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$\int_{0}^{2 x}(\sin x+|\sin x|) d x$ is equal to
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The correct answer is:
4
$\int_{0}^{2 \pi}(\sin x+|\sin x|) d x$
$=\int_{0}^{\pi}(\sin x+\sin x) d x+$
$\quad \int_{\pi}^{2 \pi}(\sin x-\sin x) d x$
$=2 \int_{0}^{\pi} \sin x d x+0=-2 \int_{0}^{\pi} \cos x d x$
$=-2(\cos \pi-\cos 0)=-2(-1-1)=4$
$=\int_{0}^{\pi}(\sin x+\sin x) d x+$
$\quad \int_{\pi}^{2 \pi}(\sin x-\sin x) d x$
$=2 \int_{0}^{\pi} \sin x d x+0=-2 \int_{0}^{\pi} \cos x d x$
$=-2(\cos \pi-\cos 0)=-2(-1-1)=4$
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