Inverting an arbitrary integral

Question

$$r(x) = \int_{x_\min}^x f(y)\, dy$$

I would like to obtain an inverse for this such that I have $x(r)$. Is this possible? I saw this post before, however my function has a $y$ involved which makes it harder for me to understand. Any hints?

Thank you for your help !

PS. I thought that I would differentiate it with respect to $y$ to get rid of the integral. Does that even make any sense?

EDIT: I must have explained the problem a little better. I have a random variable $x$ defined by the p.d.f $f(x)$ and a function of the random variable $r(x)$ defined with the p.d.f $g(r)$. I would like to obtain the p.d.f $g(r)$.

My approach was to use $$g(r) = f(x(r)) * \mid \frac{dx}{dr} \mid $$ However, I am stuck at the $ \frac{dx}{dr}$ since I find the original function $r(x)$ to be defined in an abstract way.

Answer 1

There is no way to explicitly solve this problem without knowing something about $f$. By differentiating both sides with the FTC you get

$$\frac{dr}{dx} = f(x).$$

Hence the inverse function theorem gives

$$\frac{dx}{dr}=\frac{1}{f(x(r))}.$$

But this is a nontrivial ODE, meaning that you cannot simply integrate both sides to obtain $x$.

Answer 2

Let us suppose that $r$ is invertible, and write $\chi=r^{-1}$ (as Jack d'Aurizio said, it is the case when $f$ is always positive, or always negative).

If you're looking for the derivative of $\chi$, then we simply have : $\chi'(\rho)=\frac{1}{r'(\chi(\rho))}=\frac{1}{f(\chi(\rho))}$

With a little abuse of notation : $$\frac{dx}{dr}=\frac{1}{f(x)}$$