= Solution
Take $\sigma>0$ and substitute $u=(x-\mu)/\sigma$. The density becomes the standard <Gaussian distribution> density $\varphi(u)=e^{-u^2/2}/\sqrt{2\pi}$. Its normalization follows from the supplied <Gaussian integral> after scaling by $\sqrt2$. Since $u\varphi(u)$ is an integrable <odd function>, its integral is zero, and therefore $\mathbb EX=\mu$.
For the second moment, $\varphi'=-u\varphi$. <Integration by parts> gives
$$
\int_{-\infty}^{\infty}u^2\varphi(u)du=-[u\varphi(u)]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\varphi(u)du=1.
$$
The endpoint term vanishes by Gaussian decay. Consequently
$$
\boxed{\mathbb EX=\mu,\qquad\operatorname{Var}X=\sigma^2.}
$$
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