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Choosing and finding limits of integration

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    For the given region R, find intR f(x) dA. The region has the following points:
    (-1,1), (-1,-2) and (3,-2)


    2. Relevant equations



    3. The attempt at a solution

    I'm having problems finding the boundaries for the integral. I know that we have:
    -1<=x<=3 and -2<=y<=1.

    I can also find the slope that it has and I found: y=-3/4x + 1/4.

    From there, I don't know what to do. I know i'm supposed to choose if I want to integrate first, does it matter? or how can I know which one is easier?

    I thought I could have something like: int(-1,3) int(1, -3/4x + 1/4) f(x) dA.

    Hope someone can help me with this!
     
  2. jcsd
  3. Oct 16, 2007 #2

    Dick

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    It looks like you already chosen dy as the inner variable and dx as the outer. You should probably write f(x,y) for the function and dydx for dA. Now for your problem. Why do you think the lower limit for the dy integration is 1?
     
  4. Oct 16, 2007 #3
    I just thought..the slope starts at y=1 and then it goes downwards with the -3/4x + 1/4 slope till it reaches the point (3,-2).
     
  5. Oct 16, 2007 #4

    Dick

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    The lower limit in y is the lower y value connecting the points (-1,-2) and (3,-2). Did you draw a picture of the region? Always recommended.
     
  6. Oct 16, 2007 #5
    So it would simply be -2. Yes I do have the region drawn.

    So then I would have...

    int(-1,3) int(-2, -3/4x + 1/4) f(x) dA.?
     
  7. Oct 16, 2007 #6

    Dick

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    Yes. Now write f(x,y) for f(x) (it's not just a function of x) and dA=dydx (to indicate the order of integration).
     
  8. Oct 16, 2007 #7
    I don't mean to sound dumb, but I'm not quite sure I know what you mean. Sorry. Maybe its just late :S
     
  9. Oct 17, 2007 #8

    HallsofIvy

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    The slope of what??? You haven't mentioned a line!

    Please, please, please, write what you really mean! You can't possibly mean you want to "choose if I want to integrate first"! What else can you do? You need to choose whether you want to integrate with respect to x or y first.
    Try both ways and see! Here it doesn't really matter.

    Again, what you are writing makes no sense because you haven't said which variable you are using first! Write either int(x=-1, 3) int(y= 1, (-3/4)x+ 1/4) f(x)dA, or int(-1,3)int 1, -(3/4)x+ 1/4) f(x) dydx or, even better, both: int(x=-1, 3) int(y= 1, (-3/4)x+ 1/4) f(x)dydx.

    First, explain what in the world you are talking about! You start by saying "The region has the following points: (-1,1), (-1,-2) and (3,-2)" Do you mean they are the vertices of a triangular region? If so, then you have a right triangle as your region. There are two different ways you can do this.

    If you want to do the "dy" integral first, then you choose x-values as constants to cover the entire triangle: x will go from -1 to 3. Now, draw a vertical line through the region, representing the small rectangle you would use, for each x, in setting up a "Riemann sum" for the problem. Clearly, for all x, the lower boundary is on the line y= -1. Also, for all x, the upper boundary is on the line from (-1,1) to (3, -2) which is, as you say, y= (-3/4)x+ 1/4:
    [tex]\int_{x=-1}^1 \int_{y= -1}^{(-3/4)x+ 1/4} f(x)dydx[/tex]

    If you want to do the "dx" integral first, you do just the opposite: first choose y-values as constants to cover the entire triangle: y will go from -2 to 1. Now draw a horizontal line representing the small rectangle you would use, for each y, in setting up the Riemann sum for the problem. Clearly, for all y, the left boundary is the line x= -1. Also, for all y, the right boundary is on the line from (-1,1) to (3, -2) which you can get by solving the equation above for x: x= (-4/3)y+ 1/3. The integral is:
    [tex]\int_{y= -2}^1 \int_{x= -1}^{(-4/3)y+ 1/3} f(x)dxdy[/tex]
     
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