Question

Really quick u sub question

Answer 1

\[\int\limits_{a}^{b}\sqrt{4x-1}\]

let u = sqrt 4x-1

du=4 dx
du/4 = dx and then integrate normally.

Answer 2

\[1/4\int\limits_{a}^{b}\sqrt{u} du\]

Answer 3

let u = 4x - 1
du = 4 dx
dx = 1/4 du

see your integral becomes
int 1/4 sqrt(u) du
= 1/4 int sqrt(u) du
= 1/4 int u^1/2
= ... continue it

Answer 4

I just wanted to make sure that was right. Because if I let u= square root of 4x-1 then the only term left is dx so I need only dx on the right side