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#1 2007-07-12 12:21:14
- Krizalid
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Integral Marathon
Hi y'all!!
I wanna start an
Rules are easy: person which solves an integral, may post the next one but he must have the approval of the person which proposed the previous integral if the answer is correct or not.
Let's start with an easy one
Compute
Best regards!
#2 2007-07-12 13:00:07
- mikau
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Re: Integral Marathon
sec(x) - tan(x) + x + C?
A logarithm is just a misspelled algorithm.
#4 2007-07-12 17:37:34
- mikau
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Re: Integral Marathon
Last edited by mikau (2007-07-12 17:51:38)
A logarithm is just a misspelled algorithm.
#5 2007-07-12 19:28:53
- JaneFairfax
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Re: Integral Marathon
Krizalid wrote:
but he must have the approval of the person which proposed the previous integral if the answer is correct or not
Why not just differentiate back and check? Also, is there any need to use blatantly sexist language? 
#6 2007-07-13 02:26:22
- Krizalid
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Re: Integral Marathon
mikau wrote:
sec(x) - tan(x) + x + C?
Sorry, you have to show us your solution (I dunno what I was thinking that I didn't ask you that).
JaneFairfax wrote:
Show your solution too.
JaneFairfax wrote:
Why not just differentiate back and check? Also, is there any need to use blatantly sexist language?
Well, actually when I was writing I didn't realize of that.
If you don't mind, please limit to make all type of disagreeable commentary. There're forms and forms to request the things.
#7 2007-07-13 04:44:20
- mikau
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Re: Integral Marathon
Thats right, Jane. Good job! Looks like you're next as soon as you show your solution.
and here's my solution to the first problem.
∫sin x /(1 +sin x) dx = ∫[-sin^2(x) + sin x ]/(1 - sin^2(x)) dx
= ∫ -sin^2(x)/cos^2(x) - sin(x)/cos^2(x) dx
= ∫ -tan^2(x) + sec(x)tan(x) dx
= ∫ 1 -sec^2(x) + sec(x)tan(x)dx = x - tan(x) + sec(x) + C QED.
Last edited by mikau (2007-07-13 06:26:04)
A logarithm is just a misspelled algorithm.
#8 2007-07-13 07:38:46
- Ricky
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Re: Integral Marathon
JaneFairfax wrote:
Also, is there any need to use blatantly sexist language?
http://www.xkcd.com/c145.html
Which reminds me:
http://www.xkcd.com/c263.html
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#9 2007-07-13 08:49:35
- MathsIsFun
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Re: Integral Marathon
That last comic was superb!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#10 2007-07-13 12:05:42
- mikau
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Re: Integral Marathon
i'll say!
We're still waiting for you to post the next one, Jane! 
A logarithm is just a misspelled algorithm.
#11 2007-07-14 03:23:52
- mathsyperson
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#12 2007-07-15 19:17:53
#13 2007-07-16 02:30:54
- Devantè
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Re: Integral Marathon
Doesn't this sort of thing also belong in Puzzles and Games? It is kind of a game, after all, it does follow the rule of most forum-based games.
#14 2007-07-16 05:01:45
- mathsyperson
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Re: Integral Marathon
Maybe it used to, but now it belongs in the "Jane is mean 
" section.
We don't have one of those yet, so it might as well stay here.
Why did the vector cross the road?
It wanted to be normal.
#15 2007-07-16 20:40:31
#16 2007-07-16 21:30:31
- luca-deltodesco
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Re: Integral Marathon
is that meant to be interpreted as:
or
?
The Beginning Of All Things To End.
The End Of All Things To Come.
#17 2007-07-16 22:03:11
- mathsyperson
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Re: Integral Marathon
Oops, typo. That should read "mathsy is dumb "
For some reason I read that as e^-x², and thought you were giving an impossible one. The 
didn't help either.
Why did the vector cross the road?
It wanted to be normal.
#18 2007-07-16 22:18:44
- JaneFairfax
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Re: Integral Marathon
Last edited by JaneFairfax (2007-07-16 22:22:43)
#19 2007-07-16 22:22:26
- JaneFairfax
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Re: Integral Marathon
is continuous for all real x and is therefore integrable for all real x. There is a theorem that states that all continuous functions are Riemann integrable.mathsyperson wrote:
For some reason I read that as e^-x², and thought you were giving an impossible one. The
#20 2007-07-17 01:53:38
- Ricky
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Re: Integral Marathon
Now Jane, go again.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#21 2007-07-17 08:53:20
- JaneFairfax
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Re: Integral Marathon
Yes.
I’ve also seen
I believe you can integrate an absolutely convergent series term by term within its radius of convergence, and the result would be another absolutely convergent series?
Ricky’s turn.
Last edited by JaneFairfax (2007-07-17 09:02:54)
#22 2007-07-17 10:16:01
- Ricky
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Re: Integral Marathon
Pass.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#23 2007-07-17 10:19:35
- Krizalid
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Re: Integral Marathon
I forgot to mention that don't post integrals which don't have a primitive function in elementary terms.
#24 2007-07-17 10:32:51
- mikau
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Re: Integral Marathon
Pass? :-( Who's next?
how about
this was in one of my books, i could never figure out how to do it. i think its non elementary.
Last edited by mikau (2007-07-17 10:36:11)
A logarithm is just a misspelled algorithm.
#25 2007-07-17 12:09:23
- Ricky
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Re: Integral Marathon
Not impossible but it involves complex analysis. I would suggest picking a different integral.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."