University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 85

Answer

The area of the region is $4$.
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Work Step by Step

$y=2\sin x$ and $y=\sin2x$ for $0\le x\le\pi$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $x$ in this exercise. 2) Find limits of integration: To find the limits of integration, we will find the intersection points. $$2\sin x-\sin 2x=0$$ $$2\sin x-2\sin x\cos x=0$$ $$\sin x(1-\cos x)=0$$ $$\sin x=0\hspace{1cm}\text{or}\hspace{1cm}\cos x=1$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=\pi$$ Therefore, the upper limit is $\pi$ and the lower one is $0$. 3) Find the area: The region is bounded above by $y=2\sin x$ and below by $y=\sin2x$. So according to definition, the area of the region is $$A=\int^\pi_{0}(2\sin x-\sin2x)dx$$ $$A=-2\cos x+\frac{1}{2}\cos2x\Big]^\pi_{0}$$ $$A=-2\cos\pi+\frac{1}{2}\cos2\pi+2\cos0-\frac{1}{2}\cos0$$ $$A=-2(-1)+\frac{1}{2}+2-\frac{1}{2}$$ $$A=4$$
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