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: 48

Answer

The area of the shaded region is $2$.

Work Step by Step

The shaded region is bounded above by the curve $y=(1-\cos x)\sin x$ and below by the line $y=0$ from the point $x=0$ to $x=\pi$. So the area of the region would be $$A=\int^\pi_0\Big((1-\cos x)\sin x-0\Big)dx=\int^\pi_0(1-\cos x)\sin xdx$$ We set $u=1-\cos x$, which means $$du=-(-\sin x)dx=\sin xdx$$ - For $x=\pi$, we have $u=1-\cos\pi=1-(-1)=2$ - For $x=0$, we have $u=1-\cos0=1-1=0$ Therefore, $$A=\int^2_0udu=\frac{u^2}{2}\Big]^2_0$$ $$A=\frac{1}{2}(2^2-0^2)$$ $$A=\frac{1}{2}(4)=2$$
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