:

Section 7.3 Area and the De…nite Integral 461

(b)

The sum is approximated by the integralZ 5=2

1=2

1

xdx.

6. f(x) = 3x+ 2 from x = 1 to x = 3

For n = 4 rectangles:

¢x =3¡ 14

= 0:5

(a) Using the left endpoints:

i xi f(xi)

1 1 5

2 1:5 6:5

3 2 8

4 2:5 9:5

A =4Xi=1

f(xi)¢x = 5(0:5) + 6:5(0:5) + 8(0:5) + 9:5(0:5) = 14:5

(b) Using the right endpoints:

i xi f(xi)

1 1:5 6:5

2 2 8

3 2:5 9:5

4 3 11

A = 6:5(0:5) + 8(0:5) + 9:5(0:5) + 11(0:5) = 17:5

(c) Average =14:5 + 17:5

2=32

2= 16

(d) Using the midpoints:

i xi f(xi)

1 1:25 5:75

2 1:75 7:25

3 2:25 8:75

4 2:75 10:25

A =4X1

f(xi)¢x = 5:75(0:5) + 7:25(0:5) + 8:75(0:5) + 10:25(0:5) = 16


7. f(x) = 2x+ 5 from x = 2 to x = 4


¢x =4¡ 24

= 0:5

(a) Using the left endpoints:i xi f(xi)

1 2 9

2 2:5 10

3 3 11

4 3:5 12

A =4X1

f(xi)¢x = 9 (0:5) + 10 (0:5) + 11 (0:5) + 12 (0:5) = 21

(b) Using the right endpoints:i xi f(xi)

1 2:5 10

2 3 11

3 3:5 12

4 4 13

A = 10(0:5) + 11(0:5) + 12(0:5) + 13(0:5) = 23

(c) Average =21 + 23

2=44

2= 22

(d) Using the midpoints:i xi f(xi)

1 2:25 9:5

2 2:75 10:5

3 3:25 11:5

4 3:75 12:5

A =4X1

f(xi)¢x = 9:5 (0:5) + 10:5 (0:5) + 11:5 (0:5) + 12:5 (0:5) = 22

8. f(x) = x2 from x = 1 to 5


¢x =5¡ 14

= 1


1 1 1

2 2 4

3 3 9

4 4 16

A =4Xi=1

f(xi)¢x = 1(1) + 4(1) + 9(1) + 16(1) = 30



1 2 4

2 3 9

3 4 16

4 5 25

A = 4(1) + 9(1) + 16(1) + 25(1) = 54


2=84

2= 42


13

2

9

4

25

2

25

4

37

2

49

4

49

2

81

4

A =4Xi=1

f(xi)¢x =9

4(1) +

25

4(1) +

49

4(1) +

81

4(1) = 41

9. f(x) = ¡x2 + 4 from x = ¡2 to x = 2For n = 4 rectangles:

¢x =2¡ (¡2)

4= 1


1 ¡2 ¡(¡2)2 + 4 = 02 ¡1 ¡(¡1)2 + 4 = 33 0 ¡(0)2 + 4 = 44 1 ¡(1)2 + 4 = 3

A =4Xi=1

f(xi)¢x = (0)(1) + (3)(1) + (4)(1) + (3)(1) = 10


1 ¡1 3

2 0 4

3 1 3

4 2 0

Area = 1(3) + 1(4) + 1(3) + 1(0) = 10



2= 10


i xi f(xi)

1 ¡32

7

4

2 ¡12

15

4

31

2

15

4

43

2

7

4

A =4Xi=1

f(xi)¢x =7

4(1) +

15

4(1) +

15

4(1) +

7

4(1) = 11

10. f(x) = ex ¡ 1 from x = 0 to x = 4


¢x =4¡ 04

= 1


i xi f(xi)

1 0 0

2 1 1:718

3 2 6:389

4 3 19:086

A =4Xi=1

f(xi)¢x = 0(1) + 1:718(1) + 6:389(1) + 19:086(1) ¼ 27:19


i xi f(xi)

1 1 1:718

2 2 6:389

3 3 19:086

4 4 53:598

A = 1:718(1) + 6:389(1) + 19:086(1) + 53:598(1) ¼ 80:79

(c) Average =27:19 + 80:79

2= 53:99



i xi f(xi)

11

20:649

23

23:482

35

211:182

47

232:115

A =4Xi=1

f(xi)¢x = 0:649(1) + 3:482(1) + 11:182(1) + 32:115(1) ¼ 47:43

11. f(x) = ex + 1 from x = ¡2 to x = 2


¢x =2¡ (¡2)

4= 1


i xi f(xi)

1 ¡2 e¡2 + 12 ¡1 e¡1 + 13 0 e0 + 1 = 2

4 1 e1 + 1

A =4Xi=1

f(xi)¢x =4Xi=1

f(xi)(1) =4Xi=1

f(xi) = (e¡2 + 1) + (e¡1 + 1) + 2 + e1 + 1 ¼ 8:2215 ¼ 8:22


i xi f(xi)

1 ¡1 e¡1 + 12 0 2

3 1 e+ 1

4 2 e2 + 1

Area = 1(e¡1 + 1) + 1(2) + 1(e+ 1) + 1(e2 + 1) ¼ 15:4752 ¼ 15:48

(c) Average =8:2215 + 15:4752

2= 11:84835 ¼ 11:85



1 ¡32

e¡3=2 + 1

2 ¡12

e¡1=2 + 1

31

2e1=2 + 1

43

2e3=2 + 1

A =4Xi=1

f(xi)¢x = (e¡3=2 + 1)(1) + (e¡1=2 + 1)(1) + (e1=2 + 1)(1) + (e3=2 + 1)(1) ¼ 10:9601 ¼ 10:96

12. f(x) =1

xfrom x = 1 to x = 3


¢x =3¡ 14

= 0:5


1 1 1

2 1:5 0:6667

3 2 0:5

4 2:5 0:4

A =4X1

f(xi)¢x = 1(0:5) + 0:6667(0:5) + 0:5(0:5) + 0:4(0:5) = 1:283


1 1:5 0:6667

2 2 0:5

3 2:5 0:4

4 3 0:3333

A = 0:6667(0:5) + 0:5(0:5) + 0:4(0:5) + 0:3333(0:5) = 0:95

(c) Average =1:283 + 0:95

2=2:233

2= 1:117


1 1:25 0:8

2 1:75 0:5714

3 2:25 0:4444

4 2:75 0:3636

A =4X1

f(xi)¢x = 0:8(0:5) + 0:5714(0:5) + 0:4444(0:5) + 0:3636(0:5) = 1:090


13. f(x) =2

xfrom x = 1 to x = 9


¢x =9¡ 14

= 2


1 12

1= 2

2 32

3

3 52

5= 0:4

4 72

7

A =4Xi=1

f(xi)¢x = (2)(2) +2

3(2) + (0:4)(2) +

μ2

7

¶(2) ¼ 6:7048 ¼ 6:70


1 32

3

2 52

5

3 72

7

4 92

9

Area = 2μ2

3

¶+ 2

μ2

5

¶+ 2

μ2

7

¶+ 2

μ2

9

¶=4

3+4

5+4

7+4

9¼ 3:1492 ¼ 3:15

(c) Average =6:7 + 3:15

2= 4:93


1 2 1

2 41

2

3 61

3

4 81

4

A =4Xi=1

f(xi)¢x = 1(2) +1

2(2) +

1

3(2) +

1

4(2) ¼ 4:1667 ¼ 4:17


14. (a) Width =4¡ 04

= 1; f(x) =x

2

Area = 1 ¢ fμ1

2

¶+ 1 ¢ f

μ3

2

¶+ 1 ¢ f

μ5

2

¶+ 1 ¢ f

μ7

2

¶=1

4+3

4+5

4+7

4=16

4= 4

(b)

4Z0

f(x)dx =

4Z0

x

2dx =

1

2(base)(height) =

1

2(4)(2) = 4

15.Z 5

0

(5¡ x)dx

Graph y = 5¡ x:

Z 5

0

(5¡ x)dx is the area of a triangle with base = 5¡ 0 = 5 and altitude = 5.

Area =1

2(altitude)(base) =

1

2(5)(5) = 12:5

16. (a) Area of triangle is 12 ¢base ¢height.The base is 4; the height is 2. Z 4

0

f(x)dx =1

2¢ 4 ¢ 2 = 4

(b) The larger triangle has an area of 12 ¢ 3 ¢ 3 = 92 : The smaller triangle has an area of

12 ¢ 1 ¢ 1 = 1

2 : The sum is92 +

12 =

102 = 5:

17. (a)Z 2

0

f(x)dx is the area of a rectangle with width x = 2 and length y = 4: The rectangle has area 2 ¢ 4 = 8:Z 6

2

f(x)dx is the area of one-fourth of a circle that has radius 4. The area is 14¼r2 = 1

4¼(4)2 = 4¼:

Therefore,Z 6

2

f(x)dx = 8 + 4¼:


(b)Z 2

0

f(x)dx is the area of one-fourth of a circle that has radius 2. The area is 14¼r2 = 1

4¼(2)2 = ¼:

Z 6

2

f(x)dx is the area of a triangle with base 4 and height 2. The triangle has area 12 ¢ 4 ¢ 2 = 4:

Therefore,Z 6

0

f(x)dx = 4 + ¼:

18.Z 3

¡3

p9¡ x2 dx

Graph y =p9¡ x2:

Z 3

¡3

p9¡ x2 dx is the area of a semicircle with radius 3 centered at the origin.

Area =1

2¼r2 =

1

2¼(3)2 =

9

2¼

19.Z 0

¡4

p16¡ x2 dx

Graph y =p16¡ x2:

Z 0

¡4

p16¡ x2 dx is the area of the portion of the circle in the second quadrant, which is one-fourth of a circle.

The circle has radius 4.

Area =1

4¼r2 =

1

4¼(4)2 = 4¼


20.Z 3

1

(5¡ x)dx

Graph y = 5¡ x:

Z 3

1

(5¡ x)dx is the area of a trapezoid with bases

of length 4 and 2 and height of length 2.

Area =1

2(height)(base1 + base2) =

1

2(2)(4 + 2) = 6

21.Z 5

2

(1 + 2x)dx

Graph y = 1 + 2x:

Z 5

2

(1 + 2x)dx is the area of the trapezoid with B = 11; b = 5; and h = 3: The formula for the area is

A =1

2(B + b)h;

so we have

A =1

2(11 + 5)(3) = 24:

22. (a) With n = 10;¢x = 1¡010 = 0:1; and x1 = 0 + 0:1 = 0:1, use the command seq(X2;X; 0:1; 1; 0:1) !L1. The

resulting screen is:


(b) SincenPi=1

f(xi)¢x = ¢x

μnPi=1

f(xi)

¶, use the command 0:1¤sum(L1) to approximate R 1

0x2dx. The


1R0

x2dx ¼ 0:385

(c) With n = 100;¢x = 1¡0100 = 0:01 and x1 = 0 + 0:01 = 0:01, use the command seq(X

2;X; 0:1; 1; 0:1) !L1.The resulting screen is:

Use the command 0:01¤sum(L1) to approximate R 10 x2dx. The resulting screen is:

R 10x2dx ¼ 0:33835

(d)With n = 500;¢x = 1¡0500 = 0:002; and x1 = 0+ 0:002 = 0:002, use the command seq(X

2;X; 0:1; 1; 0:1)!L1.The resulting screen is:

Use the command 0.002¤sum(L1) to approximate R 10 x2dx. The resulting screen is:

R 10 x

2dx ¼ 0:334334

(e) As n gets larger the approximation forR 10 x

2dx seems to be approaching 0.333333 or 13 . We estimateR 1

0x2dx = 1

3 .


23. (a) With n = 10, ¢x = 1¡010 = 0:1; and x1 = 0 + 0:1 = 0:1, use the command seq(X^3;X; 0:1; 0:1; 0:1) !L1.

The resulting screen is:

(b) SincenPi=1

f(xi)¢x = ¢x

μnPi=1

f(xi)

¶, use the command 0.1¤sum(L1) to approximate R 1

0x3dx. The


1Z0

x3dx ¼ 0:3025

(c)With n = 100;¢x = 1¡0100 = 0:01, and x1 = 0 + 0:01 = 0:01, use the command seq(X^3;X; 0:01; 0:1; 0:01)!L1.

The resulting screen is:

Use the command 0.01¤sum(L1) to approximate R 10x3dx. The resulting screen is:

Z 1

0

x3dx ¼ 0:255025

(d)With n = 500;¢x = 1¡0500 = 0:002, and x1 = 0+0:002 = 0:002, use the command seq (X

^3;X; 0:002; 1; 0:002)!L1.The resulting screen is:


Use the command 0.002¤sum(L1) to approximate R 10x3dx. The resulting screen is:

Z 1

0

x3dx ¼ 0:251001

(e) As ngets larger the approximation forR 10x3dx seems to be approaching 0.25 or 14 . We estimateR 1

0x3dx = 1

4 :

For Exercises 24¡34, readings on the graphs and answers may vary.24. Left endpoints:

Read values of the function from the graph for every 2 hours from midnight to 10 P.M. These values give theheights of 12 rectangles. The width of each rectangle is ¢x = 2: We estimate the area under the curve as

A =12Xi=1

f(xi)¢x

= 3:0(2) + 3:2(2) + 3:5(2) + 4:2(2) + 5:2(2) + 6:2(2) + 8:0(2) + 11:0(2) + 11:8(2) + 10:0(2) + 6:0(2) + 4:4(2)

= 153:0

Right endpoints:

Read values of the function from the graph for every 2 hours from 2 A.M. to midnight. Now we estimate thearea under the curve as

A =12Xi=1

f(xi)¢x

= 3:2(2) + 3:5(2) + 4:2(2) + 5:2(2) + 6:2(2) + 8:0(2) + 11:0(2) + 11:8(2) + 10:0(2) + 6:0(2) + 4:4(2) + 3:8(2)

= 154:6

Average:153:0 + 154:6

2=307:6

2= 153:8

The area under the curve represents the total electricity usage. We estimate this usage as about 154 millionkilowatt hours.

25. Left endpoints:

Read values of the function on the graph every 5 years from 1980 to 2000. These values give us the heights of5 rectangles. The width of each rectangle is ¢x = 5: We estimate the area under the curve as

A =5Xi=1

f(xi)¢x = 702:7(5) + 818(5) + 902:9(5) + 962:1(5) + 1084:1(5) = 22,349

Right endpoints:

Read values of the function from the graph every 5 years from 1985 to 2005. These values give the heights of5 rectangles. The width of each rectangle is ¢x = 5: We estimate the area under the curve as

A =5Xi=1

f(xi)¢x = 818(5) + 902:9(5) + 962:1(5) + 1084:1(5) + 1128:3(5) = 24,477


Average:22,349 + 24,477

2=46,8262

= 23,413

The area under the curve represents the total U.S. coal consumption. We estimate this consumption as about23,413 million short tons.

26. Left endpoints:

Read values of the function from the graph for every minute from 0 minutes to 19 minutes. These values givethe heights of 20 rectangles. The width of each rectangle is ¢x = 1: We estimate the area under the curve as20Xi=1

f(xi)¢x

= 0(1) + 2:4(1) + 2:9(1) + 3:1(1) + 3:2(1) + 3:3(1) + 3:3(1) + 3:3(1) + 3:3(1)

+ 3:3(1) + 3:3(1) + 1:1(1) + 0:7(1) + 0:6(1) + 0:5(1) + 0:4(1) + 0:3(1)

+ 0:3(1) + 0:2(1) + 0:2(1)

¼ 35:7:Right endpoints:

Read values of the function from the graph for every minute from 1 minutes to 20 minutes.Now we estimate the area under the curve as20Xi=1

f(xi)¢x

= 2:4(1) + 2:9(1) + 3:1(1) + 3:2(1) + 3:3(1) + 3:3(1) + 3:3(1) + 3:3(1) + 3:3(1)

+ 3:3(1) + 1:1(1) + 0:7(1) + 0:6(1) + 0:5(1) + 0:4(1) + 0:3(1) + 0:3(1)

+ 0:2(1) + 0:2(1) + 0:2(1)

¼ 35:9:Average:

35:7 + 35:9

2= 35:8

The area under the curve represents the total volume of oxygen inhaled. We estimate this volume as about35.8 liters.

27. (a) Left endpoints:

Read values of the function from the graph for every 14 days from 18 Feb. through 30 Apr. The values givethe heights of 6 rectangles. The width of each rectangle is ¢x = 14: We estimate the area under the curve as

A =6Xi=1

f(xi)¢x = 0(14) + 15(14) + 33(14) + 40(14) + 16(14) + 5(14) = 1526:

Right endpoints:

Read values of the function from the graph for every 14 days from 4 Mar. through 13 May. Now we estimatethe area under the curve as

A =6Xi=1

f(xi)¢x = 15(14) + 33(14) + 40(14) + 16(14) + 5(14) + 1(14) = 1540:

Average:1526 + 1540

2= 1533

There were about 1533 cases of the disease.


(b) Left endpoints:

Read values of the function from the graph for every 14 days from 18 Feb. through 30 Apr. The values givethe heights of 6 rectangles. The width of each rectangle is ¢x = 14: We estimate the area under the curve as

A =6Xi=1

f(xi)¢x = 0(14) + 10(14) + 15(14) + 10(14) + 3(14) + 1(14) = 546:

Right endpoints:

Read values of the function from the graph for every 14 days from 4 Mar. through 13 May. Now we estimatethe area under the curve as

A =6Xi=1

f(xi)¢x = 10(14) + 15(14) + 10(14) + 3(14) + 1(14) + 1(14) = 560:

Average:546 + 560

2= 553

There would have been about 553 cases of the disease.

28. Left endpoints:

Read values of the function from the graph for every 2 years from 1996 to 2002. These values give the heightsof 4 rectangles. The width of each rectangle is ¢x = 2: We estimate the area under the curve as

A =4Xi=1

f(xi)¢x = 3555(2) + 3075(2) + 3331(2) + 3650(2) = 27,222:

Right endpoints:

Read values of the function from the graph for every 5 years from 1998 to 2004. These values give the heightsof 4 rectangles. The width of each rectangle is ¢x = 2: We estimate the area under the curve as

A =4Xi=1

f(xi)¢x = 3075(2) + 3331(2) + 3650(2) + 3701(2) = 27,514

Average:27,222 + 27,514

2=54,7362

= 27,368

The area under the curve represents the number of fatal automobile accidents in California from 1996 to 2004.We estimate this number as about 27,368 collisions.

29. Read the value of the function for every 5 sec from x = 2:5 to x = 12:5: These are the midpoints of rectangleswith width ¢x = 5: Then read the function for x = 17; which is the midpoint of a rectangle with width ¢x = 4:

4Xi=1

f(xi)¢x ¼ 36(5) + 63(5) + 84(5) + 95(4) ¼ 1295

1295

3600(5280) ¼ 1900

The Porsche 928 traveled about 1900 ft.


30. Read the value for the speed every 5 sec from x = 2:5 to x = 22:5: These are the midpoints of rectangles withwidth ¢x = 5: Then read the speed for x = 26:5; which is the midpoint of a rectangle with width ¢x = 3:

6Xi=1

f(xi)¢x ¼ 28(5) + 54(5) + 72(5) + 82(5) + 92(5) + 98(3) = 1934

1934

3600(5280) ¼ 2800

The BMW 733i traveled about 2800 ft.

31. Left endpoints:

Read values of the function from the table for every number of seconds from 2.0 to 19.3. These values give theheights of 10 rectangles. The width of each rectangle varies: We estimate the area under the curve as10Xi=1

f(xi)¢x

= 30(2:9¡ 2:0) + 40(4:1¡ 2:9) + 50(5:3¡ 4:1) + 60(6:9¡ 5:3) + 70(8:7¡ 6:9) + 80(10:7¡ 8:7)+ 90(13:2¡ 10:7) + 100(16:1¡ 13:2) + 110(19:3¡ 16:1) + 120(23:4¡ 19:3)= 1876

5280

3600(1876) ¼ 2751

Right endpoints:


f(xi)¢x

= 30(2:0¡ 0) + 40(2:9¡ 2:0) + 50(4:1¡ 2:9) + 60(5:3¡ 4:1) + 70(6:9¡ 5:3) + 80(8:7¡ 6:9)+ 90(10:7¡ 8:7) + 100(13:2¡ 10:7) + 110(16:1¡ 13:2) + 120(19:3¡ 16:1) + 130(23:4¡ 19:3)= 2150

5280

3600(2150) ¼ 3153

Average:2751 + 3153

2=5904

2

= 2952

The distance traveled by the Mercedes-Benz S550 is about 2952 ft.

