Answer
$\frac{11}{9}$.
Work Step by Step
We can break the distribution into two parts: the section from 0 to $v_0$ , and the section from $v_0$ to $2v_0$. The first section is increasing linearly, so its average value is $2/3$ of the way to the right, because of expected value and integral of x^2 over integral of x being 2/3. The latter section is just uniform, so the average value is at $1.5v_0$. This section has twice the weight of the former, as it holds twice the area. Taking the weighted mean $(1.5v_0*\frac{2}{3}+\frac{2v_0}{3}*\frac{1}{3})$, we get $\frac{11}{9}v_0$, so the answer we seek is $\frac{11}{9}$.
