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 338: 50


The area of the shaded region is $2$.

Work Step by Step

The shaded region can be seen as having two parts, one half from $x=-\pi/2$ to $x=0$, and the other half from $x=-\pi$ to $x=-\pi/2$. However, since these two parts are symmetric with one another with respect to $x=-\pi/2$, we only need to calculate the area of one half and double it to get the total area of the whole shaded area. So here I would take the region from $x=-\pi/2$ to $x=0$. It is bounded above by the curve $y=\frac{\pi}{2}(\cos x)(\sin(\pi+\pi\sin x))$ and below by the line $y=0$, So the area of this region would be $$A=\int^0_{-\pi/2}\frac{\pi}{2}(\cos x)(\sin(\pi+\pi\sin x))dx$$ We set $u=\pi+\pi\sin x$, which means $$du=\pi\cos x dx=2\times\frac{\pi}{2}\cos x dx$$ $$\frac{\pi}{2}\cos xdx=\frac{1}{2}du$$ - For $x=-\pi/2$, we have $u=\pi+\pi\sin(-\pi/2)=\pi-\pi=0$ - For $x=0$, we have $u=\pi+\pi\sin0=\pi$ Therefore, $$A=\frac{1}{2}\int^{\pi}_0\sin udu=-\frac{1}{2}\cos u\Big]^{\pi}_0$$ $$A=-\frac{1}{2}(\cos\pi-\cos0)$$ $$A=-\frac{1}{2}(-1-1)=-\frac{1}{2}(-2)=1$$ The total area of the whole shaded region is $$T=2A=2$$
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