32. Left endpoints:


f(xi)¢x

= 30(3:5¡ 2:4) + 40(5:1¡ 3:5) + 50(6:9¡ 5:1) + 60(8:9¡ 6:9)+ 70(11:2¡ 8:9) + 80(14:9¡ 11:2) + 90(19:2¡ 14:9) + 100(24:4¡ 19:2)= 1671

5280

3600(1671) ¼ 2451


Right endpoints:

Read values of the function from the table for every number of seconds from 2.4 to 24.4. These values give theheights of 9 rectangles. The width of each rectangle varies: We estimate the area under the curve as

9Xi=1

f(xi)¢x

= 30(2:4¡ 0) + 40(3:5¡ 2:4) + 50(5:1¡ 3:5) + 60(6:9¡ 5:1) + 70(8:9¡ 6:9) + 80(11:2¡ 8:9)+ 90(14:9¡ 11:2) + 100(19:2¡ 14:9) + 110(24:4¡ 19:2)= 1963

5280

3600(1963) ¼ 2879

Average:

2451 + 2879

2=5330

2= 2665

The distance traveled by the Chevrolet Malibu Maxx SS is about 2665 ft.

33. (a) Read values of the function on the plain glass graph every 2 hr from 6 to 6. These are at midpoints of thewidths ¢x = 2 and represent the heights of the rectangles.

f(xi)¢x = 132(2) + 215(2) + 150(2) + 44(2) + 34(2) + 26(2) + 12(2) ¼ 1226

The total heat gain was about 1230 BTUs per square foot.

(b) Read values on the ShadeScreen graph every 2 hr from 6 to 6.

Xf(xi)¢x = 38(2) + 25(2) + 16(2) + 12(2) + 10(2) + 10(2) + 5(2) ¼ 232

The total heat gain was about 230 BTUs per square foot.

34. (a) Read the value for a plain glass window facing south for every 2 hr from 6 to 6. These are the heights, atthe midpoints, of rectangles with width ¢x = 2:

X= 10(2) + 30(2) + 80(2) + 107(2) + 79(2) + 29(2) + 10(2) ¼ 690

The heat gain is about 690 BTUs per square foot.

(b) Read the value for a window with Shadescreen facing south for every 2 hr from 6 to 6. These are theheights, at the midpoints, of rectangles with width ¢x = 2:

X= 4(2) + 10(2) + 20(2) + 22(2) + 20(2) + 10(2) + 4(2) ¼ 180

The heat gain is about 180 BTUs per square foot.


35. (a) The area of a trapezoid is

A =1

2h(b1 + b2) =

1

2(6)(1 + 2) = 9:

Car A has traveled 9 ft.

(b) Car A is furthest ahead of car B at 2 sec. Notice that from t = 0 to t = 2; v(t) is larger for car A than forcar B: For t > 2; v(t) is larger for car B than for car A:

(c) As seen in part (a), car A drove 9 ft after 2 sec. The distance of car B can be calculated as follows:

2¡ 04

=1

2= width

Distance =1

2¢ v(0:25) + 1

2v(0:75) +

1

2v(1:25) +

1

2v(1:75) =

1

2(0:2) +

1

2(1) +

1

2(2:6) +

1

2(5) = 4:4

9¡ 4:4 = 4:6The furthest car A can get ahead of car B is about 4.6 ft.

(d) At t = 3; car A travels 12(6)(2 + 3) = 15 ft and car B travels approximately 13 ft.At t = 3:5; car A travels 1

2(6)(2:5 + 3:5) = 18 ft and car B travels approximately 18.25 ft. Therefore, car Bcatches up with car A between 3 and 3.5 sec.

36. Using the left endpoints:

Distance = v0(1) + v1(1) + v2(1) + v3(1) = 0 + 8 + 13 + 17 = 38 ft

Using the right endpoints:

Distance = v1(1) + v2(1) + v3(1) + v4(1) = 8 + 13 + 17 + 18 = 56 ft

37. Using the left endpoints:

Distance = v0(1) + v1(1) + v2(1) = 10 + 6:5 + 6 = 22:5 ft

Using the right endpoints:

Distance = v1(1) + v2(1) + v3(1) = 6:5 + 6 + 5:5 = 18 ft

38. (a) Using the left endpoints:

Distance =nXi=1

f(xi)¢xi

= 0(1:84) + 12:9(1:96) + 23:8(2:58) + 26:3(0:85) + 26:3(1:73) + 26:0(0:87)

= 0 + 25:284 + 61:404 + 22:355 + 45:499 + 22:62

= 177:162

Since we multiplied the units of seconds by miles per hour, we need to divide by 3600 (the number of secondsin an hour) to get a distance in miles.

177:162

3600¼ 0:0492

The estimate of the distance is 0.0492 miles.

Section 7.4 The Fundamental Theorem of Calculus 479


Distance =nXi=1

f(xi)¢xi

= 12:9(1:84) + 23:8(1:96) + 26:3(2:58) + 26:3(0:85) + 26:0(1:73) + 25:7(0:87)

= 23:736 + 46:648 + 67:854 + 22:355 + 44:98 + 22:359

= 227:932

Divide by 3600 (the number of seconds in an hour) to get a distance in miles.

227:932

3600¼ 0:0633

The estimate of the distance is 0.0633 miles.

(c)100

1609¼ 0:0622

Johnson actually ran 0.0622 miles. The answer to part b is closer.

39. (a) Read values from the graph for every hour from 1 A.M. through 11 P.M. The values give the heights of23 rectangles. The width of each rectangle is ¢x = 1: We estimate the area under the curve as

A =23Xi=1

f(xi)¢x

= 500(1) + 550(1) + 800(1) + 1600(1) + 4000(1) + 7000(1) + 7000(1) + 5900(1) + 4500(1)

+ 3500(1) + 3100(1) + 3100(1) + 3500(1) + 3800(1) + 4100(1) + 4800(1) + 4750(1)

+ 4000(1) + 2500(1) + 2250(1) + 1800(1) + 1500(1) + 1050(1)

= 75,600

(b) Read values from the graph for every hour from 1 A.M. through 11 P.M. The values give the heights of23 rectangles. The width of each rectangle is ¢x = 1: We estimate the area under the curve as

A =23Xi=1

f(xi)¢x

= 500(1) + 400(1) + 400(1) + 700(1) + 1500(1) + 3000(1) + 4100(1) + 3900(1) + 3200(1)

+ 3600(1) + 4000(1) + 4000(1) + 4300(1) + 5200(1) + 6000(1) + 6500(1) + 6400(1)

+ 6000(1) + 4700(1) + 3100(1) + 2600(1) + 1900(1) + 1300(1)

= 77,300

7.4 The Fundamental Theorem of Calculus

1.Z 4

¡2(¡3)dp = ¡3

Z 4

¡2dp = ¡3 ¢ pj4¡2 = ¡3[4¡ (¡2)] = ¡18

2.Z 1

¡4

p2dx =

p2

Z 1

¡4dx =

p2 ¢ x

¯1¡4=p2[1¡ (¡4)] = 5p2


3.Z 2

¡1(5t¡ 3)dt = 5

Z 2

¡1t dt¡ 3

Z 2

¡1dt

=5

2t2¯¯2

¡1¡ 3t

¯2¡1

=5

2[22 ¡ (¡1)2]¡ 3[2¡ (¡1)]

=5

2(4¡ 1)¡ 3(2 + 1)

=15

2¡ 9

=15

2¡ 182

= ¡32

4.Z 2

¡2(4z + 3)dz = 4

Z 2

¡2z dz + 3

Z 2

¡2dz

= 2z2¯2¡2+ 3z

¯2¡2

= 2[22 ¡ (¡2)2] + 3[2¡ (¡2)]= 2(4¡ 4) + 3(4) = 12

5.Z 2

0

(5x2 ¡ 4x+ 2)dx

= 5

Z 2

0

x2 dx¡ 4Z 2

0

xdx+ 2

Z 2

0

dx

=5x3

3

¯20¡ 2x2

¯20+ 2x

¯20

=5

3(23 ¡ 03)¡ 2(22 ¡ 02) + 2(2¡ 0)

=5

3(8)¡ 2(4) + 2(2)

=40¡ 24 + 12

3=28

3

6.Z 3

¡2(¡x2 ¡ 3x+ 5)dx

= ¡Z 3

¡2x2 dx¡ 3

Z 3

¡2xdx+ 5

Z 3

¡2dx

= ¡13x3¯3¡2¡ 32x2¯3¡2+ 5x

¯3¡2

= ¡13[33¡(¡2)3]¡ 3

2[32¡(¡2)2]+5[3¡(¡2)]

= ¡13(27 + 8)¡ 3

2(9¡ 4) + 5(5)

= ¡353¡ 152+ 25 =

35

6

7.Z 2

0

3p4u+ 1du

Let 4u+ 1 = x; so that 4du = dx:

When u = 0; x = 4(0) + 1 = 1:When u = 2; x = 4(2) + 1 = 9:Z 2

0

3p4u+ 1du

=3

4

Z 2

0

p4u+ 1(4du)

=3

4

Z 9

1

x1=2 dx

=3

4¢ x

3=2

3=2

¯91

=3

4¢ 23(93=2 ¡ 13=2)

=1

2(27¡ 1) = 26

2= 13

8.Z 9

3

p2r ¡ 2dr =

Z 9

3

(2r ¡ 2)1=2 dr

Let u = 2r ¡ 2; so that du = 2dr:If r = 9; u = 2 ¢ 9¡ 2 = 16:If r = 3; u = 2 ¢ 3¡ 2 = 4:Z 9

3

(2r ¡ 2)1=2 dr = 1

2

Z 9

3

(2r ¡ 2)1=2 2dr

=1

2

Z 16

4

u1=2 du

=1

2¢ u

3=2

32

¯164

=1

3¢ u3=2

¯164

=1

3(163=2 ¡ 43=2)

=1

3(64¡ 8) = 56

3

9.Z 4

0

2(t1=2 ¡ t)dt = 2Z 4

0

t1=2dt¡ 2Z 4

0

t dt

= 2 ¢ t3=2

32

¯¯4

0

¡ 2 ¢ t2

2

¯¯4

0

=4

3(43=2 ¡ 03=2)¡ (42 ¡ 02)

=32

3¡ 16

= ¡163


10.Z 4

0

¡(3x3=2 + x1=2)dx

= ¡3Z 4

0

x3=2 dx¡Z 4

0

x1=2 dx

= ¡3x5=2

52

¯40¡ x

3=2

32

¯40

= ¡65(32)¡ 2

3(8)

= ¡1925¡ 163= ¡656

15

11.Z 4

1

(5ypy + 3

py)dy

= 5

Z 4

1

y3=2 dy + 3

Z 4

1

y1=2 dy

= 5

μy5=2

52

¶ ¯41+ 3

μy3=2

32

¶ ¯41

= 2y5=2¯41+ 2y3=2

¯41

= 2(45=2 ¡ 1) + 2(43=2 ¡ 1)= 2(32¡ 1) + 2(8¡ 1)= 62 + 14

= 76

12.Z 9

4

(4pr ¡ 3rpr)dr

= 4

Z 9

4

r1=2 dr ¡ 3Z 9

4

r3=2 dr

= 4r3=2

32

¯94¡ 3r

5=2

52

¯94

=8

3r3=2

¯94¡ 65r5=2

¯94

=8

3(27¡ 8)¡ 6

5(243¡ 32)

=8

3¢ 19¡ 6

5(211)

=760

15¡ 3798

15= ¡3038

15

13.Z 6

4

2

(2x¡ 7)2 dxLet u = 2x¡ 7; so that du = 2dx:When x = 6; u = 2 ¢ 6¡ 7 = 5:When x = 4; u = 2 ¢ 4¡ 7 = 1:Z 6

4

2

(2x¡ 7)2 dx =Z 5

1

u¡2du

=u¡1

¡1¯¯5

1

= ¡u¡1 ¯51

= ¡μ1

5¡ 1¶

= ¡μ¡45

¶=4

5

14.Z 4

1

¡3(2p+ 1)2

dp

= ¡3Z 4

1

(2p+ 1)¡2 dp

Let u = 2p+ 1; so that du = 2dp:

If p = 4; u = 2 ¢ 4 + 1 = 9:If p = 1; u = 2 ¢ 1 + 1 = 3:¡3Z 4

1

(2p+ 1)¡2 dp

= ¡32

Z 9

3

u¡2 du

= ¡32¢ u

¡1

¡1¯93

=3

2u

¯93

=3

18¡ 36= ¡1

3

15.Z 5

1

(6n¡2 ¡ n¡3)dn = 6Z 5

1

n¡2dn¡Z 5

1

n¡3dn

= 6 ¢ n¡1

¡1¯¯5

1

¡ n¡2

¡2¯¯5

1

=¡6n

¯¯5

1

+1

2n2

¯¯5

1

=¡65¡μ¡61

¶+

·1

2(25)¡ 1

2(1)

¸

=¡65+6

1+1

50¡ 12

=108

25


16.Z 3

2

(3x¡3 ¡ 5x¡4)dx = 3Z 3

2

x¡3 dx¡ 5Z 3

2

x¡4 dx

= 3 ¢ x¡2

¡2¯32¡ 5x

¡3

¡3¯32

= ¡32¢ 1x2

¯32+5

3¢ 1x3

¯32

= ¡32

μ1

9¡ 14

¶+5

3

μ1

27¡ 18

¶

= ¡16+3

8+5

81¡ 5

24

=5

81

17.Z ¡2

¡3

μ2e¡0:1y +

3

y

¶dy

= 2

Z ¡2

¡3e¡0:1ydy +

Z ¡2

¡3

3

ydy

= 2 ¢ e¡0:1y

¡0:1¯¯¡2

¡3+ 3 ln jyj

¯¯¡2

¡3

= ¡ 20e¡0:1y ¯¡2¡3 + 3 ln jyj¯¯¡2

¡3

= 20e0:3 ¡ 20e0:2 + 3 ln 2¡ 3 ln 3¼ 1:353

18.Z ¡1

¡2

μ¡2t+ 3e0:3t

¶dt

= ¡2Z ¡1

¡2

1

tdt+

3

0:3

Z ¡1

¡20:3e0:3tdt

= ¡ 2 ln jtj¯¯¡1

¡2+

3

0:3e0:3t

¯¯¡1

¡2

= 2 ln 2 + 10e¡0:3 ¡ 10e¡0:6¼ 3:306

19.Z 2

1

μe4u ¡ 1

(u+ 1)2

¶du

=

Z 2

1

e4udu¡Z 2

1

1

(u+ 1)2du

=e4u

4

¯¯2

1

¡ ¡1u+ 1

¯¯2

1

=e8

4¡ e

4

4+

1

2 + 1¡ 1

1 + 1

=e8

4¡ e

4

4¡ 16

¼ 731:4

20.Z 1

0:5

(p3 ¡ e4p)dp =Z 1

0:5

p3 dp¡Z 1

0:5

e4p dp

=p4

4

¯10:5¡ e

4p

4

¯10:5

=1

4¡ 1

64¡μe4

4¡ e

2

4

¶

=15

64¡ e

4

4+e2

4¼ ¡11:57

21.Z 0

¡1y(2y2 ¡ 3)5 dy

Let u = 2y2 ¡ 3; so thatdu = 4y dy and

1

4du = y dy:

When y = ¡1; u = 2(¡1)2 ¡ 3 = ¡1:When y = 0; u = 2(0)2 ¡ 3 = ¡3:

1

4

Z ¡3

¡1u5 du =

1

4¢ u

6

6

¯¡3¡1

=1

24u6¯¡3¡1

=1

24(¡3)6 ¡ 1

24(¡1)6

=729

24¡ 1

24

=728

24

=91

3

22.Z 3

0

m2(4m3 + 2)3 dm

Let u = 4m3 + 2; so that

du = 12m2 dm and1

12du = m2 dm:

Also, when m = 3;

u = 4(33) + 2 = 110;

and when m = 0;

u = 4(03) + 2 = 2:

1

12

Z 110

2

u3 du =1

12¢ u

4

4

¯1102

=1

48u4

¯1102

=146,410,000

48¡ 1648

=146,409,984

48

=9,150,624

3

¼ 3,050,208


23.Z 64

1

pz ¡ 23pz

dz

=

Z 64

1

μz1=2

z1=3¡ 2z¡1=3

¶dz

=

Z 64

1

z1=6 dz ¡ 2Z 64

1

z¡1=3 dz

=z7=6

76

¯641¡ 2z

2=3

23

=6z7=6

7

¯641¡ 3z2=3

¯641

=6(64)7=6

7¡ 6(1)

7=6

7

¡ 3(642=3 ¡ 12=3)

=6(128)

7¡ 67¡ 3(16¡ 1)

=768¡ 6¡ 315

7

=447

7¼ 63:86

24.Z 8

1

3¡ y1=3y2=3

dy

=

Z 8

1

(3y¡2=3 ¡ y¡1=3)dy

=

Z 8

1

3y¡2=3 dy ¡Z 8

1

y¡1=3 dy

=3y1=3

13

¯81¡ y

2=3

23

¯81

= 9y1=3¯81¡ 3y

2=3

2

¯81

= 9(2¡ 1)¡ 32(4¡ 1)

= 9¡ 92=9

2

25.Z 2

1

ln x

xdx


du =1

xdx:

When x = 1; u = ln 1 = 0:

When x = 2; u = ln 2:Z ln 2

0

udu =u2

2

¯ln 2

0

=(ln 2)2

2¡ 0

=(ln 2)2

2

¼ 0:2402

26.Z 3

1

pln x

xdx


du =1

xdx:

When x = 3; u = ln 3; andwhen x = 1; u = ln 1 = 0:Z ln 3

0

pudu =

Z ln 3

0

u1=2 du

=u3=2

32

¯ln 30

=2

3u3=2

¯ln 30

=2

3(ln 3)3=2 ¡ 2

3(0)3=2

=2

3(ln 3)3=2

¼ 0:7677

27.Z 8

0

x1=3px4=3 + 9 dx

Let u = x4=3 + 9; so that

du =4

3x1=3 dx and

3

4du = x1=3 dx:

When x = 0; u = 04=3 + 9 = 9:When x = 8; u = 84=3 + 9 = 25:

3

4

Z 25

9

pudu =

3

4

Z 25

9

u1=2 du

=3

4¢ u

3=2

32

¯259

=1

2u3=2

¯259

=1

2(25)3=2 ¡ 1

2(9)3=2

=125

2¡ 272

= 49


28.Z 2

1

3

x(1 + ln x)dx

Let u = 1 + ln x; so that

du =1

xdx:

When x = 2; u = 1 + ln 2; andwhen x = 1; u = 1 + ln 1 = 1:Z 1+ln 2

1

3

udu

= 3 ln juj¯1+ln 21

= 3 ln (1 + ln 2)¡ 3 ln 1= 3 ln (1 + ln 2)

¼ 1:580

29.Z 1

0

e2t

(3 + e2t)2dt

Let u = 3 + e2t; so that du = 2e2tdt:

When x = 1; u = 3 + e2¢1 = 3 + e2:When x = 0; u = 3 + e2¢0 = 4:Z 1

0

e2t

(3 + e2t)2dt =

1

2

Z 3+e2

4

u¡2du

=1

2¢ u

¡1

¡1¯¯3+e2

4

=¡12u

¯¯3+e2

4

=1

8¡ 1

2(3 + e2)

¼ 0:07687

30.Z 1

0

e2zp1 + e2z

dz

Let u = 1 + e2z; so that

du = 2e2z dz and1

2du = e2z dz:

When z = 1; u = 1 + e2; andwhen z = 0; u = 1+ e0 = 2:

1

2

Z 1+e2

2

1pudu =

1

2

Z 1+e2

2

u¡1=2 du

=1

2¢ u

1=2

12

¯1+e22

= u1=2¯1+e22

=p1 + e2 ¡p2

¼ 1:482

31. f(x) = 2x¡ 14; [6; 10]

The graph crosses the x-axis at

0 = 2x¡ 142x = 14

x = 7:

This location is in the interval. The area of theregion is¯¯Z 7

6

(2x¡ 14)dx¯¯+

Z 10

7

(2x¡ 14)dx

=¯(x2 ¡ 14x)¯7

6

¯+ (x2 ¡ 14x)¯10

7

=¯(72 ¡ 98)¡ (62 ¡ 84)¯+ (102 ¡ 140)¡ (72 ¡ 98)= j¡1j+ (¡40)¡ (¡49)= 10:

32. f(x) = 4x¡ 32; [5; 10]


0 = 4x¡ 324x = 32

x = 8:



5

(4x¡ 32)dx¯¯+

Z 10

8

(4x¡ 32)dx

=¯(2x2 ¡ 32x)¯8

5

¯+ (2x2 ¡ 32x)¯10

8

= j(128¡ 256)¡ (50¡ 160)j+ (200¡ 320)¡ (128¡ 256)= j¡18j+ (¡120) + 128= 26:

33. f(x) = 2¡ 2x2; [0; 5]

Find the points where the graph crosses the x-axisby solving 2¡ 2x2 = 0:

2¡ 2x2 = 02x2 = 2

x2 = 1

x = §1:

The only solution in the interval [0; 5] is 1.The total area isZ 1

0

(2¡ 2x2)dx+¯¯Z 5

1

(2¡ 2x2)dx¯¯

=

μ2x¡ 2x

3

3

¶¯¯1

0

+

¯¯μ2x¡ 2x3

3

¶¯¯5

1

¯¯

= 2¡ 23+

¯¯10¡ 2(53)3 ¡ 2 + 2

3

¯¯

=4

3+

¯¯¡2243

¯¯

=228

3

= 76:

