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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.