Mathematics - Area Under Curves Question with Solution | TestHub

The curve and its inverse are reflections in $y=x$; by symmetry the required area is $4A$, where $A={\int }_0^{\pi }(x+\sin x)dx-{\int }_0^{\pi }xdx={\int }_0^{\pi }\sin xdx=[-\cos x{]}_0^{\pi }=2$. Hence the area is $4\times 2=8$.