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 337: 4


$$\int^{\pi}_03\cos^2 x\sin xdx=\int^{3\pi}_{2\pi}3\cos^2 x\sin xdx=2$$

Work Step by Step

Part a) $$A=\int^{\pi}_03\cos^2 x\sin xdx$$ Set $u=\cos x$, we then have $$du=-\sin xdx$$ $$\sin xdx=-du$$ For $x=0$, we have $u=1$ and for $x=\pi$, we have $u=-1$. Therefore, $$A=-\int^{-1}_13u^2du$$ $$A=-\frac{3u^3}{3}\Big]^{-1}_1=-u^3\Big]^{-1}_1$$ $$A=-(-1-1)=2$$ Part b) $$A=\int^{3\pi}_{2\pi}3\cos^2 x\sin xdx$$ We carry out a similar substitution as in part a). For $x=2\pi$, we have $u=0$ and for $x=3\pi$, we have $u=-1$. Therefore, $$A=-\int^{-1}_{1}3u^2du=2$$
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