Explanation Given information : ∫ − ∞ ∞ x e − | x | d x Formula Used: a m n = a m / n Negative exponent rule: 1 a m = a − m Power rule of integral: ∫ a b x n ⋅ d x = [ x n + 1 n + 1 ] a b Concept Involved: If ∫ a t f ( x ) ⋅ d x exists for every t > a , then ∫ a ∞ f ( x ) ⋅ d x = lim t → ∞ ∫ a t f ( x ) ⋅ d x provided the limit exists and is finite. If ∫ t b f ( x ) ⋅ d x exists for every t < b , then ∫ − ∞ b f ( x ) ⋅ d x = lim t → − ∞ ∫ t b f ( x ) ⋅ d x provided the limit exists and is finite. If a > 0 then ∫ a ∞ 1 x p d x is convergent if p > 1 and divergent if p ≤ 1 Calculation: Description Steps Rewriting the integral by eliminating the absolute value ∫ − ∞ ∞ x e − | x | d x = ∫ − ∞ 0 x e − ( − x ) d x + ∫ 0 ∞ x e − x d x = ∫ − ∞ 0 x e x d x + ∫ 0 ∞ x e − x d x Convert the integral to a limit ∫ − ∞ 0 x e x d x + ∫ 0 ∞ x e − x d x = lim t → − ∞ ∫ t 0 x e x d x + lim t → ∞ ∫ 0 t x e − x d x Using the integration by parts to rewrite the integral L e t u = x d u = d x d v = e x d x v = e x → 1 s t I n t e g r a l lim t → − ∞ ∫ t 0 x e x d x = lim t → − ∞ [ x e x ] t 0 − lim t → − ∞ ∫ t 0 e x d x Using the integration by parts to rewrite the integral L e t u = x d u = d x d v = e − x d x v = − e − x → 2 n d I n t e g r a l lim t → ∞ ∫ 0 t x e − x d x = lim t → ∞ [ − x e − x ] 0 t − lim t → ∞ ∫ 0 t − e − x d x We need to substitute the result of integral and put everything together lim t → − ∞ ∫ t 0 x e x d x + lim t → ∞ ∫ 0 t x e − x d x = lim t → − ∞ [ x e x ] t 0 − lim t → − ∞ ∫ t 0 e x d x + lim t → ∞ [ − x e − x ] 0 t − lim t → ∞ ∫ 0 t − e − x d x = lim t → − ∞ [ x e x − e x ] t 0 + lim t → ∞ [ − x e − x − e − x ] 0 t = lim t → − ∞ ( 0 ⋅ e 0 − e 0 ) − lim t → − ∞ ( t e t − e t ) + lim t → ∞ ( − t e − t − e − t ) − lim t → ∞ ( 0 ⋅ e 0 − e 0 ) x

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 7.7, Problem 27PS

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Tofind : the value of improper integral if it converges or show that it diverges.

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Chapter 7.7, Problem 26PS

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Chapter 7 Solutions

Calculus

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Tofind : the value of improper integral if it converges or show that it diverges.

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Chapter 7 Solutions