Average Value of a Function
Define and compute the average value of f on [a,b] as (1/(b−a))∫ₐᵇ f(x)dx. Apply the Mean Value Theorem for Integrals and interpret average value in physics and economics contexts.
A lesson of Calculus, in Applications of Integration, about 10 minutes.
By the end of this lesson, you will be able to compute the average value of a function on an interval, find the point(s) guaranteed by the Mean Value Theorem for Integrals, and apply average value to physics and engineering.
Average Value Formula
The average value of $f$ on $[a,b]$:
$$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x),dx$$
Why: the average of $n$ values is $(f(x_1) + \cdots + f(x_n))/n$. As $n \to \infty$: Riemann sum over $(b-a)$ → the integral formula.
Mean Value Theorem for Integrals
If $f$ is continuous on $[a,b]$, then there exists $c \in (a,b)$ such that:
$$f(c) = f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x),dx$$
Geometrically: there is a rectangle with base $[a,b]$ and height $f(c)$ that has the same area as the region under the curve.
Example 1 — Average Value of a Polynomial
$f(x) = x^2 - 3x + 5$ on $[1, 4]$.
$f_{\text{avg}} = \frac{1}{3}\int_1^4(x^2-3x+5),dx = \frac{1}{3}\left[\frac{x^3}{3} - \frac{3x^2}{2} + 5x\right]_1^4$
$= \frac{1}{3}\left[\left(\frac{64}{3}-24+20\right) - \left(\frac{1}{3}-\frac{3}{2}+5\right)\right]$
$= \frac{1}{3}\left[\frac{64}{3} - 4 - \frac{1}{3} + \frac{3}{2} - 5\right] = \frac{1}{3}\left[21 - \frac{15}{2}\right] = \frac{1}{3}\cdot\frac{27}{2} = \frac{9}{2}$
Example 2 — Finding $c$ in MVT for Integrals
From Example 1: find $c$ where $f(c) = 9/2$.
$c^2 - 3c + 5 = 9/2 \implies c^2 - 3c + 1/2 = 0$
$c = \frac{3 \pm \sqrt{9-2}}{2} = \frac{3 \pm \sqrt{7}}{2}$
Both values: $c \approx 2.82$ and $c \approx 0.18$.
Only $c \approx 2.82$ is in $(1, 4)$ ✓.
Example 3 — Average Velocity
A car's velocity is $v(t) = t^2 + 2t$ m/s from $t = 0$ to $t = 3$. Find the average velocity.
$v_{\text{avg}} = \frac{1}{3}\int_0^3(t^2+2t),dt = \frac{1}{3}\left[\frac{t^3}{3} + t^2\right]_0^3 = \frac{1}{3}(9+9) = 6$ m/s
Example 4 — Average Temperature
Temperature over 12 hours: $T(t) = 50 + 5\sin\left(\frac{\pi t}{12}\right)$ (°F). Find the average.
$T_{\text{avg}} = \frac{1}{12}\int_0^{12}\left[50 + 5\sin\left(\frac{\pi t}{12}\right)\right]dt$
$= \frac{1}{12}\left[50t - \frac{60}{\pi}\cos\left(\frac{\pi t}{12}\right)\right]_0^{12}$
$= \frac{1}{12}\left[\left(600 - \frac{60}{\pi}\cos\pi\right) - \left(0 - \frac{60}{\pi}\cos 0\right)\right]$
$= \frac{1}{12}\left[600 + \frac{60}{\pi} + \frac{60}{\pi}\right] = 50 + \frac{10}{\pi} \approx 53.18°F$
Summary
Average value: $\dfrac{1}{b-a}\int_a^b f(x),dx$