# section 5.5

## Table of Contents

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 $[0,4]$