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

Answer

$A = 2-2\ln 2$

Work Step by Step

From the graph, we can see that on the interval $\big[-\frac{\pi}{3}, 0 \big]$, $2\sin(x) \leq \tan(x) dx$ and on the interval $\big[0, \frac{\pi}{3} \big]$, $2\sin(x) \geq \tan(x) dx$; therefore, the area between the two curves is $$A = \displaystyle\int_{-\frac{\pi}{3}}^{0} \tan(x) - 2\sin(x)dx + \displaystyle\int_{0}^{\frac{\pi}{3}} 2\sin(x) - \tan(x)dx$$ $$A=\Big[\ln |\cos (x)| -2\cos (x)\Big]^{0}_{-\frac{\pi}{3}} + \Big[-2\cos (x) + \ln |\cos (x)|\Big]^{\frac{\pi}{3}}_0$$ $$A=\big[ -2(1)+\ln(\frac{1}{2})-2(-\frac{1}{2})\big] - \big[-2(\frac{1}{2})+\ln(\frac{1}{2})+2(1)\big]$$ $$A = 2 + \ln(\frac{1}{4})$$ $$A = 2-2\ln 2$$
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