Question

Let $a>0$ and $f\left(x\right)$ is monotonic increasing such that $f\left(0\right)=0$ and $f\left(a\right)=b$, then ${\int }_0^{a}f\left(x\right)dx+{\int }_0^b{f}^{-1}\left(x\right)dx$ is equal to

• A. $a+b$
• B. $ab+b$
• C. $ab+a$
• D. $ab$

Answer 1

The correct option is D $ab$

Let $I={\int }_0^{a}f\left(x\right)dx+{\int }_0^b{f}^{-1}\left(x\right)dx$
Let $I_1={\int }_0^{a}f\left(x\right)dx$ and
$I_2={\int }_0^b{f}^{-1}\left(x\right)dx$
Refer to the figure. $I_1+I_2$ will be equal to the area of the rectangle with length $a$ and breadth $b$.
Hence, $I=ab$