Solution
= Solution
By the density formula for <expected value>,
$$
\mathbb E[X]=\int_0^\infty x^2e^{-x^2/2}\,dx.
$$
For the standard <normal distribution>, the integrand in the second <moment> is even, so
$$
\mathbb E[Z^2]=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^2e^{-z^2/2}\,dz
=\frac2{\sqrt{2\pi}}\int_0^\infty z^2e^{-z^2/2}\,dz.
$$
Comparing these two <integrals> gives
$$
\boxed{\mathbb E[X]=\sqrt{\frac\pi2}\,\mathbb E[Z^2],\qquad\beta=\sqrt{\frac\pi2}.}
$$