Clearly,$\mathrm{f}:[0,2\mathrm{\pi }]\to [0,2\mathrm{\pi }]$ given by

f(x) =x + sin x is a bijection. So, its inverse exists, the graph of ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)$  is the mirror image of the graph of f(x) in the line y = x.

$\therefore$ Required area = 4 (Area of one loop)

                              $\begin{array}{l}=4\left[{\int }_0^{\mathrm{\pi }} (\mathrm{x}+\sin \mathrm{x})\mathrm{dx}-{\int }_0^{\mathrm{\pi }} \mathrm{xdx}\right]\\ =4{\int }_0^{\mathrm{\pi }} \sin \mathrm{xdx}\\ =-4[\cos \mathrm{x}{]}_0^{\mathrm{\pi }}\\ =-4[\cos \mathrm{\pi }-\cos 0]\\ =8\text{ sq units }\end{array}$