Question

How do you find the point c in the interval `0<=x<=2` such that f(c) is equation to the average value of `f(x)=x^(2/3)`?

Answer 1

Solve the equation `f(x) = 1/(2-0) int_0^2 f(x) dx`. Discard any solutions outside `[0,2]`

Explanation:

`1/(2-0) int_0^2 f(x) dx = 1/2 int_0^2 x^(2/3) dx = (3(2^(2/3)))/5`

So you need to solve

`x^(2/3) = (3(2^(2/3)))/5`.

I get `x = (6sqrt15)/25`