Solution
= Solution
Use the <substitution rule> $u=x^2$, $du=2x\,dx$. The convergent <improper integral> is
$$
\boxed{\int_0^\infty xe^{-x^2}\,dx=\frac12\int_0^\infty e^{-u}\,du=\frac12.}
$$
= Solution
Use the <substitution rule> $u=x^2$, $du=2x\,dx$. The convergent <improper integral> is
$$
\boxed{\int_0^\infty xe^{-x^2}\,dx=\frac12\int_0^\infty e^{-u}\,du=\frac12.}
$$