34. f(x) = 9¡ x2; [0; 6]


0 = 9¡ x2x2 = 9

x = §3:

In the interval, the graph crosses at x = 3:The area of the region is

Z 3

0

(9¡ x2)dx+¯¯Z 6

3

(9¡ x2)dx¯¯

=

μ9x¡ x

3

3

¶ ¯30+

¯¯μ9x¡ x

3

3

¶ ¯63

¯¯

= (27¡ 9) + j(54¡ 72)¡ (27¡ 9)j= 18 + j¡36j= 18 + 36

= 54:

35. f(x) = x3; [¡1; 3]

The solution

x3 = 0

x = 0

indicates that the graph crosses the x-axis at 0 inthe given interval [¡1; 3]:


The total area is¯¯Z 0

¡1x3 dx

¯¯+

Z 3

0

x3 dx

=

¯¯x44

¯¯0

¡1+

¯¯x44

¯¯3

0

=

¯¯μ0¡ 1

4

¶¯¯+

μ34

4¡ 0¶

=1

4+81

4=82

4

=41

2:

36. f(x) = x3 ¡ 2x; [¡2; 4]


0 = x3 ¡ 2x= x(x2 ¡ 2)

x = 0; x =p2; and x = ¡p2:

These locations are all in the interval.The area of the region is¯¯Z ¡p2

¡2(x3 ¡ 2x)dx

¯¯+ ¯¯

Z 0

¡p2(x3 ¡ 2x)dx

¯¯

+

¯¯Z

p2

0

(x3 ¡ 2x)dx¯¯+ ¯¯

Z 4

p2

(x3 ¡ 2x)dx¯¯

=

¯¯μx44¡ x2

¶ ¯¯¡p2

¡2

¯¯+ ¯¯

μx4

4¡ x2

¶¯¯0

¡p2

¯¯

+

¯¯μx4

4¡ x2

¶¯¯p2

0

¯¯+ ¯¯

μx4

4¡ x2

¶¯¯4

p2

¯¯

= j(1¡ 2)¡ (4¡ 4)j+ j0¡ (1¡ 2)j+ j(1¡ 2)¡ 0 + (64¡ 16)¡ 1¡ 2j= j¡1j+ j1j+ j¡1j+ j49j= 1 + 1 + 1 + 49

= 52:

37. f(x) = ex ¡ 1; [¡1; 2]

Solveex ¡ 1 = 0:

ex = 1

x ln e = ln 1

x = 0

The graph crosses the x-axis at 0 in the given in-terval [¡1; 2]:The total area is¯

¯Z 0

¡1(ex ¡ 1)dx

¯¯+

Z 2

0

(ex ¡ 1)dx

=¯(ex ¡ x)

0

¡1+ (ex ¡ x)

¯20

= j(1¡ 0)¡ (e¡1 + 1)j+ (e2 ¡ 2)¡ (1¡ 0)

=¯1¡ e¡1 ¡ 1¯+ e2 ¡ 2¡ 1

=1

e+ e2 ¡ 3

¼ 4:757:38. f(x) = 1¡ e¡x; [¡1; 2]


0 = 1¡ e¡xe¡x = 1

¡x ln e = ln 1¡x = 0x = 0:


The area of the region is

¯¯Z 0

¡1(1¡ e¡x)dx

¯¯+

Z 2

0

(1¡ e¡x)dx

=¯(x+ e¡x )

0

¡1+¯(x+ e¡x )

2

0

=¯(1)¡ (¡1 + e1)¯+ (2 + e¡2)¡ (e0)

= j2¡ ej+ 2+ e¡2 ¡ 1= j¡0:718j+ 1+ e¡2= 0:718 + 1:135

= 1:854:

39. f(x) =1

x¡ 1e; [1; e2]


0 =1

x¡ 1e

1

x=1

ex = e:

This location is in the interval. The area of theregion is

Z e

1

μ1

x¡ 1e

¶dx+

¯¯Z e2

e

μ1

x¡ 1e

¶dx

¯¯

=

¯¯ln jxj ¡ x

e

¯e1+¯³ln jxj ¡ x

e

´ ¯e2e

¯¯

= 0¡μ¡1e

¶+ j(2¡ e)¡ 0j

=1

e+ j2¡ ej

= e¡ 2 + 1e:

40. f(x) = 1¡ 1

x; [e¡1; e]


0 = 1¡ 1

x1

x= 1

x = 1:


e¡1

μ1¡ 1

x

¶dx

¯¯+

Z e

1

μ1¡ 1

x

¶dx

=¯(x¡ ln jxj)j1e¡1

¯+ (x¡ ln jxj)je1

=¯(1¡ 0)¡ [e¡1 ¡ (¡1)]¯+ (e¡ 1)¡ (1¡ 0)

=

¯¯1¡

μ1

e+ 1

¶¯¯+ (e¡ 1)¡ 1

= e¡ 2 + 1e:

41. y = 4¡ x2; [0; 3]From the graph, we see that the total area isZ 2

0

(4¡ x2) dx+¯¯Z 3

2

(4¡ x2)dx¯¯

=

μ4x¡ x

3

3

¶¯¯2

0

+

¯¯μ4x¡ x3

3

¶¯¯3

2

¯¯

=

·μ8¡ 8

3

¶¡ 0¸

+

¯¯·(12¡ 9)¡

μ8¡ 8

3

¶¸¯¯

=16

3+

¯¯3¡ 163

¯¯

=16

3+7

3

=23

3


42. f(x) = x2 ¡ 2x; [¡1; 2]

From the graph, we see that the total area is

Z 0

¡1(x2 ¡ 2x)dx+

¯¯Z 2

0

(x2 ¡ 2x)dx¯¯

=

μx3

3¡ x2

¶¯¯0

¡1+

¯¯ μx3

3¡ x2

¶¯¯2

0

¯¯

= ¡μ¡13¡ 1¶+

¯¯83 ¡ 4

¯¯

=4

3+

¯¯¡43

¯¯ = 4

3+4

3=8

3:

43. y = ex ¡ e; [0; 2]

From the graph, we see that the total area is¯¯Z 1

0

(ex ¡ e)dx¯¯+

Z 2

1

(ex ¡ e)dx

=¯(ex ¡ xe)

¯10

¯+ (ex ¡ xe)

¯2

1

=¯(e1 ¡ e)¡ (e0 + 0)¯+ (e2 ¡ 2e)¡ (e1 ¡ e)= j¡1j+ e2 ¡ 2e= 1 + e2 ¡ 2e¼ 2:952:

44. y =lnx

x;·1

e; e

¸

From the graph, we can see that the total area is¯¯Z 1

1e

μlnx

x

¶dx

¯¯+ Z e

1

μlnx

x

¶dx

=

¯¯ (lnx)2

2

¯¯1

1e

¯¯+ (lnx)2

2

¯¯e

1

=

¯¯0¡ 12

¯¯+ 12 ¡ 0

=1

2+1

2

= 1

45.Z c

a

f(x)dx =

Z b

a

f(x)dx+

Z c

b

f(x)dx

46. The equation for Exercise 45 isZ c

a

f(x)dx =

Z b

a

f(x)dx+

Z c

b

f(x)dx:

(a) If b is replaced by d; we getZ c

a

f(x)dx =

Z d

a

f(x)dx+

Z c

d

f(x)dx:

This is a correct statement.

(b) If b is replaced by g; we getZ c

a

f(x)dx =

Z g

a

f(x)dx+

Z c

g

f(x)dx:

This is a correct statement.

47.Z 16

0

f(x)dx =

Z 2

0

f(x)dx+

Z 5

2

f(x)dx

+

Z 8

5

f(x)dx+

Z 16

8

f(x)dx

=1

2¢ 2(1 + 3) + ¼(3

2)

4¡ ¼(3

2)

4¡ 12(3)(8)

= 4 +9

4¼ ¡ 9

4¼ ¡ 12

= ¡8

48. Prove:Z b

a

k f(x)dx = k

Z b

a

f(x)dx:

Assume F (x) is an antiderivative of f(x):Then kF (x) is an antiderivative of kf(x):Z b

a

k f(x)dx = kF (x)¯ba

= kF (b)¡ kF (a)= k[F (b)¡ F (a)]

= k

Z b

a

f(x)dx

49. Prove:Z b

a

f(x)dx =

Z c

a

f(x)dx+

Z b

c

f(x)dx:

Let F (x) be an antiderivative of f(x):Z c

a

f(x)dx+

Z b

c

f(x)dx

= F (x)¯ca+ F (x)

¯bc

= [F (c)¡ F (a)] + [F (b)¡ F (c)]= F (c)¡ F (a) + F (b)¡ F (c)= F (b)¡ F (a)

=

Z b

a

f(x)dx


50. Prove:Z b

a

f(x)dx = ¡Z a

b

f(x)dx:

Assume F (x) is an antiderivative of f(x):Z b

a

f(x)dx = [F (x)]¯ba

= F (b)¡ F (a)= (¡1) [F (a)¡ F (b)]

= ¡1Z a

b

f(x)dx

= ¡Z a

b

f(x)dx

51.Z 4

¡1f(x)dx

=

Z 0

¡1(2x+ 3)dx

Z 4

0

³¡x4¡ 3´dx

= (x2 + 3x)¯0¡1+

μ¡x

2

8¡ 3x

¶ ¯40

= ¡(1¡ 3) + (¡2¡ 12)= 2¡ 14= ¡12

52.Z 1

0

ex2

dx = 1:46265;

Z 2

0

ex2

dx = 16:45263

(a) Since ex2

is symmetric about the y-axis,Z 1

¡1ex

2

dx = 2

Z 1

0

ex2

dx

= 2(1:46265)

= 2:92530:

(b)Z 2

1

ex2

dx =

Z 2

0

ex2

dx¡Z 1

0

ex2

dx

= 16:45263¡ 1:46265= 14:98998

53. (a) g(t) = t4 and c = 1, use substitution.

f(x) =

Z x

c

g(t)dt

=

Z x

1

t4dt

=t5

5

¯¯x

1

=x5

5¡ (1)

5

5

=x5

5¡ 15

(b) f 0(x) =d

dx(f(x))

=d

dx

μx5

5¡ 15

¶

=1

5¢ ddx(x5)¡ d

dx

μ1

5

¶

=1

5¢ 5x4 ¡ 0

= x4

Since g(t) = t4, then g(x) = x4 and we see f 0(x) =g(x):

(c) Let g(t) = et2

and c = 0, then f(x) =Z x

0

et2

dt.

f(1) =

Z 1

0

et2

dt and f(1:01) =Z 1:01

0

et2

dt.

Use the fnInt command in the Math menu of your

calculator to …ndZ 1

0

ex2

dx andZ 1:01

0

ex2

dx. The

resulting screens are:

f(1) ¼ 1:46265f(1:01) ¼ 1:49011

Usef(1 + h)¡ f(1)

hto approximate f 0(1) with

h = 0:01

f(1 + h)¡ f(1)h

=f(1:01)¡ f(1)

0:01

¼ 1:49011¡ 1:462650:01

= 2:746

So f 0(1) ¼ 2:746, and g(1) = e12 = e ¼ 2:718:


54. (a)Z 5

¡5x(x2 + 3)7dx

Let u = x2 + 3; so that du = 2xdx:

When x = 5; u = 52 + 3 = 28:When x = ¡5; u = (¡5)2 + 3 = 28:

Z 5

¡5x(x2 + 3)7dx =

1

2

Z 28

28

u7du

=1

2¢ u

8

8

¯¯28

28

=1

2

μ288

8¡ 28

8

8

¶

= 0

55. P 0(t) = (3t+ 3)(t2 + 2t+ 2)1=3

(a)Z 3

0

3(t+ 1)(t2 + 2t+ 2)1=3 dt

Let u = t2 + 2t+ 2; so that

du = (2t+ 2)dt and1

2du = (t+ 1)dt:

When t = 0; u = 02 + 2 ¢ 0 + 2 = 2:When t = 3; u = 32 + 2 ¢ 3 + 2 = 17:

3

2

Z 17

2

u1=3 du =3

2¢ u

4=3

43

¯172

=9

8u4=3

¯172

=9

8(17)4=3 ¡ 9

8(2)4=3

¼ 46:341

Total pro…ts for the …rst 3 yr were

9000

8(174=3 ¡ 24=3) ¼ $46; 341:

(b)Z 4

3

3(t+ 1)(t2 + 2t+ 2)1=3 dt

Let u = t2 + 2t+ 2; so that

du = (2t+ 2)dt = 2(t+ 1)dt and

3

2du = 3(t+ 1)dt:

When t = 3; u = 32 + 2 ¢ 3 + 2 = 17:

When t = 4; u = 42 + 2 ¢ 4 + 2 = 26:3

2

Z 26

17

u1=3 du =9

8u4=3

¯2617

=9

8(26)4=3 ¡ 9

8(17)4=3

¼ 37:477Pro…t in the fourth year was

9000

8(264=3 ¡ 174=3) ¼ $37; 477:

(c) limt!1 P 0(t)

= limt!1 (3t+ 3)(t2 + 2t+ 2)1=3

=1The annual pro…t is slowly increasing without bound.

56. H 0(x) = 20 ¡ 2x is the rate of change of thenumber of hours it takes a worker to produce thexth item.

(a) The total number of hours required to producethe …rst 5 items isZ 5

0

(20¡ 2x)dx = (20x¡ x2)¯50

= 100¡ 25 = 75:

It would take 75 hr to produce 5 items.

(b) The total number of hours required to producethe …rst 10 items isZ 10

0

(20¡ 2x)dx = (20x¡ x2)¯100

= (200¡ 100)¡ (0) = 100:

It would take 100 hr to produce the …rst 10 items.

57. P 0(t) = 140t5=2

Z 4

0

140t5=2 dt = 140 ¢ t7=2

72

¯40

= 40t7=2¯40

= 5120

Since 5120 is above the total level of acceptablepollution (4850), the factory cannot operate for 4years without killing all the …sh in the lake.


58. The tanker is leaking oil at a rate in barrels perhour of

L0(t) =80 ln (t+ 1)

t+ 1:

(a)Z 24

0

80 ln (t+ 1)

t+ 1dt

Let u = ln (t+ 1); so that du = 1t+1 dt:

When t = 24; u = ln 25:When t = 0; u = ln 1 = 0:

80

Z ln 25

0

udu = 80u2

2

¯ln 250

= 40u2¯ln 250

= 40(ln 25)2 ¡ 40(0)2¼ 414

About 414 barrels will leak on the …rst day.

(b)Z 48

24

80 ln (t+ 1)

t+ 1dt

Let u = ln (t+1), so that the limits of integrationwith respect to u are ln 25 and ln 49:

80

Z ln 49

ln 25

udu = 40u2¯ln 49ln 25

= 40(ln 49)2 ¡ 40(ln 25)2¼ 191

About 191 barrels will leak on the second day.

(c) limt!1 L

0(t) = limt!1

80 ln (t+ 1)

t+ 1= 0

The number of barrels of oil leaking per day isdecreasing to 0.

59. Growth rate is 0:6 + 4(t+1)3 ft/yr.

(a) Total growth in the second year isZ 2

1

·0:6 +

4

(t+ 1)3

¸dt

=

·0:6t+

4

¡2(t+ 1)2¸¯¯2

1

=

·0:6(2)¡ 2

(2 + 1)2

¸

¡·0:6(1)¡ 2

(1 + 1)2

¸

=44

45¡ 1

10

¼ 0:8778 ft.

(b) Total growth in the third year is

Z 3

2

·0:6 +

4

(t+ 1)3

¸dt

=

·0:6t+

4

¡2(t+ 1)2¸¯¯3

2

=

·0:6(3)¡ 2

(3 + 1)2

¸

¡·0:6(2)¡ 2

(2 + 1)2

¸

=67

40¡ 4445

¼ 0:6972 ft.

60. Total growth after 3.5 days is

Z 3:5

0

R0(x)dx =Z 3:5

0

150e0:2xdx

= 150 ¢ e0:2x

0:2

¯¯3:5

0

= 750e 0:2x¯3:50

= 750e0:7 ¡ 750e0¼ 760:3:

61. R0(t) =5

t+ 1+

2pt+ 1

(a) Total reaction from t = 1 to t = 12 is

Z 12

1

μ5

t+ 1+

2pt+ 1

¶dt

=£5 ln(t+ 1) + 4

pt+ 1

¤¯121

= (5 ln 13 + 4p13)¡ (5 ln 2 + 4p2)

¼ 18:12:

(b) Total reaction from t = 12 to t = 24 is

Z 24

12

μ5

t+ 1+

2pt+ 1

¶dt

=£5 ln(t+ 1) + 4

pt+ 1

¤¯2412

= (5 ln 25 + 4p25)¡ (5 ln 13 + 4p13)

¼ 8:847:


62. F (T ) =Z T

0

f(x)dx

=

Z T

0

kbx dx

=

Z T

0

ke(ln b)x dx

= k

Z T

0

e(ln b)x dx

=k

ln b¢ e(ln b)x

¯T0

=k

ln b

£e(ln b)T ¡ 1¤

=k

ln b

£bT ¡ 1¤

63. (b)Z 60

0

n(x)dx

(c)Z 10

5

p5x+ 1dx

Let u = 5x+ 1: Then du = 5dx:When x = 5; u = 26; when x = 10; u = 51:

1

5

Z 51

26

u1=2 du

=1

5¢ u

3=2

32

¯5126

=2

15u3=2

¯5126

=2

15

¡513=2 ¡ 263=2¢

¼ 30.89 million

64. w0(t) = (3t+ 2)1=3

w(t) =

Z 3

0

(3t+ 2)1=3dt =1

3¢ (3t+ 2)

4=3

43

¯¯3

0

=(3t+ 2)4=3

4

¯¯3

0

=1

4(114=3 ¡ 24=3)

¼ 5:486 mg

65. v = k(R2 ¡ r2)

(a) Q(R) =Z R

0

2¼vr dr

=

Z R

0

2¼k(R2 ¡ r2)r dr

= 2¼k

Z R

0

(R2r ¡ r2)dr

= 2¼k

μR2r2

2¡ r

4

4

¶ ¯R0

= 2¼k

μR4

2¡ R

4

4

¶

= 2¼k

μR4

4

¶

=¼kR4

2

(b) Q(0:4) =¼k(0:4)4

2

= 0:04k mm/min

66.Z 9

3

(0:1762x2 ¡ 3:986x+ 22:68)dx

=

μ0:1762

3x3 ¡ 3:986

2x2 + 22:68x

¶¯¯9

3

= 85:5036¡ 51:6888= 33:8148

The total increase in the length of a ram’s hornduring the period is about 33.8 cm.

67. E(t) = 753t¡0:1321

(a) Since t is the age of the beagle in years,to convert the formula to days, let T = 365t;or t = T

365 :

E(T ) = 753

μT

365

¶¡0:1321¼ 1642T¡0:1321

Now, replace T with t:

E(t) = 1642t¡0:1321


(b) The beagle’s age in days after one year is 365days and after 3 years she is 1095 days old.Z 1095

365

1642t¡0:1321dt

= 16421

0:8679t0:8679

¯¯1095

365

¼ 1892(1; 0950:8679 ¡ 3650:8679)¼ 505,155

The beagle’s total energy requirements are about505,000 kJ/W0:67:

68.Z 100

0

0:85e0:0133xdx

Let u = 0:0133x; so that du = 0:0133dx; ordx = 1

0:0133du

When x = 100; u = 1:33:When x = 0; u = 0:

Z 1:33

0

0:85euμ

1

0:0133

¶du =

0:85

0:0133eu¯¯1:33

0

=0:85

0:0133(e1:33 ¡ e0)

¼ 177:736

The total mass of the column is about 178 g.

69. (a) f(x) = 40:1 + 2:03x¡ 0:741x2Z 9

0

(40:1 + 2:03x¡ 0:741x2)dx

=

μ40:1x+

2:03

2x2 ¡ 0:741

3x3¶ ¯9

0

= (40:1x+ 1:015x2 ¡ 0:247x3)¯90

¼ 263This integral shows that the total population aged0 to 90 was about 263 million.

(b)Z 5:5

3:5

(40:1 + 2:03x¡ 0:741x2)dx

= (40:1x+ 1:015x2 ¡ 0:247x3¯5:53:5

¼ 210:1591¡ 142:1936= 67:9655

The number of baby boomers is about 68 million.

70.Z 5

2:5

(0:058x3 ¡ 1:08x2 + 4:81x+ 6:26)dx

=

μ0:058

4x4 ¡ 1:08

3x3 +

4:81

2x2 + 6:26x

¶¯¯5

2:5

= (0:0145x4 ¡ 0:36x3 + 2:405x2 + 6:26x)¯52:5

¼ 55:4875¡ 25:6227= 29:8648

In 2000, approximately 30% of the population hadan income between $25,000 and $50,000.

71. c0(t) = kert

(a) c0(t) = 1:2e0:04t

(b) The amount of oil that the company will sell in

the next ten years is given by the integralZ 10

0

1:2e0:04tdt:

(c)Z 10

0

1:2e0:04tdx =1:2e0:04t

0:04

¯100

= 30e0:04t¯100

= 30e0:4 ¡ 30¼ 14:75

This represents about 14:75 billion barrels of oil.

