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. Z 6 dk =6 Z 1 dk =6 Z k 0 dy =6 ¢ 1 1 k 0+1 + C =6k + C 6. Z 9 dy =9 Z 1 dy =9 Z y 0 dy = 9 1 y 0+1 + C =9y + C 7. Z (2z + 3) dz =2 Z z dz +3 Z z 0 dz =2 ¢ 1 1+1 z 1+1 +3 ¢ 1 0+1 z 0+1 + C = z 2 +3z + C 8. Z (3x ¡ 5) dx =3 Z x dx ¡ 5 Z x 0 dx =3 ¢ 1 2 x 2 ¡ 5 ¢ 1 1 x + C = 3x 2 2 ¡ 5x + C 9. Z (6t 2 ¡ 8t + 7) dt =6 Z t 2 dt ¡ 8 Z t dt +7 Z t 0 dt = 6t 3 3 ¡ 8t 2 2 +7t + C =2t 3 ¡ 4t 2 +7t + C 10. Z (5x 2 ¡ 6x + 3) dx =5 Z x 2 dx ¡ 6 Z x dx +3 Z x 0 dx = 5x 3 3 ¡ 6x 2 2 +3x + C = 5x 3 3 ¡ 3x 2 +3x + C 11. Z (4z 3 +3z 2 +2z ¡ 6) dz =4 Z z 3 dz +3 Z z 2 dz +2 Z z dz ¡ 6 Z z 0 dz = 4z 4 4 + 3z 3 3 + 2z 2 2 ¡ 6z + C = z 4 + z 3 + z 2 ¡ 6z + C 12. Z (16y 3 +9y 2 ¡ 6y + 3) dy = 16 Z y 3 dy +9 Z y 2 dy ¡ 6 Z y dy +3 Z dy = 16y 4 4 + 9y 3 3 ¡ 6y 2 2 +3y + C =4y 4 +3y 3 ¡ 3y 2 +3y + C 13. Z (5 p z + p 2) dz =5 Z z 1=2 dz + p 2 Z dz = 5z 3=2 3 2 + p 2z + C =5 μ 2 3 ¶ z 3=2 + p 2z + C = 10z 3=2 3 + p 2z + C 14. Z (t 1=4 + ¼ 1=4 ) dt = t 1=4+1 1 4 +1 + ¼ 1=4 t + C = t 5=4 5 4 + ¼ 1=4 t + C = 4t 5=4 5 + ¼ 1=4 t + C 443
95
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c:swp2507calwaism2001CWA CH 7 PDF FINAL€¦ · dy = 16y4 4 + 9y3 3 ¡ 6y2 2 +3y +C =4y4 +3y3 ... +4 e¡0:4x + 0:1 ¶ dx = ¡3 Z dx x +4 Z e¡0:4x dx +e0:1 Z dx = ¡3lnjxj+ 4e¡0:4x
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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
446 Chapter 7 INTEGRATION
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:
Section 7.1 Antiderivatives 447
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
448 Chapter 7 INTEGRATION
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
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.
452 Chapter 7 INTEGRATION
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
454 Chapter 7 INTEGRATION
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
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
áex22
+ 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.
Section 7.2 Substitution 459
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.
460 Chapter 7 INTEGRATION
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.
(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:
Section 7.3 Area and the De…nite Integral 469
(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
470 Chapter 7 INTEGRATION
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:
Section 7.3 Area and the De…nite Integral 471
(b) SincenPi=1
f(xi)¢x = ¢x
μnPi=1
f(xi)
¶, use the command 0:1¤sum(L1) to approximate R 1
0x2dx. The
resulting screen is:
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 .
472 Chapter 7 INTEGRATION
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
resulting screen is:
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:
Section 7.3 Area and the De…nite Integral 473
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
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
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
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
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
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
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
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
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.
476 Chapter 7 INTEGRATION
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:
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
Read values of the function from the table for every number of seconds from 2.0 to 23.4. These values give theheights of 11 rectangles. The width of each rectangle varies: We estimate the area under the curve as11Xi=1
The distance traveled by the Mercedes-Benz S550 is about 2952 ft.
32. Left endpoints:
Read values of the function from the table for every number of seconds from 2.4 to 19.2. These values give theheights of 8 rectangles. The width of each rectangle varies: We estimate the area under the curve as8Xi=1
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
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.
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:
(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:
(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.
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
(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
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
496 Chapter 7 INTEGRATION
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
To …nd the points of intersection of the graphs,substitute for y:
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
500 Chapter 7 INTEGRATION
17. y = x3 ¡ x2 + x+ 1; y = 2x2 ¡ x+ 1Find the points of intersection.
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)
Find the points of intersection.
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:
Section 7.5 The Area Between Two Curves 501
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
Find the points of intersection.
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:
The area between the curves isZ 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:
The area between the curves isZ 1
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
502 Chapter 7 INTEGRATION
To …nd the points of intersection of the graphs,substitute for y:
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
To …nd the points of intersection of the graphs,substitute for y:
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):
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:
The area between the curves is given by
¡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.
Section 7.5 The Area Between Two Curves 503
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
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.
(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.
(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
(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.
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:)
Section 7.5 The Area Between Two Curves 509
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
The area between the curves is given by
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:
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
516 Chapter 7 INTEGRATION
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.
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
518 Chapter 7 INTEGRATION
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
Section 7.6 Numerical Integration 519
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
(a) Trapezoidal rule:
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
520 Chapter 7 INTEGRATION
(b) Simpson’s rule:Z 9
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
(a) Trapezoidal rule:Z 7
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.
Section 7.6 Numerical Integration 521
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):
522 Chapter 7 INTEGRATION
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]
The total milk consumed is about 1231 kg.
(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
The total milk consumed is about 1224 kg.
Section 7.6 Numerical Integration 523
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
The total milk consumed is about 1250 kg.
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
524 Chapter 7 INTEGRATION
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.
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:
Chapter 7 Review Exercises 527
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
528 Chapter 7 INTEGRATION
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
Chapter 7 Review Exercises 529
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
Find the points of intersection.
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
530 Chapter 7 INTEGRATION
51. f(x) = 5¡ x2; g(x) = x2 ¡ 3; x = 0; x = 4
Find the points of intersection.
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:
Chapter 7 Review Exercises 531
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.
532 Chapter 7 INTEGRATION
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
Chapter 7 Review Exercises 533
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:
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.
536 Chapter 7 INTEGRATION
(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
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.