Solução_Calculo_Stewart_6e

F.
254 ¤ CHAPTER 5 INTEGRALS
TX.10
75. First Figure Let u = √ x,sox = u 2 and dx =2udu.Whenx =0, u =0;whenx =1, u =1. Thus,
A 1 = 1
0 e√x dx = 1
0 eu (2udu)=2 1
0 ueu du.
Second Figure A 2 = 1
0 2xex dx =2 1
0 ueu du.
Third Figure
Let u =sinx,sodu =cosxdx.Whenx =0, u =0;whenx = π 2 , u =1.Thus,
A 3 = π/2
0
e sin x sin 2xdx= π/2
0
e sin x (2 sin x cos x) dx = 1
0 eu (2udu)=2 1
0 ueu du.
Since A 1 = A 2 = A 3 , all three areas are equal.
77. The rate is measured in liters per minute. Integrating from t =0minutes to t =60minutes will give us the total amount of oil
that leaks out (in liters) during the first hour.
60
r(t) dt = 60
0 0 100e−0.01t dt [u = −0.01t, du = −0.01dt]
= 100 −0.6
e u (−100 du) =−10,000 e u −0.6
= −10,000(e −0.6 − 1) ≈ 4511.9 ≈ 4512 liters
0 0
79. The volume of inhaled air in the lungs at time t is
V (t)= t
f(u) du = t 1
sin 2π
u du = 2πt/5 1
sin v 5
0 0 2 5 0 2
= 5
4π
− cos v
2πt/5
0
= 5
4π
− cos
2π
5 t +1 = 5
4π
1 − cos
2π
5 t liters
2π dv substitute v = 2π 5 u, dv = 2π 5 du
81. Let u =2x. Thendu =2dx,so 2
f(2x) dx = 4
f(u) 1
du
= 1 4 f(u) du = 1 (10) = 5.
0 0 2 2 0 2
83. Let u = −x. Thendu = −dx,so
b f(−x) dx = −b
a
f(u)(−du) = −a
−a −b
f(u) du = −a
f(x) dx
−b
From the diagram, we see that the equality follows from the fact that we are
reflecting the graph of f, and the limits of integration, about the y-axis.
85. Let u =1− x. Thenx =1− u and dx = −du,so
1
0 xa (1 − x) b dx = 0
(1 − 1 u)a u b (−du) = 1
0 ub (1 − u) a du = 1
0 xb (1 − x) a dx.
87.
x sin x
1+cos 2 x = x · sin x
t
2 − sin 2 = xf(sin x),wheref(t) = . By Exercise 86,
x 2 − t2 π
0
x sin x
π
1+cos 2 x dx = xf(sin x) dx = π 2
0
π
0
f(sin x) dx = π 2
Let u =cosx. Thendu = − sin xdx.Whenx = π, u = −1 and when x =0, u =1.So
π π
2 0
sin x
1+cos 2 x dx = − π −1
2 1
du
1+u = π 1
2 2 −1
= π 2 [tan−1 1 − tan −1 (−1)] = π 2
π
0
sin x
1+cos 2 x dx
du
1+u = π
tan −1 u 1
2 2
−1
π
4 − − π
= π2
4 4