Explanation Given: ∫ − 3 1 + 16 x 2 d x Concept Used: ∫ a f ( x ) d x = a ∫ f ( x ) d x d d x x n = n x n − 1 d d x ( tan − 1 x ) = 1 1 + x 2 ∫ 1 1 + x 2 d x = tan − 1 x + C , where C is an arbitrary constant. Calculation: To solve the integrals ∫ − 3 1 + 16 x 2 d x First write it as ∫ − 3 1 + 16 x 2 d x = ∫ − 3 1 + ( 4 x ) 2 d x First use substitution by putting 4 x = u ⇒ d ( 4 x ) = d u ⇒ 4 d x = d u ⇒ d x = d u 4 Thus given integral changes as ∫ − 3 1 + ( 4 x ) 2 d x = ∫ − 3 1 + u 2 d u 4 = − 3 4 ∫ 1 1 + u 2 d u Now use fundamental formulae of integrals to solve it further, as ∫ − 3 1 + ( 4 x ) 2 d x = − 3 4 ∫ 1 1 + u 2 d u = − 3 4 t a n − 1 u + C Where C is an arbitrary constant

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 4.4, Problem 40S

To determine

To solve the integrals

∫−31+16 x 2dx

And also check the answer by differentiating.

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Chapter 4.4, Problem 39S

Chapter 4.4, Problem 41S

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To solve the integrals ∫ − 3 1 + 16 x 2 d x And also check the answer by differentiating.

Solution Summary: The author explains how to solve the integrals displaystyle 'int' and check the answer by differentiating.

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To solve the integrals

∫−31+16 x 2dx

And also check the answer by differentiating.

Want to see the full answer?

Chapter 4 Solutions

To solve the integrals ∫ − 3 1 + 16 x 2 d x And also check the answer by differentiating.

Solution Summary: The author explains how to solve the integrals displaystyle 'int' and check the answer by differentiating.

Videos

To solve the integrals ∫−31+16 x 2dx And also check the answer by differentiating.

Want to see the full answer?

Chapter 4 Solutions

To solve the integrals

∫−31+16 x 2dx

And also check the answer by differentiating.