Explanation Definition used: A function F is called an antiderivative of f on an interval I if F ′ ( x ) = f ( x ) for all x in I . Theorem used: If F is an antiderivative of f on the interval I , then the most general antiderivative of f on I is F ( x ) + C where C is an arbitrary constant.” Calculation: The given function is f ( x ) = 4 − 3 ( 1 + x 2 ) − 1 , F ( 1 ) = 0 . Find the general antiderivative of F as follows. F ( x ) = 4 − 3 ( 1 + x 2 ) − 1 = 4 − 3 1 + x 2 = 4 − 3 1 ( 1 + x 2 ) = 4 x − 3 tan − 1 x + C ( The antiderivative of 1 1 + x 2 is tan − 1 x + C The antiderivative of k ( constant ) is k x + C ) Substitute x = 1 in F ( x ) = 4 x − 3 tan − 1 x + C to find the value of C

Single Variable Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112785

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4.7, Problem 16E

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The antiderivative of F of f for the condition F(1)=0.

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Chapter 4.7, Problem 15E

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Single Variable Essential Calculus: Early Transcendentals

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