Explanation Given: The integral function is ∫ 0 0.6 x 2 9 − 25 x 2 d x The region lies between x = 0 and x = 0.6 . Calculation: Consider x = 3 5 sin θ (1) Differentiate both sides of the Equation (1). d x = 3 5 cos θ d θ Show the integral function as follows: ∫ 0 0.6 x 2 9 − 25 x 2 d x (2) Substitute ( 3 5 sin θ ) for x and ( 3 5 cos θ d θ ) for dx in Equation (2). ∫ 0 0.6 x 2 9 − 25 x 2 d x = ∫ 0 0.6 ( 3 5 sin θ ) 2 9 − 25 ( 3 5 sin θ ) 2 ( 3 5 cos θ d θ ) = ∫ 0 0.6 9 25 sin 2 θ 3 5 cos θ 9 − 25 × 9 25 sin 2 θ d θ = 27 125 ∫ 0 0.6 sin 2 θ cos θ 9 − 9 sin 2 θ d θ = 27 125 ∫ 0 0.6 sin 2 θ cos θ 9 ( 1 − sin 2 θ ) d θ = 27 125 ∫ 0 0.6 sin 2 θ cos θ 3 cos 2 θ d θ = 27 125 × 3 ∫ 0 0.6 sin 2 θ cos θ cos θ d θ = 9 125 ∫ 0 0.6 sin 2 θ d θ = 9 125 ∫ 0 0.6 1 2 ( 1 − cos 2 θ ) d θ = 9 250 ∫ 0 0

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

2nd Edition

ISBN: 9781285101552

Author: James Stewart

Publisher: Cenage

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Chapter 6.2, Problem 53E

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To evaluate: The integral function ∫00.6x29−25x2 dx.

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Chapter 6.2, Problem 61E

Chapter 6.2, Problem 56E

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Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

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The integral function ∫ 0 0.6 x 2 9 − 25 x 2 d x .

Solution Summary: The author evaluates the integral function with a value of 0.05652.

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To evaluate: The integral function ∫00.6x29−25x2 dx.

Want to see the full answer?

Chapter 6 Solutions