Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 7 REVIEW ¤ 349
71.
(c)Ifwewanttheerrortobelessthan0.00001,wemusthave|E S| ≤ 3.8π5
180n 4 ≤ 0.00001,
so n 4 ≥
3.8π 5
≈ 646,041.6 ⇒ n ≥ 28.35. Sincen must be even for Simpson’s Rule, we must have n ≥ 30
180(0.00001)
to ensure the desired accuracy.
x 3
x 5 +2 ≤ x3
x 5 = 1 x 2 for x in [1, ∞). ∞
convergent by the Comparison Theorem.
73. For x in 0, π 2
1
1
∞
x dx is convergent by (7.8.2) with p =2> 1. Therefore, 2
, 0 ≤ cos 2 x ≤ cos x.Forx in π
2 ,π , cos x ≤ 0 ≤ cos 2 x.Thus,
area = π/2
(cos x − cos 2 x) dx + π
0 π/2 (cos2 x − cos x) dx
= sin x − 1 x − 1 sin 2x π/2
2 4
+ 1
x + 1 sin 2x − sin x π
=
0 2 4
1 − π π/2 4 − 0 + π − π
− 1 2 4
=2
75. Using the formula for disks, the volume is
V = π/2
π [f(x)] 2 dx = π π/2
(cos 2 x) 2 dx = π π/2
1 (1 + cos 0 0 0 2 2x)2 dx
= π 4
π/2
0
(1 + cos 2 2x +2cos2x) dx = π 4
= π 4
3
2 x + 1 2
77. By the Fundamental Theorem of Calculus,
1 sin 4x +2 1
sin 2x π/2
= π 4 2 0 4
π/2
0 1+
1
3π
4
2 (1 + cos 4x)+2cos2x dx
+ 1
8 · 0+0 − 0 = 3π2
16
1
x 3
dx is
x 5 +2
∞
0
f 0 (x) dx = lim
t→∞
t
0 f 0 (x) dx = lim
t→∞
[f(t) − f(0)] = lim
t→∞
f(t) − f(0) = 0 − f(0) = −f(0).
79. Let u =1/x ⇒ x =1/u ⇒ dx = −(1/u 2 ) du.
∞
0
∞
Therefore,
0
ln x
0
1+x dx = 2
∞
ln x
∞
1+x dx = − 2
ln (1/u)
− du 0
− ln u
0
=
1+1/u 2 u 2 ∞ u 2 +1 (−du) = ∞
0
ln x
dx =0.
1+x2
ln u
∞
1+u du = − 2
0
ln u
1+u 2 du