How to prove that $f(x)x - \int_{0}^{x}{f(t) \,dt} = \int_{f(0)}^{f(x)}{f^{-1}(t) \,dt},$ for all invertible functions.

Question

A while ago, I found that:

$$f(x)x - \int_{0}^{x}{f(t) \,dt} = \int_{f(0)}^{f(x)}{f^{-1}(t) \,dt}.$$

I managed to prove it for a few functions, and I believe that it may be the case for all invertible functions.

However, I would like to be able to prove it for all invertible functions, if, of course, it is true.

Answer 1

If $f(a)=c$ and $f(b)=d$, then

$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$

Now, let $a=0$, $b=x$, $c=f(0)$, and $d=f(x)$.