Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 46


$$\int^{3\pi/2}_{0}|\sin x|dx=3$$

Work Step by Step

$$A=\int^{3\pi/2}_{0}|\sin x|dx$$ From $(0,\pi]$, $(\sin x)\geq0$, therefore $|\sin x|=\sin x$. From $(\pi,\frac{3\pi}{2})$, $x\lt0$, therefore, $|\sin x|=-\sin x$ So, $A$ would become $$A=\int^{\pi}_{0}(\sin x)dx+\int^{3\pi/2}_{\pi}(-\sin x)dx$$ $$A=(-\cos x)\Bigg]^{\pi}_{0}+(\cos x)\Bigg]^{3\pi/2}_{\pi}$$ $$A=-(\cos\pi-\cos0)+(\cos\frac{3\pi}{2}-\cos\pi)$$ $$A=-(-1-1)+(0+1)$$ $$A=3$$
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