Pre-Calc Homework Solutions 339

Pre-Calc Homework Solutions 339 - Section 8.3 sin 1 0 b 339...

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Unformatted text preview: Section 8.3 sin 1 0 b 339 28. On [0, ], 0 sin 0 1 , so 32. 0 dx 4 1 x 1 x b on [4, ) 1 dx x b d d lim b d 1 lim b 4 lim 2 b x 4 lim [2 b b 4] x 1 0 0 d b Since this integral diverges, the given integral diverges. lim b 2 0 33. 0 b 2 2 ) 1 1 x3 dx x3 1 1 on [1, x3 b ) lim ( 2 b lim b 1 dx x3 1 2 x 2 1 2 b 2 b 1 2 0 2 lim b lim b 1 2 1 2 Since this integral converges, the given integral converges. x2 x2 x2 0 Since this integral converges, the given integral converges. 2 1 2 29. 2xe 2xe 0 dx 0 2xe b dx x2 b 0 2xe dx x2 dx 34. 0 1 0 dx 1 dx 1 x 2 dx 1 b dx 1 x2 x) (1 x)] dx x)(1 x) 1 x) 2(1 x) 1 ln 1 2 b 0 x 2 0 x 2 1 dx lim b 0 2xe e x2 b2 lim b1 0 b lim b 1/2[(1 (1 1 2(1 lim [ e b 0 1] x2 0 1 lim b1 0 dx b 2xe x2 0 dx lim b b 2xe e x2 dx lim b1 1 ln 1 2 1 1 ln 2 1 1 1 ln 2 1 x x x b b x 0 lim b lim b b1 b2 lim [ 1 b e 1 lim b1 0 The integral converges. 4 Since this integral diverges, the given integral diverges. x 30. 0 e x x 4 dx lim b0 b e x dx 4 2 35. 0 1 dx 1 dx 1 x x 1 0 dx 1 b 0 2 dx 1 x x dx 1 1 lim b0 2e 2 x b lim b1 0 x b lim [ 2e b0 2e b] lim b1 ln 1 x 0 2e 2 2 lim ( ln 1 b1 b 0) The integral converges. 1 t dt 0 Since this integral diverges, the given integral diverges. 0 on [0, ]. 31. 0 1 sin t t dt b on (0, ] since sin t lim b0 t t lim b0 2 t b lim [2 b0 2 b] 2 Since this integral converges, the given integral converges. ...
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