The area bounded by $f\left(x\right)=x+\sin x$ and its inverse between $x=0$ and $\mathrm{x}=2\pi$ is

The given function is $f\left(x\right)=x+\sin x$

The area of the region bounded by the curve $f\left(x\right)=x+\sin x$ and its inverse is equal to double the are of the region bouned by $f\left(x\right)=x+\sin x$ and the line $y=x$

The area of the shaded region in the following figure is the area required. 

Hence the required area is four times the area of the region bouned by the curve $y=x+\sin x$ and the lines $x=0,x=\mathrm{\pi }, \mathrm{y}=\mathrm{x}$

$\begin{array}{rcl}Required area& =& 4{\int }_0^{\mathrm{\pi }}\left(x+\sin x-x\right) dx\\ & =& 4{\int }_0^{\mathrm{\pi }}\sin x dx\\ & =& 4{{\left(-\cos x\right)}^{\mathrm{\pi }}}_0\\ & =& 8\end{array}$

Therefore, the area of the region bouned by the curve  $f\left(x\right)=x+\sin x$ and its inverse between $x=0$ and $\mathrm{x}=2\pi$ is $\boxed{\mathbf{8}\mathbf{ }\mathbf{\text{square units}}}$