# Thread: [SOLVED] area under the curve

1. ## [SOLVED] area under the curve

I'm having trouble proving an equation.

Take $f(x)$ to be a continuous strictly increasing function with $f(0)=0$ and let ${f}^{-1}(x)$ be its inverse. Then:

$\int_{0}^{a}f(x)dx+\int_{0}^{f(a)}{f}^{-1}(y)dy=af(a)$

This should be visually obvious. The graph of $f(x)$ divides the rectangle with vertices $(0,0)$ and $(a,f(a))$ in two.

In the proof, I would like not to assume differentiability (since the equation is obviously true non-smooth functions too)

Thanks.

2. Here's what I mean by visually obvious:

File:Young.png - Wikipedia, the free encyclopedia

Just put b=f(a) in that figure.

any ideas?

3. $\int_0^{f(a)}f^{-1}(y)dy =
$

$
( \: y=f(t) \: \: t=f^{-1}(y) \: )
$

$
= \: \int_0^{f^{-1}(f(a))}f^{-1}(f(t))df(t) \:= \: \int_0^{a}tdf(t) \: = \: tf(t)| _0^a - \int_0^a f(t)dt

$

4. I forgot that using Stieltjes integrals one is free to do change of variable and integration by parts without assuming differentiability.

Thanks.