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Reversing integration

  1. Nov 14, 2008 #1
    b]1. The problem statement, all variables and given/known data[/b]

    Given Domain :[tex]\int[/tex][tex]\int[/tex]f(x,y)dydx
    0[tex]\leq[/tex]x[tex]\leq[/tex]1
    x-1[tex]\leq[/tex]y[tex]\leq[/tex]2-2x
    reiterate the integrals so the order is reversed

    2. Relevant equations



    3. The attempt at a solution
    not really sure how to complete,
    [tex]\int[/tex][tex]\int[/tex]f(xy)dxdy
    y+1[tex]\leq[/tex]x[tex]\leq[/tex](2-y)/2
    -1[tex]\leq[/tex]y[tex]\leq[/tex]2

    Is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 14, 2008 #2

    HallsofIvy

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    It should be obvious that it is not correct. The first integral is a number. This integral will, after integrating, be a function of y. In order to be a number, the limits of integration on the "outside" integeral, with respect to y, must be numbers, not functions of y.

    Draw a picture! In the original "outside" integral, x ranges from 0 to -1. For each x, y ranges from x-1 up to 2- 2x. Those are straight lines and you should see that the area is a triangle with vertices (0, -1), (2, 0) and (0, 2).

    Now look at it from the side. To cover that entire area, y needs to vary from -1 to 1: those will be the limits on the "outer", dy, integral. For each y, x varies form 0 up to x= a function of y, given by the line making the right boundary. It looks to me like you will need to separate that into two integrals: y form -1 to 0 and then from 0 to 1.
     
  4. Nov 14, 2008 #3
    I have the verticies of the triangle at (2,0), (1,0), and (0,-1)?

    wouldn't that mean dy would range from -1 to 2?

    the lower part of the triangle dy from -1 to 0, x goes from 0 to y+1
    ther upper part, dy ranges from 0 to 2, x goes from 0 to (y-2)/2

    do you put an addition sign in between the the two integrals?
     
  5. Nov 14, 2008 #4

    HallsofIvy

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    Then draw the graph more carefully. The two lines y= x- 1 and y= 2- 2x intersect at (1, 0). y= x-1 intersects x=0 at y= -1 and y= 2- 2x intersects x= 0 at y= 1. The vertices are, as I said, (0, -1), (0, 1), and (1, 0).
     
  6. Nov 14, 2008 #5

    gabbagabbahey

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    Really? :wink:
     
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