1. Determine whether the improper integral converges or diverges.

http://img194.imageshack.us/img194/3...erintegral.png

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- Feb 9th 2010, 06:25 PMBelowzero78Improper Integral
1. Determine whether the improper integral converges or diverges.

http://img194.imageshack.us/img194/3...erintegral.png - Feb 9th 2010, 06:29 PMVonNemo19
Where are you stuck?

- Feb 9th 2010, 06:32 PMBelowzero78
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 PMBelowzero78
anyone know how to solve this?

- Feb 9th 2010, 06:50 PMVonNemo19
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 PMBelowzero78
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 PMVonNemo19
=$\displaystyle \lim_{x\to{21}^-}\int_{20}^{21}\frac{1}{\sqrt[3]{-(x-21)}}dx+\lim_{x\to21^+}\int_{21}^{22}\frac{1}{\sqr t[3]{x-21}}$

- Feb 9th 2010, 07:24 PMBelowzero78
does it converge to 3? This is what i got as an answer.

- Feb 9th 2010, 07:39 PMBelowzero78
Anyone?