Why do I get a negative sign in ∫₀^∞ e^(-ax²)x dx?

[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