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 2008-04-26, 04:31 #1 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 3×5×61 Posts What Integration Technique? Alright, I'm in dire need of some assistance. This integral was a homework problem (second semester calculus) and although I have a fancy calculator and can determine the annoyingly clean answer, I have no idea how to integrate this function by hand. A pasted picture of the integral is provided in png format for your aesthetic viewing pleasure. Thank you. Attached Thumbnails
 2008-04-26, 06:38 #2 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 64110 Posts What section in the book is this problem from?
 2008-04-26, 06:48 #3 Chris Card     Aug 2004 2×5×13 Posts I think you need to look at symmetries that the integrand possesses, which are easier to see after a simple substitution. Chris PS I make the answer = 1
 2008-04-26, 16:36 #4 frmky     Jul 2003 So Cal 5×11×41 Posts Ah, yes. One of those integrals that you don't actually DO the integral, you determine the solution purely by symmetry upon clever substitutions. This one had both Mathematica and me stumped. Thanks Chris! Greg Last fiddled with by frmky on 2008-04-26 at 16:37
 2008-04-27, 00:20 #5 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 3×5×61 Posts Yes, I knew the answer was one-- my TI-89 was kind enough to provide me with that answer. Alright, then looking at the graph, how can you determine the solution based upon symmetry? My calculus text book mentions NOTHING about using symmetry to solve the integral. This is one of those ridiculous Putnam Exam Challenge problems. Thanks.
 2008-04-27, 03:20 #6 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts f(3+y)+f(3-y)=1
 2008-04-27, 17:41 #7 frmky     Jul 2003 So Cal 5·11·41 Posts As hinted above, shift the origin by 3 to get a symmetric integral over -1 to 1. Call this integral 1. Then replace x by -x (and of course dx by -dx) and see what you get as an equivalent integral. Call this integral 2. Now add integral 1 and integral 2 to get an integral that is easier to do. Call this integral 3. Now, since integral 1 and 2 are equal, they are each 1/2 of integral 3.
 2008-04-27, 18:38 #8 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Don't think we can go any further without contravening Rool2 of the Homework forum. David May as well suggest considering the area under the curve though. Last fiddled with by davieddy on 2008-04-27 at 18:46
 2008-04-27, 19:16 #9 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts I assume that an indefiniite integral was impossible (or at least difficult).
2008-04-27, 19:53   #10
frmky

Jul 2003
So Cal

1000110011112 Posts

Quote:
 Originally Posted by davieddy I assume that an indefiniite integral was impossible (or at least difficult).
I think so. I suspect the square roots were there to ensure it since, interestingly, the answer does not depend on the power of the logs.

Greg

2008-04-28, 02:55   #11
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by frmky As hinted above, shift the origin by 3 to get a symmetric integral over -1 to 1. Call this integral 1. Then replace x by -x (and of course dx by -dx) and see what you get as an equivalent integral. Call this integral 2. Now add integral 1 and integral 2 to get an integral that is easier to do. Call this integral 3. Now, since integral 1 and 2 are equal, they are each 1/2 of integral 3.
If "technique" is required, I would show that integrating
f(x) from 2 to 3 was the same as integrating f(3-y) from y=0 to 1.
More trivially, integrating f(x) from 3 to 4 is the same as
integrating f(3+y) from y=0 to 1.

Summing these I would then conclude that the answer was
the integral of (f(3+y) + f(3-y)) from y=0 to 1.

Last fiddled with by davieddy on 2008-04-28 at 03:34

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