Solução_Calculo_Stewart_6e

F.
250 ¤ CHAPTER 5 INTEGRALS
TX.10
61. Since m 0 (x) =ρ(x), m = 4
ρ(x) dx =
4
9+2 √ 4
x dx = 9x + 4 0 0
3 x3/2 =36+ 32 3
0
− 0=
140
3
=46 2 3 kg.
63. Let s be the position of the car. We know from Equation 2 that s(100) − s(0) = 100
0
v(t) dt. We use the Midpoint Rule for
0 ≤ t ≤ 100 with n =5. Note that the length of each of the five time intervals is 20 seconds = 20 hour = 1 hour.
3600 180
So the distance traveled is
100
v(t) dt ≈ 1
1
247
[v(10) + v(30) + v(50) + v(70) + v(90)] = (38 + 58 + 51 + 53 + 47) = ≈ 1.4 miles.
0 180 180 180
65. From the Net Change Theorem, the increase in cost if the production level is raised
from 2000 yards to 4000 yards is C(4000) − C(2000) = 4000
2000 C0 (x) dx.
4000
2000 C0 (x) dx = 4000
2000 3 − 0.01x +0.000006x
2
dx = 3x − 0.005x 2 +0.000002x 3 4000
=60,000 −2,000 = $58,000
2000
67. (a) We can find the area between the Lorenz curve and the line y = x by subtracting the area under y = L(x) from the area
under y = x. Thus,
area between Lorenz curve and line y = x
coefficient of inequality = =
area under line y = x
=
1
0
[x − L(x)] dx
[x 2 /2] 1 =
0
1
0
[x − L(x)] dx
1/2
1
0
[x − L(x)] dx
1
0 xdx
=2 1
[x − L(x)] dx
0
(b) L(x) = 5
12 x2 + 7 x ⇒ L(50%) = L
1
12 2 =
5
+ 7 = 19 =0.39583, so the bottom 50% of the households receive
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at most about 40% of the income. Using the result in part (a),
coefficient of inequality =2 1
[x − L(x)] dx =2 1 0 0 x −
5
12 x2 − 7 x 12
dx =2 1
5
x − 5 x2 dx
0 12 12
=2 1 5
(x −
0 12 x2 ) dx = 5 1
6 2 x2 − 1 x3 1
= 5 1 − 1
3 0 6 2 3 =
5 1
6 6 =
5
36
5.5 The Substitution Rule
1. Let u = −x. Thendu = − dx, so dx = − du. Thus, e −x dx = e u (−du) =−e u + C = −e −x + C. Don’t forget that it
is often very easy to check an indefinite integration by differentiating your answer. In this case,
d
dx (−e−x + C) =−[e −x (−1)] = e −x , the desired result.
3. Let u = x 3 +1.Thendu =3x 2 dx and x 2 dx = 1 du,so 3
x 2 √u
x 3 +1dx =
1 du = 1 u 3/2
3
3 3/2 + C = 1 3 · 2
3 u3/2 + C = 2 9 (x3 +1) 3/2 + C.
5. Let u =cosθ. Thendu = − sin θdθand sin θdθ = −du, so
cos 3 θ sin θdθ= u 3 (−du) =− u4
4 + C = − 1 4 cos4 θ + C.
7. Let u = x 2 .Thendu =2xdxand xdx= 1 2 du,so x sin(x 2 ) dx = sin u 1
2 du = − 1 2 cos u + C = − 1 2 cos(x2 )+C.