## Calculus: Early Transcendentals 8th Edition

$$\int^{3\pi/2}_{0}|\sin x|dx=3$$
$$A=\int^{3\pi/2}_{0}|\sin x|dx$$ 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 $$A=\int^{\pi}_{0}(\sin x)dx+\int^{3\pi/2}_{\pi}(-\sin x)dx$$ $$A=(-\cos x)\Bigg]^{\pi}_{0}+(\cos x)\Bigg]^{3\pi/2}_{\pi}$$ $$A=-(\cos\pi-\cos0)+(\cos\frac{3\pi}{2}-\cos\pi)$$ $$A=-(-1-1)+(0+1)$$ $$A=3$$