Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then
$$
E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x).
$$
The medians of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.

Why do the medians solve
$$
\min_{c \in \mathbb{R}} E(|X-c|) \, ?
$$

For every real valued random variable $X$,
$$
\mathrm E(|X-c|)=\int_{-\infty}^c\mathrm P(X\leqslant t)\,\mathrm dt+\int_c^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt
$$ 
hence the function $u:c\mapsto \mathrm E(|X-c|)$ is differentiable almost everywhere and, where $u'(c)$ exists, $u'(c)=\mathrm P(X\leqslant c)-\mathrm P(X\geqslant c)$. Hence $u'(c)\leqslant0$ if $c$ is smaller than every median, $u'(c)=0$ if $c$ is a median, and $u'(c)\geqslant0$ if $c$ is greater than every median.

The formula for $\mathrm E(|X-c|)$ is the integrated version of the relations  $$(x-y)^+=\int_y^{+\infty}[t\leqslant x]\,\mathrm dt$$ and $|x-c|=((-x)-(-c))^++(x-c)^+$, which yield, for every $x$ and $c$,
$$
|x-c|=\int_{-\infty}^c[x\leqslant t]\,\mathrm dt+\int_c^{+\infty}[x\geqslant t]\,\mathrm dt
$$
