Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 10



Work Step by Step

The area $(A)$ can be calculated as: $A=\int_{\pi/3}^{\pi} [\sin x -\sin 2x] \ dx \\= [-\cos x+\dfrac{\cos 2x}{2}]_{\pi/3}^{\pi}\\=-[\cos \pi-\cos (\pi/3)] +\dfrac{1}{2} (\cos 2 (\pi-\dfrac{\pi}{3})]\\= 1+\dfrac{1}{2}-(-\dfrac{1}{2}-\dfrac{1}{4}) \\=\dfrac{9}{4}$
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