I really need help on improper integrals, please respond?

Question

anonymous · Answer

ok

anonymous · Answer

Okay, so I'm really confused as to what to do. Let's do an example problem, $$\int\limits_{0}^{1} dx/3x-2$$

anonymous · Answer

this isn't an improper integral

anonymous · Answer

It's dx over 3x-2, i'm looking at my book right now.

anonymous · Answer

ok

anonymous · Answer

okay. so I don't know what to do.

anonymous · Answer

well we can't have a zero denominator, which means we can't have x=2/3.  This lies within our limits of integration which is why the integral is "improper".

anonymous · Answer

We need to split this integral

anonymous · Answer

what are the steps?

anonymous · Answer

$$\int\limits_{0}^{2/3}\frac{dx}{3x-2}+\int\limits_{2/3}^{1}\frac{dx}{3x-2}$$

anonymous · Answer

do you know what to do next?

anonymous · Answer

We evaluate the one-sided limits of each term separately:$$\int\limits_{0}^{2/3}\frac{dx}{3x-2}=\lim_{t \rightarrow 2/3^{-1}}\int\limits_{0}^{t}\frac{dx}{3x-2}$$

anonymous · Answer

I get:$$\lim_{t \rightarrow (2/3)^{-1}}\frac{1}{3}(\ln \left| 3t-2 \right|-\ln \left| 0-2 \right|)$$

anonymous · Answer

I'll let you take it from here

anonymous · Answer

Yes, it diverges so you wouldn't continue. Therefore the answer you just be that.

anonymous · Answer

yeppers

anonymous · Answer

dont forget to medal me to close this question