Explanation Given: The given function is ∫ 1 ∞ 1 x 1 3 + x 3 d x Calculation: You have to check the convergence of ∫ 1 ∞ 1 ( x + x 2 ) 1 3 d x using the comparison test The comparison test is given as: If f ( x ) and g ( x ) are two functions in interval [ a , b ) such that g ( x ) ≥ f ( x ) ≥ 0 then If ∫ a b g ( x ) converges then ∫ a b f ( x ) also converges If ∫ a b f ( x ) diverges then ∫ a b g ( x ) also diverges. To check the convergence of a given function you have to chose a function that is larger or smaller than the given function and then check the convergence or divergence of the chosen function and then according to the comparison test check the convergence of the original function. The given function is ∫ 1 ∞ 1 ( x + x 2 ) 1 3 d x . You can choose a function that is greater or smaller than the given function. Consider a function ∫ 1 ∞ 1 2 1 3 x 2 3 d x Now for x > 1 ; 1 2 1 3 x 2 3 < 1 ( x + x 2 ) 1 3 Therefore ∫ 1 ∞ d x 2 1 3 x 2 3 < ∫ 1 ∞ 1 ( x + x 2 ) 1 3 d x Now you have to show that ∫ 1 ∞ d x 2 1 3 x 2 3 is convergent in the given interval...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.7, Problem 74E

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Chapter 8.7, Problem 73E

Chapter 8.7, Problem 75E

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Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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Whether or not the integral converges using comparison test:

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Chapter 8 Solutions