Solução_Calculo_Stewart_6e

F.
SECTION TX.105.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ¤ 249
64
1+ 3√
x
64
1
64
x1/3
39. √ dx =
+ dx = x −1/2 + x (1/3) − (1/2) 64
dx = (x −1/2 + x −1/6 ) dx
1 x 1 x1/2 x 1/2 1
1
64
=
2x 1/2 + 6 5 x5/6 =
16 + 192
5 − 2+
6
5 =14+
186
= 256
5 5
1
√
1/ 3
t 2
√
− 1
1/ 3
41.
0 t 4 − 1 dt = t 2
√
− 1
1/ 3
0 (t 2 +1)(t 2 − 1) dt = 1
0 t 2 +1 dt = arctan t 1/ √
3
=arctan 1/ √
3 − arctan 0
0
= π − 0= π 6 6
2
43. (x − 2 |x|) dx = 0
[x − 2(−x)] dx + 2
[x − 2(x)] dx = 0
3xdx+ 2
(−x) dx =3 1
x2 0
− 1
x2 2
−1 −1 0 −1 0 2 −1 2 0
=3
0 − 1 2 − (2 − 0) = −
7
= −3.5
2
45. The graph shows that y = x + x 2 − x 4 has x-intercepts at x =0and at
x = a ≈ 1.32. So the area of the region that lies under the curve and above the
x-axis is
a (x + 0 x2 − x 4 ) dx = 1
2 x2 + 1 3 x3 − 1 x5 a
5 0
= 1
2 a2 + 1 3 a3 − 1 a5 − 0 ≈ 0.84
5
47. A = 2
0 2y − y
2
dy = y 2 − 1 y3 2
=
3
4 − 8 0 3 − 0=
4
3
49. If w 0 (t) is the rate of change of weight in pounds per year, then w(t) represents the weight in pounds of the child at age t. We
know from the Net Change Theorem that 10
w 0 (t) dt = w(10) − w(5), so the integral represents the increase in the child’s
5
weight (in pounds) between the ages of 5 and 10.
51. Since r(t) istherateatwhichoilleaks,wecanwriter(t) =−V 0 (t),whereV (t) is the volume of oil at time t. [Note that the
minus sign is needed because V is decreasing, so V 0 (t) is negative, but r(t) is positive.] Thus, by the Net Change Theorem,
120
r(t) dt = − 120
V 0 (t) dt = − [V (120) − V (0)] = V (0) − V (120), which is the number of gallons of oil that leaked
0 0
from the tank in the first two hours (120 minutes).
53. By the Net Change Theorem, 5000
1000 R0 (x) dx = R(5000) − R(1000), so it represents the increase in revenue when
production is increased from 1000 units to 5000 units.
55. In general, the unit of measurement for b
f(x) dx is the product of the unit for f(x) and the unit for x. Sincef(x) is
a
measured in newtons and x is measured in meters, the units for 100
f(x) dx are newton-meters. (A newton-meter is
0
abbreviated N·m and is called a joule.)
57. (a) Displacement = 3
(3t − 5) dt = 3
0 2 t2 − 5t 3
= 27 − 15 = − 3 m
0 2 2
(b) Distance traveled = 3
|3t − 5| dt = 5/3
(5 − 3t) dt + 3
(3t − 5) dt
0 0 5/3
= 5t − 3 t2 5/3
+ 3
2 0 2 t2 − 5t 3
= 25 − 3 · 25 + 27 − 15 − 3 · 25 − 25
5/3 3 2 9 2 2 9 3 =
41
m 6
59. (a) v 0 (t) =a(t) =t +4 ⇒ v(t) = 1 2 t2 +4t + C ⇒ v(0) = C =5 ⇒ v(t) = 1 2 t2 +4t +5m/s
(b) Distance traveled = 10
|v(t)| dt = 10
1
0 0 2 t2 +4t +5 dt = 10 1
0 2 t2 +4t +5 dt = 1
6 t3 +2t 2 +5t 10
0
= 500
3 +200+50=4162 m 3