Explanation Given: The improper integral is ∫ 0 1 1 x p d x . Definition used: Improper integral “If f is continuous on ( a , b ] with lim c → a + f ( x ) = ± ∞ , then ∫ a b f ( x ) d x = lim c → a + ∫ a b f ( x ) d x .” Theorem used: Fundamental theorem of calculus “If f is continuous on [ a , b ] and F is anti-derivative of f , then ∫ a b f ( x ) d x = F ( b ) − F ( a ) .” Calculation: The improper integral is ∫ 0 1 1 x p d x . The integrand f ( x ) = 1 x p is defined on ( 0 , 1 ] . In order to find the value p such that ∫ 0 1 1 x p d x is convergence, first calculate the integral with p > 0 and p ≠ 1 . Use the definition and theorem mentioned above to calculate the value of the improper integral. That is, ∫ 0 1 1 x p d x = lim c → 0 + ∫ c 1 1 x p d x = lim c → 0 + [ x 1 − p 1 − p ] c 1 = 1 1 − p lim c → 0 + [ ( 1 ) 1 − p − c 1 − p ] = 1 1 − p lim c → 0 + [ 1 − c 1 − p ] To find the value of p, following three cases are required. Case 1: when p > 1 Since p > 1 , this implies p − 1 > 0 and use it in the above calculated integral

MyLab Math with Pearson eText -- Standalone Access Card -- for Calculus: Early Transcendentals (3rd Edition)

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ISBN: 9780134856926

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 8.9, Problem 94E

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To find: The value of p such that the improper integral ∫011xpdx exists.

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Chapter 8.9, Problem 93E

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MyLab Math with Pearson eText -- Standalone Access Card -- for Calculus: Early Transcendentals (3rd Edition)

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The value of p such that the improper integral ∫ 0 1 1 x p d x exists.

Solution Summary: The author explains the value of p such that the improper integral exists, and the integrand is defined on (0,1).