Theorem 9.2.7. Verifying the normal probability variance.
Using a similar change of variable as in the previous theorem 9.2.6,
\begin{equation*}
E[(X-\mu)^2] = \int_{-\infty}^{\infty} (x-\mu)^2 \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx = \sigma^2
\end{equation*}