The area enclosed by the curve $\left|y\right|=\sin 2x$ when $\mathrm{x}\in [0,2\pi ]$ is

The given curve is $\left|y\right|=\sin 2x$ and the given interval is $x\in \left[0,2\mathrm{\pi }\right]$

The rough sketch to the required area is as below

The required area is the area of shaded region.

The area of the shaded region is four times the area of the region bounded by the curve $y=\sin 2x$, the lines $x=0,x=\frac{\mathrm{\pi }}{2}$ with $x-$ axis.

Hence, the area of the shaded region is $A=4{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2xdx$

we know that $\int \sin 2x dx=-\frac{\cos 2x}{2}+c$

Hence, 

$\begin{array}{rcl}A& =& 4{\left(-\frac{\cos 2x}{2}\right)}_0^{\frac{\mathrm{\pi }}{2}}\\ & =& 4\left(\frac{1}{2}\right)\\ & =& 2\end{array}$

Therefore, the area of the region bounded by the curve $\left|y\right|=\sin 2x$ in the interval $x\in \left[0,2\mathrm{\pi }\right]$ is $\boxed{2 \mathrm{sq} \mathrm{units}}$