# Weibull integral

1. Mar 20, 2010

### longrob

Can someone explain this..
$$P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{v}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{v}{\eta}\right)^{\beta}\right)dv=\intop_{0}^{x}e^{-x}dx\hphantom{}\; where\phantom{\:}x=\left(\frac{v}{\eta}\right)^{\beta}$$

Thanks !

2. Mar 20, 2010

### Galileo

It just seems like a change of variable, but beware your notation.
In the first integral you use v both as the integration variable and in the upper limit of integration. In the second, x plays both those roles too. This is confusing.
I'd call the integration variable v' in the first one, then just make a change of variable substitution.

3. Mar 20, 2010

### arildno

In line with Galileo's comment, this is how you should write it:
$$P(v)=\frac{\beta}{\eta}\intop_{0}^{v}\left(\frac{V}{\eta}\right)^{\beta-1}\exp\left(-\left(\frac{V}{\eta}\right)^{\beta}\right)dV=\intop_{0}^{x(v)}e^{-X}dX\hphantom{}\; where\phantom{\:}x(v)=\left(\frac{v}{\eta}\right)^{\beta}$$

4. Mar 20, 2010

Thanks !!