anonymous
  • anonymous
how to find the area between y=sinx, y=cosx, y=0 between {0,pi/2} ans: 2-sqrt2
Mathematics
katieb
  • katieb
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goformit100
  • goformit100
anonymous
  • anonymous
Cant you find the middle point? Then solve using two integrals?
anonymous
  • anonymous
i tried that.

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anonymous
  • anonymous
I mean they should intersect at pi/6 ( where they have the same value) --> unit circle I believe ( I don't remember so much)
anonymous
  • anonymous
sqrt2/2
anonymous
  • anonymous
and this is calc II
anonymous
  • anonymous
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)
anonymous
  • anonymous
I think im totally wrong though x.x I was just trying to help. Sorry if im mistaken
anonymous
  • anonymous
what are you trying to do?
anonymous
  • anonymous
calculus or regular
anonymous
  • anonymous
calc 2
anonymous
  • anonymous
finding the area between 0, sinx and cosx on the interval of 0,pi/2
anonymous
  • anonymous
|dw:1363164661036:dw|
anonymous
  • anonymous
pretty much.
anonymous
  • anonymous
so you'll need to split the integral up
anonymous
  • anonymous
|dw:1363164746737:dw|
anonymous
  • anonymous
so you need the intersection point
anonymous
  • anonymous
already found that.
anonymous
  • anonymous
the reason why you need two? If you do high - low , they switch at the intersection
anonymous
  • anonymous
alright so what are you asking... are you done?
anonymous
  • anonymous
[|dw:1363618349617:dw|
anonymous
  • anonymous
\[ A = \int_y\int_xdxdy=\left[\int_{x=0}^{\pi\over4}\int_{y=0}^{\sin x}+\int_{x={\pi\over4}}^{\pi\over2}\int_{y=0}^{\cos x}\right]dxdy\\ A_1 = \int_{x=0}^{\pi\over4}\left[\int_{y=0}^{\sin x}dy\right]dx=\int_0^{\pi\over4}\sin x\,dx=1-{1\over\sqrt{2}}\\ A_2= \int_{x={\pi\over4}}^{\pi\over2}\left[\int_{y=0}^{\cos x}dy\right]dx=\int_{\pi\over4}^{\pi\over2}\cos x\,dx=1-{1\over\sqrt{2}}\\ A=A_1+A_2=2-{2\over\sqrt{2}}\\ A=2-\sqrt{2} \]

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