Answer
$0$
Work Step by Step
We have
$$\int_{2\pi }^{4 \pi } \sin xd x= -\cos x|_{2\pi }^{4 \pi }\\
= -\cos (4 \pi )+ \cos (2\pi )=0.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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