Verified answer

There's a neat trick for this integral.

Let I = ∫(0 to ∞) e^(-x^2) dx.

Then I^2 = [∫(0 to ∞) e^(-x^2) dx] [∫(0 to ∞) e^(-y^2) dy]

= ∫(0 to ∞)∫(0 to ∞) e^(-x^2) e^(-y^2) dx dy

= ∫(0 to ∞)∫(0 to ∞) e^(-(x^2 + y^2)) dx dy

Now change to polar coordinates:

= ∫(0 to ∞)∫(0 to 2π) e^(-r^2) r dθ dr

= 2π ∫(0 to ∞) r e^(-r^2) dr; let u = r^2, du = 2r dr

= 2π (1/2) ∫(0 to ∞) e^(-u) du

= π [-e^(-u)][0 to ∞]

= π (0 + 1)

and hence I = √π.