Averages with Functions - At A Glance

Let f (x) = 4 – x². Is the average value of f on [-2, 2]

• strictly between 0 and 2?

• equal to 2?

• strictly between 2 and 4?

• equal to 4?

The graph of f looks like this:

We want a rectangle on [-2, 2] whose area is the same as the area between f and the x-axis on [-2, 2]. The height of that rectangle is the average value of f on [-2, 2]. Since this is an example, we'll try out all the options offered in the problem and see which one works.

• If the average value of f were strictly between 0 and 2, we would get a picture like this.

This rectangle doesn't cover as much area as it needs to, so the average value can't be between 0 and 2.

• If the average value of f were equal to 2, the rectangle still doesn't cover quite enough area. The area in the rectangle but not under f doesn't quite make up for the area under f but not in the rectangle.

• Having the average value of f strictly between 2 and 4 could work.

• The average value of f can't be 4, because that's too big. The rectangle would cover all the area between f and the x-axis on [0 ,2], and then some!

This is like the story of Goldilocks and the 3 Bears. The area under m can't be too big or too small; it needs to be just right.

The average value of f on [a, b] is the constant m for which the weighted area between m and the x-axis on [a, b] equals the weighted area between f and the x-axis on [a, b]. In symbols, it's the value of m for which

.

We know what happens when we integrate a constant m from a to b: we get m(b – a). So the average value of f on [a, b] is the value of m for which

If we divide both sides by b – a we get the formula

.

This is the formula that shows up in most textbooks. We like the equation

better because it helps us remember what the average value m is, but we admit the formula

is quite handy when it comes to actually calculating average values.

Calculate the average value of  on [0, 3].

Using the handy formula, the average value of f (x) on [0, 3] is

.

The graph of f(x) on [0,3] is a quarter of a circle of radius 3. This means the integral of f from 0 to 3 is

So the average value of f on [0,3] is

When finding average values, it's a good idea to do a sanity check. If the function mostly accumulates area above the x-axis on an interval, its average value on that interval should be positive. If the function mostly accumulates area below the x-axis on an interval, its average value on that interval should be negative.