Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 307: 25



Work Step by Step

Consider $f(x)=\int_{0}^{\pi} (2\sin x-\sin 2x) dx$ Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ This implies that $[-2 \cos x+\dfrac{\cos 2x}{2}]_0^{\pi}=-2[\cos \pi-\cos 0]+\dfrac{(\cos 2 \pi-\cos 0)}{2} $ Thus $f(x)=(-2) \cdot (-1)+(-2) \cdot (1)+\dfrac{1}{2}-\dfrac{1}{2}=4$
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.