This is a relatively straightforward question:
I need to find $\text{E}|x|$, given $x ~ \sim \mathcal{N}(\mu, \sigma^2)$. Presumably, this would be found as
$$\text{E}|x| = \int_{\mathbb{R}} |x| \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right)$$
I'm not sure how to integrate this. I think there's some trick with transforming it and using other probability distributions, and that the integration is unnecessary, but I can't think of anything.
Let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ be the pdf of standard normal distribution for all $x\in\mathbb R$.
Then,
\begin{align} E\left[|X|\right]&=\int_{-\infty}^\infty |x|\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\,dx \\&=-\int_{-\infty}^0 \frac{x}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\,dx+\int_0^\infty \frac{x}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\,dx \\&=-\sigma \int_{-\infty}^{-\mu/\sigma}t\phi(t)\,dt-\mu\int_{-\infty}^{-\mu/\sigma}\phi(t)\,dt+\sigma\int_{-\mu/\sigma}^\infty t\phi(t)\,dt+\mu\int_{-\mu/\sigma}^\infty \phi(t)\,dt\quad\small \left[x-\mu=\sigma t\right] \end{align}
Now recall that $\phi'(t)=-t\phi(t)$, which reduces the above to
\begin{align} E\left[|X|\right]&=2\sigma \phi\left(\frac{-\mu}{\sigma}\right)+\mu\left[1-2\Phi\left(\frac{-\mu}{\sigma}\right)\right]\quad,\,\small \Phi'=\phi \end{align}
You can rewrite this by using $\phi(-x)=\phi(x)$ and $\Phi(-x)=1-\Phi(x)$.