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rosho

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

  • one year ago
  • one year ago

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  1. goformit100
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    @marcleclair

    • one year ago
  2. MarcLeclair
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    Cant you find the middle point? Then solve using two integrals?

    • one year ago
  3. rosho
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    i tried that.

    • one year ago
  4. MarcLeclair
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    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
  5. rosho
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    sqrt2/2

    • one year ago
  6. rosho
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    and this is calc II

    • one year ago
  7. MarcLeclair
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    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
  8. MarcLeclair
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    I think im totally wrong though x.x I was just trying to help. Sorry if im mistaken

    • one year ago
  9. Outkast3r09
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    what are you trying to do?

    • one year ago
  10. Outkast3r09
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    calculus or regular

    • one year ago
  11. rosho
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    calc 2

    • one year ago
  12. rosho
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    finding the area between 0, sinx and cosx on the interval of 0,pi/2

    • one year ago
  13. Outkast3r09
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    |dw:1363164661036:dw|

    • one year ago
  14. rosho
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    pretty much.

    • one year ago
  15. Outkast3r09
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    so you'll need to split the integral up

    • one year ago
  16. Outkast3r09
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    |dw:1363164746737:dw|

    • one year ago
  17. Outkast3r09
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    so you need the intersection point

    • one year ago
  18. rosho
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    already found that.

    • one year ago
  19. Outkast3r09
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    the reason why you need two? If you do high - low , they switch at the intersection

    • one year ago
  20. Outkast3r09
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    alright so what are you asking... are you done?

    • one year ago
  21. electrokid
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    [|dw:1363618349617:dw|

    • one year ago
  22. electrokid
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    \[ A = \int_y\int_xdxdy=\left[\int_{x=0}^{\pi\over4}\int_{y=0}^{\sin x}+\int_{x={\pi\over4}}^{\pi\over2}\int_{y=0}^{\cos x}\right]dxdy\\ A_1 = \int_{x=0}^{\pi\over4}\left[\int_{y=0}^{\sin x}dy\right]dx=\int_0^{\pi\over4}\sin x\,dx=1-{1\over\sqrt{2}}\\ A_2= \int_{x={\pi\over4}}^{\pi\over2}\left[\int_{y=0}^{\cos x}dy\right]dx=\int_{\pi\over4}^{\pi\over2}\cos x\,dx=1-{1\over\sqrt{2}}\\ A=A_1+A_2=2-{2\over\sqrt{2}}\\ A=2-\sqrt{2} \]

    • one year ago
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