# Improper Integral

• Feb 9th 2010, 06:25 PM
Belowzero78
Improper Integral
1. Determine whether the improper integral converges or diverges.

http://img194.imageshack.us/img194/3...erintegral.png
• Feb 9th 2010, 06:29 PM
VonNemo19
Where are you stuck?
• Feb 9th 2010, 06:32 PM
Belowzero78
Well , I first let u=x-2, so du=dx. Also, when x=22, u=20, and when x=20,u=18.

In another question exactly the same just different numbers, i did my own procedure and it worked out, but for this one, im not sure if I have to split up the integral into two parts since last time the denominator of the integral u approached 0 (unbounded near u=0).
• Feb 9th 2010, 06:49 PM
Belowzero78
anyone know how to solve this?
• Feb 9th 2010, 06:50 PM
VonNemo19
Note that

$\displaystyle |x-21|=\left\{\begin{array}{cc}-(x-21),&\mbox{ if } x\leq 21\\x-21, & \mbox{ if } x>21\end{array}\right.$

See where I'm goin here? You've gotta break it up.
• Feb 9th 2010, 06:53 PM
Belowzero78
yeah, i understand that, but im not sure what are the upper and lower bounds of the integral with a negative sign since x less than or equal to 21.
• Feb 9th 2010, 07:02 PM
VonNemo19
=$\displaystyle \lim_{x\to{21}^-}\int_{20}^{21}\frac{1}{\sqrt[3]{-(x-21)}}dx+\lim_{x\to21^+}\int_{21}^{22}\frac{1}{\sqr t[3]{x-21}}$
• Feb 9th 2010, 07:24 PM
Belowzero78
does it converge to 3? This is what i got as an answer.
• Feb 9th 2010, 07:39 PM
Belowzero78
Anyone?