Section 5.5 Average Value of a Function
Recall Averages
If we want to find the average of a dataset: $$ { 3, 5, 2, 9, 0, 8, 1 } $$ we add together all the elements and divide by the total number of them: $$ \dfrac{3+ 5+ 2+ 9+ 0+ 8+ 1}{7} = 4 $$ … turns out our average of my set of made up numbers is 4. Nice.
Continuous averages
… what instead of discrete numbers, I have a continuous set of values. We still want to add together all the values and divide by then size of that set. How can we do this? Integration is repeated addition! Length of the interval is the size of the set!
Average Value of a Function
If $f$ is a continuous function on an interval $[a, b]$, we define the average value of $f$ on that interval to be: $$ f_{\text{ave}} = \dfrac{1}{b-a} \int_a^b f(x)\, dx $$
Notation note The text uses $f_{\text{ave}}$ but I like to abbreviate average by “avg,” so if you see me write $f_{\text{avg}}$, I mean the exact same thing. It’s my fave way of writing it.
Example
Find the average value of $\cos x$ on the interval $[0, \pi/2]$.
Solution:
Click to reveal the answer.
$\begin{align*} f_\text{ave} &= \dfrac{1}{b-a} \int_a^b f(x) \, dx \\ &= \dfrac{1}{\frac\pi2 - 0} \int_0^{\pi/2} \cos x\, dx \\ &= \frac2\pi (\sin x) \Big|_{x=0}^{\pi/2} \\ &= \frac2\pi (\sin (\pi/2) - \sin 0) \\ &= \frac2\pi \end{align*}$Example
Find the average value of the function $\sin x$ between $0$ and $\pi$.
Solution:
Click to reveal the answer.
$\begin{align*} f_\text{ave} &= \dfrac{1}{b-a} \int_a^b f(x) \, dx \\ &= \dfrac{1}{\frac\pi - 0} \int_0^{\pi} \sin x\, dx \\ &= \frac1\pi (-\cos x) \Big|_{x=0}^{\pi} \\ &= \frac1\pi (-\cos (\pi) - -\cos 0) \\ &= \frac1\pi (-(-1) + 1)\\ &= \frac{2}{\pi} \end{align*}$Example
Find the average value of $f(x) = x^2$ between -2 and 2.
Solution:
Click to reveal the answer.
$\begin{align*} f_\text{ave} &= \dfrac{1}{b-a} \int_a^b f(x) \\, dx \\ &= \dfrac{1}{2 - (-2)} \int_{-2}^2 x^2 \\, dx \\ &= \dots\\ &= \frac43 \end{align*}$Mean Value Theorem for Integrals
We had a Mean Value Theorem for derivatives which told us that there is a point where the function attains its mean/average value. Here we get the same for integrals:
If $f$ is continuous on $[a,b]$ then there exists a number $c$ in $[a, b]$ so that $$ f(c) = \dfrac{1}{b-a} \int_a^b f(x)\, dx $$
or equivalently, $$ \int_a^b f(x) \, dx = f(c) (b-a) $$
Example
Find the values $c$ which satisfy the mean value theorem of integrals for $f(x) = x^2$ on the interval $[-2, 2]$.
Solution: Note: Above we found $f_{\text{ave}} = \frac43$.
Click to reveal the answer.
$f(c) = f_{\text{ave}} = \frac{4}{3}$ so that $c^2 = \frac43$ or $c= \pm \frac{2\sqrt{3}}{3}$Example
Find all values $c$ which satisfy the mean value theorem for integrals for the function $f(x) = \sqrt{x}$ on $[0,4]$