How to find the value of the integral Suppose $f$ is continuous, $f(0) = 0,\;\; f(1) = 1, \;\; f’(x) > 0,$ and $\int\limits_{0}^{1}f(x)\,dx=\dfrac{1}{3}$.  Find the value of the integral $\int\limits_{0}^{1}f^{-1}(y)\,dy.$
 A: Edit: As Andre Nicolas says, this can be made very visual. The graph of $f$ splits the square $[0,1]^2$ into two regions. The area below the curve is the integral of $f$ by integration over $x\in[0,1]$. The area above the curve is the integral of $f^{-1}$ by integration over $y\in[0,1]$. So their sum is the area of the square, $1$. 
Now here is an analytical argument in case $f$ is $C^1$. By assumptions, $f$ is a $C^1$ diffeomorphism of $(0,1)$ onto itself. So we can perform the change of variable $y=f(x)$, $dy=f'(x)dx$. This yields
$$
\int_0^1f^{-1}(y)dy=\int_{0}^1f^{-1}(f(x))f'(x)dx=\int_0^1xf'(x)dx.
$$
Now integrate by parts.
A: Here is an algebraic answer.
\begin{align}
I & = \underbrace{\int_0^1 f^{-1}(y) dy = \int_{f^{-1}(0)}^{f^{-1}(1)}xd(f(x))}_{f^{-1}(y) = x \text{ i.e. } y = f(x)} = \overbrace{\int_0^1 x d(f(x))}^{f^{-1}(x) > 0 \implies \text{one-to-one}}\\
& = \left.x f(x) \right \vert_{x=0}^{x=1} - \int_0^1 f(x)dx = 1 \cdot f(1) - 0 \cdot f(0) - \int_0^1 f(x)dx = 1 - \dfrac13 = \dfrac23
\end{align}
A: Hint: Draw a picture. Identify the region the first integral computes the area of. Do the same for the second integral.
