Question

What is the average value of the function `f(x) = x - (x^2)` on the interval `[0,2]`?

Answer 1

The average value of `f` on `[a,b}` is `1/(b-a) int_a^b f(x) dx`.

Explanation:

For this function on this interval, I get `-1/3`

Ave`= 1/(2-0) int_0^2 (x-x^2) dx`

`= 1/2[x^2/2-x^3/3]_0^2`

`= 1/2[(4/2-8/3)-(0)]`

`= 1/2(-2/3) = -1/3`