Answer
$-\frac{1}{1+\pi}$
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 $$A'(\pi)= \frac{\cos \pi}{1+\pi} =-\frac{1}{1+\pi}
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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