Explanation Integral Formula used: ∫ d x x a 2 − x 2 = − 1 a sech − 1 ( x a ) + c , for 0 < x < a Calculation: Consider the integral ∫ 1 6 1 4 d t t 1 − 4 t 2 , ∫ 1 6 1 4 d t t 1 − 4 t 2 = ∫ 1 6 1 4 2 d t 2 t 1 − 4 t 2 Let x = 2 t . Differentiate with respect to t , d d t ( x ) = d d t ( 2 t ) d x d t = 2 d d t ( t ) d x d t = 2 d t = d x 2 Obtain the limits in term of x . Substitute t = 1 6 in x = 2 t , x = 2 × 1 6 = 1 3 Thus, the lower limit is 1 3 . Substitute t = 1 4 in x = 2 t , x = 2 × 1 4 = 1 2 Thus, the upper limit is 1 2 . Substitute x = 2 t and d t = d x 2 in ∫ 1 6 1 4 d t t 1 − 4 t 2 = ∫ 1 6 1 4 2 d t 2 t 1 − 4 t 2 and evaluate the integral in the variable t

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 6.10, Problem 62E

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To evaluate: The definite integral ∫1614dtt1−4t2.

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Chapter 6.10, Problem 61E

Chapter 6.10, Problem 63E

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Calculus: Early Transcendentals (2nd Edition)

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The definite integral ∫ 1 6 1 4 d t t 1 − 4 t 2 .

Solution Summary: The author evaluates the definite integral displaystyle 'underset'16overset

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To evaluate: The definite integral ∫1614dtt1−4t2.

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Chapter 6 Solutions