Main Concept

Given a continuous function $f\left(x\right)$, its average value over [a , b] is defined as $\frac{1}{b - a}{\int }_a^b f\left(x\right) \ⅆx$.

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 ${\int }_a^{b}v\left(t\right) \ⅆt$.  Therefore, your average driving speed was $\frac{1}{b - a}{\int }_a^b v\left(t\right) \ⅆt$, 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 $\left(b-a\right)$ on both sides equals:

 $f\left(t\'\right)\left(b - a\right) \= {\int }_a^{b}f\left(t\right) \ⅆt$ .

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.

More MathApps

MathApps/Calculus