Question

The area bounded by the curve $f\left(x\right)=x+\sin x$ and its inverse between the ordinates $x=0$ to $x=2\pi$ is

• A. $4$ sq. units
• B. $8$ sq. units
• C. $4\pi$ sq. units
• D. $8\pi$ sq. units

Answer 1

The correct option is B $8$ sq. units

$f\left(x\right)=x+\sin x$
For $x\in (0,\pi ),\sin x>0$
$\Rightarrow f\left(x\right)>x$
For $x\in (\pi ,2\pi ),\sin x<0$
$\Rightarrow f\left(x\right)<x$

We know that, $f^{-1}\left(x\right)$ is the mirror image of $f\left(x\right)$ with respect to the line $y=x$


So, the area between $f\left(x\right)$ and $f^{-1}\left(x\right)$ is double the area between $f\left(x\right)$ and $y=x$ from $x=0$ to $x=2\pi$
Which is equivalent to $4$ times the area between $f\left(x\right)$ and $y=x$ from $x=0$ to $x=\pi$

So, the required area is,
$A=4\underset{0}{\overset{\pi }{\int }}\left[\right(x+\sin x)-x]dx$
$=4\underset{0}{\overset{\pi }{\int }}\sin xdx$
$=4[-\cos x{]}_0^{\pi }$
$=4\times 2$
$=8\text{sq. units}$