Question

How do you find the integral of `int dx/(sqrt(2x-1)` from 1/2 to 2?

Answer 1

Try the substitution `u = 2x-1` in hopes of getting `int u^(-1/2) du`.

Explanation:

With `u = 2x-1`, we get `du=2dx` so the indefinite integral becomes

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

The limits of integration change from `x=1/2` to `u=2(1/2)-1 = 0`
and from `x=2` to `u=2(2)-1=3`

The new problem is to evaluate

`1/2int_0^3 u^(-1/2) du`

`= [u^(1/2)]_0^3 = sqrt3`