Explanation Given: Let J n = ∫ 0 ∞ x n e − α x d x n ≥ 1 α > 0 Calculation: Consider the given integration : J n = ∫ 0 ∞ x n e − α x d x Let u = x n and d v = e − α x Then, d u = n x n − 1 d x and v = − 1 α e − α x Using integration by parts. Then J n = ∫ 0 ∞ x n e − α x d x = − 1 α [ x n e − α x ] 0 ∞ + n α ∫ 0 ∞ x n − 1 e − α x d x = lim R → ∞ − 1 α [ x n e − α x ] 0 R + n α J n − 1 = lim R → ∞ − 1 α R n e − α R + 0 + n α J n − 1 = lim R → ∞ − R n α e α R + n α J n − 1 = lim R → ∞ − n R n − 1 α e α R + n α J n − 1 { using L'hopital’s rule} = lim R → ∞ − n ( n − 1 ) ( n − 2 ) ...... ( 3 ) ( 2 ) ( 1 ) α e α R + n α J n − 1 { using L'hopital’s rule} = 0 + n α J n − 1 Thus, J n = n α J n − 1 Now, to prove J 0 = 1 α . Again, consider the given integration: J n = ∫ 0 ∞ x n e − α x d x .Then J 0 = ∫ 0 ∞ x 0 e − α x d x = lim R → ∞ ∫ 0 R e − α x d x = lim R → ∞ [ − 1 α e − α x ] 0 R = lim R → ∞ − 1 α ( e − α R − e − α × 0 ) = − 1 α ( 0 − 1 ) = 1 α Thus, J 0 = 1 α Now, to compute the value of J 4

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 7.7, Problem 93E

To determine

To prove:

the given function Jn=nαJn−1 and J0=1α. Again, to compute J4. Also, prove Jn=nαn+1.

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Chapter 7.7, Problem 92E

Chapter 7.7, Problem 94E

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Chapter 7 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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To prove: the given function J n = n α J n − 1 and J 0 = 1 α . Again, to compute J 4 . Also, prove J n = n α n + 1 .

Solution Summary: The author explains how to prove the given function J_n=

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To prove:

the given function Jn=nαJn−1 and J0=1α. Again, to compute J4. Also, prove Jn=nαn+1.

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Chapter 7 Solutions

To prove: the given function J n = n α J n − 1 and J 0 = 1 α . Again, to compute J 4 . Also, prove J n = n α n + 1 .

Solution Summary: The author explains how to prove the given function J_n=

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To prove: the given function Jn=nαJn−1 and J0=1α. Again, to compute J4. Also, prove Jn=nαn+1.

Want to see the full answer?

Chapter 7 Solutions

To prove:

the given function Jn=nαJn−1 and J0=1α. Again, to compute J4. Also, prove Jn=nαn+1.