Pre-Calc Homework Solutions 335

Pre-Calc Homework Solutions 335 - Section 8.3 Section 8.3...

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Unformatted text preview: Section 8.3 Section 8.3 Exercises 1. (a) The integral is improper because of an infinite limit of integration. (b) 0 335 4. (a) The integral is improper because of two infinite limits of integration. (b) 2x dx 2x dx 2x dx 2 1)2 (x2 1)2 (x 2 1)2 0 (x 0 2x dx 2x dx lim 2 1)2 (x 2 1)2 b b (x 0 dx x2 1 b lim b b 0 dx x2 1 1 0 1 b 0 lim tan lim (tan b x b 0) lim b b (x 2 1) (b 2 1 0 b lim [ 1 0 1) 1 1 2 The integral converges. (c) 2x dx (x 2 1)2 b 0 lim b b 2x dx (x 2 1)2 lim ( x 2 lim [ (b2 b 1) 1) 0 1 1 b 0 2 2x dx (x 2 1)2 1] 1 2. (a) The integral is improper because the integrand has an infinite discontinuity at x 0. 1 1 1 The integral converges. (b) 0 dx x 1 lim b0 b dx x 1 (c) 0 5. (a) The integral is improper because the integrand has an infinite discontinuity at 0. b) 2 ln 2 lim b0 2 x b lim (2 b0 2 (b) 0 x 2 1/x ln 2 e dx lim b0 b x e1/x 2 1/x e dx ln 2 b The integral converges. (c) 2 3. (a) The integral involves improper integrals because the integrand has an infinite discontinuity at x 0. 1 lim b0 b0 lim [ e1/ln 2 The integral diverges. (c) No value dx x1/3 e1/b] (b) 8 0 8 dx x1/3 dx x1/3 0 8 dx x1/3 b 8 1 0 6. (a) The integral is improper because the integrand has an infinite discontinuity at x 0. /2 /2 lim b0 dx x1/3 b 8 (b) 0 cot d lim b0 b /2 cot b d lim b0 3 2/3 x 2 3 2/3 b 2 1 b lim b0 cos d sin b lim b0 1 0 6 6 /2 lim b0 b0 ln sin ln sin b ) dx x1/3 lim b0 dx x1/3 1 b lim (0 The integral diverges. (c) No value 7. 1 lim b0 3 2/3 x 2 3 2 lim b0 3 2/3 b 2 dx x 1.001 b 3 2 1 8 lim b 1 dx x1.001 lim 6 3 2 9 2 b b 1000 x 0.001 0.001 b 1 dx x1/3 lim ( 1000b 1000) 1000 The integral converges. (c) 9 2 ...
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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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