University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 342: 25



Work Step by Step

We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ where $C$ is a constant of proportionality and $\int \sin x dx=\cos x +c$ $\int_{0}^{\pi} (2\sin x-\sin 2x) dx=[-2 \cos x+\dfrac{\cos 2x}{2}]_0^{\pi}$ or, $=-2[\cos \pi-\cos 0]+\dfrac{1}{2} (\cos 2 \pi-\cos 0)$ or $=(-2)(-1)+2=4$
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.