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MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Cant you find the middle point? Then solve using two integrals?

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0I mean they should intersect at pi/6 ( where they have the same value) > unit circle I believe ( I don't remember so much)

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Yeah and then integrate both int ( pi/6  sinx ) + int ( cos x  pi/6) no? Yeah I'm taking cal II too ( well in quebec, might be different curiculum)

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0I think im totally wrong though x.x I was just trying to help. Sorry if im mistaken

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0what are you trying to do?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0calculus or regular

rosho
 one year ago
Best ResponseYou've already chosen the best response.0finding the area between 0, sinx and cosx on the interval of 0,pi/2

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363164661036:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0so you'll need to split the integral up

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363164746737:dw

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0so you need the intersection point

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0the reason why you need two? If you do high  low , they switch at the intersection

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.0alright so what are you asking... are you done?

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0[dw:1363618349617:dw

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0\[ A = \int_y\int_xdxdy=\left[\int_{x=0}^{\pi\over4}\int_{y=0}^{\sin x}+\int_{x={\pi\over4}}^{\pi\over2}\int_{y=0}^{\cos x}\right]dxdy\\ A_1 = \int_{x=0}^{\pi\over4}\left[\int_{y=0}^{\sin x}dy\right]dx=\int_0^{\pi\over4}\sin x\,dx=1{1\over\sqrt{2}}\\ A_2= \int_{x={\pi\over4}}^{\pi\over2}\left[\int_{y=0}^{\cos x}dy\right]dx=\int_{\pi\over4}^{\pi\over2}\cos x\,dx=1{1\over\sqrt{2}}\\ A=A_1+A_2=2{2\over\sqrt{2}}\\ A=2\sqrt{2} \]
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