Troubleshooting the Integral of e^-ax^2: Where Did That Negative Sign Come From?

[genericusrnme]

Hey guys

I've been reading through a few books and I can't seem to work this out;
\int _0^{\infty }e^{-a x^2}xdx = \frac{1}{2a}
I keep getting
\int _0^{\infty }e^{-a x^2}xdx = -\frac{1}{2a}
I do the old variable switch
u = x^2
du=2 x dx
Which leaves me with
\int_0^{\infty } \frac{e^{-a u}}{2} \, du
Which them leads me to
\text{Lim} u\to 0\frac{e^{- a u}}{-2 a}=-\frac{1}{2a}
Where am I picking up this unwanted - from?
I can't see where I've gone wrong

 

[tiny-tim]

hey genericusrnme!

Where am I picking up this unwanted - from?

0 is the lower limit … so there's an extra minus

 

[genericusrnme]

man I feel stupid now :L

thanks tiny tim :D