Explanation Given information: ∫ 1 / 2 ∞ d x x ( ln x ) 3 Formula: ∫ a b 1 x n d x = [ x − n + 1 − n + 1 ] a b Calculation: Test the below expression using the direct comparison test. 1 x ( ln x ) 3 Write the function f ( x ) as below. f ( x ) = 1 x ( ln x ) 3 Write the function g ( x ) as below. g ( x ) = 1 x The following term cannot be tested using this direct comparison test as the following condition exists as given below. f ( x ) < g ( x ) on [ 1 2 , 1 ] f ( x ) > g ( x ) on [ 1 , ∞ ] This does not satisfy the condition of continuity within the given limit [ 1 2 , ∞ ] and the condition 0 ≤ f ( x ) ≤ g ( x ) . The discontinuity exists in the function at the value of 1 for x , where it will reach infinity. Hence, use the limit comparison test. Test the below expression using the limit comparison test. 1 x ( ln x ) 3 Apply also the following test for the convergence as below. lim x → ∞ f ( x ) g ( x ) = lim x → ∞ 1 x ( ln x ) 3 1 x = lim x → ∞ 1 ( ln x ) 3 = 1 ∞ = 0 Therefore, the term is not within the range ( 0 , ∞ ) . The value should fall in between the range 0 < x < ∞ and the functions f ( x ) and g ( x ) are not continuous. Hence, the integral term f ( x ) cannot be tested with this limit comparison test. Express the integral f ( x ) having discontinuity at x equal to 1 as below. ∫ 1 / 2 ∞ d x x ( ln x ) 3 = lim b → 1 − ∫ 1 / 2 b d x x ( ln x ) 3 + lim b → ∞ ∫ 1 b d x x ( ln x ) 3 (1) Assume the term u as ln x

Student Solutions Manual Single Variable For University Calculus: Early Transcendentals

4th Edition

ISBN: 9780135166130

Author: Joel R. Hass, Maurice D. Weir, George B. Thomas Jr., Przemyslaw Bogacki

Publisher: PEARSON

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Chapter 8.7, Problem 37E

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Whether an integral converges or diverges by testing it either using the direct comparison test, or the Limit comparison test by adopting any one of the appropriate method.

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Chapter 8.7, Problem 36E

Chapter 8.7, Problem 38E

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Student Solutions Manual Single Variable For University Calculus: Early Transcendentals

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