University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 25

Answer

$-1$

Work Step by Step

$\int_{\pi/2}^{\pi} (\frac{sin(2x)}{2\cdot sinx}) dx=\int_{\pi/2}^{\pi} (\frac{2sinxcosx}{2\cdot sinx}) dx=\int_{\pi/2}^{\pi} (cosx) dx=[sinx]_{\pi/2}^{\pi}=[sin\pi]-[sin\frac{\pi}{2}]=0-1=-1$
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