The normalization condition of the univariate Gaussian distribution is given by: $$\begin{align} \int_{-\infty}^{\infty}\mathcal{N}\left(\textit{x}\mid\mu, \sigma^{2}\right)dx &= 1\\ \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left\lbrace-\frac{1}{2\sigma^2}\left(x-\mu\right)^2\right\rbrace dx &= 1\\ \int_{-\infty}^{\infty} \exp\left\lbrace-\frac{1}{2\sigma^2}\left(x-\mu\right)^2\right\rbrace dx &= \left(2\pi\sigma^2\right)^{1/2} \end{align}$$ Differentiating both sides of $(3)$ with respect to $\sigma^2$, $$\begin{align} \int_{-\infty}^{\infty} \exp\left\lbrace-\frac{1}{2\sigma^2}\left(x-\mu\right)^2\right\rbrace \left\lbrace \frac{\left(x-\mu\right)^2}{2}\right\rbrace \left\lbrace \frac{1}{\left(\sigma^2\right)^2} \right\rbrace dx &= \left(\frac{1}{2}\right) \left(2\pi\sigma^2\right)^{-1/2} \left(2\pi\right)\\ \int_{-\infty}^{\infty} \exp\left\lbrace-\frac{1}{2\sigma^2}\left(x-\mu\right)^2\right\rbrace \left(x-\mu\right)^2 dx &= \sigma^2 \, \sqrt{2\pi\sigma^2}\\ \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left\lbrace-\frac{1}{2\sigma^2}\left(x-\mu\right)^2\right\rbrace \left(x-\mu\right)^2 dx &= \sigma^2 \end{align}$$ It follows directly from $(6)$ that $$\begin{align} \mathbb{E}\left[\left(x-\mu\right)^2\right] &= var\left[x\right] = \sigma^2 \end{align}$$ Expanding the left hand side of equation $(7)$, $$\begin{align*} \mathbb{E}\left[x^2\right] - 2\mu\mathbb{E}\left[x\right] + \mu^2 &= \sigma^2\\ \mathbb{E}\left[x^2\right] - 2\mathbb{E}\left[x\right]^2 + \mathbb{E}\left[x\right]^2 &= \sigma^2\\ \mathbb{E}\left[x^2\right] - \mathbb{E}\left[x\right]^2 &= \sigma^2 \end{align*}$$