1 Average value of a function

Suppose a time-varying function [maths rendering] is defined on the interval [maths rendering] . The area, [maths rendering] , under the graph of [maths rendering] is given by the integral    [maths rendering] . This is illustrated in Figure 5.

Figure 5

{a. the area under the curve from $t=a$ to $t=b$. b. the area under the curve and the area of the rectangle are equal}

On Figure 3 we have also drawn a rectangle with base spanning the interval [maths rendering] and which has the same area as that under the curve. Suppose the height of the rectangle is [maths rendering] . Then

area of rectangle = area under curve [maths rendering]

The value of [maths rendering] is the mean value of the function across the interval [maths rendering] .

Key Point 2

The mean value of a function [maths rendering] in the interval [maths rendering] is [maths rendering]

The mean value depends upon the interval chosen. If the values of [maths rendering] or [maths rendering] are changed, then the mean value of the function across the interval from [maths rendering] to [maths rendering] will in general change as well.

Example 2

Find the mean value of [maths rendering] over the interval [maths rendering] .

Solution

Using Key Point 2 with [maths rendering] and [maths rendering] and [maths rendering]

[maths rendering]

Task!

Find the mean value of [maths rendering] over the interval [maths rendering] .

Use Key Point 2 with [maths rendering] and [maths rendering] to write down the required integral:

[maths rendering]

Now evaluate the integral:

[maths rendering]