Question

What is the average value of the function `f(x) = x^2` on the interval `[0,3]`?

Answer 1

The average value is `3`.

Explanation:

The average value of a function `f` on an interval `[a,b]` is

`1/(b-a) int_a^b f(x) dx`

So the value we seek is

`1/(3-0)int_0^3 x^2 dx`

`= 1/3 x^3/3]_0^3`

`= (3)^3/9 - (0)^3/9 = 3`