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anonymous
 4 years ago
\[\int\limits_{0}^{2\Pi}\left senxcosx \rightdx\]
anonymous
 4 years ago
\[\int\limits_{0}^{2\Pi}\left senxcosx \rightdx\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is this from 0 to 2pi?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sinx, sorry (I speak spanish, in spanish is seno , not sine :D)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm getting 4squareroot of 2...I'll be right back...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits_{0}^{\pi/4}+\int\limits\limits\limits_{\pi/4}^{\pi/2}+\int\limits\limits\limits_{\pi/2}^{3\pi/4}+\int\limits\limits\limits_{3\pi/4}^{2\pi}..\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328592298972:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There are a couple of ways you can approach the problem...I first approached it using a graph and added the areas between the funcitons...or

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits\limits_{0}^{\pi/4}(sinx+cosx)dx+\int\limits\limits\limits\limits_{\pi/4}^{\pi/2}(sinxcosx)dx+\int\limits\limits\limits\limits_{\pi/2}^{3\pi/4}+\int\limits\limits\limits\limits_{3\pi/4}^{2\pi}..\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now that looks right....cinar....to start...but from pi/4 to 5pi/4 for the second term, then from 5pi/4 to 2pi for the last.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328592868525:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328592924702:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both representations are correct as far as area is concerned. but integrate from o to pi/4, then pi/4 to 5pi/4, then 5pi/4 to 2pi...
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