I have the type 2 improper integral $\lim_{t \to 1^-} \int^t_0 \dfrac{1}{\sqrt{|x-1|}} \: dx$.
My calculations are as follows:
$\lim_{t \to 1^-} \int^t_0 \dfrac{1}{\sqrt{1-x}} \: dx$
Let $u = 1 - x$.
$\implies du = -dx$
Upper limit of integration becomes $u = 1 - t$
Lower limit of integration becomes $u = 1$
$\lim_{t \to 1^-} \int^{1 - t}_1 \dfrac{-1}{\sqrt{u}} \: du$ = $\lim_{t \to 1^-} \left( \dfrac{-2}{\sqrt{u}} \right)^{1 - t}_1$
When I changed the limits of integration due to substitution, did I also have to change the limit value?
$= \lim_{t \to 1^-} \left( \dfrac{-2}{\sqrt{1-t}} + 2\right)$
$= -\infty + 2 $
Therefore, this improper integral is divergent.
I would greatly appreciate it if people could please take the time to review my calculations and check if they're correct. I would also appreciate a response with regards to my question about changing the limit value when I changed the limits of integration.