To my knowledge, there are two ways of finding average (what average I am not too sure, forgot what my teacher said. But I think it has something to do with the mean value theorem)
Say you have a velocity-versus-time graph:
The 1st way of finding the average (acceleration?) is by using the slope of secant line: $$m_\text{secant}=\frac{v(b)-v(a)}{b-a}$$
Well, since (I just found out) the secant method is for finding average acceleration, when using the integral method, we need to first take the derivative of v(t), whose graph is shown, to get a(t). $$a_\text{avg}=\frac{1}{b-a} \int_a^b a(t) \, dt$$
If we assume that $$ v'(t)=a(t),\quad t\in [a,b], $$ then there is no difference, since $$ \int_{a}^{b} a(t) dt=\int_{a}^{b} v'(t) dt=v(b)-v(a), $$ giving $$ a_{avg}=\frac{1}{b-a} \int_{a}^{b} a(t) dt=\frac{v(b)-v(a)}{b-a}=m_{secant}. $$