integrals

[leemaj]

ok, so there is this argument in math today about integrals of stuff. Here is what my teacher and i agree on:
http://www.geocities.com/leemaj13/calc.x

The book says that since the graph diverges, then the integral is divergent and so it cant be done. what do you guys think? this might be a good thing to ask your professors too.

[coleslaw]

Over the given integration range, the integral IS divergent and therefore cannot be solved using traditional integration techniques.

[sho.gun]

well I must say... if you drew all that in paint, you did a pretty good job.

as for the question... uhmm.. ehh... :idea:


FALSE!

*clap* *clap* *clap*

[Kevster]

Originally posted by coleslaw
Over the given integration range, the integral IS divergent and therefore cannot be solved using traditional integration techniques.

:stupid:

That's correct in my book (Zill).

[hapoo]

simply put, the integral is a method to find the area under a graph, now since your area has and infinate hole in it (hence infinate area) theres really no integer answer for it. But since your range cancels the infinate parts out through semmetry the only thing remaining is the ln2, that answer is correct

[coleslaw]

Originally posted by hapoo
now since your area has and infinate hole in it (hence infinate area) theres really no integer answer for it.That's not always the case. There can be a finite area within an infinite boundary.

[Tse How]

yah, u cant take ln of a negative number, that should be obvious right off the bat.

[helius]

Split the integral into two parts, i.e. integrate the thing from (-1, +1) and (+1, +2). 1/x is an odd function, therefore the first part will be zero. You're left with the integrating it in the range of (+1, +2).

Well, okay. I'll admit to not having done any calculus in over 4 years but any teacher who'd agree with the "-inf + ln(2) + inf" line ought to be slapped hard. :)

And also, yes. You can take the log of a negative number.

[Napoleon54]

Haha, nice proof.

According to your proof, the region from -1 to 0 has no area. That doesn't sound right.

Additionally, I don't belive +infinity and -infinity cancel. Infinity is not a number, it's a concept, an approximation for something that we can not measure; it has no definite value and thus can not be manipulated, added, subtracted, etc...

[poiselle]

While it is cancelling a negative and a positive infinity by adding them is done often it is not technically correct. The integral can't be evaluated over that region because of the singularity. I had a professor that tried cancelling infinities and my buddy the math and physics major almost had a heart attack. Your method makes sense conceptually, but math does not always make sense conceptually. Your teacher should have known better though.

[leemaj]

Originally posted by hapoo
simply put, the integral is a method to find the area under a graph, now since your area has and infinate hole in it (hence infinate area) theres really no integer answer for it. But since your range cancels the infinate parts out through semmetry the only thing remaining is the ln2, that answer is correct

yea, thats exactly what i think

[coleslaw]

Originally posted by leemaj
yea, thats exactly what i think You can think that all you want, but you're still wrong! :P

[Markel]

I can't be of much help for several reasons:
1) I can't view the image in the initial post
2) I've forgotten almost everything I learned in calculus/analysis

However, in analysis 3 I do remember integrating over infinite discontinuities. That stuff made my head hurt (and almost kept me from graduating on time :eek: ).

[leemaj]

Originally posted by coleslaw
That's not always the case. There can be a finite area within an infinite boundary.

yea but this is an infinate area over a finite boundry...and the infinite areas are exactly equal due to the symmetry in the graph.

[hapoo]

Originally posted by coleslaw
That's not always the case. There can be a finite area within an infinite boundary.


Thats true (either way they cancel out though), in which case we're looking at the divergence. The problem can be considered of being in the form of a p-series (with a power of 1). Recalling the p-series rule we know that any series with a power less than 2 diverges. Hence the equation diverges -> area goes to infinity.

[leemaj]

Originally posted by hapoo



Thats true (either way they cancel out though), in which case we're looking at the divergence. The problem can be considered of being in the form of a p-series (with a power of 1). Recalling the p-series rule we know that any series with a power less than 2 diverges. Hence the equation diverges -> area goes to infinity.

yea, but the area also goes to negative infinity, and we are wondering if they cancel out since their values equal each other because the graph is symmetrical