I have been looking at
$$
v_{avg} = \frac{v_{i} + v_{f}}{2},
$$
when the acceleration is constant, where $v_i$ is equal to the initial velocity and $v_f$ is equal to the final velocity.
How can you derive this using calculus?
$$
v_{avg} = \frac{\Delta x}{\Delta t} = \frac{1}{t_{f} – t_{i}} \int_{t_{i}}^{t_{f}} v(t) dt
$$
We know that
$$
v(t) = at + v_0,
$$
thus by substituting this into the previous integral yields
$$
v_{avg} = \frac{1}{t_{f} – t_{i}} \left[a\frac{t_f^2}{2} + v_{0} t_f – a\frac{t_i^2}{2} – v_{0} t_i\right].
$$
But there's nothing else, which I can think of. Idea?
Best Answer
There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.