Answer
$2\pi-1.$
Work Step by Step
We have
$$
\int_{\pi/2}^{2pi} x \sin x d x=R_2-R_3-R_4=\pi-1-\pi-1-2\pi+1=-2\pi-1.
$$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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: 40
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