Pre-Calc Homework Solutions 256

Pre-Calc Homework Solutions 256 - 256 Section 6.3 (c) Using...

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Unformatted text preview: 256 Section 6.3 (c) Using the result from part (b): Let u x3 3x 2 dx x3 ex x 3e x (x 3 (d) x n e cos t dt C C 2e tcos t for t 25. First, we evaluate e Let u du e t t cos t dt. dv cos t dt sin t t dv v e x dx ex e t du x 3 e x dx e t t dt v sin t e dv 3x 2 e x dx 3(x 2 3x 2 2x 6x ... 2)e x 6)e x ( 1)n 2 n cos t dt e t e sin t dt C C C Let u du sin t dt cos t t e t dt t v d n x dx d n x dx 2 1 2 d n x x e dx n e cos t dt 2 e tcos t dt e tcos t dt t e sin t e t(sin t 1 t e (sin t 2 e cos t cos t) cos t) or [x n ... nx n n(n 1 1)x n 1)!x ( 1)n (n ( 1)n(n!)]e x C (e) Use mathematical induction or argue based on tabular integration. Alternately, show that the derivative of the answer to Now we find the average value of y 0 t 2 . 1 2 1 2 0 2 0 part (d) is x ne x: 2e tcos t dt e tcos t dt [x n 2 cos t) 0 d xn dx Average value nx n 1 n(n 1 1)x n 2 ... ( 1)n nx n 1 (n!)x ( 1)nn! e x 2 C n(n 1)x n 1 e t(sin t 2 1 [e 2 ( 1) 2 1 2 e 2 ... d dx ( 1)n 1(n!)x nx n 1 ( 1)nn!]e x n(n 1)x n 2 e0( 1)] ex [x n ... [xn nx n ... ( 1)n 1(n!)x 1 ( 1)nn!] 2 0.159 dv e x dx ex n(n 1)x n 26. (a) Let u du xe x dx x dx xe x xe x (x ( 1)n 1(n!)x [nx n n(n 1 ( 1)nn!]e x 1)x n 2)x n 3 2 v e x dx ex 1)e x n(n 1)(n C C 27. Let w ... x ne x ( 1)n 1n!]e x x. Then dw x dx w dw dx 2 x , so dx 2 x dw 2w dw. (b) Using the result from part (a): Let u du x 2 e x dx x2 2x dx x2 ex x 2e x (x 2 dv v e x dx ex sin Let u (sin w)(2w dw) dv v sin w dw cos w 2 w sin w dw du 2x e x dx 2(x 2x 1)e x 2)e x C C w sin w dw w cos w w cos w cos w dw sin w C sin x dx 2 w sin w dw 2w cos w 2 x cos 2 sin w x 2 sin C x C ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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