\[\int\limits_{0}^{2\Pi}\left| senx-cosx \right|dx\]

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\[\int\limits_{0}^{2\Pi}\left| senx-cosx \right|dx\]

Mathematics
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Is this from 0 to 2pi?
yes
Absolute value

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give me just a sec
what senx is?
sinx, sorry (I speak spanish, in spanish is seno , not sine :D)
I'm getting 4squareroot of 2...I'll be right back...
\[\int\limits\limits\limits_{0}^{\pi/4}+\int\limits\limits\limits_{\pi/4}^{\pi/2}+\int\limits\limits\limits_{\pi/2}^{3\pi/4}+\int\limits\limits\limits_{3\pi/4}^{2\pi}..\]
|dw:1328592298972:dw|
There 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
\[\int\limits\limits\limits\limits_{0}^{\pi/4}(-sinx+cosx)dx+\int\limits\limits\limits\limits_{\pi/4}^{\pi/2}(sinx-cosx)dx+\int\limits\limits\limits\limits_{\pi/2}^{3\pi/4}+\int\limits\limits\limits\limits_{3\pi/4}^{2\pi}..\]
now 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.
|dw:1328592868525:dw|
|dw:1328592924702:dw|
both 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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