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Find the area between three curves

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Sketch the region enclosed by the curves and compute its area as an integral along the x or y axis.
    y+x=4 y-x=0 y+3x=2

    2. Relevant equations

    top function - bottom function dx OR right function-left function dy

    3. The attempt at a solution
    I originally had chosed to integrate with respect to y, I solved all equations for y as follows:
    y=4-x y=x y=2-3x

    I set all equations equal to each other to find three intersection x values: These are
    x=2, x=-2, and x=1/2

    This makes since considering the way I sketched the graph: see attachment.

    photo (18).jpg


    Now the Area is shaded in orange and labeled A: to set up the integral I did:
    1/2
    (4-x)-(2-3x) dx = 2x+x2 evaluated from -2 to 1/2
    -2

    My answer to this integral was (5/4)

    Then I took this integral:
    2
    (4-x)-(x) = 4x-x2 evaluated from 1/2 to 2
    1/2

    My answer to this integral was (7/4)

    It was olvious to me at this point that something went wrong because (5/4)-(7/4)=-(1/2) and areas cannot be negative.

    This is the first time I've attempted to find the area between three curves. I am not sure how to arrange the integrals and which one is considered "top" or "bottom" because the graph that I drew shows the lower functions switching, which I tried to reflect in my calculations.

    I'm really working hard to wrap my head around this stuff, so any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 26, 2012 #2

    SammyS

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    The lines y=4-x , and y=2-3x intersect at x = -1 not at x = -2.

    By the way, the lines y=4-x, and y=x are perpendicular to each other, so it's easy to check the answer.
     
  4. Sep 26, 2012 #3
    Thanks, I wouldn't of caught that.
    So I've edited my limits of integration:
    1/2
    ((4-x)-(2-3x)) = 4.25
    -1

    2
    ((4-x)-(x))= (7/4)
    1/2

    4.25-(7/4) = 2.5

    Alas, I am still incorrect. Is there something else I am missing?

    Thanks again.
     
  5. Sep 26, 2012 #4

    SammyS

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    Be careful with your signs !

    You integrations are correct in the previous post.

    You are messing up when plugging in the limits of integration.
     
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