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: 10

Answer

$$A = 1-\frac{\pi}{4} $$
1517854933

Work Step by Step

Find where the functions intersect: $$\sin x = \frac{2x}{\pi} \rightarrow x = \frac{\pi}{2}$$ Set the integral, with the top bound being the intersection point and the lower bound being zero in this case: $$\displaystyle\int_{0}^{\frac{\pi}{2}} \sin x - \frac{2x}{\pi} dx = \displaystyle\int_{0}^{\frac{\pi}{2}} \sin x dx - \frac{2}{\pi} \displaystyle \int_{0}^{\frac{\pi}{2}} x dx$$ Take the integral $$- \cos x - \frac{2}{\pi} \times \frac{x^{2}}{2} = -\cos x - \frac{x^{2}}{\pi}$$ Evaluate the end points and simplify: $$- \cos x - \frac{x^{2}}{\pi} = \Bigg(-\cos\frac{\pi}{2} - \frac{ \bigg( \frac{\pi}{2} \bigg) ^{2}}{\pi}\Bigg)+1=0-\frac{\pi}{4}+1=1-\frac{\pi}{4}$$
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