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. |