| By Cantwait4colleg (Cantwait4colleg) on Monday, May 10, 2004 - 10:40 pm: Edit |
Stupid question...
Integral of (1/x)dx from 0 to 1...
| By Miseryxsignals (Miseryxsignals) on Monday, May 10, 2004 - 10:52 pm: Edit |
integral of 1/x = lnx.
soo... ln(1) - ln(0). And you cant do the ln(0), Sorry.
| By Athlonmj (Athlonmj) on Monday, May 10, 2004 - 10:54 pm: Edit |
Yes you can, it's an improper integral so substitute a for 0 and take the limit as it approaches 0.
| By Efilsiertaeht (Efilsiertaeht) on Monday, May 10, 2004 - 11:05 pm: Edit |
approaches 0, not infinity.
| By Sandwraith (Sandwraith) on Monday, May 10, 2004 - 11:05 pm: Edit |
either way the point is that the value of the integral is not a finite answer
| By Mike105 (Mike105) on Monday, May 10, 2004 - 11:07 pm: Edit |
ò0 1(1/x)dx = ¥
| By Miseryxsignals (Miseryxsignals) on Monday, May 10, 2004 - 11:13 pm: Edit |
haha oops. Shows what a year w/o math can do to you. SOrry about that, thanks for catching my mistake! 
| By Ubercollegeman (Ubercollegeman) on Monday, May 10, 2004 - 11:40 pm: Edit |
That is an improper integral. If you want the full way to do it..
lim a->0+ ln(1) - ln(a) =
lim a->0+ 0 - lna = -(-infty) = infty.
^not good mathematical notation, but whatever!
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