Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 356: 33


$ f(t)=4\cdot\tan^{-1}t-\pi$

Work Step by Step

$\left[\begin{array}{cc} \text{}f' & \text{particular antiderivative,}f\\ 4\cdot\dfrac{1}{1+t^{2}} & 4\cdot\tan^{-1}t\\ & \end{array}\right]$ $f(t)=4\cdot\tan^{-1}t+C$ Given that $f(1)=0$, we find C: $0=4\cdot\tan^{-1}1+C$ $0=4\displaystyle \cdot\dfrac{\pi}{4}+C$ $ C=-\pi$ $ f(t)=4\cdot\tan^{-1}t-\pi$
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