Question

What is `f(x) = int x/sqrt(x-2) dx` if `f(3)=-1`?

Answer 1

`f(x) = 2/3(x -2)^(3/2) + 4(x- 2)^(1/2) - 17/3`

Explanation:

Let `u = x - 2`. Then `du = dx`, and `x = u + 2`.

`f(x) = int x/sqrt(u)dx`

`f(x) = int (u +2)/sqrt(u) du`

`f(x) = int u/sqrt(u) + 2/sqrt(u) du`

`f(x) = int u^(1/2) + 2u^(-1/2) du`

`f(x) = 2/3u^(3/2) + 4u^(1/2) + C`

`f(x) = 2/3(x - 2)^(3/2) +4(x- 2)^(1/2) + C`

Now we solve for `C`.

`-1 = 2/3(3- 2)^(3/2) + 4(3 -2)^(1/2) + C`

`-1 = 2/3 + 4 + C`

`-17/3 = C`

Therefore the function is

`f(x) = 2/3(x -2)^(3/2) + 4(x- 2)^(1/2) - 17/3`

Hopefully this helps!