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I haven't done calc in a couple years, but shouldn't you have a negative sign before (your answer) since you are integrating from 2pi to 0, not 0 to 2pi.
If it helps, my TI-89 simplifies your definite integral to: -16*pi^5/15
So you expanded the polynomial correctly (not counting the extra ^2 which you meant to remove) and you integrated them correctly. Now all you need to do is find a value for that expression at theta=0, find a value of the expression at theta=2pi, and subtract.
mspencer on
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True, those shouldn't be there.
Aye - yet it gives me the wrong answer. Guess I am screwing up somewhere further along the line. Ah well. Hand it in and let the Gods decide.
So you expanded the polynomial correctly (not counting the extra ^2 which you meant to remove) and you integrated them correctly. Now all you need to do is find a value for that expression at theta=0, find a value of the expression at theta=2pi, and subtract.
XBL Michael Spencer || Wii 6007 6812 1605 7315 || PSN MichaelSpencerJr || Steam Michael_Spencer || Ham NOØK
QRZ || My last known GPS coordinates: FindU or APRS.fi (Car antenna feed line busted -- no ham radio for me X__X )
Also check out the integrator:
http://integrals.wolfram.com/index.jsp
This is a good tool to double check your indefinite integrals.