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

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) We evaluate the integral $I = \int_{0}^{2\pi} (\sin x + |\sin x|) \, dx$. 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: $I = \int_{0}^{\pi} (\sin x + \sin x) \, dx + \int_{\pi}^{2\pi} (\sin x - \sin x) \, dx$ $I = \int_{0}^{\pi} 2\sin x \, dx + \int_{\pi}^{2\pi} 0 \, dx$ $I = 2 [-\cos x]_{0}^{\pi} + 0$ $I = -2 (\cos \pi - \cos 0)$ $I = -2 (-1 - 1) = -2(-2) = 4$.

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

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