Pre-Calc Homework Solutions 338

# Pre-Calc Homework Solutions 338 - 338 Section 8.3 0 23...

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Unformatted text preview: 338 Section 8.3 0 23. Integrate u du d e d 0 e d by parts. dv v e lim b b 25. 0 e x dx lim b- e x dx 0 0 b b e lim b- x dx ex 0 e d e e x dx e x e x dx e x lim (1 b eb) 1 b- x b 0 dx lim b 0 dx 1) lim b e e d 0 e e d 0 e C 0 e d lim ( e b b 1 lim b e e b e eb) c x dx 1 1 2 lim ( 1 b be b ce b 1 lim c c Note that lim be b b lim c c ec 26. Integrate x ln x dx by parts. u du ln x 1 dx x lim c 1 ec 0 and lim e b lim e c 0. dv v 1 2 x ln x 2 1 x dx 1 2 x 2 1 x dx 2 1 2 x ln x 2 1 2 x 4 24. Integrate 2e u du 2e 2 sin 2 cos sin d d sin d by parts. dv v 2e e e d x ln x dx 1 C x ln x 0 lim b0 b x ln x dx 1 2 x ln x 2 1 4 1 2 x 4 1 b sin 2e cos d lim b0 Integrate 2e u du 2e Thus, 2e sin d 2 cos 2 sin cos d by parts. dv e e d lim b0 1 2 b ln b 2 1 2 b 4 d v 2e 1 4 Note that lim b 2 ln b b0 lim b0 ln b 1/b 2 lim b0 1/b 2/b 3 cos d cos 2e sin d lim b0 /2 b2 2 0. b 27. 0 tan d b /2 0 cos lim sin d b 2e 2 2e 2e 2e 0 sin 2e e lim b 0 2e sin sin b cos 2e e sin sin 2e cos cos d sin C1 C d lim b /2 ln cos 0 sin sin sin d d d lim [ ln cos b b /2 0] The integral diverges. b 2e e b lim b b e e b cos 0 lim ( e sin b cos b 1) 1 ...
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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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