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rosho
 2 years ago
how to find the area between y=sinx, y=cosx, y=0 between {0,pi/2}
ans: 2sqrt2
rosho
 2 years ago
how to find the area between y=sinx, y=cosx, y=0 between {0,pi/2} ans: 2sqrt2

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

MarcLeclair
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0what are you trying to do?

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

rosho
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1363164661036:dw

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

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

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

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

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

electrokid
 2 years 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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