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FDWKCalcSM_ch9 - Section 9.1 377 55 Continued xe x dx = xe...

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Section 9.1 377 55. Continued xe dx xe e dx xe e C x x x x x = − + = − + Area = = + →∞ →∞ lim lim b x x b b b b xe e be e 0 1 1 1 1 1 = − + = − + = →∞ →∞ lim lim b b b b b e e 56. (a) xe dx x 2 0 (b) lim b x b xe dx →∞ 2 0 (c) Note that xe dx x 2 can be found by parts: xe dx x e e dx x x x = 2 2 2 2 2 ( ) ( ) = − + 2 4 2 2 xe e C x x . Area = lim lim b x b b x x b xe dx xe e →∞ →∞ = 2 0 2 2 0 2 4 = + ( ) = →∞ lim . b b b be e 2 4 0 4 4 2 2 57. (a) π π x dy dy y 2 2 0 0 1 = + ( ) (b) lim ( ) b b dy y →∞ + π 1 2 0 (c) Volume = + = + →∞ →∞ lim ( ) lim ( ) b b b b dy y y π π 1 1 2 1 0 0 = + + = →∞ lim b b π π 1 1 1 58. Note that xe dx x can be found by parts: xe dx x e e dx xe e C x x x x x = = − + ( ) ( ) . So xe dx xe dx xe e x k x k x x k k →∞ →∞ = = lim lim 0 0 0 = + →∞ lim k k k k e e 1 1 By L’Hopital’s rule, lim lim . k k k k k e e →∞ →∞ = = 1 0 Therefore, xe dx k e e x k k k →∞ = + lim 1 1 0 = + = 0 0 1 1 The integral converges to 1. Chapter 9 Infinite Series Section 9.1 Power Series (pp. 473– 483) Exploration 1 Finding Power Series for Other Functions 1. 1 2 3 + + + − + x x x x n ( ) . 2. x x x x x n n + + + − + + 2 3 4 1 1 ( ) . 3. 1 2 4 8 2 2 3 + + + + + + x x x x n ( ) . 4. 1 1 1 1 1 1 2 3 + + + − + ( ) ( ) ( ) ( ) ( ) . x x x x n n This geometric series converges for –1 < x 1 < 1, which is equivalent to 0 < x < 2. The interval of convergence is (0, 2). 5. 1 3 1 3 1 1 3 1 1 3 1 1 3 2 3 + + + ( ) ( ) ( ) x x x + n n x ( ) . 1 This geometric series converges for –1 < x 1 < 1, which is equivalent to 0 < x < 2. The interval of convergence is (0, 2). Exploration 2 Finding a Power Series for tan –1 x 1. 1 1 2 4 6 2 + + + − + x x x x n n ( ) . 2. tan = + 1 2 1 1 x t dt x 0 = + + + − + = + ( ( ) ) 1 1 3 5 2 4 0 6 2 3 5 t t t t dt t t t t x n n 7 2 1 0 3 5 7 1 2 1 3 5 + + − + + = + + ( ) n n x t n x x x x x n n n 7 2 1 7 1 2 1 + + − + + ( ) . 3. The graphs of the first four partial sums appear to be converging on the interval ( 1, 1).
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378 Section 9.1 4. When x = 1, the series becomes 1 1 3 1 5 1 7 1 2 1 + + + + + ( ) . n n This series does appear to converge. The terms are getting smaller, and because they alternate in sign they cause the partial sums to oscillate above and below a limit. The two calculator statements shown below will cause the succes- sive partial sums to appear on the calculator each time the ENTER button is pushed. The partial sums will appear to be approaching a limit of π /4 (which is tan ( )), 1 1 although very slowly.
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