Find the DC component of a Full wave rectifier (DC Power supply)
What you need to do is compute the average of a function. The definition of the average of a function over the interval \$[a,b]\$ is:

$$
\overline{f} = \frac{1}{b-a}\int_a^bf(x)\,dx
$$

The sine is a periodic function, so the average must be computed over a period. The absolute value of a sine is still a periodic function, and its period is T=\$\frac{\pi}{\omega_0}\$, so \$a=0\$ and \$b=T\$. Let's throw in the formula what you've got:

$$
V_{dc}=\overline{V_r(t)} = \frac{1}{T-0}\int_0^T|V_msin(\omega_0t)|\,dx=...
$$

But in that interval the sine is always positive so we can take away the absolute value:

$$
...= \frac{1}{T-0}\int_0^T V_msin(\omega_0t)\,dx=\frac{V_m}{T}\cdot\Bigg[-\frac{cos(\omega_0t)}{\omega_0}\Bigg]_0^T = \frac{V_m}{T\omega_0}\cdot\Bigg(1-cos(\omega_0T)\Bigg) = \frac{V_m\omega_0}{\pi\omega_0}\cdot\Bigg(1-cos(\omega_0\cdot\frac{\pi}{\omega_0})\Bigg) = 2\cdot\frac{V_m}{\pi}
$$

And there you go.