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)=sqrt(2x)`?

Answer 1

Please see below.

Explanation:

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

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

So we need to find the average value, then solve the equation `f(x) = "average value"` on the interval `[a,b]`. If the equation has more than one solution then `c` can be any one of the solutions in `[a,b]`

In this question we get

Solve: `sqrt(2x) = 1/(2-0) int_0^2 sqrt(2x) dx`

Evaluating the integral gets us:

Solve: `sqrt(2x) = 4/3`.

The only solution is `x=8/9`