Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises: 16


$$A = 4\pi $$

Work Step by Step

$y=\cos x$ is below the entire interval; subtract from the other function in the integral to find the area. $$\displaystyle\int_{0}^{2\pi}(2-\cos x - \cos x)dx = \displaystyle\int_{0}^{2\pi} (2-2\cos x)dx$$ $$= \left[2x-2\sin x \right]^{2\pi}_0$$ $$=4\pi - 0 - (0-0)$$ $$ = 4\pi$$
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