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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) → π/ 2. So the final answer is 1 6 ( π/ 2) = π 12 . (b) Let’s 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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This note was uploaded on 09/17/2011 for the course MAT 1332 taught by Professor Munteanu during the Winter '07 term at University of Ottawa.
 Winter '07
 MUNTEANU
 Improper Integrals, Integrals

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