20080426, 04:31  #1 
"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. 
20080426, 06:38  #2 
Cranksta Rap Ayatollah
Jul 2003
641_{10} Posts 
What section in the book is this problem from?

20080426, 06:48  #3 
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 
20080426, 16:36  #4 
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 20080426 at 16:37 
20080427, 00:20  #5 
"Kyle"
Feb 2005
Somewhere near M52..
3×5×61 Posts 
Yes, I knew the answer was one my TI89 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.

20080427, 03:20  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
f(3+y)+f(3y)=1

20080427, 17:41  #7 
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. 
20080427, 18:38  #8 
"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 20080427 at 18:46 
20080427, 19:16  #9 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
I assume that an indefiniite integral was
impossible (or at least difficult). 
20080427, 19:53  #10 
Jul 2003
So Cal
100011001111_{2} Posts 

20080428, 02:55  #11  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
f(x) from 2 to 3 was the same as integrating f(3y) 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(3y)) from y=0 to 1. Last fiddled with by davieddy on 20080428 at 03:34 

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