LA-8.7-improper_integrals

LA-8.7-improper_inte - MthSc 108-Learning Activity Section 8.7 Improper Integrals 1 1 x e dx x Solution e x x 1 dx = lim a → e x x dx a 1

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Unformatted text preview: MthSc 108-Learning Activity Section 8.7: Improper Integrals 1. 1 x e dx x- ∫ Solution: e- x x 1 ∫ dx = lim a → + e- x x dx a 1 ∫ Dealing with only the indefinite integral, let u = - x ⇒ du = - 1 2 x dx , so e- x x dx ∫ = - 2 e u du ∫ = - 2 e u + C = - 2 e- x + C ⇒ lim a → o + e- x x dx = lim a → o +- 2 e- x a 1 = a 1 ∫ lim a → o +- 2 e- 1 + 2 e- a = - 2 e + 2 e = 2- 2 e So the improper integral converges to 2- 2 e . 2. 2 2 2 1 dt t ∞- ∫ Solution 1: 2 t 2- 1 dt 2 ∞ ∫ = lim b →∞ 2 t 2- 1 2 b ∫ = lim b →∞ 2 t- 1 ( ) t + 1 ( ) dt 2 b ∫ Dealing with only the indefinite integral, we must find a PFD. 2 t- 1 ( ) t + 1 ( ) = A t- 1 + B t + 1 ⇒ 2 = A + B ( ) t + A- B ⇒ A + B = A- B = 2 ⇒ A = 1, B = - 1 ⇒ 2 t- 1 ( ) t + 1 ( ) ∫ = 1 t- 1 dt ∫- 1 t + 1 dt ∫ = ln t- 1- ln t + 1 + C = ln t- 1 t + 1 + C ⇒ lim b →∞ 2 t- 1 ( ) t + 1 ( ) dt 2 b ∫ = lim b →∞ ln t- 1 t + 1 2 b = lim b →∞ ln b- 1 b...
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This note was uploaded on 01/30/2011 for the course MTHSC 108 taught by Professor Any during the Spring '08 term at Clemson.

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LA-8.7-improper_inte - MthSc 108-Learning Activity Section 8.7 Improper Integrals 1 1 x e dx x Solution e x x 1 dx = lim a → e x x dx a 1

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