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 Primeinator 2008-04-26 04:31

What Integration Technique?

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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.

 Orgasmic Troll 2008-04-26 06:38

What section in the book is this problem from?

 Chris Card 2008-04-26 06:48

I think you need to look at symmetries that the integrand possesses, which are easier to see after a simple substitution.

Chris

PS
[SPOILER]I make the answer = 1[/SPOILER]

 frmky 2008-04-26 16:36

Ah, yes. [SPOILER]One of those integrals that you don't actually DO the integral, you determine the solution purely by symmetry upon clever substitutions.[/SPOILER] This one had both Mathematica and me stumped. Thanks Chris!

Greg

 Primeinator 2008-04-27 00:20

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.

 davieddy 2008-04-27 03:20

[spoiler]f(3+y)+f(3-y)=1[/spoiler]

 frmky 2008-04-27 17:41

[SPOILER]As hinted above, shift the origin by 3 to get a symmetric integral over -1 to 1. Call this integral 1. [/SPOILER]

[SPOILER]Then replace x by -x (and of course dx by -dx) and see what you get as an equivalent integral. Call this integral 2. [/SPOILER]

[SPOILER]Now add integral 1 and integral 2 to get an integral that is easier to do. Call this integral 3. [/SPOILER]

[SPOILER] Now, since integral 1 and 2 are equal, they are each 1/2 of integral 3. [/SPOILER]

 davieddy 2008-04-27 18:38

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.

 davieddy 2008-04-27 19:16

I assume that an indefiniite integral was
impossible (or at least difficult).

 frmky 2008-04-27 19:53

[QUOTE=davieddy;132252]I assume that an indefiniite integral was impossible (or at least difficult).[/QUOTE]

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

 davieddy 2008-04-28 02:55

[quote=frmky;132248][spoiler]As hinted above, shift the origin by 3 to get a symmetric integral over -1 to 1. Call this integral 1. [/spoiler]

[spoiler]Then replace x by -x (and of course dx by -dx) and see what you get as an equivalent integral. Call this integral 2. [/spoiler]

[spoiler]Now add integral 1 and integral 2 to get an integral that is easier to do. Call this integral 3. [/spoiler]

[spoiler] Now, since integral 1 and 2 are equal, they are each 1/2 of integral 3. [/spoiler][/quote]

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.

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