If $\int_0^{2 \pi}|x \sin x| d x=k \pi$, then $k=$
If $\int_0^{2 \pi}|x \sin x| d x=k \pi$, then $k=$
$1$
$2$
$3$
$4$
Solution
$\int_0^{2 \pi}|x \sin x| d x=k \pi$ ...(i)
Let $\mathrm{I}=$\int_0^{2 \pi}|x \sin x| d x$
$=\int_0^{2 \pi} x|\sin x| d x$\{\because x>0\}$
$=\left[\int_0^\pi x \sin x d x-\int_\pi^{2 \pi} x \sin x d x\right]$
$\begin{aligned} & =[-x \cos x]_0^\pi+\left[\int_0^\pi 1 \cdot \cos x d x\right] \\ & -[-x \cos x]_\pi^{2 \pi}-\left[\int_\pi^{2 \pi} \cos x d x\right]\end{aligned}$
$\begin{aligned} & =\pi+\int_0^{\frac{\pi}{2}} \cos x d x-\int_{\frac{\pi}{2}}^\pi \cos x d x+[2 \pi+\pi]^\pi \\ & -\int_\pi^{3 \pi / 2} \cos x d x+\int_{3 \pi / 2}^{2 \pi} \cos x d x\end{aligned}$
$=4 \pi+1-1+1-1=4 \pi$
$\Rightarrow I=4 \pi \Rightarrow \int_0^{2 \pi}|x \sin x| d x=4 \pi$....(ii)
From (i) and (ii), $k=4$