Question

How do you find the average value of `f(x)=4x^(1/2)` as x varies between `[0,3]`?

Answer 1

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

Explanation:

`1/(3-0) int_0^3 4x^(1/2)dx = 4/3 int_0^3 x^(1/2) dx`

`= 4/3[ {: 2/3 x^(3/2)]_0^3]`

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

`= 4/3[2/3 (3sqrt3)]`

`= (8sqrt3)/3`