Finding a function's average value over a particular interval

What is the average value of a function?

In the same way that we can find the average of set of numbers, we can also find the average value of a function over a specific interval.

The formula we use to find the average value of a function $f(x)$ over the interval $[a,b]$ is

$f_{avg}=\frac{1}{b-a}\int^b_af(x)\ dx$

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Think about the average value of a function as the average height the function attains above the $x$-axis. If the function were $y=3$, then the height of the function is always $3$ everywhere, so the average height of the function would also be $3$. When the function gets more complicated, we can use the average value formula to find its average height on $[a,b]$.

How to calculate a function’s average value over a particular interval

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Using the average value integration formula

Example

Calculate the average value of the function over the interval.

$f(x)=x^3-2x^2+e^{2x}$

on $[3,7]$

We’ll use the formula for average value

$f_{avg}=\frac{1}{b-a}\int^b_af(x)\ dx$

and get

$f_{avg}=\frac{1}{7-3}\int^7_3x^3-2x^2+e^{2x}\ dx$

$f_{avg}=\frac{1}{4}\int^7_3x^3-2x^2+e^{2x}\ dx$

Think about the average value of a function as the average height the function attains above the x-axis.

Next we can break the integral apart by term.

$f_{avg}=\frac{1}{4}\int^7_3x^3\ dx+\frac{1}{4}\int^7_3-2x^2\ dx+\frac{1}{4}\int^7_3e^{2x}\ dx$

$f_{avg}=\frac{1}{4}\int^7_3x^3\ dx-\frac{2}{4}\int^7_3x^2\ dx+\frac{1}{4}\int^7_3e^{2x}\ dx$

Integrate.

$f_{avg}=\frac14\left(\frac{x^4}{4}\right)\bigg|^7_3-\frac24\left(\frac{x^3}{3}\right)\bigg|^7_3+\frac14\left(\frac{e^{2x}}{2}\right)\bigg|^7_3$

$f_{avg}=\frac{x^4}{16}-\frac{x^3}{6}+\frac{e^{2x}}{8}\bigg|^7_3$

Now we can evaluate on the interval.

$f_{avg}=\left[\frac{(7)^4}{16}-\frac{(7)^3}{6}+\frac{e^{2(7)}}{8}\right]-\left[\frac{(3)^4}{16}-\frac{(3)^3}{6}+\frac{e^{2(3)}}{8}\right]$

$f_{avg}=150,367$

The average value of the function $f(x)=x^3-2x^2+e^{2x}$ over the interval $[3,7]$ is $150,367$.

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