The average value of a function

The mean of finitely many numbers is their sum divided by their count; the mean of a function is its integral divided by the length of the interval. We find the average value of $f(x) = x^2$ on $0 \leq x \leq 2$ .

\frac{1}{2 - 0} \int_0^2 x^2 \, dx = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}

The average value is $\dfrac{4}{3} \approx 1.333$ . The values of $f$ run from $0$ to $4$ , but levelled out they come to this height.

Replacing it with a rectangle

The operation replaces the region by a rectangle of the same area.

Figure	Dimensions	Area
Under the parabola	the interval $[0, 2]$	$\dfrac{8}{3}$
Rectangle	width $2$ , height $\dfrac{4}{3}$	$\dfrac{8}{3}$

The crests and troughs of the curve balance out exactly at that height.

The point where the average is attained

Solving $c^2 = \dfrac{4}{3}$ gives $c = \dfrac{2}{\sqrt{3}} \approx 1.1547$ , inside the interval. That such a $c$ always exists is the mean value theorem for integrals¹.

f(c) = \frac{1}{b - a} \int_a^b f(x) \, dx

For a continuous function the average lies between the minimum and the maximum, and every intermediate value is attained, so the average itself is reached somewhere.

It changes with the interval

Interval	Average value
$0 \leq x \leq 1$	$\dfrac{1}{3}$
$1 \leq x \leq 2$	$\dfrac{7}{3}$
$0 \leq x \leq 2$	$\dfrac{4}{3}$

Averaging the first two returns $\dfrac{4}{3}$ . When an interval is split into pieces of equal length, the average of the averages is the original average.

Applications

Integrating a velocity over time and dividing by the time gives the average speed
The root mean square of a current comes from the average of its square
A representative value obtained by levelling out measurements

The parabola on the graph is $y = x^2$ , the horizontal line is the average value $y = \dfrac{4}{3}$ , and the large dots are the two ends of the interval together with the point where the average is attained.

Mean value theorem, Wikipedia