$\begin{aligned} & \text { Let } I=\int_0^\pi|\sin 2 x| d x \\\\ & =2 \int_0^{\pi / 2}|\sin 2 x| d x \quad[\because \sin 2 x \text { is periodic function }] \\\\ & =2 \int_0^{\pi / 2} \sin 2 x \,d x[\sin 2 x \text { is positive in range }(0, \pi / 2)] \\\\ & =2\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\\\ & =-[\cos \pi-\cos 0]=-(-1-1)=2 \end{aligned}$