Question

Suppose f is continuous $f\left(0\right)=0,f\left(1\right)=1,f^{\prime }\left(x\right)>0$ and ${\int }_0^{1}f\left(x\right)dx=\frac{1}{3}$. Find the value of the definite integral ${\int }_0^1{f}^{-1}\left(y\right)dy$.

Answer 1

${\int }_0^{1}f\left(x\right)dx=\frac{1}{3},f\left(x\right)=y;f^{-1}\left(y\right)=x\to 1$

$={\int }_0^{1}ydx=\frac{1}{3}$

Integral $ydx$ is the area under graph of $f\left(x\right)$ from $0$ to $1$

Area can also be found by integrating $xdy$ from $f\left(0\right)$ to $f\left(1\right)$

Here $f\left(0\right)=0$ and $f\left(1\right)=1$

${\int }_{f\left(0\right)}^{f\left(1\right)}xdy={\int }_0^1{f}^{-1}\left(y\right)dy=$Area $=\frac{1}{3}$