Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.2 Definite Integrals - 5.2 Exercises - Page 359: 37

Answer

$\pi.$

Work Step by Step

We have $$ \int_{0}^{\pi} x \sin x d x=R_1+R_2=1+\pi-1=\pi. $$
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