int_{0}^{2pi} |sin x| , dx = | vedclass

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(D) We need to evaluate the integral $I = \int_{0}^{2\pi} |\sin x| \, dx$. Since $|\sin x| = \sin x$ for $x \in [0, \pi]$ and $|\sin x| = -\sin x$ for

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$4$

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(D) We need to evaluate the integral $I = \int_{0}^{2\pi} |\sin x| \, dx$. Since $|\sin x| = \sin x$ for $x \in [0, \pi]$ and $|\sin x| = -\sin x$ for $x \in [\pi, 2\pi]$,we split the integral: $I = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx$ $I = [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{2\pi}$ $I = (-(\cos \pi - \cos 0)) + (\cos 2\pi - \cos \pi)$ $I = (-(-1 - 1)) + (1 - (-1))$ $I = (2) + (2) = 4$.

$\int_{0}^{2\pi} |\sin x| \, dx = $

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