Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 279: 87



Work Step by Step

By making use of Theorem 1 (FTC II), we have $$ A'(x)=\frac{d}{d x} \int_{2}^{x} \frac{\cos t}{1+t} d t=\frac{\cos x}{1+x} $$ Hence $$A'(\pi)= \frac{\cos \pi}{1+\pi} =-\frac{1}{1+\pi} $$
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