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Order of integration

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the double integration by changing the order of integration

    2. Relevant equations


    \int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy


    3. The attempt at a solution

    I've been trying to sketch the graph with

    y/2<= x <= 1

    and

    0 <= y <= 2

    Is the problem trying to get the {y/2} in the limit of \int_{y/2}^{1} cos(x^2)dx to be the equivalent value of x ?

    I'm not if my graph that I sketched should be a triangle from the origin out to x= 1 y=2

    or should it be a curve from (0,1) to (1,cos(1^2))

    regards
    Brendan
     
  2. jcsd
  3. Apr 23, 2009 #2

    djeitnstine

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    Gold Member

    if x= y/2, y = 2x
     
  4. Apr 24, 2009 #3
    I see that you have made the integral.

    \int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy

    into

    \int_{0}^{2} \int_{2x}^{1} cos(x^2)dxdy

    by solving y/2 for x = 2x.

    Can you use the same rational for

    \int_{1}^{3} \int_{0}^{ln(x)} x dydx

    solving y = ln(x) for x therefore x = e^(y()

    than the integral would be:

    int_{1}^{3} \int_{0}^{e^(y)} x dydx


    regards
    Brendan
     
  5. Apr 24, 2009 #4

    djeitnstine

    User Avatar
    Gold Member

    I see that you have made the integral.

    \int_{0}^{2} \int_{y/2}^{1} cos(x^2)dxdy

    into

    \int_{0}^{2} \int_{2x}^{1} cos(x^2)dxdy

    __________________________________

    This is incorrect. When solving for y in y/2 = x (y=2x) It is necessary to sketch the curves so that you can see the boundaries clearly. It seems to me you have a triangle with boundaries y=2x, x=1, and y=0. If you solve for 'y' like that its necessary to change the order of integration and write new limits.

    Now think about what you did when you exchanged y for x in your limits...when you try to integrate the function you can't...the whole reason why you change order in this case is so that you may perform the double integrals.

    In the second case, it is again similar to the triangle but with the upper boundary ln{x}
     
  6. Apr 25, 2009 #5
    I've sketched the graph of
    \int_{1}^{3} \int_{0}^{ln(x)} x dydx

    And the boundaries are the:

    x axis <= y <= ln(x)
    1 <= x <=3

    Is the integral now

    \int_{x}^{ln(x)} \int_{1}^{3} x dxdy ?

    regards
     
  7. Apr 25, 2009 #6

    djeitnstine

    User Avatar
    Gold Member

    You should restudy double integration. That double integral there makes no sense. Just a hint that allows you to know whether what you're doing is right or wrong, you should change the function in terms of your second integrating variable.

    \int_{1}^{3} \int_{0}^{ln(x)} x dydx


    this double integral goes from the lower y limit to the upper, then from the left x limit to the right. In order to change that order you have to change the function y=ln{x} in terms of x....e^y = x

    so when doing dxdy we start from dx and are going from the left limit to the right limit, as you see on your graph the limits now become \int_{e^y}^{3} dx and for dy its from lower to upper... \int_{0}^{ln(3)} dy

    putting it all together \int_{0}^{ln(3)}\int_{e^y}^{3} x dxdy (note that I said before the function should be in terms of the second integrating variable...)
     
  8. Apr 25, 2009 #7
    Thanks for your reply,

    \int_{0}^{ln(3)}\int_{e^y}^{3} x dxdy


    with the X limits \int_{e^y}^{3} when I graphed the original function those limits were 1 <= x <=3

    Why do we change 1 to be e^y and not change 3
     
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