# Inverse integral numerically

1. Oct 23, 2007

### jimbo_durham

I have a complicated integral which i need to compute numerically. I can do this in C++ using a version of Simpson's rule. I also need to compute the inverse of this integral (presumably this is what it is called) ie i have

d=integral(f(x)dx)

and i need to be able to compute x given a value of d.

The function f(x) is too complicated to simply rearrange, so i need an iterative way of guessing x, running it in the program, seeing what value of d is given, and then improving my guess of x until i reach my starting d value.

This seems on the surface to be a simple iteration problem, however i cannot find an efficent way of doing this.

Can anyone tell me of a nice way of solving this problem (and even if it is called an 'inverse integration' or not?) and if there is a name given to the numerical method of solving it.

It has been mentioned that there is a 'press' method however i can find no mention of this. Prehaps it involves finding the roots of an equation which takes the value of the integral evaluated at some guessed x, and the roots give the point where the guessed x corresponds to the known d, thus solving the problem? does this ring any bells?

Last edited: Oct 23, 2007
2. Oct 23, 2007

### arildno

Well, one way would be:
Let:
$$y(x)=y(0)+\int_{x_{0}}^{x}f(t)dt$$
Or, that is:
$$y=y_{0}+\int_{x(0)}^{x(y)}f(t)dt$$
Differentiating the latter expression with respect to y yields:
$$1=f(x(y))\frac{dx}{dy}$$
That is, you may solve the following differential equation numerically:
$$\frac{dx}{dy}=\frac{1}{f(x)}, x(y_{0})=x(y_{0})$$

3. Oct 23, 2007

### jimbo_durham

i am not sure how to apply that, i think it is worth me giving you the integral;

$$d_{M}=c_{1}\cdot sinh\left[c_{2}\int^{b}_{a}\left[\left(1+z\right)^{2}\cdot\left(1+c_{3}\cdot z\right)-z\cdot\left(2+z\right)\cdot c_{4}\right]^{-\frac{1}{2}}dz\right]$$

i am trying to find z given a value of $$d_{M}$$.

4. Oct 23, 2007

### jimbo_durham

note in formula, $$a, b, c$$ are known constants