Solução_Calculo_Stewart_6e

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F. TX.10 CHAPTER 11 PROBLEMS PLUS ¤ 511 (f) From part (c) we have π =16arctan 1 − 4arctan 1 5 239 , so from parts (d) and (e) we have 16(0.197395560) − 4(0.004184077)
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 .
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Solução_Calculo_Stewart_6e

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