Why is average velocity the midpoint of initial and final velocity under constant acceleration? Since average velocity is defined as$^1$
$$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf x}-\vec{\mathbf x}_0}{t-t_0},$$
where $\vec{\mathbf x}$ denotes position, why is this quantity equal to
$$\frac{\vec{\mathbf v}+\vec{\mathbf v}_0}{2},$$
where $\vec{\mathbf v}=\frac{d\vec{\mathbf x}}{dt}$ and $\vec{\mathbf v}_0=\left.\frac{d\vec{\mathbf x}}{dt}\right|_{t=t_0}$, when acceleration is constant?
What in particular about constant acceleration allows average velocity to be equal to the midpoint of velocity?
$^1$: Resnick, Halliday, Krane, Physics (5th ed.), equation 2-7.
 A: 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:
$$ v_{av} = v_0 + \frac{1}{2}(v - v_0) = \frac{v_0 + v}{2} $$
A: 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
$$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf v}(t_1)+\vec{\mathbf v}(t_0)}{2}.$$
