Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 26

Answer

$0$

Work Step by Step

We have $$\int_{2\pi }^{4 \pi } \sin xd x= -\cos x|_{2\pi }^{4 \pi }\\ = -\cos (4 \pi )+ \cos (2\pi )=0.$$
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