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. |