(d)Z T

0

1:2e0:04t dt = 30e0:04t¯T0

= 30e0:04T ¡ 30Solve

20 = 30e0:04T ¡ 30:50 = 30e0:04T

5

3= e0:04T

ln5

3= 0:04T ln e

T =ln 5

3

0:04

¼ 12:8

The oil will last about 12.8 years.

(e)Z T

0

1:2e0:02t dt = 60e0:02t¯T0

= 60e0:02T ¡ 60


Solve

20 = 60e0:02T ¡ 60:80 = 60e0:02T

4

3= e0:02T

ln4

3= 0:02T ln e

T =ln 4

3

0:02¼ 14:4

The oil will last about 14.4 years.

72. c0(t) = 1:2e0:04t

c(T ) =

Z T

0

1:2e0:04t

=1:2

0:04e0:04t

¯T0

= 30(e0:04T ¡ e0)= 30

¡e0:04T ¡ 1¢

In 5 yr,

c(5) = 30(e0:04(5) ¡ 1)= 30(e0:2 ¡ 1)¼ 6:64 billion barrels.

7.5 The Area Between Two Curves

1. x = ¡2; x = 1; y = 2x2 + 5; y = 0

Z 1

¡2[(2x2 + 5)¡ 0] =

μ2x3

3+ 5x

¶¯¯1

¡2

=

μ2

3+ 5

¶¡μ¡163¡ 10

¶

= 21

2. x = 1; x = 2; y = 3x3 + 2; y = 0

Z 2

1

[(3x3 + 2)¡ 0]dx =μ3x4

4+ 2x

¶ ¯21

= (12 + 4)¡μ3

4+ 2

¶

=53

4

3. x = ¡3; x = 1; y = x3 + 1; y = 0

To …nd the points of intersection of the graphs,substitute for y:

x3 + 1 = 0

x3 = ¡1x = ¡1

The region is composed of two separate regionsbecause y = x3 + 1 intersects y = 0 at x = ¡1:Let f(x) = x3 + 1; g(x) = 0:In the interval [¡3;¡1]; g(x) ¸ f(x):In the interval [¡1; 1]; f(x) ¸ g(x):

Section 7.5 The Area Between Two Curves 495

Z ¡1

¡3[0¡ (x3 + 1)]dx+

Z 1

¡1[(x3 + 1)¡ 0]dx

=

μ¡x44¡ x

¶¯¯¡1

¡3+

μx4

4+ x

¶¯¯1

¡1

=

μ¡14+ 1

¶¡μ¡814+ 3

¶+

μ1

4+ 1

¶¡μ1

4¡ 1¶

= 20

4. x = ¡3; x = 0; y = 1¡ x2; y = 0

To …nd the points of intersection of the graphs in[¡3; 0]; substitute for y:

1¡ x2 = 0x2 = 1

x = ¡1 or x = 1

The region is composed of two separate regionsbecause y = 1¡ x2 intersects y = 0 at x = ¡1:Let f(x) = 1¡ x2; g(x) = 0:In the interval [¡3;¡1]; g(x) ¸ f(x):In the interval [¡1; 0]; f(x) ¸ g(x):Z ¡1

¡3[0¡ (1¡ x2)]dx+

Z 0

¡1[(1¡ x2)¡ 0]dx

=

μ¡x+ x

3

3

¶¯¯¡1

¡3+

μx¡ x

3

3

¶¯¯0

¡1

=

μ1¡ 1

3

¶¡ (3¡ 9) + 0¡

μ¡1 + 1

3

¶

=22

3

5. x = ¡2; x = 1; y = 2x; y = x2 ¡ 3

Find the points of intersection of the graphs ofy = 2x and y = x2 ¡ 3 by substituting for y:

2x = x2 ¡ 30 = x2 ¡ 2x¡ 30 = (x¡ 3)(x+ 1)

The only intersection in [¡2; 1] is at x = ¡1:In the interval [¡2;¡1]; (x2 ¡ 3) ¸ 2x:In the interval [¡1; 1]; 2x ¸ (x2 ¡ 3):

Z ¡1

¡2[(x2 ¡ 3)¡ (2x)]dx+

Z 1

¡1[(2x)¡ (x2 ¡ 3)]dx

=

Z ¡1

¡2(x2 ¡ 3¡ 2x)dx

+

Z 1

¡1(2x¡ x2 + 3)dx

=

μx3

3¡ 3x¡ x2

¶ ¯¡1¡2

+

μx2 ¡ x

3

3+ 3x

¶ ¯1¡1

= ¡13+ 3¡ 1¡

μ¡83+ 6¡ 4

¶+ 1¡ 1

3+ 3

¡μ1 +

1

3¡ 3¶

=5

3+ 6 =

23

3


6. x = 0; x = 6; y = 5x; y = 3x+ 10

To …nd the intersection of y = 5x and y = 3x+10;substitute for y:

5x = 3x+ 10

2x = 10

x = 5

If x = 5; y = 5(5) = 25:The region is composed of two separate regionsbecause y = 5x and y = 3x+10 intersect at x = 5(that is, (5; 25)):Let f(x) = 3x+ 10; g(x) = 5x:

In the interval [0; 5]; f(x) ¸ g(x):In the interval [5; 6]; g(x) ¸ f(x):Z 5

0

(3x+ 10¡ 5x)dx+Z 6

5

[5x¡ (3x+ 10)]dx

=

Z 5

0

(¡2x+ 10)dx+Z 6

5

(2x¡ 10x)¯65

=

μ¡2x22

+ 10x

¶ ¯50+

μ2x2

2¡ 10x

¶ ¯65

= (¡x2 + 10x)¯50+ (x2 ¡ 10x)

¯65

= ¡25 + 50 + (36¡ 60)¡ (25¡ 50)= 26

7. y = x2 ¡ 30y = 10¡ 3x

Find the points of intersection.

x2 ¡ 30 = 10¡ 3xx2 + 3x¡ 40 = 0(x+ 8)(x¡ 5) = 0x = ¡8 or x = 5

Let f(x) = 10¡ 3x and g(x) = x2 ¡ 30:

The area between the curves is given by

Z 5

¡8[f(x)¡ g(x)]dx

=

Z 5

¡8[(10¡ 3x)¡ (x2 ¡ 30)]dx

=

Z 5

¡8(¡x2 ¡ 3x+ 40)dx

=

μ¡x33¡ 3x

2

2+ 40x

¶ ¯5¡8

=¡533¡ 3(5)

2

2+ 40(5)

¡·¡(¡8)3

3¡ 3(¡8)

2

2+ 40(¡8)

¸

=¡1253

¡ 752+ 200¡ 512

3+192

2+ 320

¼ 366:1667:

8. y = x2 ¡ 18; y = x¡ 6

Find the intersection points.

x2 ¡ 18 = x¡ 6x2 ¡ x¡ 12 = 0

(x¡ 4)(x+ 3) = 0

The curves intersect at x = ¡3 and x = 4:


Z 4

¡3[(x¡ 6)¡ (x2 ¡ 18)]dx

=

Z 4

¡3(x¡ 6¡ x2 + 18)dx

=

Z 4

¡3(¡x2 + x+ 12)dx

=

μ¡x33+x2

2+ 12x

¶ ¯4¡3

=

μ¡643+ 8 + 48

¶¡μ9 +

9

2¡ 36

¶

=

μ¡643+ 56

¶¡μ¡27 + 9

2

¶

=¡643+ 83¡ 9

2

=343

6

¼ 57:167

9. y = x2; y = 2x


x2 = 2x

x2 ¡ 2x = 0x(x¡ 2) = 0x = 0 or x = 2

Let f(x) = 2x and g(x) = x2:The area between the curves is given byZ 2

0

[f(x)¡ g(x)]dx =Z 2

0

(2x¡ x2)dx

=

μ2x2

2¡ x

3

3

¶ ¯20

= 4¡ 83=4

3:

10. y = x2; y = x3

Find the intersection points.

x2 = x3

x2 ¡ x3 = 0x2(1¡ x) = 0

The curves intersect at x = 0 and x = 1:In the interval [0; 1]; x2 > x3:

Z 1

0

(x2 ¡ x3)dx =μx3

3¡ x

4

4

¶ ¯10

=1

3¡ 14

=1

12

11. x = 1; x = 6; y =1

x; y =

1

2


1

x=1

2

x = 2

The region is composed of two separate regionsbecause y = 1

x intersects y =12 at x = 2:

Let f(x) = 1x ; g(x) =

12 :

In the interval [1; 2]; f(x) ¸ g(x):In the interval [2; 6]; g(x) ¸ f(x):


Z 2

1

μ1

x¡ 12

¶dx+

Z 6

2

μ1

2¡ 1

x

¶dx

=³ln jxj ¡ x

2

´¯21+³x2¡ ln jxj

´¯62

= (ln 2¡ 1)¡μ0¡ 1

2

¶+ (3¡ ln 6)¡ (1¡ ln 2)

= 2 ln 2¡ ln 6 + 32

¼ 1:095

12. x = 0; x = 4; y =1

x+ 1; y =

x¡ 12

Find the intersection points.1

x+ 1=x¡ 12

x2 ¡ 1 = 2x2 ¡ 3 = 0

In the interval [0; 4]; the only intersection point isat x =

p3:Z p

3

0

μ1

x+ 1¡ x¡ 1

2

¶dx+

Z 4

p3

μx¡ 12

¡ 1

x+ 1

¶dx

=

μln jx+ 1j ¡ x

2

4+x

2

¶ ¯p30

+

μx2

4¡ x2¡ ln jx+ 1j

¶ ¯4p3

= ln (p3 + 1)¡ 3

4+

p3

2

+

((4¡ 2¡ ln 5)¡

"3

4¡p3

2¡ ln (p3 + 1)

#)

= ln (p3 + 1)¡ 3

4+

p3

2+ 2

¡ ln 5¡ 34+

p3

2+ ln (

p3 + 1)

= ln (p3 + 1) + ln (

p3 + 1)¡ ln 5 + 1

2+p3

= ln(p3 + 1)2

5+1

2+p3

¼ 2:633

13. x = ¡1; x = 1; y = ex; y = 3¡ ex

To …nd the point of intersection, set ex = 3 ¡ exand solve for x:

ex = 3¡ ex2ex = 3

ex =3

2

ln ex = ln3

2

x ln e = ln3

2

x = ln3

2

The area of the region between the curves fromx = ¡1 to x = 1 isZ ln 3=2

¡1[(3¡ ex)¡ ex]dx+

Z 1

ln 3=2

[ex¡ (3¡ ex)]dx

=

Z ln 3=2

¡1(3¡ 2ex)dx+

Z 1

ln 3=2

(2ex ¡ 3)dx

= (3x¡ 2ex)¯ln 3=2¡1

+ (2ex ¡ 3x)¯1ln 3=2

=

·μ3 ln

3

2¡ 2eln 3=2

¶¡ [3(¡1)¡ 2e¡1]

¸

+

·2e1 ¡ 3(1)¡

μ2eln 3=2 ¡ 3 ln 3

2

¶¸

=

·μ3 ln

3

2¡ 3¶¡μ¡3¡ 2

e

¶¸

+

·2e¡ 3¡

μ3¡ 3 ln 3

2

¶¸

= 6 ln3

2+2

e+ 2e¡ 6 ¼ 2:605:


14. x = ¡1; x = 2; y = e¡x; y = ex

The total area between the curves from x = ¡1to x = 2 isZ 0

¡1(e¡x ¡ ex)dx+

Z 2

0

(ex ¡ e¡x)dx

= (¡e¡x ¡ ex)¯0¡1+ (ex + e¡x)

¯20

= [(¡1¡ 1)¡ (¡e¡ e¡1)]+ [(e2 + e¡2)¡ (1 + 1)]= e2 + e¡2 + e+ e¡1 ¡ 4¼ 6:611:

15. x = ¡1; x = 2; y = 2e2x; y = e2x + 1


2e2x = e2x + 1

e2x = 1

2x = 0

x = 0

The region is composed of two separate regionsbecause y = 2e2x intersects y = e2x + 1 at x = 0:Let f(x) = 2e2x; g(x) = e2x + 1:In the interval [¡1; 0]; g(x) ¸ f(x):

In the interval [0; 2]; f(x) ¸ g(x):Z 0

¡1(e2x + 1¡ 2e2x)dx+

Z 2

0

[2e2x ¡ (e2x + 1)]dx

=

μ¡e

2x

2+ x

¶¯¯0

¡1+

μe2x

2¡ x

¶¯¯2

0

=

μ¡12+ 0

¶¡μ¡e

¡2

2¡ 1¶+

μe4

2¡ 2¶¡μ1

2¡ 0¶

=e¡2 + e4

2¡ 2

¼ 25:37

16. x = 2; x = 4; y =x¡ 14

; y =1

x¡ 1

To …nd the points of intersection of the graphs in[2; 4]; substitute for y:

x¡ 14

=1

x¡ 1(x¡ 1)(x¡ 1) = 4x2 ¡ 2x+ 1 = 4x2 ¡ 2x¡ 3 = 0

x = ¡1 or x = 3The region is composed of two separate regionsbecause y = x¡1

4 intersects y = 1x¡1 at x = 3:

Let f(x) = x¡14 ; g(x) =

1x¡1 .

In the interval [2; 3]; g(x) ¸ f(x):In the interval [3; 4]; f(x) ¸ g(x):Z 3

2

μ1

x¡ 1 ¡x¡ 14

¶dx+

Z 4

3

μx¡ 14

¡ 1

x¡ 1¶dx

=

·ln jx¡1j¡ x(x¡2)

8

¸¯¯3

2

+

·x(x¡2)8

¡ln jx¡1j¸¯¯4

3

=

μln 2¡ 3

8

¶¡ 0 + (1¡ ln 3)¡

μ3

8¡ ln 2

¶

= 2 ln2¡ ln 3 + 14

¼ 0:5377


17. y = x3 ¡ x2 + x+ 1; y = 2x2 ¡ x+ 1Find the points of intersection.

x3 ¡ x2 + x+ 1 = 2x2 ¡ x+ 1x3 ¡ 3x2 + 2x = 0x(x2 ¡ 3x+ 2) = 0x(x¡ 2)(x¡ 1) = 0

The points of intersection are at x = 0; x = 1;

and x = 2:

Area between the curves isZ 1

0

[(x3 ¡ x2 + x+ 1)¡ (2x2 ¡ x+ 1)]dx

+

Z 2

1

[(2x2 ¡ x+ 1)¡ (x3 ¡ x2 + x+ 1)]dx

=

Z 1

0

(x3¡3x2+2x)dx+Z 1

0

(¡x3+3x2¡2x)dx

=

μx4

4¡ x3 + x2

¶¯¯1

0

+

μ¡x44+ x3 ¡ x2

¶¯¯2

1

=

·μ1

4¡ 1 + 1

¶¡ (0)

¸

+

·(¡4 + 8¡ 4)¡

μ¡14+ 1¡ 1

¶¸

=1

4+1

4

=1

2:

18. y = 2x3 + x2 + x+ 5;y = x3 + x2 + 2x+ 5

To …nd the points of intersection, substitute for y:

2x3 + x2 + x+ 5 = x3 + x2 + 2x+ 5

x3 ¡ x = 0x(x2 ¡ 1) = 0

The points of intersection are at x = 0; x = ¡1;and x = 1:The area of the region between the curves is

Z 0

¡1[(2x3 + x2 + x+ 5)¡ (x3 + x2 + 2x+ 5)]dx

+

Z 1

0

[(x3+x2+2x+5)¡ (2x3+x2+x+5)]dx

=

Z 0

¡1(x3 ¡ x)dx+

Z 1

0

(¡x3 + x)dx

=

μx4

4¡ x

2

2

¶ ¯0¡1+

μ¡x

4

4+x2

2

¶ ¯10

=

·0¡

μ1

4¡ 12

¶¸+

·μ¡14+1

2

¶¡ 0¸

=1

4+1

4=1

2:

19. y = x4 + ln (x+ 10);y = x3 + ln (x+ 10)


x4 + ln (x+ 10) = x3 + ln (x+ 10)

x4 ¡ x3 = 0x3(x¡ 1) = 0x = 0 or x = 1

The points of intersection are at x = 0 and x = 1:The area between the curves isZ 1

0

[(x3 + ln(x+ 10))¡ (x4 + ln(x+ 10))]dx

=

Z 1

0

(x3 ¡ x4)dx

=

μx4

4¡ x

5

5

¶¯¯1

0

=

μ1

4¡ 15

¶¡ (0) = 1

20:


20. y = x5 ¡ 2 ln (x+ 5);y = x3 ¡ 2 ln (x+ 5)To …nd the points of intersection, substitute for y:

x5 ¡ 2 ln (x+ 5) = x3 ¡ 2 ln (x+ 5)x5 ¡ x3 = 0

x3(x2 ¡ 1) = 0The points of intersection are at x = 0 and x = 1and x = ¡1:In the interval [¡1; 0];

x5 ¡ 2 ln (x+ 5) > x3 ¡ 2 ln (x+ 5):

In the interval [0; 1];

x5 ¡ 2 ln (x+ 5) < x3 ¡ 2 ln (x+ 5):

The area between the curves isZ 0

¡1[(x5 ¡ 2 ln (x+ 5))¡ (x3 ¡ 2 ln (x+ 5))]dx

+

Z 1

0

[(x3 ¡ 2 ln (x+ 5))¡ (x5 ¡ 2 ln (x+ 5))]dx

=

Z 0

¡1(x5 ¡ x3)dx+

Z 1

0

(x3 ¡ x5)dx

=

μx6

6¡ x

4

4

¶ ¯0¡1+

μx4

4¡ x

6

6

¶ ¯10

=

·0¡

μ1

6¡ 14

¶¸+

·μ1

4¡ 16

¶¡ 0¸

=1

12+1

12=1

6:

21. y = x4=3; y = 2x1=3


x4=3 = 2x1=3

x4=3 ¡ 2x1=3 = 0x1=3(x¡ 2) = 0x = 0 or x = 2

The points of intersection are at x = 0 and x = 2:


0

(2x1=3 ¡ x4=3)dx = 2x4=3

43

¡ x7=3

73

¯¯2

0

=3

2x4=3 ¡ 3

7x7=3

¯¯2

0

=

·3

2(2)4=3 ¡ 3

7(2)7=3

¸¡ 0

=3(24=3)

2¡ 3(2

7=3)

7

¼ 1:62:22. y =

px; y = x

px

To …nd the points of intersection, substitute for y:px = x

px

xpx¡px = 0px(x¡ 1) = 0

The points of intersection are at x = 0 and x = 1:In [0; 1];

px > x

px:


0

(px¡ xpx)dx =

Z 1

0

(x1=2 ¡ x3=2)dx

=

μx3=2

3=2¡ x

5=2

5=2

¶ ¯10

=

μ2

3x3=2 ¡ 2

5x5=2

¶ ¯10

=

·2

3(1)¡ 2

5(1)

¸¡ 0

=4

15:

23. x = 0; x = 3; y = 2e3x; y = e3x + e6



2e3x = e3x + e6

e3x = e6

3x = 6

x = 2

The region is composed of two separate regionsbecause y = 2e3x intersects y = e3x+ e6 at x = 2:Let f(x) = 2e3x; g(x) = e3x + e6:In the interval [0; 2]; g(x) ¸ f(x):In the interval [2; 3]; f(x) ¸ g(x):Z 2

0

(e3x + e6 ¡ 2e3x)dx+Z 3

2

[2e3x ¡ (e3x + e6)]dx

=

μ¡e

3x

3+ e6x

¶¯¯2

0

+

μe3x

3¡ e6x

¶¯¯3

2

=

μ¡e

6

3+2e6

¶¡μ¡13+0

¶+

μe9

3¡3e6

¶¡μe6

3¡2e6

¶

=e9 + e6 + 1

3

¼ 2836

24. x = 0; x = 3; y = ex; y = e4¡x


ex = e4¡x

x = 4¡ x2x = 4

x = 2

The region is composed of two separate regionsbecause y = ex intersects y = e4¡x at x = 2:Let f(x) = ex; g(x) = e4¡x:In the interval [0; 2]; g(x) ¸ f(x):In the interval [2; 3]; f(x) ¸ g(x):

Z 2

0

(e4¡x ¡ ex)dx+Z 3

2

(ex ¡ e4¡x)dx

= (¡e4¡x ¡ ex)¯20+ (ex + e4¡x)

¯32

= (¡e2 ¡ e2)¡ (¡e4 ¡ 1) + (e3 + e)¡ (e2 + e2)= e4 + e3 ¡ 4e2 + e+ 1¼ 48:85

25. Graph y1 = ex and y2 = ¡x2¡ 2x on your graph-ing calculator. Use the intersect command to …ndthe two intersection points. The resulting screensare:

These screens show that ex = ¡x2¡ 2x when x ¼¡1:9241 and x ¼ ¡0:4164:In the interval [¡1:9241;¡0:4164],

ex < ¡x2 ¡ 2x:


¡0:4164Z¡1:9241

[(¡x2 ¡ 2x)¡ ex]dx:

Use the fnInt command to approximate this de…-nite integral.The resulting screen is:

The last screen shows that the area is approxi-mately 0.6650.


