## anonymous 4 years ago I really need help on improper integrals, please respond?

1. anonymous

ok

2. anonymous

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

3. anonymous

this isn't an improper integral

4. anonymous

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

5. anonymous

ok

6. anonymous

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

7. anonymous

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".

8. anonymous

We need to split this integral

9. anonymous

what are the steps?

10. anonymous

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

11. anonymous

do you know what to do next?

12. anonymous

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

13. anonymous

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

14. anonymous

I'll let you take it from here

15. anonymous

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

16. anonymous

yeppers

17. anonymous

dont forget to medal me to close this question