If int_0^{2 pi} |x sin x| , dx = k pi,then k | vedclass

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(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$. In the inte

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$4$

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(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,and in the interval $[\pi, 2 \pi]$,$\sin x \le 0$. Thus,$I = \int_0^{\pi} x \sin x \, dx - \int_{\pi}^{2 \pi} x \sin x \, dx$. Using integration by parts,$\int x \sin x \, dx = -x \cos x + \sin x$. Evaluating the first integral: $\int_0^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi} = (-\pi(-1) + 0) - (0 + 0) = \pi$. Evaluating the second integral: $\int_{\pi}^{2 \pi} x \sin x \, dx = [-x \cos x + \sin x]_{\pi}^{2 \pi} = (-2 \pi(1) + 0) - (-\pi(-1) + 0) = -2 \pi - \pi = -3 \pi$. Therefore,$I = \pi - (-3 \pi) = 4 \pi$. Given $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,we have $4 \pi = k \pi$,which implies $k = 4$.

If $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,then $k =$

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