Given a continuous function $f\left(x\right)$, its average value over [a , b] is defined as $\frac{1}{ba}{\int}_{a}^{b}f\left(x\right)DifferentialD;x$.
Suppose that you are driving a car in a straight line and that v(t) is your velocity at time, t. The distance you traveled over the time interval from t = a to t = b can be written using the definite integral ${\int}_{a}^{b}v\left(t\right)DifferentialD;t$. Therefore, your average driving speed was $\frac{1}{ba}{\int}_{a}^{b}v\left(t\right)DifferentialD;t$, which represents the average value of the velocity function.
A geometric way to interpret the average value of a function is in terms of area. According to the Mean Value Theorem for integrals, given a continuous function f(x) defined over the closed interval $\left[a\,b\right]$, there exists a point $t\'$ in the interior $\left(a\,b\right)$ such that: ${}$$f\left(t\'\right)\,$the instantaneous value of $f$ at $t\'$ is equal to the average value of $f\left(x\right)$ over the whole interval $\left[a\,b\right]$. Thus, multiplying by $\left(ba\right)$ on both sides equals:
$f\left(t\'\right)\left(ba\right)equals;{\int}_{a}^{b}f\left(t\right)DifferentialD;t$ .
In other words, the area is defined by a rectangle, whose height is the average value of a function and whose width is an interval equal to the area under the entire function, occurring over the same interval.
