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September 1, 2014

September 1, 2014

Posted by **sarah** on Saturday, February 9, 2008 at 5:34pm.

1/(x-1)

from 0 to 2

this is improper at one, so I split it up into two integrals

ln(x-1) from 0-1 and

ln(x-1) from 1-2

I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1))

and then the same thing for the second part

I didn't know if this was right though, or what the answer would be

- calculus -
**drwls**, Sunday, February 10, 2008 at 10:56amYou used a correct procedure, but both integrals you broke it up into involve the log of zero or a negative number, both of which are "improper". I don't see how one can get a finite integral this way.

Here is another way that might work:

Define the integral of the sum of two limits. One is the integral from 0 to 1-a and the other is the integral from 1+a to 2, as value of a goes to zero. Consider a as always positive. You may find that two two terms always cancel each other whatever a is, so that the limit will be zero.

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