improper integration

**davecs77** · July 15th 2007, 05:59 PM

I am trying to do this
integral sign going from 0 to 4 (0 on the bottom, 4 on the top) 1/(4-x) dx
i got stuck.
Thanks a lot!!!

**Krizalid** · July 15th 2007, 06:30 PM

$\int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$

**davecs77** · July 15th 2007, 06:36 PM

Originally Posted by Krizalid

$\int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$

Yeah i was able to get that far, but not so sure how to solve it.

**Jhevon** · July 15th 2007, 06:52 PM

Originally Posted by davecs77

Yeah i was able to get that far, but not so sure how to solve it.

If you got that far, what's the problem?

$\lim_{a \to 4} \int_0^a \frac {1}{4 - x}~dx$

$= \lim_{a \to 4} \left[ - \ln (4 - x) \right]_{0}^{a}$

$= \lim_{a \to 4} \left( - \ln (4 - a) + \ln 4 \right)$

Now take the limit

Thread: improper integration

LinkBack

Thread Tools

Display

improper integration

Similar Math Help Forum Discussions

[SOLVED] Improper Integration

Improper Integration

improper integration by parts

Improper Integration

Improper Integration

Search Tags