Solução_Calculo_Stewart_6e

F.
512 ¤ CHAPTER 11 PROBLEMS PLUS
TX.10
(b) The total volume is
π ∞
0
e −x/5 sin 2 xdx= ∞
n=1
V n = 250π
101
∞
[e −(n−1)π/5 − e −nπ/5 ]= 250π [telescoping sum].
101
n=1
Another method: If the volume in part (a) has been written as V n = 250π
101 e−nπ/5 (e π/5 − 1), then we recognize ∞
as a geometric series with a = 250π
101 (1 − e−π/5 ) and r = e −π/5 .
15. If L is the length of a side of the equilateral triangle, then the area is A = 1 2 L · √3
2 L = √ 3
4 L2 and so L 2 = 4
Let r be the radius of one of the circles. When there are n rows of circles, the figure shows that
L = √ 3 r + r +(n − 2)(2r)+r + √ 3 r = r 2n − 2+2 √ 3 ,sor =
The number of circles is 1+2+···+ n =
A n =
=
n(n +1)
πr 2 n(n +1)
= π
2
2
n(n +1)
2
A n
A = n(n +1) π
√ 2
n + 3 − 1 2 √ 3
=
π
L
2 n + √ 3 − 1 .
n(n +1)
, and so the total area of the circles is
2
L 2
4 n + √ 3 − 1 2
4A/ √ 3
4 n + √ 3 − 1 2 = n(n +1) πA
√ 2
n + 3 − 1 2 √ 3
1+1/n π
√ 2
1+ 3 − 1 /n 2 √ 3 → π
2 √ 3
as n →∞
⇒
√
3
A.
V n
n=1
17. As in Section 11.9 we have to integrate the function x x by integrating series. Writing x x =(e ln x ) x = e x ln x and using the
Maclaurin series for e x ,wehavex x =(e ln x ) x = e x ln x
= ∞ (x ln x) n
= ∞ x n (ln x) n
. As with power series, we can
n!
n!
integrate this series term-by-term:
1
with u =(lnx) n , dv = x n dx,sodu =
1
0
x n (ln x) n dx = lim
0
t→0 + 1
t
=0− n
n +1
x x dx = ∞
n=0
n(ln x)n−1
x
1
0
n=0
x n (ln x) n
dx = ∞
n!
dx and v = xn+1
n +1 :
n=0
n=0
1
1
n!
x
x n (ln x) n n+1
1
dx = lim
t→0 + n +1 (ln x)n t
1
0
x n (ln x) n−1 dx
0
x n (ln x) n dx. We integrate by parts
1
− lim
t→0 + t
(where l’Hospital’s Rule was used to help evaluate the first limit). Further integration by parts gives
1
0
1
0
1
0
x n (ln x) k dx = −
k
n +1
1
x n (ln x) n dx = (−1)n n!
(n +1) n 1
x x dx = ∞
n=0
1
n!
1
0
0
0
x n (ln x) k−1 dx and, combining these steps, we get
x n (ln x) n dx = ∞
x n dx = (−1)n n!
(n +1) n+1 ⇒
n=0
1 (−1) n n!
n! (n +1) = ∞
n+1
n=0
(−1) n
(n +1) = ∞
n+1
n=1
n
n +1 xn (ln x) n−1 dx
(−1) n−1
n n .