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

Question

4975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-14T02:02:46

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

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

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

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

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