Average Value of a Function

Average Value and the Mean Value Theorem for Integrals

By the end of this lesson, you will be able to compute the average value of a function and apply the Mean Value Theorem for Integrals.

Average Value of a Function

The average value of $f$ on $[a, b]$ is:

$$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x), dx$$

Why this formula? The average of a finite set of values is $\frac{\text{sum}}{n}$. For a continuous function, the integral replaces the sum and $b-a$ replaces $n$.

Geometric interpretation: $f_{\text{avg}}$ is the height of the rectangle with the same base $[a,b]$ and the same area as the region under $f(x)$.

The Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, there exists at least one $c \in [a, b]$ such that:

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

Interpretation: a continuous function achieves its average value at least once in the interval.

Examples

Example 1: Average value of $f(x) = x^2$ on $[0, 3]$:

$f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2, dx = \frac{1}{3}\cdot\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot 9 = 3$

The average value is 3.

Example 2: Temperature $T(t) = 60 + 10\sin\left(\frac{\pi t}{12}\right)$ (°F) over a 24-hour period. Find the average temperature.

$T_{\text{avg}} = \frac{1}{24}\int_0^{24}\left(60 + 10\sin\frac{\pi t}{12}\right), dt$

$= \frac{1}{24}\left[60t - \frac{120}{\pi}\cos\frac{\pi t}{12}\right]_0^{24}$

$= \frac{1}{24}\left[\left(1440 - \frac{120}{\pi}(1)\right) - \left(0 - \frac{120}{\pi}(1)\right)\right] = \frac{1}{24}(1440) = 60°F$

(The average is just the midpoint of the sine oscillation, which makes intuitive sense.)

Summary

Average value: $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x), dx$

MVT for Integrals: if $f$ is continuous, some $c$ satisfies $f(c) = f_{\text{avg}}$.

Geometric: $f_{\text{avg}}$ = height of rectangle with equal area under the curve.