ass 3 answ - Professor Jason Levy, University of Ottawa,...

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Unformatted text preview: Professor Jason Levy, University of Ottawa, MAT 1332C, Winter 2011 Assignment 3 Solutions 1. For each of the following improper integrals, determine whether it converges, and evaluate it if if does. (a) Z t 2 4 + t 6 dt (b) Z 1 e- t sin(2 t ) dt Solution: (a) We express the improper integral as a limit, and use two substitu- tions, u = t 3 ( du = 3 t 2 dt ), and then v = u/ 2 ( dv = du/ 2): Z t 2 4 + t 6 dt = lim T Z T t 2 4 + t 6 dt = lim T Z T 3 du/ 3 4 + u 2 = lim T 1 3 * 4 Z T 3 du 1 + ( u/ 2) 2 = lim T 1 12 Z T 3 / 2 2 dv 1 + ( v ) 2 = lim T 1 6 arctan( v ) T 3 / 2 = lim T 1 6 (arctan( T 3 / 2)- 0) . As T , T 3 / 2 also , and arctan( T 3 / 2) / 2. So the final answer is 1 6 ( / 2) = 12 . (b) Lets first evaluate the indefinite integral, as this is a little tricky: Write I for R e- t sin(2 t ) dt . Then I = Z e- t sin(2 t ) dt =- e- t sin(2 t ) + 2 Z e- t cos(2 t ) dt by parts, with u = sin(2 t ) , dv = e- t dt =- e- t sin(2 t )- 2 e- t cos(2 t )- 4 Z e- t sin(2 t ) dt by parts, with u = cos(2 t ) , dv = e- t dt =- e- t sin(2 t )- 2 e- t cos(2 t )- 4 I, Notes: (1) Really the equation should be I =- e- t sin(2 t )- 2 e- t cos(2 t )- 4 I + C , C any constant. The + C is because of the inherent ambiguity in an indefinite...
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ass 3 answ - Professor Jason Levy, University of Ottawa,...

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