26. Graph y1 = lnx and y2 = x3 ¡ 5x2 + 6x¡ 1 onyour graphing calculator. Use the intersectcommand to …nd the two intersection points. Theresulting screens are:

These screens show that lnx = x3 ¡ 5x2 + 6x¡ 1when x ¼ 1:4027 and x ¼ 3:4482.In the interval [1:4027; 3:4482],

lnx > x3 ¡ 5x2 + 6x¡ 1:The area between the curves is given by

3:4482Z1:4027

[lnx¡ (x3 ¡ 5x2 + 6x¡ 1)]dx:

Use the fnInt command to approximate this de…-nite integral. The resulting screen is:

The last screen shows that the area is approxi-mately 3.3829.

27. (a) It is pro…table to use the machine untilS0(x) = C0(x):

150¡ x2 = x2 + 114x

2x2 +11

4x¡ 150 = 0

8x2 + 11x¡ 600 = 0

x =¡11§p121¡ 4(8)(¡600)

16

=¡11§ 139

16

x = 8 or x = ¡9:375

It will be pro…table to use this machine for 8 years.Reject the negative solution.

(b) Since 150¡x2 > x2+ 114 x; in the interval [0; 8];

the net total savings in the …rst year areZ 1

0

·(150¡ x2)¡

μx2 +

11

4x

¶¸dx

=

Z 1

0

μ¡2x2 ¡ 11

4x+ 150

¶dx

=

μ¡2x33

¡ 11x2

8+ 150x

¶¯¯1

0

= ¡23¡ 118+ 150

¼ $148:(c) The net total savings over the entire period ofuse areZ 8

0

·(150¡ x2)¡

μx2 +

11

4x

¶¸dx

=

μ¡2x33

¡ 11x2

8+ 150x

¶¯¯8

0

=¡2(83)3

¡ 11(82)

8+ 150(8)

=¡10243

¡ 7048+ 1200

¼ $771:

28. (a) S0(x) = ¡x2 + 4x+ 8; C 0(x) = 3

25x2

S0(x) = C0(x)

¡x2 + 4x+ 8 = 3

25x2

¡25x2 + 100x+ 200 = 3x2

0 = 28x2 ¡ 100x¡ 2000 = 7x2 ¡ 25x¡ 500 = (7x+ 10)(x¡ 5)

x = ¡107

or x = 5


Since time would not be negative, 5 is the onlysolution. It will pay to use the device for 5 yr.

(b) The total savings over 5 yr is given byZ 5

0

(¡x2 + 4x+ 8)dx

=

μ¡x33+ 2x2 + 8x

¶ ¯50

=¡1253

+ 90

= 48:33:

The total cost over 5 yr is given by

Z 5

0

3

25x2 dx =

x3

25

¯50= 5:

Net savings = $48:33 million¡ $5 million= $43:33 million

29. (a) E0(x) = e0:1x and I 0(x) = 98:8¡ e0:1x

To …nd the point of intersection, where pro…t willbe maximized, set the functions equal to each otherand solve for x:

e0:1x = 98:8¡ e0:1x2e0:1x = 98:8

e0:1x = 49:4

0:1x = ln 49:4

x =ln 49:4

0:1

x ¼ 39

The optimum number of days for the job to lastis 39.

(b) The total income for 39 days isZ 39

0

(98:8¡ e0:1x)dx

=

μ98:8x¡ e

0:1x

0:1

¶¯¯39

0

=³98:8x¡ 10e0:1x

¯390

= [98:8(39)¡ 10e3:9]¡ (0¡ 10)= $3369:18:

(c) The total expenditure for 39 days is

Z 39

0

e0:1x dx =e0:1x

0:1

¯390

= 10e0:1x¯390

= 10e3:9 ¡ 10= $484:02:

(d) Pro…t = Income¡ Expense= 3369:18¡ 484:02= $2885:16

30. (a) R0(t) = 104¡ 0:4et=2;C 0(t) = 0:3et=2

It will no longer be pro…table when C0(t) > R0(t):Find t when C 0(t) > R0(t):

0:3et=2 > 104¡ 0:4et=20:7et=2 > 104

et=2 >104

0:7

ln et=2 > ln

μ104

0:7

¶

t > 2 ln

μ104

0:7

¶

t > 10

It will no longer be pro…table to use the processafter 10 yr.

(b) The total net savings is

Z 10

0

[(104¡ 0:4et=2)¡ 0:3et=2]dt

=

Z 10

0

(104¡ 0:7et=2)dt

=

μ104t¡ 0:7e

t=2

1=2

¶ ¯100

= (104t¡ 1:4et=2)¯100

= [(104t¡ 1:4et=2)¡ (0¡ 1:4)]= 1041:4¡ 1:4e5¼ 834:

The net total savings will be $834,000.


31. S(q) = q5=2+2q3=2+ 50; q = 16 is the equilibriumquantity.

Producers’ surplus =Z q0

0

[p0¡S(q)]dq; where p0 isthe equilibrium price and q0 is equilibrium supply.

p0 = S(16) = (16)5=2 + 2(16)3=2 + 50

= 1202

Therefore, the producers’ surplus isZ 16

0

[1202¡ (q5=2 + 2q3=2 + 50)]dq

=

Z 16

0

(1152¡ q5=2 ¡ 2q3=2)dq

=

μ1152q ¡ 2

7q7=2 ¡ 4

5q5=2

¶¯¯16

0

= 1152(16)¡ 27(16)7=2 ¡ 4

5(16)5=2

= 18; 432¡ 32; 7687

¡ 40965

= 12; 931:66:

The producers’ surplus is 12,931.66.

32. S(q) = 100+ 3q3=2+ q5=2; equilibrium quantity isq = 9:

Producers’ surplus =Z q0

0

[p0 ¡ S(q)]dq

p0 = S(9) = 424Z 9

0

[424¡ (100 + 3q3=2 + q5=2)]dq

=

Z 9

0

(324¡ 3q3=2 ¡ q5=2)]dq

=

μ324q ¡ 6

5q5=2 ¡ 2

7q7=2

¶ ¯90

=

·μ324(9)¡ 6

5(9)5=2 ¡ 2

7(9)7=2

¶¡ 0¸

= 2916¡ 14585

¡ 43747

= 1999:54

The producers’ surplus is 1999.54.

33. D(q) =200

(3q + 1)2; q = 3 is the equilibrium quantity.

Consumers’ surplus =Z q0

0

jD(q)¡ p0j dq

p0 = D(3) = 2

Therefore, the consumers’ surplus is

Z 3

0

·200

(3q + 1)2¡ 2¸dq =

Z 3

0

200

(3q + 1)2dq¡

Z 3

0

2dq:

Let u = 3q + 1, so that

du = 3dq and1

3du = dq:

Z 3

0

200

(3q + 1)2dq¡

Z 3

0

2dq =1

3

Z 10

1

200

u2du¡

Z 3

0

2dq

=200

3

Z 10

1

u¡2du¡Z 3

0

2dq

=200

3¢ u

¡1

¡1¯¯10

1

¡ 2q¯¯3

0

= ¡2003u

¯¯10

1

¡ 6

= ¡20030

+200

3¡ 6

= 54

34. D(q) =32,000(2q + 8)3

; q = 6 is the equilibrium quantity.

Consumers’ surplus =Z q0

0

jD(q)¡ p0j dq

p0 = D(6) =32,000203

= 4

Therefore, the consumers’ surplus is

Z 6

0

·32,000(2q + 8)3

¡ 4¸dq =

Z 6

0

32,000(2q + 8)3

dq¡Z 6

0

4dq:

Let u = 2q + 8, so that

du = 2dq and1

2du = dq:


Z 6

0

32,000(2q + 8)3

dq ¡Z 6

0

4dq =1

2

Z 20

8

32,000u3

du¡Z 6

0

4dq

= 16,000Z 20

8

u¡3du¡Z 6

0

4dq

= 16,000 ¢ u¡2

¡2¯¯20

8

¡ 4q¯¯6

0

= ¡ 8000u2

¯¯20

8

¡ 24

= ¡8000400

+800064

¡ 24

= 81

35. S(q) = q2 + 10qD(q) = 900¡ 20q ¡ q2

(a) The graphs of the supply and demand func-tions are parabolas with vertices at (¡5;¡25) and(¡10; 1900); respectively.

(b) The graphs intersect at the point where they-coordinates are equal.

q2 + 10q = 900¡ 20q ¡ q22q2 + 30q ¡ 900 = 0q2 + 15q ¡ 450 = 0(q + 30)(q ¡ 15) = 0

q = ¡30 or q = 15

Disregard the negative solution.The supply and demand functions are inequilibrium when q = 15:

S(15) = 152 + 10(15) = 375

The point is (15; 375):

(c) Find the consumers’ surplus.Z q0

0

[D(q)¡ p0)]dq

p0 = D(15) = 375

Z 15

0

[(900¡ 20q ¡ q2)¡ 375]dq

=

Z 15

0

(525¡ 20q ¡ q2)dq

=

μ525q ¡ 10q2 ¡ 1

3q3¶¯¯15

0

=

·525(15)¡ 10(15)2 ¡ 1

3(15)3

¸¡ 0

= 4500

The consumer’s surplus is $4500.(d) Find the producers’ surplus.Z q0

0

[p0 ¡ S(q)]dq

p0 = S(15) = 375Z 15

0

[375¡ (q2 + 10q)]dq

=

Z 15

0

(375¡ q2 ¡ 10q)dq

=

μ375q ¡ 1

3q3 ¡ 5q2

¶ ¯150

=

·375(15)¡ 1

3(15)3 ¡ 5(15)2

¸¡ 0

= 3375

The producer’s surplus is $3375.

36. S(q) = (q + 1)2

D(q) =1000

q + 1

(a) The graph of the supply function is a parabolawith vertex at (¡1; 0). The graph of the demandfunction is the graph of a rational function withvertical asymptote of x = ¡1 and horizontal as-ymptote of y = 0:

(b) Find the equilibrium point by setting the twofunctions equal.

(q + 1)2 =1000

q + 1

(q + 1)3 = 1000

q3 + 3q2 + 3q + 1 = 1000

q3 + 3q2 + 3q ¡ 999 = 0(q ¡ 9)(q2 + 12q + 111) = 0


Since q2 + 12q + 111 has no real roots, q = 9 isthe only root. At the equilibrium point where thesupply and demand are both 9 items, the price is

S(9) = (9 + 1)2 = 100:

The equilibrium point is (9; 100):

(c) The consumers’ surplus is given byZ 9

0

μ1000

q + 1¡ 100

¶dq = (1000 ln jq + 1j ¡ 100q )

9

0

= 1000 ln(9 + 1)¡ 100(9)¡ 0¼ 1402:59

Here the consumers’ surplus is 1402.59.

(d) The producers’ surplus is given byZ 9

0

[100¡ (q + 1)2]dq =Z 9

0

(99¡ q2 ¡ 2q)dq

= (99q ¡ 13q3 ¡ q2)

¯¯9

0

= 99(9)¡ 13(9)3 ¡ (9)2 ¡ 0

= 567

Here the producers’ surplus is 567.

37. (a) S(q) = q2 + 10q; S(q) = 264 is the price thegovernment set.

264 = q2 + 10q

0 = q2 + 10q ¡ 2640 = (q ¡ 12)(q + 22)q = 12 or q = ¡22

Only 12 is a meaningful solution here. Thus, 12units of oil will be produced.

(b) The consumers’ surplus is given byZ 12

0

(900¡ 20q ¡ q2 ¡ 264)dq

=

Z 12

0

(636¡ 20q ¡ q2)dq

=

μ636q ¡ 10q2 ¡ 1

3q3¶¯¯12

0

= 636(12)¡ 10(12)2 ¡ 13(12)3 ¡ 0

= 5616

Here the consumer’ surplus is $5616. In this case,the consumers’ surplus is 5616 ¡ 4500 = $1116

larger.

(c) The producers’ surplus is given byZ 12

0

[264¡ (q2 + 10q)]dq

=

Z 12

0

(264¡ q2 ¡ 10q)dq

=

μ264q ¡ 1

3q3 ¡ 5q2

¶¯¯12

0

= 264(12)¡ 13(12)3 ¡ 5(12)2 ¡ 0

= 1872

Here the producers’ surplus is $1872. In this case,the producers’ surplus is 3375 ¡ 1872 = $1503

smaller.

(d) For the equilibrium price, the total consumers’and producers’ surplus is

4500 + 3375 = $7875

For the government price, the total consumers’and producers’ surplus is

5616 + 1872 = $7488:

The di¤erence is

7875¡ 7488 = $387:

39. (a) The pollution level in the lake is changing atthe rate f(t) ¡ g(t) at any time t: We …nd theamount of pollution by integrating.Z 12

0

[f(t)¡ g(t)]dt

=

Z 12

0

[10(1¡ e¡0:5t)¡ 0:4t]dt

=

μ10t¡ 10 ¢ 1

¡0:5e¡0:5t ¡ 0:4 ¢ 1

2t2¶¯¯12

0

= (20e¡0:5t + 10t¡ 0:2t2)¯120

= [20e¡0:5(12) + 10(12)¡ 0:2(12)2]¡ [20e¡0:5(0) + 10(0)¡ 0:2(0)2]= (20e¡6 + 91:2)¡ (20)= 20e¡6 + 71:2 ¼ 71:25

After 12 hours, there are about 71.25 gallons.


(b) The graphs of the functions intersect at about25.00. So the rate that pollution enters the lakeequals the rate the pollution is removed at about25 hours.

(c)Z 25

0

[f(t)¡ g(t)]dt

= (20e¡0:5t + 10t¡ 0:2t2)¯250

= [20e¡0:5(25) + 10(25)¡ 0:2(25)2]¡ 20= 20e¡12:5 + 105¼ 105

After 25 hours, there are about 105 gallons.

(d) For t > 25; g(t) > f(t); and pollution is beingremoved at the rate g(t) ¡ f(t): So, we want tosolve for c, where

cZ0

[f(t)¡ g(t)]dt = 0:

(Altternatively, we could solve for c in

cZ25

[g(t)¡ f(t)dt = 105:

One way to do this with a graphing calculator isto graph the function

y =

xZ0

[f(t)¡ g(t)]dt

and determine the values of x for which y = 0:The …rst window shows how the function can bede…ned.

A suitable window for the graph is [0; 50] by [0; 110]:

Use the calculator’s features to approximate wherethe graph intersects the x-axis. These are at 0and about 47.91. Therefore, the pollution will beremoved from the lake after about 47.91 hours.

40. (a) The pollution level in the lake is changing atthe rate f(t) ¡ g(t) at any time t: We …nd theamount of pollution by integrating.Z 12

0

[f(t)¡ g(t)]dt

=

Z 12

0

[15(1¡ e¡0:05t)¡ 0:3t]dt

=

μ15t¡ 15 1

¡0:05e¡0:05t ¡ 0:31

2t2¶¯12

0

= (300e¡0:05t + 15t¡ 0:15t2)¯120

= [300e¡0:05(12) + 15(12)¡ 0:15(12)2]¡ [300e¡0:05(0) + 15(0)¡ 0:15(0)2]= (300e¡0:6 + 158:4)¡ (300)= 300e¡0:6 ¡ 141:6¼ 23:04

After 12 hours, there are about 23.04 gallons.

(b) The graphs of the functions intersect at about44.63. So the rate that pollution enters the lakeequals the rate the pollution is removed at about44.63 hours.

(c)Z 44:63

0

[f(t)¡ g(t)]dt

= (300e¡0:05t + 15t¡ 0:15t2)¯44:630

= [300e¡0:05(44:63) + 15(44:63)¡ 0:15(44:63)2]¡ 300= (300e¡2:2315 + 370:674465)¡ 300= 300e¡2:2315 + 70:674465¼ 102:88

After 44.63 hours, there are about 102.88 gallons.

(d) For t > 44:63; g(t) > f(t); and pollution isbeing removed at the rate g(t)¡f(t): So, we wantto solve for c; where

cZ0

[f(t)¡ g(t)]dt = 0

(Alternatively, we could solve for c in

cZ44:63

[g(t)¡ f(t)]dt = 102:88:)


One way to do this with a graphing calculator isto graph the function

y =

xZ0

[f(t)¡ g(t)]dt

and determine the values of x for which y = 0:The …rst window shows how the function can bede…ned.

A suitable window for the graph is [0:75] by [0; 110]:

Use the calculator’s features to approximate wherethe graph intersects the x-axis. These are at 0and abaout 73.47. Therefore, the pollution willbe removed from the lake after about 73.47 hours.

41. I(x) = 0:9x2 + 0:1x

(a) I(0:1) = 0:9(0:1)2 + 0:1(0:1)= 0:019

The lower 10% of income producers earn 1.9% oftotal income of the population.

(b) I(0:4) = 0:9(0:4)2 + 0:1(0:4) = 0:184

The lower 40% of income producers earn 18.4% oftotal income of the population.

(c)The graph of I(x) = x is a straight line throughthe points (0; 0) and (1; 1). The graph ofI(x) = 0:9x2 + 0:1x is a parabola with vertex¡¡ 1

18 ;¡ 1360

¢. Restrict the domain to 0 · x · 1:

(d) To …nd the points of intersection, solve

x = 0:9x2 + 0:1x:

0:9x2 ¡ 0:9x = 00:9x(x¡ 1) = 0x = 0 or x = 1


Z 1

0

[x¡ (0:9x2 + 0:1x)]dx

=

Z 1

0

(0:9x¡ 0:9x2)dx

=

μ0:9x2

2¡ 0:9x

3

3

¶¯¯1

0

=0:9

2¡ 0:93= 0:15:

42. y =px; y =

x

2

To …nd the points of intersection, substitute for y:

px =

x

2x

2¡px = 0

x¡ 2px = 0px(px¡ 2) = 0

x = 0 or x = 4

Area =Z 4

0

³px¡ x

2

´dx =

Z 4

0

³x1=2 ¡ x

2

´dx

=

μ2

3x3=2 ¡ 1

4x2¶ ¯4

0=16

3¡ 4 = 4

3


7.6 Numerical Integration

1.Z 2

0

(3x2 + 2)dx

n = 4; b = 2; a = 0; f(x) = 3x2 + 2

i xi f(xi)

0 0 2

11

22:75

2 1 5

33

28:75

4 2 14

(a) Trapezoidal rule:

Z 2

0

(3x2 + 2)dx

¼ 2¡ 04

·1

2(2) + 2:75 + 5 + 8:75 +

1

2(14)

¸

= 0:5(24:5)

= 12:25

(b) Simpson’s rule:

Z 2

0

(3x2 + 2)dx

¼ 2¡ 03(4)

[2 + 4(2:75) + 2(5) + 4(8:75) + 14]

=2

12(72)

= 12

(c) Exact value:

Z 2

0

(3x2 + 2)dx = (x3 + 2x)¯20

= (8 + 4)¡ 0= 12

2.Z 2

0

(2x2 + 1)dx

n = 4; b = 2; a = 0; f(x) = 2x2 + 1

i xi f(xi)

0 0 1

11

21:5

2 1 3

33

25:5

4 2 9

(a) Trapezoidal rule:Z 2

0

(2x2 + 1)dx

¼ 2¡ 04

·1

2(1) + 1:5 + 3 + 5:5 +

1

2(9)

¸

= 0:5(15)

= 7:5

(b) Simpson’s rule:Z 2

0

(2x2 + 1)dx

¼ 2¡ 03(4)

[1 + 4(1:5) + 2(3) + 4(5:5) + 9]

=2

12(44)

¼ 7:333(c) Exact value:Z 2

0

(2x2 + 1)dx =

μ2x3

3+ x

¶¯¯2

0

=

μ16

3+ 2

¶¡ 0

=22

3¼ 7:333

3.Z 3

¡1

3

5¡ x dx

n = 4; b = 3; a = ¡1, f(x) = 3

5¡ xi xi f(xi)

0 ¡1 0:5

1 0 0:6

2 1 0:75

3 2 1

4 3 1:5

Section 7.6 Numerical Integration 511


¡1

3

5¡ x dx

¼ 3¡ (¡1)4

·1

2(0:5) + 0:6 + 0:75 + 1 +

1

2(1:5)

¸

= 1(3:35)

= 3:35


¡1

3

5¡ x dx

¼ 3¡ (¡1)3(4)

[0:5+4(0:6)+2(0:75)+4(1)+1:5]

=1

3

μ99

10

¶

=33

10¼ 3:3

(c) Exact value:

Z 3

¡1

3

5¡ x dx = ¡3 ln j5¡ xj¯¯3

¡1

= ¡3(ln j2j ¡ ln j6j)= 3 ln 3 ¼ 3:296

4.Z 5

1

6

2x+ 1dx

n = 4; b = 5; a = 1; f(x) = 62x+1

i xi f(xi)

0 1 2

1 26

5

2 36

7

3 42

3

4 56

11


1

6

2x+ 1dx

¼ 5¡ 14

·1

2(2) +

6

5+6

7+2

3+1

2

μ6

11

¶¸

= 1

μ1 +

6

5+6

7+2

3+3

11

¶

¼ 3:997


1

6

2x+ 1dx

¼ 5¡ 13(4)

