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 - Page 434: 21


$$A = (2-2\ln 2)$$

Work Step by Step

From the graph, we can see that $2sin(x) \geq \tan(x) dx$ on the interval $\big[0, \frac{\pi}{3} \big]$; therefore, the area between the two curves is $$A = \displaystyle\int_{0}^{\frac{\pi}{3}} 2\sin(x) - \tan(x)dx$$ $$A=2\Big[-2\cos (x) + \ln |\cos (x)|\Big]^{\frac{\pi}{3}}_0$$ $$A=2\big[ -1+ \ln (\frac{1}{2})\big] - 2\big[ -2\times1+\ln 1\big]$$ $$A = 2-2\ln 2$$
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