Hello All,
I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.
Upper limit 2, lower limit 0 (x-1)^-1/3
Any help you can give me would be greatly appreciated.
Thanks!
Hello All,
I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.
Upper limit 2, lower limit 0 (x-1)^-1/3
Any help you can give me would be greatly appreciated.
Thanks!
Put:
$\displaystyle I_{a,b}= \int_0^a (x-1)^{-1/3}\; dx+ \int_b^2 (x-1)^{-1/3}\; dx,\ a<1,\ b>1$
For the integral to converge requires that:
$\displaystyle I=\lim_{a\to 1_-, b \to 1_+}I_{a,b}$
exists, which is eqivalent to requiring that:
$\displaystyle \lim_{a \to 1_-}\int_0^a (x-1)^{-1/3}\; dx$
and
$\displaystyle \lim_{b \to 1_+}\int_b^2 (x-1)^{-1/3}\; dx$
both exist.
CB
Where it converges? I'm not sure I know what you mean. Both the constituent limits exist so the improper integral converges (the only point that there is a problem is around x=1 where the integrand misbehaves, which is why that is the point about which we break up the integral)
CB