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 - Page 320: 17


$-\frac{\sqrt 2}{4}+\frac{1}{2}$

Work Step by Step

$\int_{0}^{\pi/8} (sin(2x)) dx=\frac{1}{2}\cdot\int_{0}^{\pi/8} (2sin(2x)) dx=[-\frac{1}{2}\cdot cos(2x)]_{0}^{\pi/8}=[-\frac{1}{2}\cdot cos(2\cdot\frac{\pi}{8})]-[-\frac{1}{2}\cdot cos(2\cdot0)]=[-\frac{1}{2}\cdot cos(\frac{\pi}{4})]-[-\frac{1}{2}\cdot cos(0)]=[-\frac{1}{2}\cdot(\frac{\sqrt 2}{2})]+\frac{1}{2}\cdot1=-\frac{\sqrt 2}{4}+\frac{1}{2}$
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