Question

Question #7bb57

Answer 1

Solve `sqrt(x+1) = 1/3 int_0^3 sqrt(x+1) dx`

Explanation:

The mean value theorem for integrals says that if `f` is continuous on `[a,b]` then there is a `c` in `(a,b)` such that `f(c)` equals the average value of `f` on `[a,b]`.

To find these `c`'s, solve the equation

`f(x) = 1/(b-a) int_a^b f(x) dx`. Retain only solutions in `(a,b)`.

In this question, we get

`sqrt(x+1) = 1/3 {:(x+1)^(3/2)]_0^3 = 7/3`

So `x = 40/9`