·2+4

μ6

5

¶+2

μ6

7

¶+4

μ2

3

¶+6

11

¸

=1

3

μ2 +

24

5+12

7+8

3+6

11

¶¼ 3:909

(c) Exact value:Z 5

1

6

2x+ 1dx = 3 ln j2x+ 1j

¯51

= 3(ln j11j ¡ ln j3j)

= 3 ln11

3¼ 3:898

5.Z 2

¡1(2x3 + 1)dx

n = 4; b = 2; a = ¡1; f(x) = 2x3 + 1

i xi f(x)

0 ¡1 ¡1

1 ¡ 1

4

31

32

21

2

5

4

35

4

157

32

4 2 17


¡1(2x3 + 1)dx

¼ 2¡ (¡1)4

·1

2(¡1) + 31

32+5

4+157

32+1

2(17)

¸

= 0:75(15:125)

¼ 11:34(b) Simpson’s rule:Z 2

¡1(2x3 + 1)dx

¼ 2¡ (¡1)3(4)

·¡1+4

μ31

32

¶+2

μ5

4

¶+4

μ157

32

¶+17

¸

=1

4(42)

= 10:5


(c) Exact value:Z 2

¡1(2x3 + 1)dx

=

μx4

2+ x

¶¯¯2

¡1

= (8 + 2)¡μ1

2¡ 1¶

=21

2

= 10:5

6.Z 3

0

(2x3 + 1)dx

n = 4; b = 3; a = 0; f(x) = 2x3 + 1

i x f(x)

0 0 1

13

4

59

32

23

2

31

4

39

4

761

32

4 3 55


0

(2x3 + 1)dx

¼ 3¡ 04

·1

2(1)+

59

32+31

4+761

32+1

2(55)

¸

=3

4

μ491

8

¶


0

(2x3 + 1)dx

=3¡ 03(4)

·1+4

μ59

32

¶+2

μ31

4

¶+4

μ761

32

¶+55

¸

=1

4(174)

= 43:5

(c) Exact value:

Z 3

0

(2x3 + 1)dx =

μx4

2+ x

¶¯¯3

0

=

μ81

2+ 3

¶¡ 0

=87

2

= 43:5

7.Z 5

1

1

x2dx

n = 4; b = 5; a = 1; f(x) =1

x2

i xi f(xi)

0 1 1

1 2 0:25

2 3 0:1111

3 4 0:0625

4 5 0:04


1

1

x2dx

¼ 5¡ 14

·1

2(1)+0:25+0:1111+0:0625+

1

2(0:04)

¸

¼ 0:9436


1

1

x2dx

¼ 5¡ 112

[1+4(0:25)+2(0:1111)+4(0:0625)+0:04)]

¼ 0:8374

(c) Exact value:

Z 5

1

x¡2 dx = ¡x¡1¯51

= ¡15+ 1

=4

5= 0:8


8.Z 4

2

1

x3dx

n = 4; b = 4; a = 2; f(x) =1

x3

i xi f(xi)

0 2 0:125

1 2:5 0:064

2 3 0:03703

3 3:5 0:02332

4 4 0:015625


2

dx

x3¼ 4¡ 2

4

·1

2(0:125) + 0:064 + 0:03703

+ 0:02332 +1

2(0:015625)

¸

¼ 1

2(0:19466) ¼ 0:0973


2

dx

x3¼ 4¡ 23(4)

[0:125 + 4(0:064) + 2(0:03703)

+ 4(0:02332) + 0:015625]

¼ 1

6(0:56397)


2

dx

x3=

Z 4

2

x¡3 dx =x¡2

¡2¯42=¡12x2

¯42

=¡132+1

8=3

32= 0:09375

9.Z 1

0

4xe¡x2

dx

n = 4; b = 1; a = 0; f(x) = 4xe¡x2

i xi f(xi)

0 0 0

11

4e¡1=16

21

22e¡1=4

33

43e¡9=16

4 1 4e¡1


0

4xe¡x2

dx

¼ 1¡ 04

·1

2(0) + e¡1=16 + 2e¡1=4

+3e¡9=16 +1

2(4e¡1)

¸

=1

4(e¡1=16 + 2e¡1=4 + 3e¡9=16 + 2e¡1)


0

4xe¡x2

dx

¼ 1¡ 03(4)

[0 + 4(e¡1=16) + 2(2e¡1=4)

+ 4(3e¡9=16) + 4e¡1]

=1

12(4e¡1=16 + 4e¡1=4 + 12e¡9=16 + 4e¡1)

¼ 1:265(c) Exact value:

Z 1

0

4xe¡x2

dx = ¡2e¡x2¯10

= (¡2e¡1)¡ (¡2)= 2¡ 2e¡1 ¼ 1:264

10.Z 4

0

xp2x2 + 1dx

n = 4; b = 4; a = 0; f(x) = xp2x2 + 1

i xi f(xi)

0 0 0

1 1p3

2 2 6

3 3 3p19

4 4 4p33


0

xp2x2 + 1dx

¼ 4¡ 04

·1

2(0)+

p3+6+3

p19 +

1

2(4p33)

= 1(p3 + 6 + 3

p19 + 2

p33)

¼ 32:30



0

xp2x2 + 1dx

¼ 4¡ 03(4)

[0 + 4(p3) + 2(6) + 4(3

p19) + 4

p33]

=1

3(4p3 + 12 + 12

p19 + 4

p33)


0

xp2x2 + 1dx =

(2x2 + 1)3=2

6

¯¯4

0

=333=2 ¡ 1

6

¼ 31:4311. y =

p4¡ x2

n = 8; b = 2; a = ¡2; f(x) = p4¡ x2

i xi y

0 ¡2:0 0

1 ¡1:5 1:32289

2 ¡1:0 1:73205

3 ¡0:5 1:93649

4 0 2

5 0:5 1:93649

6 1:0 1:73205

7 1:5 1:32289

8 2:0 0


¡2

p4¡ x2 dx

¼ 2¡ (¡2)8

¢·1

2(0) + 1:32289 + 1:73205 + ¢ ¢ ¢+ 1

2(0)

¸¼ 5:991


¡2

p4¡ x2 dx

¼ 2¡ (¡2)3(8)

¢[0+4(1:32289)+2(1:73205)+4(1:93649)+2(2)+ 4(1:93649) + 2(1:73205) + 4(1:32289) + 0]

¼ 6:167(c) Area of semicircle =

1

2¼r2

=1

2¼(2)2

¼ 6:283Simpson’s rule is more accurate.

12. 4x2 + 9y2 = 36

y2 =36¡ 4x2

9

y = §13

p36¡ 4x2

An equation of the semiellipse is

y =1

3

p36¡ 4x2:

n = 12; b = ¡3; a = 3i xi y

0 ¡3 0

1 ¡2:5 1:1055

2 ¡2 1:4907

3 ¡1:5 1:7321

4 ¡1 1:8856

5 ¡0:5 1:9720

6 0 2

7 0:5 1:9720

8 1 1:8856

9 1:5 1:7321

10 2 1:4907

11 2:5 1:1055

12 3 0



A

2=6

12[1

2(0) + 1:1055 + 1:4907 + 1:7321

+ 1:8856 + 1:972 + 2 + 1:972 + 1:8856

+ 1:7321 + 1:4907 + 1:1055 + 2(0)]

¼ 9:186(b) Simpson’s rule:

A

2=

6

3(12)[0 + 4(1:1055) + 2(1:4907)

+ 4(1:7321) + 2(1:8856) + 4(1:972)

+ 2(2) + 4(1:972) + 2(1:8856) + 4(1:7321)

+ 2(1:4907) + 4(1:1055) + 0]

=1

6(55:982)

¼ 9:330(c) The trapezoidal rule gives the area of the re-gion as 9.1859. Simpson’s rule gives the area of theregion as 9.3304. The actual area is 3¼ ¼ 9:4248:Simpson’s rule is a better approximation.

13. Since f(x) > 0 and f 00(x) > 0 for all x between aand b, we know the graph of f(x) on the intervalfrom a to b is concave upward. Thus, the trape-zoid that approximates the area will have an areagreater than the actual area. Thus,

T >

Z b

a

f(x)dx:

The correct choice is (b).

14. (a) f(x) = x2; [0; 3]

T >

Z b

a

f(x)dx

By looking at the graph of y = x2 and dividingthe area between 0 and 3 into an even number oftrapezoids, you can see that each trapezoid has anarea greater than the actual area [case (b)].

(b) f(x) =px; [0; 9]

T <

Z b

a

f(x)dx

By looking at the graph of y =px and divid-

ing area between 0 and 9 into an even number oftrapezoids, you can see that each trapezoid has anarea less than the actual area [case (a)].

(c) You can’t say which is larger because sometrapezoids are greater than the given area andsome are less than the given area [case (c)].

15. (a)Z 1

0

x4 dx =

μ1

5

¶x5¯¯1

0

=1

5

= 0:2

(b) n = 4; b = 1; a = 0; f(x) = x4

Z 1

0

x4 dx ¼ 1¡ 04

·1

2(0) +

1

256+1

16+81

256+1

2(1)

¸

=1

4

μ226

256

¶

¼ 0:220703

n = 8; b = 1; a = 0; f(x) = x4

Z 1

0

x4 dx ¼ 1¡ 08

·1

2(0) +

1

4096+

1

256+

81

4096

+1

16+625

4096+81

256+2401

4096+1

2(1)

¸

=1

8

μ6724

4096

¶

¼ 0:20520

n = 16; b = 1; a = 0; f(x) = x4

Z 1

0

x4 dx ¼ 1¡ 016

·1

2(0) +

1

65,536+

1

4096

+81

65,536+

1

256+

625

65,536

+81

4096+

2401

65; 536+1

16

+6561

65,536+625

4096+14; 641

65,536

+81

256+28; 561

65,536+2401

4096

+50; 625

65,536+1

2(1)

¸

¼ 1

16

μ211; 080

65,536

¶

¼ 0:201302


n = 32; b = 1; a = 0; f(x) = x4Z 1

0

x4 dx

¼ 1¡ 032

·1

2(0) +

1

1,048,576+

1

65,536

+81

1,048,576+

1

4096+

625

1,048,576

+81

65,536+

2401

1,048,576+

1

256+

6561

1,048,576

+625

65,536+

14,6411,048,576

+81

4096+

28,5611,048,576

+2401

65,536+

50,6251,048,576

+1

16+

83,5211,048,576

+6561

65,536+130,3211,048,576

+625

4096+194,4811,048,576

+14,64165,536

+279,8411,048,576

+81

256+390,6251,048,576

+28,56165,536

+531,4411,048,576

+2401

4096+707,2811,048,576

+50,62565,536

+923,5211,048,576

+1

2(1)

¸

¼ 1

32

μ6,721,8081,048,576

¶¼ 0:200325

To …nd error for each value of n; subtract as indicated.

n = 4: (0:220703¡ 0:2) = 0:020703n = 8: (0:205200¡ 0:2) = 0:005200n = 16: (0:201302¡ 0:2) = 0:001302n = 32: (0:200325¡ 0:2) = 0:000325(c) p = 1

41(0:020703) = 4(0:020703)

= 0:082812

81(0:005200) = 8(0:005200)

= 0:0416

Since these are not the same, try p = 2:

p = 2:

42(0:020703) = 16(0:020703)

= 0:331248

82(0:005200) = 64(0:005200) = 0:3328

162(0:001302) = 256(0:001302)

= 0:333312

322(0:000325) = 1024(0:000325)

= 0:3328

Since these values are all approximately the same,the correct choice is p = 2:

16. As n changes from 4 to 8, for example, the errorchanges from 0.020703 to 0.005200.

0:020703a = 0:005200

a ¼ 1

4

Similar results would be obtained using other val-ues for n:The error is multiplied by 1

4 :

17. (a)Z 1

0

x4 dx =1

5x5¯¯1

0

=1

5

= 0:2

(b) n = 4; b = 1; a = 0; f(x) = x4

Z 1

0

x4 dx ¼ 1¡ 03(4)

·0 + 4

μ1

256

¶+ 2

μ1

16

¶

+ 4

μ81

256

¶+ 1

¸

=1

12

μ77

32

¶

¼ 0:2005208

n = 8; b = 1; a = 0; f(x) = x4

Z 1

0

x4 dx ¼ 1¡ 03(8)

·0 + 4

μ1

4096

¶+ 2

μ1

256

¶

+ 4

μ81

4096

¶+ 2

μ1

16

¶+ 4

μ625

4096

¶

+ 2

μ81

256

¶+ 4

μ2401

4096

¶+ 1

¸

=1

24

μ4916

1024

¶

¼ 0:2000326


n = 16; b = 1; a = 0; f(x) = x4Z 1

0

x4 dx

¼ 1¡ 03(16)

·0 + 4

μ1

65,536

¶+ 2

μ1

4096

¶

+ 4

μ81

65,536

¶+ 2

μ1

256

¶+ 4

μ625

65,536

¶

+ 2

μ81

4096

¶+ 4

μ2401

65,536

¶+ 2

μ1

16

¶

+ 4

μ6561

65,536

¶+ 2

μ625

4096

¶+ 4

μ14; 641

65,536

¶

+ 2

μ81

256

¶+ 4

μ28,56165,536

¶+ 2

μ2401

4096

¶

+ 4

μ50; 625

65,536+ 1

¶¸

=1

48

μ157; 288

16; 384

¶¼ 0:2000020

n = 32; b = 1; a = 0; f(x) = x4Z 1

0

x4 dx

¼ 1¡ 03(32)

·0 + 4

μ1

1,048,576

¶+ 2

μ1

65,536

¶

+4

μ81

1,048,576

¶+2

μ1

4096

¶+4

μ625

1,048,576

¶

+ 2

μ625

65,536

¶+ 4

μ14; 641

1,048,576

¶+ 2

μ81

4096

¶

+4

μ28,5611,048,576

¶+2

μ2401

65,536

¶+4

μ50; 625

1,048,576

¶

+ 2

μ1

16

¶+ 4

μ83,5211,048,576

¶+ 2

μ6561

65,536

¶

+4

μ130,3211,048,576

¶+2

μ625

4096

¶+4

μ194,4811,048,576

¶

+ 2

μ14,64165,536

¶+ 4

μ279,8411,048,576

¶+ 2

μ81

256

¶

+ 4

μ390,6251,048,576

¶+ 2

μ28,56165,536

¶+ 4

μ531,4411,048,576

¶

+ 2

μ2401

4096

¶+ 4

μ707,2811,048,576

¶+ 2

μ50,62565,536

¶

+ 4

μ923,5211,048,576

¶+ 1

¸

=1

96

μ50,033,168262,144

¶¼ 0:2000001

To …nd error for each value of n; subtract as indicated.

n = 4: (0:2005208¡ 0:2) = 0:0005208n = 8: (0:2000326¡ 0:2) = 0:0000326n = 16: (0:2000020¡ 0:2) = 0:0000020n = 32: (0:2000001¡ 0:2) = 0:0000001(c) p = 1:

41(0:0005208) = 4(0:0005208) = 0:0020832

81(0:0000326) = 8(0:0000326) = 0:0002608

Try p = 2:

42(0:0005208) = 16(0:0005208) = 0:0083328

82(0:0000326) = 64(0:0000326) = 0:0020864

Try p = 3:

43(0:0005208) = 64(0:0005208) = 0:0333312

83(0:0000326) = 512(0:0000326) = 0:0166912

Try p = 4:

44(0:0005208) = 256(0:0005208) = 0:1333248

84(0:0000326) = 4096(0:0000326) = 0:1335296

164(0:0000020) = 65536(0:0000020) = 0:131072

324(0:0000001) = 1048576(0:0000001) = 0:1048576

These are the closest values we can get; thus,p = 4:

18. As n changes from 4 to 8, the error changes from0.0005208 to 0.0000326.

0:0005208a = 0:0000326

a ¼ 1

16

Similar results would be obtained using other val-ues for n:The error is multiplied by 1

16 :

19. Midpoint rule:

n = 4; b = 5; a = 1; f(x) =1

x2;¢x = 1

i xi f(xi)

13

2

4

9

25

2

4

25

37

2

4

49

49

2

4

81


Z 5

1

1

x2dx ¼

4Xi=1

f(xi)¢x

=4

9(1) +

4

25(1) +

4

49(1) +

4

81(1)

¼ 0:7355

Simpson’s rule:

m = 8; b = 5; a = 1; f(x) =1

x2

i xi f(xi)

0 1 1

13

2

4

9

2 21

4

35

2

4

25

4 31

9

57

2

4

49

6 41

16

79

2

4

81

8 51

25

Z 5

1

1

x2dx

¼ 5¡ 13(8)

·1 + 4

μ4

9

¶+ 2

μ1

4

¶+ 4

μ4

25

¶

+ 2

μ1

9

¶+ 4

μ4

49

¶+ 2

μ1

16

¶

+ 4

μ4

81

¶+1

25

¸

¼ 1

6(4:82906)

¼ 0:8048From #7 part a, T ¼ 0:9436, when n = 4. Toverify the formula evaluate 2M+T

3 :

2M + T

3¼ 2(0:7355) + 0:9436

3

¼ 0:8048

20. Midpoint rule: n = 4; b = 4; a = 2; f(x) =1

x3;

¢x =1

2

i xi f(xi)

19

4

64

729

211

4

64

1331

313

4

64

2197

415

4

64

3375

Z 4

2

1

x3dx

¼4Xi=1

f(xi)¢x

=64

729

μ1

2

¶+

64

1331

μ1

2

¶+

64

2197

μ1

2

¶+

64

3375

μ1

2

¶

¼ 0:09198Simpson’s rule:

n = 8; b = 4; a = 2; f(x) =1

x3

i xi f(xi)

0 21

8

19

4

64

729

25

2

8

125

311

4

64

1331

4 31

27

513

4

64

2197

67

2

8

343

715

4

64

3375

8 41

64


Z 4

2

1

x3dx

¼ 4¡ 23(8)

·1

8+ 4

μ64

729

¶+ 2

μ8

125

¶+ 4

μ64

1331

¶

+ 2

μ1

27

¶+ 4

μ64

2197

¶+ 2

μ8

343

¶

+ 4

μ64

3375

¶+1

64

¸

¼ 1

12(1:125223) ¼ 0:09377

From #8 part a, T ¼ 0:0973, when n = 4. To ver-ify the formula evaluate 2M+T

3 .

2M + T

3¼ 2(0:09198) + 0:0973

3

¼ 0:09377

21. (a)

(b) A =7¡ 16

·1

2(0:4) + 0:6 + 0:9 + 1:1

+ 1:3 + 1:4 +1

2(1:6)

¸

= 6:3

(c) A =7¡ 13(6)

[0:4 + 4(0:6) + 2(0:9) + 4(1:1)

+ 2(1:3) + 4(1:4) + 1:6]

¼ 6:27

22. (a)

(b) A =7¡ 16

·1

2(9) + 9:2 + 9:5 + 9:4

+ 9:8 + 10:1 +1

2(10:5)

¸

= 57:75

(c) A =7¡ 13(6)

[9:0 + 4(9:2) + 2(9:5)

+ 4(9:4) + 2(9:8) + 4(10:1) + 10:5]

=1

3(172:9)

= 57:63

23. y = e¡t2

+1

t+ 1

The total reaction is

Z 9

1

μe¡t

2

+1

t+ 1

¶dt:

n = 8; b = 9; a = 1; f(t) = e¡t2

+ 1t+1

i xi f(xi)

0 1 0:8679

1 2 0:3516

2 3 0:2501

3 4 0:20004 5 0:1667

5 6 0:1429

6 7 0:1250

7 8 0:1111

8 9 0:1000


Z 9

1

μe¡t

2

+1

t+ 1

¶dt

¼ 9¡ 18

·1

2(0:8679) + 0:3516 + 0:2501

+ ¢ ¢ ¢+ 1

2(0:1000)

¸

¼ 1:831



1

μe¡t

2

+1

t+ 1

¶dt

¼ 9¡ 13(8)

[0:8679 + 4(0:3516) + 2(0:2501)

+ 4(0:2000) + 2(0:1667) + 4(0:1429)

+ 2(0:1250) + 4(0:1111) + 0:1000]

=1

3(5:2739)

¼ 1:758

24. y =2

t+ 2+ e¡t

2=2

The total growth is

Z 7

1

μ2

t+ 2+ e¡t

2=2

¶dt:

n = 12; b = 7; a = 1; f(t) = 2t+2 + e

¡t2=2

i xi f(xi)

0 1 1:273

1 1:5 0:8961

2 2 0:6353

3 2:5 0:4884

4 3 0:4111

5 3:5 0:3658

6 4 0:3337

7 4:5 0:3077

8 5 0:2857

9 5:5 0:2667

10 6 0:2500

11 6:5 0:2353

12 7 0:2222


1

μ2

t+ 2+ e¡t

2=2

¶dt

=7¡ 112

·1

2(1:273) + 0:8961 + 0:6353

+ 0:4884 + 0:4111 + 0:3658 + 0:3337

+ 0:3077 + 0:2857 + 0:2667 + 0:2500

+0:2353 +1

2(0:2222)

¸

=1

2(5:2234)

¼ 2:612

(b) Simpson’s rule

Z 7

1

μ2

t+ 2+ e¡t

2=2

¶dt

=7¡ 13(12)

[1:273+4(0:8961)+2(0:6353)+4(0:4884)

+2(0:4111)+4(0:3658)+2(0:3337)+4(0:3077)

+2(0:2857)+4(0:2667)+2(0:2500)+4(0:2353)

+ 0:2222]

=1

6(15:5668)

¼ 2:594

25. Note that heights may di¤er depending on thereadings of the graph. Thus, answers may vary.n = 10; b = 20; a = 0

i xi f(xi)

0 0 01 2 5

2 4 3

3 6 2

4 8 1:5

5 10 1:2

6 12 1

7 14 0:58 16 0:39 18 0:210 20 0:2

Area under curve for Formulation A

=20¡ 010

·1

2(0) + 5 + 3 + 2 + 1:5 + 1:2

+ 1 + 0:5 + 0:3 + 0:2 +1

2(0:2)

¸

= 2(14:8)

¼ 30 mcgh/ml

This represents the total amount of drug availableto the patient for each ml of blood.


