= Solution
Let $I=\int_0^\infty e^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant <double integral>. In <polar coordinates> the <Jacobian determinant> is $r$, so
$$
I^2=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta=\frac\pi4,\qquad \boxed{I=\frac{\sqrt\pi}2.}
$$
Using <integration by parts>, with the boundary term $xe^{-x^2}$ vanishing at both endpoints,
$$
\boxed{\int_0^\infty x^2e^{-x^2}dx=\frac12\int_0^\infty e^{-x^2}dx=\frac{\sqrt\pi}4.}
$$
The substitution $u=x^2$ also gives the two <antiderivatives>
$$
\boxed{\int xe^{-x^2}dx=-\frac12e^{-x^2}+C,\qquad \int x^3e^{-x^2}dx=-\frac12(x^2+1)e^{-x^2}+C.}
$$
These values also follow from the <Gaussian moment recurrence> and can be checked by <differentiation>.
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