1 Average value of a function

Suppose a time-varying function $f\left(t\right)$ is defined on the interval $a\le t\le b$ . The area, $A$ , under the graph of $f\left(t\right)$ is given by the integral    $A={\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)dt$ . This is illustrated in Figure 5.

Figure 5

On Figure 3 we have also drawn a rectangle with base spanning the interval $a\le t\le b$ and which has the same area as that under the curve. Suppose the height of the rectangle is $m$ . Then

area of rectangle = area under curve $\Rightarrow m\left(b-a\right)={\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)dt\Rightarrow m=\frac{1}{b-a}{\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)dt$

The value of $m$ is the mean value of the function across the interval $a\le t\le b$ .

The mean value depends upon the interval chosen. If the values of $a$ or $b$ are changed, then the mean value of the function across the interval from $a$ to $b$ will in general change as well.

Example 2

Find the mean value of $f\left(t\right)=t^2$ over the interval $1\le t\le 3$ .

Solution

Using Key Point 2 with $a=1$ and $b=3$ and $f\left(t\right)=t^2$

$\text{mean value}=\frac{1}{b-a}{\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)dt=\frac{1}{3-1}{\int \; <!--nolimits\; -->}_1^3{t}^{2}dt=\frac{1}{2}{\left[\frac{t^3}{3}\right]}_1^3=\frac{13}{3}$

Task!

Find the mean value of $f\left(t\right)=t^2$ over the interval $2\le t\le 5$ .

Use Key Point 2 with $a=2$ and $b=5$ to write down the required integral:

Answer

Now evaluate the integral:

Answer