question_answer

The area bounded by the curve \[f(x)=x+\sin x\] and its inverse function between the ordinates \[x=0\] and \[x=2\pi \] is

A)  \[4\pi \] sq. units            

B)  \[8\pi \] sq. units

C)  4 sq. units         

D)  8 sq. units

Correct Answer: D

Solution :

\[\because \,f:\,\left[ 0,\,2\pi  \right]\to \left[ 0,\,2\pi  \right]\] is given by \[f(x)=x+\sin x\] is a bijective function. The graph of \[{{f}^{-1}}(x)\] is the mirror image of the graph of \[f(x)\] in the line \[y=x\]. \[\therefore \] Required Area = 2 (Area of one loop) \[=2\times 2\left[ \int\limits_{0}^{\pi }{(x+\sin x)}dx-\int\limits_{0}^{\pi }{xdx} \right]\] \[=4\left[ \int\limits_{0}^{\pi }{\sin xdx} \right]\] = 8 sq. units.