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I don't get this integral at all.

  1. Aug 17, 2013 #1
    How did he do that?

    I follow a regular "do the integral inside first" then "do the integral outside second" after I have initially just put the '3' completely outside but he seems to get the 3 outside first in the form of 6 and then shrinking the whole of it. (???)

    2AxNuSQ.png

    I would normally be OK with not getting it but in this case he gets a result of 1/4 and I get 1/12 with my method.

    The freakiest thing? I saw a postgraduate mathematician getting 1/12 as well. How did the Professor get 1/4?

    The question did have a prerequisite of finding fx(x) (the above is from fx,y(x,y)), but I doubt it's related..
     
  2. jcsd
  3. Aug 17, 2013 #2

    haruspex

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    Can't answer that without seeing your working.
     
  4. Aug 17, 2013 #3
    Wait. Stop the Presses.

    I figured it out.

    I left the '3' outside and forgot to use it at all in the end (funny that the postgrad did the same).

    Now I still wonder what his trick was. Does anyone know? The regular "start from the inside going outside" is confirmed now to give the same result with his shortened version.
     
  5. Aug 17, 2013 #4
    Something tells me, he may have used the fact of equivalency of fx(x) and fy(y) (notice the 'x' and 'y' are interchangeable..
     
    Last edited: Aug 17, 2013
  6. Aug 17, 2013 #5

    HallsofIvy

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    If you are getting "1/12" I wonder if you are not factoring out the "3" and then forgetting to multiply it again?
    (Okay, you say that's what you did.)

    I was also, at first, puzzled by the change from the first integral to the second. Surely "[itex]3xy(x+ y- x^2- y^2[/itex]" is NOT the same as [itex]6 xy(x- x^2)[/itex]. But you can use a form of "symmetry". we can write [itex]x+ y- x^2- y^2[/itex] as [itex]x- x^2+ y- y^2[/itex] and then recognize that, since the integrand is symmetric in x and y, we can use, instead, [itex]x- x^+ x- x^2= 2(x- x^2)[/itex].
     
  7. Aug 17, 2013 #6
    Thanks. That should point me to the right direction if I choose to search for it in depth.

    And yes, I did make a mistake of forgetting to use the '3' at all in the end.
     
  8. Aug 17, 2013 #7
    Wait a second. If I understand this correctly, was it only a matter of replacing all instances of y with x (after noticing the symmetry)?

    In that case is 6IntegralIntegral(x^3-x^4)dxdx just 6Integral(x^2-x^3)dx?
     
  9. Aug 17, 2013 #8

    D H

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    [tex]
    \begin{aligned}
    \int_0^1\int_0^1\,3xy(x+y-x^2-y^2)\,dx\,dy
    =&\phantom{+}\int_0^1\int_0^1\,3xy(x-x^2)\,dx\,dy \\
    &+\int_0^1\int_0^1\,3xy(y-y^2)\,dx\,dy \\[8pt]
    =&\phantom{+}
    \int_0^1\int_0^1\,3xy(x-x^2)\,dx\,dy \\
    &+ \int_0^1\int_0^1\,3xy(y-y^2)\,dy\,dx\qquad\text{change order} \\[8pt]
    =&\phantom{+}
    \int_0^1\int_0^1\,3xy(x-x^2)\,dx\,dy \\
    &+ \int_0^1\int_0^1\,3xy(x-x^2)\,dx\,dy\qquad\text{rename dummy variables} \\[8pt]
    =&
    \int_0^1\int_0^1\,6xy(x-x^2)\,dx\,dy
    \end{aligned}
    [/tex]
     
  10. Aug 17, 2013 #9
    Is that a proof? Because the shortcut of just replacing all 'y's with 'x's seems to work arithmetically.
     
  11. Aug 18, 2013 #10

    D H

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    When is that ever valid? Certainly not here! When this replacement happens to yield the correct result, it's just luck.

    I could have bypassed the middle two steps by saying that ##\int_0^1\int_0^1 3xy(x-x^2)\,dx\,dy## and ##\int_0^1\int_0^1 3xy(y-y^2)\,dx\,dy## are obviously equal to one another. (Beware the mathematician who uses the word "obviously".) I added those middle two steps to explicitly show this equality.
     
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