Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

108 D CHAPTER 11 PROBLEMS PLUS
15. If Lis the length of a side of the equilateral triangle, then the area is A= ~L · 4L = 4£ 2 and so £ 2 =faA.
Let r be the radius of one of the circles. When there are n rows of circles, the figure shows that
L
L = v'3r+r+(n-2)(2r)+r+v'3r=r(2n-2+2../3),sor = ( y'3 )'
2 n+ 3-1
The number of circles is 1 + 2 + · · · + n = n( n 2
+ 1 ) , and so the total area of the circles is
n(n + 1) 2 n(n + 1) £ 2
An = 1fT = · 7r 2
2 2 4(n+v'3- 1)
n(n+ 1) 4A/J3 n(n+1) 1rA
= 2 1f 4(n+v'3 - 1) 2 = {n+../3-1) 2 2../3 =>
An n(n + 1) 1r
A = (n+../3-1) 2 2.;3
1 + 1/n 1r 1r
= -> -- asn ->oo
[1+(J3-1)/n) 2 2v'3 2../3
r 2r 2r
~-----L------~
17. As in Section 11.9 we have to integrate the function x"' by integrating series. Writing x"' = ( e 1 " "')"' = e"' 1 ""' and using the
Maclaurin series for e"', we have x"' = ( eln "')"' = e"' 1 " "' = E ( x In x) n 1
= E x" (~ x) n . As with power series, we can
n = O n. n=O n.
1
·1 oo 11
x"(ln xt oo 1 1 1
dx = 2: I
integrate this series term-by-term: x"' dx = 2:
1
0 n=O 0 n. n=O n. 0
n(ln x)"- 1 x"+ 1
withu = (lnx)",dv =x" dx,sodu= · dxand v = --: .
x n+ 1
f \"(lnx)"dx = lim /. 1 x"(ln xfdx= lim [x"+l (lnx)"]
} 0 t-o+ t . t-o+ n + 1 t t-o+ it n +
= o- _ n _ t x"(Inxf- 1 dx
n+ 1 } 0
(where !'Hospital's Rule was used to help evaluate the first limit). Further integration by parts gives
1
x"(lnxf dx. We integrate by parts
- lim ( 1 ___!!:__ 1
x"(1nxf- 1 dx
(
1
x"(ln x)k dx = -~ 1
} 0 n + 1 0
1
x"(ln x)k-
1
dx and, combining these steps, we get
1x"(ln x)"dx = (-1)"n!11 x"dx= (-1)"n! =?
0
(n+1)" 0
· (n+ 1)n+l
1 "' d = ~ ..!_ 1 1 "(In ·)"' d = ~ ..!_ (- 1t n! = ~ (-1)" = ~ (- 1)"-1
X X L... I X X X L... I ( 1 ) +1 , 0 n=O n. 0 . n=O n. n + n n=O L... ( n + 1 " n=l n"
) +1 L... ·
00 x2n+l 1
19. By Table 1 in Section 11. 10, tan- 1 x = E ( - lt- 2
' --for JxJ < 1. In particular, for x = r.;• we
n=O n+ 1 v3
7r ( 1 )
00
(1/../3) 2 "+ 1 00
(1)" i 1
have 6 = tan-1 y'3 = n~O ( - .1t 2n + 1 = n~O ( -1)" 3 y'3 2n + 1' SO
6 00 (-1)" 00 (-1)" ( 00 ' (-1)" ) .
·rr - - - 2 3 - 2 3 1 =>
- ../3 n~O (2n + 1)3" - y'3 n~O (2n + 1)3" - y'3 + f1 (2n + 1)3n
00 (-1)" 7r
E =--1.
n=1 (2n + 1)3" 2 y'3
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