26. n = 10; b = 20; a = 0

i xi y

0 0 0

1 2 2:0

2 4 2:9

3 6 3:0

4 8 2:5

5 10 2:0

6 12 1:75

7 14 1:0

8 16 0:75

9 18 0:50

10 20 0:25

A =20¡ 010

·1

2(0) + 2 + 2:9 + 3 + 2:5 + 2

+ 1:75 + 1:0 + 0:75 + 0:5 +1

2(0:25)

¸

= 33:05 (This answer may vary dependingupon readings from the graph.)

The area under the curve, about 33 mcg/ml; rep-resents the total amount of drug available to thepatient for each ml of blood.

27. As in Exercise 25, readings on the graph mayvary, so answers may vary. The area both underthe curve for Formulation A and above the mini-mum e¤ective concentration line in on the interval£12 ; 6¤:

Area under curve for Formulation A on£12 ; 1¤;

with n = 1

=1¡ 1

2

1

·1

2(2 + 6)

¸

=1

2(4) = 2

Area under curve for Formulation A on [1; 6]; withn = 5

=6¡ 15

·1

2(6) + 5 + 4 + 3 + 2:4 +

1

2(2)

¸= 18:4

Area under minimum e¤ective concentration line£12 ; 6¤= 5:5(2) = 11:0

Area under the curve for Formulation A and aboveminimum e¤ective concentration line

= 2 + 18:4¡ 11:0

¼ 9 mcgh/ml

This represents the total e¤ective amount of drugavailable to the patient for each ml of blood.

28. The area both under the curve for FormulationB and above the minimum e¤ective concentrationline is on the interval (2; 10):n = 8; b = 10; a = 2

i xi y

0 2 2:0

1 3 2:4

2 4 2:9

3 5 2:8

4 6 3:0

5 7 2:6

6 8 2:5

7 9 2:2

8 10 2:0

Let AB = area under Formulation B curve be-tween t = 2 and t = 10:

AB =10¡ 28

·1

2(2) + 2:4 + 2:9 + 2:8 + 3

+ 2:6 + 2:5 + 2:2 +1

2(2)

¸

AB = 20:4

Let AME = area under minimum e¤ective concen-tration curve between t = 2 and t = 10:

AME = (10¡ 2)(2) = 16

So the area between AB and AME between t = 2and t = 10 is 20:4¡ 16 = 4:4:This area, about 4.4 mcgh/ml; represents the to-tal e¤ective amount of the drug available to thepatient for each ml of blood.

Notice that between t = 0 and t = 12; the graphfor Formulation B is below the line.Thus, no area exists under the curve for Formula-tion B and above the minimum e¤ective concen-tration line in the intervals (0; 2) and (10; 12):


29. y = b0wb1e¡b2w

(a) If t = 7w then w =t

7:

y = b0

μt

7

¶b1e¡b2t=7

(b) Replacing the constants with the given values,we have

y = 5:955

μt

7

¶0:233e¡0:027t=7dt

In 25 weeks, there are 175 days.

175Z0

5:955

μt

7

¶0:233e¡0:027t=7dt

n = 10; b = 175; a = 0;

f(t) = 5:955

μt

7

¶0:233e¡0:027t=7

i ti f(ti)

0 0 0

1 17:5 6:89

2 35 7:57

3 52:5 7:78

4 70 7:77

5 87:5 7:65

6 105 7:46

7 122:5 7:23

8 140 6:97

9 157:5 6:70

10 175 6:42

Trapezoidal rule:Z 175

0

5:955

μt

7

¶0:233e¡0:027t=7dt

¼ 175¡ 010

·1

2(0) + 6:89 + 7:57 + 7:78 + 7:77

+ 7:65 + 7:46 + 7:23 + 6:97 + 6:70 +1

2(6:42)

¸

= 17:5(69:23)

= 1211.525

The total milk consumed is about 1212 kg.

Simpson’s rule:Z 175

0

5:955

μt

7

¶0:233e¡0:027t=7dt

¼ 175¡ 03(10)

[0 + 4(6:89) + 2(7:57) + 4(7:78)

+ 2(7:77) + 4(7:65) + 2(7:46) + 4(7:23)

+ 2(6:97) + 4(6:70) + 6:42]


(c) Replacing the constants with the givenvalues, we have

y = 8:409

μt

7

¶0:143e¡0:037t=7:

In 25 weeks, there are 175 days.

Z 175

0

8:409

μt

7

¶0:143e¡0:037t=7dt

n = 10; b = 175; a = 0;

f(t) = 8:409

μt

7

¶0:143e¡0:037t=7

i ti f(ti)

0 0 0

1 17:5 8:74

2 35 8:80

3 52:5 8:50

4 70 8:07

5 87:5 7:60

6 105 7:11

7 122:5 6:63

8 140 6:16

9 157:5 5:71

10 175 5:28

Trapezoidal rule:Z 175

0

8:409

μt

7

¶0:143e¡0:037t=7dt

¼ 175¡ 010

·1

2(0) + 8:74 + 8:80 + 8:50

+ 8:07 + 7:60 + 7:11 + 6:63

+ 6:16 + 5:71 +1

2(5:28)

¸

= 17:5(69:96)

= 1224.30



Simpson’s rule:Z 175

0

8:409

μt

7

¶0:143e¡0:037t=7dt

¼ 175¡ 03(10)

[0 + 4(8:74) + 2(8:80) + 4(8:50)

+ 2(8:07) + 4(7:60) + 2(7:11) + 4(6:63)

+ 2(6:16) + 4(5:71) + 5:28]

=35

6(214:28)

= 1249.97


30. For the period Feb. 18 through May 13, there aresix 14-day intervals or 84 days.

n = 6; b = 84; a = 0; f(t) as listed

(a) i ti f(ti)

0 Feb. 18 01 Mar. 4 122 Mar. 18 303 Apr. 1 404 Apr. 15 185 Apr. 29 86 May 13 3

Simpson’s rule:Z b

a

f(t)dt

¼ 84¡ 03(6)

[0 + 4(12) + 2(30) + 4(40) + 2(18)

+ 4(8) + 3]

=14

3(339)

= 1,582

There were about 1,582 cases.

(b) i ti f(ti)

0 Feb. 18 01 Mar. 4 102 Mar. 18 143 Apr. 1 114 Apr. 15 25 Apr. 29 16 May 13 1

Simpson’s rule:Z b

a

f(t)dt

¼ 84¡ 03(6)

[0 + 4(10) + 2(14) + 4(11) + 2(2)

+ 4(1) + 1]

¼ 14

3(121)

¼ 564:67There were about 565 cases.

31. (a)

(b)7¡ 16

1

2

·(4) + 7 + 11 + 9 + 15 + 16 +

1

2(23)

¸

= 71:5

(c)7¡ 13(6)

[4 + 4(7) + 2(11) + 4(9)

+ 2(15) + 4(16) + 23]

= 69:0

32. (a)

(b) A =7¡ 16

·1

2(12) + 16 + 18 + 21 + 24

+ 27 +1

2(32)

¸

A = 1(128)

= 128

(c) A =7¡ 13(6)

[12 + 4(16) + 2(18) + 4(21)

+ 2(24) + 4(27) + 32]

A = 128


33. We need to evaluateZ 36

12

(105e0:01px + 32)dx:

Using a calculator program for Simpson’s rule withn = 20; we obtain 3413.18 as the value of this inte-gral. This indicates that the total revenue betweenthe twelfth and thirty-sixth months is about 3413.

34. We need to evaluateZ 182

7

3:922t0:242e¡0:00357tdt

Using a calculator program for Simpson’s rule withn = 20; we obtain 1400.88 as the value of this inte-gral. This indicates that the total amount of milkconsumed by a calf from 7 to 182 days is about1400 kg.

35. Use a calculator program for Simpson’s rule withn = 20 to evaluate each of the integrals in thisexercise.

(a)Z 1

¡1

μ1p2¼e¡x

2=2

¶dx ¼ 0:6827

The probability that a normal random variable iswithin 1 standard deviation of the mean is about0.6827.

(b)Z 2

¡2

μ1p2¼e¡x

2=2

¶dx ¼ 0:9545

The probability that a normal random variable iswithin 2 standard deviations of the mean is about0.9545.

(c)Z 3

¡3

μ1p2¼e¡x

2=2

¶dx ¼ 0:9973

The probability that a normal random variable iswithin 3 standard deviations of the mean is about0.9973.

Chapter 7 Review Exercises

5.Z(2x+ 3)dx =

2x2

2+ 3x+C

= x2 + 3x+C

6.Z(5x¡ 1)dx = 5x2

2¡ x+C

7.Z(x2 ¡ 3x+ 2)dx

=x3

3¡ 3x

2

2+ 2x+C

8.Z(6¡ x2)dx = 6x¡ x

3

3+C

9.Z3pxdx = 3

Zx1=2 dx

=3x3=2

32

+C

= 2x3=2 +C

10.Z p

x

2dx =

Z1

2x1=2 dx

=12x

3=2

32

+C

=x3=2

3+C

11.Z(x1=2 + 3x¡2=3)dx

=x3=2

32

+3x1=3

13

+C

=2x3=2

3+ 9x1=3 +C

12.Z(2x4=3 + x¡1=2)dx

=2x7=3

73

+x1=2

12

+C

=6x7=3

7+ 2x1=2 +C

13.Z ¡4x3dx =

Z¡4x¡3 dx

=¡4x¡2¡2 +C

= 2x¡2 +C

14.Z

5

x4dx =

Z5x¡4 dx

=5x¡3

¡3 +C

= ¡ 5

3x3+C

15.Z¡3e2x dx = ¡3e2x

2+C

16.Z5e¡x dx = ¡5e¡x +C

Chapter 7 Review Exercises 525

17.Zxe3x

2

dx =1

6

Z6xe3x

2

dx

Let u = 3x2; so that du = 6xdx:

=1

6

Zeu du

=1

6eu +C

=e3x

2

6+C

18.Z2xex

2

dx = ex2

+C

19.Z

3x

x2 ¡ 1 dx = 3μ1

2

¶Z2xdx

x2 ¡ 1Let u = x2 ¡ 1; so that du = 2xdx:

=3

2

Zdu

u

=3

2ln juj+C

=3 ln

¯x2 ¡ 1¯2

+C

20.Z ¡x2¡ x2 dx = ¡

1

2

Z ¡2xdx2¡ x2

Let u = 2¡ x2; so thatdu = ¡2xdx:

=1

2

Zdu

u

=1

2ln juj+C

=1

2ln¯2¡ x2¯+C

21.Z

x2 dx

(x3 + 5)4=1

3

Z3x2 dx

(x3 + 5)4

Let u = x3 + 5; so that

du = 3x2 dx:

=1

3

Zdu

u4

=1

3

Zu¡4 du

=1

3

μu¡3

¡3¶+C

=¡(x3 + 5)¡3

9+C

22.Z(x2 ¡ 5x)4(2x¡ 5)dx


du = (2x¡ 5)dx:Z(x2 ¡ 5x)4(2x¡ 5)dx

=

Zu4 du

=u5

5+C

=(x2 ¡ 5x)5

5+C

23.Z

x3

e3x4dx =

Zx3e¡3x

4

= ¡ 1

12

Z¡12x3e¡3x4 dx

Let u = ¡3x4; so that du = ¡12x3 dx:

= ¡ 1

12

Zeu du

= ¡ 1

12eu +C

=¡e¡3x412

+C

24.Ze3x

2+4 xdx

Let u = 3x2 + 4 so that

du = 6xdx:

Ze3x

2+4 xdx =1

6

Z(6x)(e3x

2

)dx

=1

6

Zeu du

=1

6eu + C

=e3x

2+4

6+C


25.Z(3 lnx+ 2)4

xdx

Let u = 3 lnx+ 2 so that

du =3

xdx:

Z(3 lnx+ 2)4

xdx =

1

3

Z3(3 lnx+ 2)4

xdx

=1

3

Zu4du

=1

3¢ u

5

5+C

=(3 lnx+ 2)5

15+ C

26.Z p

5 lnx+ 3

xdx

Let u = 5 lnx+ 3 so that

du =5

xdx:

Z p5 lnx+ 3

xdx =

1

5

Z5p5 lnx+ 3

xdx

=1

5

Zu1=2du

=1

5¢ 2u

3=2

3+C

=2(5 lnx+ 3)3=2

15+C

27. f(x) = 3x+ 1; x1 = ¡1; x2 = 0; x3 = 1;x4 = 2; x5 = 3

f(x1) = ¡2; f(x2) = 1; f(x3) = 4;f(x4) = 7; f(x5) = 10

5Xi=1

f(xi)

= f(1) + f(2) + f(3) + f(4) + f(5)

= ¡2 + 1 + 4+ 7+ 10= 20

28. (a)Z 4

0

f(x)dx = 0; since the area above the x-

axis from 0 to 2 is identical to the area below thex-axis from 2 to 4.

(b)Z 4

0

f(x)dx can be computed by calculating

the area of the rectangle and triangle that makeup the region shown in graph.

Area of rectangle = (length)(width)= (3)(1) = 3

Area of triangle =1

2(base)(height)

=1

2(1)(3) =

3

2Z 4

0

f(x)dx = 3 +3

2=9

2= 4:5

29. f(x) = 2x+ 3; from x = 0 to x = 4

¢x =4¡ 04

= 1

i xi f(xi)

1 0 3

2 1 5

3 2 7

4 3 9

A =4Xi=1

f(xi)¢x

= 3(1) + 5(1) + 7(1) + 9(1)

= 24

30.Z 4

0

(2x+ 3)dx

Graph y = 2x+ 3:

Z 4

0

(2x+ 3)dx is the area of a trapezoid with

B = 11; b = 3; h = 4: The formula for thearea is

A =1

2(B + b)h:

A =1

2(11 + 3)(4)

A = 28;

so Z 4

0

(2x+ 3)dx = 28:


31. (a) Since s(t) represents the odometer reading,the distance traveled between t = 0 and t = T

will be s(T )¡ s(0):

(b)Z T

0

v(t)dt = s(T ) ¡ s(0) is equivalent to theFundamental Theorem of Calculus with a = 0;

and b = T because s(t) is an antiderivative of v(t):

32. The Fundamental Theorem of Calculus states thatZ b

a

f(x)dx = F (x)¯ba= F (b)¡ F (a);

where f is continuous on [a; b] and F is any anti-derivative of f:

33.Z 2

1

(3x2 + 5)dx =

μ3x3

3+ 5x

¶¯¯2

1

= (23 + 10)¡ (1 + 5)= 18¡ 6= 12

34.Z 6

1

(2x2 + x)dx

=

μ2x3

3+x2

2

¶¯¯6

1

=

·2(6)3

3+(6)2

2

¸¡·2(1)3

3+(1)2

2

¸

= 144 + 18¡ 23¡ 12

= 162¡ 23¡ 12

=965

6

¼ 160:83

35.Z 5

1

(3x¡1 + x¡3)dx =μ3 ln jxj+ x

¡2

¡2¶¯¯5

1

=

μ3 ln 5¡ 1

50

¶¡μ3 ln 1¡ 1

2

¶

= 3 ln 5 +12

25¼ 5:308

36.Z 3

1

(2x¡1 + x¡2)dx =μ2 ln jxj+ x

¡1

¡1¶¯¯3

1

=

μ2 ln 3¡ 1

3

¶¡ (2 ln 1¡ 1)

= 2 ln 3 +2

3¼ 2:864

37.Z 1

0

xp5x2 + 4dx

Let u = 5x2 + 4; so that

du = 10xdx and1

10du = xdx:

When x = 0; u = 5(02) + 4 = 4:When x = 1; u = 5(12) + 4 = 9:

=1

10

Z 9

4

pudu =

1

10

Z 9

4

u1=2 du

=1

10¢ u

3=2

3=2

¯94=1

15u3=2

¯94

=1

15(9)3=2 ¡ 1

15(4)3=2

=27

15¡ 8

15

=19

15

38.Z 2

0

x2(3x3 + 1)1=3 dx =(3x3 + 1)4=3

12

¯¯2

0

=254=3

12¡ 1

4=3

12

=254=3 ¡ 112

¼ 6:008

39.Z 2

0

3e¡2xdx =¡3e¡2x2

¯¯2

0

=¡3e¡42

+3

2

=3(1¡ e¡4)

2¼ 1:473

40.Z 5

1

5

2e0:4xdx =

5

2¢ 52

Z 5

1

0:4e0:4xdx

=5

2¢ 52¢ e2x=5

¯¯5

1

=25

4(e2 ¡ e0:4)

=25(e2 ¡ e0:4)

4¼ 36:86


41.Z 1=2

0

xp1¡ 16x4 dx

Let u = 4x2: Then du = 8xdx:When x = 0; u = 0; and when x = 1

2 ; u = 1:

Thus,

Z 1=2

0

xp1¡ 16x4 dx = 1

8

Z 1

0

p1¡ u2 du:

Note that this integral represents the area of rightupper quarter of a circle centered at the originwith a radius of 1.

Area of circle = ¼r2 = ¼(12) = ¼Z 1

0

p1¡ u2 du = ¼

4

1

8

Z 1

0

p1¡ u2 du = 1

8¢ ¼4=¼

32

42.Z e5

1

p25¡ (lnx)2

xdx

Let u = lnx: Then du = 1x dx:

When x = e5; u = ln(e5) = 5:When x = 1; u = ln(1) = 0:

Thus,

Z e5

1

p25¡ (lnx)2

xdx =

Z 5

0

p25¡ u2 du:

Note that this integral represents the area of aright upper quarter of a circle centered at the ori-gin with a radius of 5.

Area of circle = ¼r2 = ¼(5)2 = 25¼Z 5

0

p25¡ u2 du = 25¼

4

43.Z p

7

1

2xp36¡ (x2 ¡ 1)2 dx

Let u = x2 ¡ 1: Then du = 2xdx:When x =

p7; u = (

p7)2 ¡ 1 = 6:

When x = 1; u = (p1)2 ¡ 1 = 0:

Thus,

Z p7

1

2xp36¡ (x2 ¡ 1)2 dx =

Z 6

0

p36¡ u2du:

Note that this integral represents the area of aright upper quarter of a circle centered at the ori-gin with a radius of 6.

