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

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goformit100 Group TitleBest ResponseYou've already chosen the best response.0
@marcleclair
 one year ago

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

rosho Group TitleBest ResponseYou've already chosen the best response.0
i tried that.
 one year ago

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

rosho Group TitleBest ResponseYou've already chosen the best response.0
and this is calc II
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Yeah 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)
 one year ago

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

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

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

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

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

rosho Group TitleBest ResponseYou've already chosen the best response.0
pretty much.
 one year ago

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

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

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

rosho Group TitleBest ResponseYou've already chosen the best response.0
already found that.
 one year ago

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

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

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

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