int_0^{2 pi} (sin x + |sin x|) , dx = | vedclass

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(C) We need to evaluate the integral $I = \int_0^{2 \pi} (\sin x + |\sin x|) \, dx$. Since the function $|\sin x|$ changes behavior at $x = \pi$,we sp

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$4$

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(C) We need to evaluate the integral $I = \int_0^{2 \pi} (\sin x + |\sin x|) \, dx$. Since the function $|\sin x|$ changes behavior at $x = \pi$,we split the integral: $I = \int_0^{\pi} (\sin x + |\sin x|) \, dx + \int_{\pi}^{2 \pi} (\sin x + |\sin x|) \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. In the interval $[\pi, 2 \pi]$,$\sin x \le 0$,so $|\sin x| = -\sin x$. Thus,$I = \int_0^{\pi} (\sin x + \sin x) \, dx + \int_{\pi}^{2 \pi} (\sin x - \sin x) \, dx$. $I = \int_0^{\pi} 2 \sin x \, dx + \int_{\pi}^{2 \pi} 0 \, dx$. $I = 2 [-\cos x]_0^{\pi} + 0$. $I = 2 [-\cos(\pi) - (-\cos(0))] = 2 [-(-1) - (-1)] = 2 [1 + 1] = 4$.

$\int_0^{2 \pi} (\sin x + |\sin x|) \, dx =$

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