Why is average velocity the midpoint of initial and final velocity under constant acceleration?

Sticking to one dimension for simplicity, at a constant acceleration, $a$, the distance travelled in a time $t$ is simply:

$$ s = v_0 t + \frac{1}{2}at^2 $$

So the average velocity, $v_{av} = s/t$, is:

$$ v_{av} = v_0 + \frac{1}{2}at $$

But acceleration $\times$ time is just the change in velocity i.e. $at = v - v_0$ so:

$$ v_{av} = v_0 + \frac{1}{2}(v - v_0) = \frac{v_0 + v}{2} $$

Note that $\vec{\mathbf v}_\mathrm{av}$ is defined as the average value of $\vec{\mathbf v}$: $$\vec{\mathbf v}_\mathrm{av}:=\frac{1}{t_1-t_0}\int_{t_0}^{t_1}\vec{\mathbf v}(t)\,\mathrm dt.$$ Since $\vec{\mathbf x}$ is the antiderivative of $\vec{\mathbf v}$, this equals $$\frac{\vec{\mathbf x}(t_1)-\vec{\mathbf x}(t_0)}{t_1-t_0}.$$ However, when acceleration is constant, and thus $\vec{\mathbf v}$ is a line (that is, $\vec{\mathbf v}(t)=\vec{\mathbf a}t+\vec{\mathbf v}_0$), then by plugging into the average value integral, you obtain the equality $$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf v}(t_1)+\vec{\mathbf v}(t_0)}{2}.$$