Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 253: 68



Work Step by Step

$f’’(\theta)=cos(\theta),$ $f’(\frac{\pi}{2})=1,$ $f(\frac{\pi}{2})=6$ $f’(\theta)=\int cos(\theta)d\theta$ $f’(\theta)=sin(\theta)+C_1$ Substitute in $f’(\frac{\pi}{2})=1$ to solve for $C_1$ $1=sin(\frac{\pi}{2})+C_1$ $C_1=0$ $f’(\theta)=sin(\theta)$ $f(\theta)=\int sin(\theta)d\theta$ $f(\theta)=-cos(\theta)+C_2$ Substitute in $f(\frac{\pi}{2})=6$ to solve for $C_2$ $6=-cos(\frac{\pi}{2})+C_2$ $C_2=6$ $f(\theta)=-cos(\theta)+6$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.