Area of circle = ¼r2 = ¼(6)2 = 36¼Z 6

0

p36¡ u2 du = 36¼

4= 9¼

44. f(x) =p4x¡ 3; [1; 3]

Area =Z 3

1

p4x¡ 3dx

=

Z 3

1

(4x¡ 3)1=2dx

=2

3¢ 14¢ (4x¡ 3)3=2

¯¯3

1

=1

6(9)3=2 ¡ 1

6(1)3=2

=1

6(26)

=13

3

45. f(x) = (3x+ 2)6; [¡2; 0]

Area =Z 0

¡2(3x+ 2)6dx

=(3x+ 2)7

21

¯¯0

¡2

=27

21¡ (¡4)

7

21

=5504

7

46. f(x) = xex2

; [0; 2]

Area =Z 2

0

xex2

dx

=ex

2

2

¯¯2

0

=e4

2¡ 12

=e4 ¡ 12

¼ 26:80

47. f(x) = 1 + e¡x; [0; 4]Z 4

0

(1 + e¡x)dx = (x¡ e¡x)¯40

= (4¡ e¡4)¡ (0¡ e0)= 5¡ e¡4¼ 4:982


48. f(x) = 5¡ x2; g(x) = x2 ¡ 3

Points of intersection:

5¡ x2 = x2 ¡ 32x2 ¡ 8 = 0

2(x2 ¡ 4) = 0x = §2

Since f(x) ¸ g(x) in [¡2; 2]; the area between thegraphs isZ 2

¡2[f(x)¡ g(x)]dx =

Z 2

¡2[(5¡ x2)¡ (x2 ¡ 3)]dx

=

Z 2

¡2(¡2x2 + 8)dx

=

μ¡2x33

+ 8x

¶¯¯2

¡2

= ¡23(8) + 16 +

2

3(¡8)¡ 8(¡2)

=¡323+ 32

=64

3:

49. f(x) = x2 ¡ 4x; g(x) = x¡ 6


x2 ¡ 4x = x¡ 6x2 ¡ 5x+ 6 = 0

(x¡ 3)(x¡ 2) = 0x = 2 or x = 3

Since g(x) ¸ f(x) in the interval [2; 3]; the areabetween the graphs isZ 3

2

[g(x)¡ f(x)]dx

=

Z 3

2

[(x¡ 6)¡ (x2 ¡ 4x)]dx

=

Z 3

2

(¡x2 + 5x¡ 6)dx

=

μ¡x33+5x2

2¡ 6x

¶¯¯3

2

=¡273+5(9)

2¡ 6(3)¡ ¡8

3

¡ 5(4)2+ 6(2)

= ¡193+25

2¡ 6 = 1

6:

50. f(x) = x2 ¡ 4x; g(x) = x+ 6; x = ¡2; x = 4

Points of intersection:

x2 ¡ 4x = x+ 6x2 ¡ 5x¡ 6 = 0

(x+ 1)(x¡ 6) = 0x = ¡1 or x = 6

Thus, the area isZ ¡1

¡2[x2¡4x¡(x+6)]dx+

Z 4

¡1[x+6¡(x2¡4x)]dx

=

μx3

3¡ 5x

2

2¡ 6x

¶¯¯¡1

¡2+

μ¡x

3

3+5x2

2+ 6x

¶¯¯4

¡1

=

μ19

6+2

3

¶+

μ128

3+19

6

¶

=149

3


51. f(x) = 5¡ x2; g(x) = x2 ¡ 3; x = 0; x = 4


5¡ x2 = x2 ¡ 38 = 2x2

4 = x2

§2 = x

The curves intersect at x = 2 and x = ¡2:Thus, the area isZ 2

0

[(5¡ x2)¡ (x2 ¡ 3)]dx

+

Z 4

2

[(x2 ¡ 3)¡ (5¡ x2)]dx

=

Z 2

0

(¡2x2 + 8)dx+Z 4

2

(2x2 ¡ 8)dx

=

μ¡2x33

+ 8x

¶¯¯2

0

+

μ2x3

3¡ 8x

¶¯¯4

2

=¡163+ 16 +

μ128

3¡ 32

¶¡μ16

3¡ 16

¶

=32

3+128

3¡ 32¡ 16

3+ 16

= 32:

52.Z 3

1

ln x

xdx

Trapezoidal Rule:

n = 4; b = 3; a = 1; f(x) = ln xx

i x1 f(xi)

0 1 0

1 1:5 0:27031

2 2 0:34657

3 2:5 0:36652

4 3 0:3662

Z 3

1

ln x

xdx ¼ 3¡ 1

4

·1

2(0) + 0:27031 + 0:34657

+ 0:36652 +1

2(0:3662)

¸

= 0:5833

Exact value:Z 3

1

ln x

xdx

=1

2(ln x)2

¯31

=1

2(ln 3)2 ¡ 1

2(ln 1)2

¼ 0:6035

53.Z 10

2

xdx

x¡ 1Trapezoidal Rule:

n = 4; b = 10; a = 2; f(x) = xx¡1

i xi f(xi)

0 2 2

1 44

3

2 66

5

3 88

7

4 1010

9Z 10

2

x

x¡ 1 dx

¼ 10¡ 24

·1

2(2) +

4

3+6

5+8

7+1

2

μ10

9

¶¸¼ 10:46

Exact Value:

Let u = x¡ 1; so that du = dx and x = u+ 1:ThenZ 10

2

x

x¡ 1 dx =Z 9

1

u+ 1

udu

=

Z 9

1

μ1 +

1

u

¶du

=

Z 9

1

du+

Z 9

1

1

udu

= u9

1+ ln juj

¯91

= (9¡ 1) + (ln 9¡ ln 1)= 8 + ln 9 ¼ 10:20:


54.Z 1

0

expex + 4dx

Trapezoidal Rule:

n = 4; b = 1; a = 0; f(x) = expex + 4

i xi f(xi)

0 0 2:236

1 0:25 2:952

2 0:5 3:919

3 0:75 5:236

4 1 7:046Z 1

0

expex + 4dx

=1¡ 04

·1

2(2:236) + 2:952

+ 3:919 + 5:236 +1

2(7:046)

¸¼ 4:187

Exact value:Z 1

0

expex + 4dx =

Z 1

0

ex(ex + 4)1=2dx

=2

3(ex + 4)3=2

¯¯1

0

=2

3(e+ 4)3=2 ¡ 2

3(5)3=2

¼ 4:155

55.Z 3

1

ln x

xdx

Simpson’s rule:

n = 4; b = 3; a = 1; f(x) = ln xx

i xi f(xi)

0 1 0

1 1:5 0:27031

2 2 0:34657

3 2:5 0:36652

4 3 0:3662Z 3

1

ln x

xdx

¼ 3¡ 13(4)

[0 + 4(0:27031) + 2(0:34657)

+ 4(0:36652) + 0:3662]

¼ 0:6011This answer is close to the value of 0.6035 obtainedfrom the exact integral in Exercise 52.

56.Z 10

2

xdx

x¡ 1Simpson’s Rule:

i xi f(x1)

0 2 2

1 44

3

2 66

5

3 88

7

4 1010

9

Z 10

2

x

x¡ 1 dx

¼ 10¡ 23(4)

·2+4

μ4

3

¶+2

μ6

5

¶+4

μ8

7

¶+10

9

¸

¼ 10:28

This answer is close to the answer of 10.20 ob-tained from the exact integral in Exercise 53.

57.Z 1

0

expex + 4dx

Simpson’s rule:

n = 4; b = 1; a = 0; f(x) = expex + 4

i xi f(xi)

0 0 2:236

1 0:25 2:952

2 0:5 3:919

3 0:75 5:236

4 1 7:046

Z 1

0

expex + 4dx

=1¡ 03(4)

[2:236 + 4(2:952) + 2(3:919)

+ 4(5:236) + 7:046

¼ 4:156

This answer is close to the answer of 4.155 ob-tained from the exact integral in Exercise 54.


58. (a)Z 5

1

·px¡ 1¡

μx¡ 12

¶¸dx

=

Z 5

1

μpx¡ 1¡ x

2+1

2dx

¶

=

μ2

3(x¡ 1)3=2 ¡ x

2

4+x

2

¶ ¯51

=

μ16

3¡ 254+5

2

¶¡μ0¡ 1

4+1

2

¶

=16

3¡ 6 + 2 = 4

3

(b) n = 4; b = 5; a = 1; f(x) =px¡ 1¡ x

2+1

2

i xi f(xi)

0 1 0

1 2 0:5

2 3 0:41421

3 4 0:23205

4 5 0Z 5

1

μpx¡ 1¡ x

2+1

2

¶dx

=

μ5¡ 14

¶·1

2(0) + 0:5 + 0:41421

+ 0:23205 +1

2(0)

¸= 1:146

(c)Z 5

1

μpx¡ 1¡ x

2+1

2

¶dx

=

μ5¡ 13(4)

¶[0 + 4(0:5) + 2(0:41421)

+ 4(0:23205) + 0]

=

μ1

3

¶(3:75662)

= 1:252

59.Z 2

¡2[x(x¡ 1)(x+ 1)(x¡ 2)(x+ 2)]2dx

(a) Trapezoidal Rule:

n = 4; b = ¡2; a = 2;f(x) = [x(x¡ 1)(x+ 1)(x¡ 2)(x+ 2)]2

i xi f(xi)

0 ¡2 0

1 ¡1 0

2 0 0

3 1 0

4 2 0

Z 2

¡2[x(x¡ 1)(x+ 1)(x¡ 2)(x+ 2)]2dx

¼ 2¡ (¡2)4

·1

2(0) + 0 + 0 + 0 +

1

2(0)

¸

= 0

(b) Simpson’s Rule

n = 4; b = 2; a = 2;

f(x) = [x(x¡ 1)(x+ 1)(x¡ 2)(x+ 2)]2

i xi f(xi)

0 ¡2 0

1 ¡1 0

2 0 0

3 1 0

4 2 0

Z 2

¡2[x(x¡ 1)(x+ 1)(x¡ 2)(x+ 2)]2dx

¼ 2¡ (¡2)3(4)

[0 + 4(0) + 2(0) + 4(0) + 0]

= 0

60.Z 2

0

f(2x)dx =1

2

Z 4

0

f(x)dx

=1

2

Z 2

0

f(x)dx+1

2

Z 4

2

f(x)dx

=1

2(3) +

1

2(5)

= 4

The answer is c.

61. C0(x) = 3p2x¡ 1; 13 units cost $270.

C(x) =

Z3(2x¡ 1)1=2 dx

=3

2

Z2(2x¡ 1)1=2 dx

Let u = 2x¡ 1; so thatdu = 2dx:

=3

2

Zu1=2 du

=3

2

μu3=2

3=2

¶+C

= (2x¡ 1)3=2 +CC(13) = [2(13)¡ 1]3=2 +C


Since C(13) = 270;

270 = 253=2 +C

270 = 125 +C

C = 145:

Thus,C(x) = (2x¡ 1)3=2 + 145:

62. C0(x) = 82x+1 ; …xed cost is $18.

C(x) =

Z8

2x+ 1dx

= 4 ln j2x+ 1j+ kIf x = 0; C(x) = 18:

Thus18 = ln j1j+ kk = 18:

Thus,C(x) = 4 ln j2x+ 1j+ 18:

63. Read values for the rate of investment income ac-cumulation for every 2 years from year 1 to year9. These are the heights of rectangles with width¢x = 2:

Total accumulated income= 11; 000(2) + 9000(2) + 12; 000(2) + 10; 000(2)

+ 6000(2) ¼ $96; 00064. Total amount

=

Z t

0

100,000e0:03t dt

=100,000e0:03t

0:03

¯t0

=10,000,000

3(e0:03t ¡ 1)

Set this expression equal to 4,000,000.10,000,000

3(e0:03t ¡ 1) = 4,000,000

e0:03t ¡ 1 = 1:20:03t = ln 2:2

t ¼ 26:3It will take him 26.3 years to use up the supply.

65. S0(x) = 3p2x+ 1+ 3

S(x) =

Z 4

0

(3p2x+ 1+ 3)dx

= [(2x+ 1)3=2 + 3x]¯40

= (27 + 12)¡ (1 + 0)= 38

Total sales = $38,000.

66. (a) f(x) =Z(0:1908x+ 1:148)dx

=0:1908x2

2+ 1:148x+C

= 0:0954x2 + 1:148x+C

Since the productivity in 1992 (x = 2) was 100,

100 = 0:0954(2)2 + 1:148(2) +C

100 = 2:6776 +C

C = 97:3224

Thus,

f(x) = 0:0954x2 + 1:148x+ 97:3224:

(b) 2005 corresponds to x = 15:

f(15) = 0:0954(15)2 + 1:148(15) + 97:3224

¼ 136:0The productivity was approximately 136, whichdi¤ers from the actual value by 0.2.

67. S(q) = q2 + 5q + 100D(q) = 350¡ q2S(q) = D(q) at the equilibrium point.

q2 + 5q + 100 = 350¡ q22q2 + 5q ¡ 250 = 0

(¡2q + 25)(q ¡ 10) = 0q = ¡25

2or q = 10

Since the number of units produced would not benegative, the equilibrium point occurs when q =10:

Equilibrium supply= (10)2 + 5(10) + 100 = 250

Equilibrium demand= 350¡ (10)2 = 250

(a) Producers’ surplus

=

Z 10

0

[250¡ (q2 + 5q + 100)]dx

=

Z 10

0

(¡q2 ¡ 5q + 150)dx

=

μ¡q33¡ 5q

2

2+ 150q

¶¯¯10

0

=¡10003

¡ 5002+ 1500

=$2750

3¼ $916:67


(b) Consumers’ surplus =Z 10

0

[(350¡ q2)¡ 250]dx =Z 10

0

(100¡ q2)dx =μ100q ¡ q

3

3

¶¯¯10

0

= 1000¡ 10003

=$2000

3¼ $666:67

68. S0(x) = 225¡ x2; C0(x) = x2 + 25x+ 150S0(x) = C0(x)

225¡ x2 = x2 + 25x+ 1502x2 + 25x¡ 75 = 0(2x¡ 5)(x+ 15) = 0

x =5

2= 2:5

The company should use the machinery for 2.5 years.Z 2:5

0

[(225¡ x2)¡ (x2 + 25x+ 150)]dx =Z 2:5

0

(¡2x3 ¡ 25x+ 75)dx =μ¡2x3

3¡ 25x

2

2+ 75x

¶ ¯2:50

=¡2(2:53)

3¡ 25(2:5

2)

2+ 75(2:5) ¼ 98:95833 ¼ 99; 000

The net savings are about $99,000.

69. (a)Total amount =1

2(2:394)+2:366+2:355+2:282+2:147+2:131+2:118+2:097+2:073+1:983+

1

2(1:869)

¼ 21:684This calculation yields a total of 21.684 billion barrels.

70. The inventory must be replenished when

19¡ S(T ) = 1:19¡ (e3T ¡ 1) = 1

19 = e3T

3T = ln 19

T =1

3ln 19

Therefore, the inventory carrying cost from 0 to T week is

15

Z T

0

[19¡ S(t)]dt = 15Z T

0

[19¡ (e3t ¡ 1)]dt

= 15

Z T

0

(20¡ e3t)dt

= 15

μ20t¡ 1

3e3t¶¯¯1

0

= 15

μ20T ¡ 1

3e3T +

1

3

¶

= 15

μ20

3ln 19¡ 19

3+1

3

¶

¼ 15(13:63)¼ 204:

The correct choice is (c).


71. f(t) = 100¡ tp0:4t2 + 1The total number of additional spiders in the …rstten months isZ 10

0

(100¡ tp0:4t2 + 1)dt;where t is the time in months.

=

Z 10

0

100dt¡Z 10

0

tp0:4t2 + 1dt:

Let u = 0:4t2 + 1; so that

du = 0:8tdt and 10:8 du = t dt:

When t = 10; u = 41:When t = 0; u = 1:

=

Z 10

0

100dt¡ 1

0:8

Z 41

1

u1=2 du

= 100t

¯¯10

0

¡ 54¢ u

3=2

32

¯¯41

1

= 1000¡ 5

6u3=2

¯¯41

1

¼ 782The total number of additional spiders in the …rst10 months is about 782.

72. The total number of infected people over the …rst

four months isZ 4

0

100t

t2 + 1dt; where t is time in

months.Let u = t2 + 1; so that

du = 2t dt and 50 du = 100t dt:

If t = 4; u = 17; soZ 4

0

100t

t2 + 1dt

= 50

Z 17

1

1

udu = 50 ln juj

¯171

= 50 ln 17¡ 50 ln j1j= 50 ln 17

¼ 141:66:Approximately 142 people are infected.

73. (a) The total area is the area of the triangle on[0; 12] with height 0.024 plus the area of the rec-tangle on [12; 17:6] with height 0.024.

A =1

2(12¡ 0)(0:024) + (17:6¡ 12)(0:024)

= 0:144 + 0:1344

= 0:2784

(b) On [0; 12] we de…ned the function f(x) withslope 0:024¡0

12¡0 = 0:002 and y-intercept 0.

f(x) = 0:002x

On [12; 17:6]; de…ne g(x) as the constant value.

g(x) = 0:024:

The area is the sum of the integrals of these twofunctions.

A =

Z 12

0

0:002xdx+

Z 17:6

12

0:024dx

= 0:001x2¯120+ 0:024x

¯17:612

= 0:001(122 ¡ 02) + 0:024(17:6¡ 12)= 0:144 + 0:1344

= 0:2784

74. Since answers are found by estimating values onthe graph, exact answers may vary slightly; how-ever when rounded to the nearest hundred, all an-swers should be the same. Sample solution:

(a) Left endpoints:

Read the values of the function from the graph,using the open circles for the functional values.The values of x and f(x) are listed in the table.

x 0 2 5 15 30 45 60f(x) 30 50 60 105 85 70 55

The values give the heights of 6 rectangles. Thewidth of each rectangle is found by subtractingsubsequent values of x: We estimate the area un-der the curve as6Xi=1

f(xi)4 xi = 30(2) + 50(3) + 60(10) + 105(15)

+ 85(15) + 70(15)

= 4710.

Right endpoints:

We estimate the area under the curve as6Xi=1

f(xi)4 xi = 50(2) + 60(3) + 105(10) + 85(15)

+ 70(15) + 55(15)

= 4480.

Average:

4710 + 44802

= 4595 ¼ 4600 pM.


(b) Read the values of the function from the graph, using the closed circles for the functional values. Thevalues of x and g(x) are listed in the table.

x 0 2 5 15 30 45 60g(x) 20 42 42 70 52 40 20

The values give the heights of 6 rectangles. The width of each rectangle is found by subtracting subsequentvalues of x: We estimate the area under the curve as

6Xi=1

g(xi)4 xi = 20(2) + 42(3) + 42(10) + 70(15) + 52(15) + 40(15)

= 3016.

Right endpoints:

We estimate the area under the curve as

6Xi=1

g(xi)4 xi = 42(2) + 42(3) + 70(10) + 52(15) + 40(15) + 20(15)

= 2590.

Average:3016 + 2590

2= 2803 ¼ 2800 pM.

(c)4600¡ 2800

2800¼ 0:6428

The area under the curve is about 64% more for the fasting sheep.

75. (a) Total amount =1

2(271,553) + 278,325 + 274,690 + 290,525 + 289,890

+ 309,569 + 317,567 + 335,869 + 331,055 +1

2(331,208)

¼ 2,728,871This calculation yields a total of about $2,728,871.

76. v(t) = t2 ¡ 2t

s(t) =

Z t

0

(t2 ¡ 2t)dt

s(t) =t3

3¡ t2 + s0

If t = 3; s = 8:

8 = 9¡ 9 + s08 = s0

Thus,

s(t) =t3

3¡ t2 + 8:

77. For each month, subtract the average temperature from 65± (if it falls below 65±F), then multiply this numbertimes the number of days in the month. The sum is the total number of heating degree days. Readings mayvary, but the sum is approximately 4800 degree-days. (The actual value is 4868 degree-days.)

Extended Application/Estimating Depletion Dates for Mineralsl 537

Extended Application: EstimatingDepletion Dates for Minerals

1. 2,300,000 ¥ 17,100 ¼ 135The reserves would last about 135 yr.

2. 2,300,000 =17,1000:02

(e0:02T1 ¡ 1)2,300,000(0:02)

17,100= e0:02T1 ¡ 1

2:6901 + 1 = e0:02T1

3:6901 = e0:02T1

ln 3:6901 = 0:02T1

T1 =ln 3:6901

0:02

T1 ¼ 65:3The reserves would last about 65.3 yr.

3. 15,000,000 =63,0000:06

(e0:06T1 ¡ 1)15,000,000(0:06)

63,000+ 1 = e0:06T1

15:286 ¼ e0:06T1ln 15:286 ¼ 0:06T1

T1 ¼ ln 15:286

0:06

= 45:4

The depletion time for bauxite is about 45.4 yr.

4. 2; 000; 000 =2200

0:04(e0:04T1 ¡ 1)

2; 000; 000(0:04)

2200+ 1 = e0:04T1

37:36 = e0:04T1

ln 37:36 = 0:04T1T1 ¼ 90:5

The depletion time for bituminous coal is about90.5 yr.

5. k(t) =0:5

t+ 25

(a) For t = 0;

k(t) =0:5

0 + 25= 0:02:

This gives a growth rate of 2% for 1970.

For t = 25;

k(t) =0:5

25 + 25= 0:01:

This gives a growth rate of 1% for 1996.

(b) Use the form of the function k(t) =a

t+ b;

where a and b are both constants. Since k(0) =

0:03; k(t) =a

t+ b, where

0:03 =a

0 + b=a

b. Or a = 0:03b:

Also, since k(25) = 0:02,

0:02 =a

25 + b: Or a = 0:02(25 + b):

Solve:0:03b = 0:02(25 + b)

0:03b = 0:5 + 0:02b

0:01b = 0:5

b = 50

Find a using substitution.

a = 0:03b

a = 0:03(50)

a = 1:5

The function that satis…es these conditions is

k(t) =1:5

t+ 50:

(a) Total consumption = 17,100Z T

0

ek(t)¢tdt

= 17,100Z T

0

e1:5t=(t+50)dt

(b) Use the fnInt command on a graphing calculator toevaluate

17,100Z T

0

e1:5t=(t+50)dt

for di¤erent values of T:

For T = 70 the integral is about 2,158,000.For T = 71 the integral is about 2,199,000.For T = 72 the integral is about 2,240,000.For T = 73 the integral is about 2,282,000.For T = 74 the integral is about 2,324,000.

We would estimate that starting in 1970 the petroleumreserves would last for about 73 years, that is, until2043.

c:swp2507calwaism2001cwa ch 7 pdf final€¦ · dy = 16y4 4 + 9y3 3 ¡ 6y2 2 +3y +c =4y4 +3y3 ......

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