# Inserting limits into an integrated term (Quick question)

1. Apr 9, 2015

### rwooduk

1. The problem statement, all variables and given/known data
Derive the momentum for a charged particle going through matter.

2. Relevant equations
None.

3. The attempt at a solution
I understand the derivation but there's one step im not clear about, and i'm probably being really stupid but this:

if the -infinity term is squared then doesnt it become positive and the sum equal zero?

2. Apr 9, 2015

### Dick

Yes, you are correct and that's a really sloppy derivation. What really happens is that when you factor the x^2 out of the square root you get |x|. And x/|x| is 1 at +infinity and -1 at -infinity. Does that help?

Last edited: Apr 9, 2015
3. Apr 9, 2015

### rwooduk

Thanks for the reply, It's hard to visualise what you mean, is there any chance you could Latex it? If not, it's fine, I'll just remember that it goes to (2/b^2) and not zero.

thanks again!

4. Apr 9, 2015

### Dick

What would be the point of memorizing that something like that? $\sqrt{x^2+b^2}=|x| \sqrt{1+\frac{b^2}{x^2}}$. Notice the factor in front is not $x$, it's $|x|$!

5. Apr 10, 2015

### rwooduk

Ahh, i see now, yes that makes sense! Thanks very much for this!