Explanation Given information: 1 1 − x 2 Formula: ∫ a b 1 c x + d d x = [ ln | c x + d | ] a b Calculation: Test the below expression using the direct comparison test. 1 1 − x 2 = 1 ( 1 − x ) ( 1 + x ) Write the function f ( x ) as below. f ( x ) = 1 ( 1 − x ) ( 1 + x ) 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 [ 0 , 0.618 ] f ( x ) > g ( x ) on [ 0.618 , 1 ] f ( x ) < g ( x ) on [ 1 , 2 ] This does not satisfy the condition of continuity within the given limit [ 0 , 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 1 − x 2 = 1 ( 1 − x ) ( 1 + x ) Apply also the following test for the convergence as below. lim x → ∞ f ( x ) g ( x ) = lim x → ∞ 1 ( 1 − x 2 ) 1 x = lim x → ∞ x ( 1 − x 2 ) Apply l’Hospital’s rule as the term lim x → ∞ x ( 1 − x 2 ) yields lim θ → ∞ ∞ ∞ . lim x → ∞ f ( x ) g ( x ) = lim x → ∞ 1 ( − 2 x ) = lim θ → ∞ 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 as f ( x ) having discontinuity at x equal to 1 as below. ∫ 0 2 d x ( 1 − x ) ( 1 + x ) = lim b → 1 − ∫ 0 b d x ( 1 − x ) ( 1 + x ) + lim b → 2 − ∫ 1 b d x ( 1 − x ) ( 1 + x ) (1) Write the integrand term in the form of partial fraction as below. 1 1 − x 2 = 1 ( 1 − x ) ( 1 + x ) = A ( 1 − x ) + B ( 1 + x ) Simplify it as below. 1 1 − x 2 = A ( 1 − x ) + B ( 1 + x ) (2) 1 = A ( 1 + x ) + B ( 1 − x ) 1 = x ( A − B ) + A + B (3) Equate the constant term of the equation (3) as below

University Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780321999580

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

Publisher: PEARSON

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Chapter 8.7, Problem 43E

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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 42E

Chapter 8.7, Problem 44E

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University Calculus: Early Transcendentals (3rd Edition)

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