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Math Help - is this possible?

  1. #1
    Member i_zz_y_ill's Avatar
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    is this possible?

     \int^{\infty}_{0} \int^{x}_{0}\frac {e^{-{\frac {x}{y}-y}}}{y} dy dx = \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx


    in ref to http://www.mathhelpforum.com/math-he...ob-153251.html i dont think it is
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  2. #2
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    You are right. They are not the same The first integral takes x going from 0 to infinity and, for each x, y going from 0 to x. That integral is over the region in the first quadrant bounded by the x-axis, y= 0, and by the line y= x, that is, the area in the first quadrant below the line y= x. The second integral take y going from 0 to infinity and, for each y, x going from 0 to y- the region in the first quadrant bounded by the y-axis, x= 0, and by the line y= x. That is the region in the first quadrant above the line y= x.

    What would be correct is either
    \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy
    or
    \int_0^\infty\int_x^\infty f(x,y)dydx= \int_0^\infty\int_0^y f(x,y) dxdy
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  3. #3
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    [ going from 0 to infinity and, for each y, x going from 0 to y- the region in the first quadrant bounded by the y-axis, x= 0, and by the line y= x. That is the region in the first quadrant above the line y= x.

    \int_0^\infty\int_x^\infty f(x,y)dydx= \int_0^\infty\int_0^y f(x,y) dxdy[/QUOTE]

    which region is this then? and do you not mean for each x?

    bare with me a second,,i believe youve answered my question giving exp(-1) as the answer as opposed to 1-exp(-1) which gets me somewhere,,just a sec
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  4. #4
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    so as the question in the original post said,,it should be the region under y=x(as i knew) and therefore \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy which gives exp(-1)


    ahh now i see that \int_0^\infty\int_0^y f(x,y) dxdy is above the line,,,but

    why do we start with dydx and where does y= \infty come from in  \int_0^\infty\int_y^\infty f(x,y)dxdy

    my only thoughts as to why we start dydx can be that the region is bounded by x= \infty and this needs to be the last integral calculated with the upper limit \infty to give a numeric answer although i cant convince myself this is a valid way to tackle the question

    ok cool so the answer is \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy but how were the limits changed and what procedure would youuse to get the rhs without the lhs
    Last edited by bluesblues; August 11th 2010 at 04:14 AM.
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  5. #5
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    ok its cool i think i got it now,,,i can start dxdy if i wanted or dydx doesnt matter the answer is exp(-1)
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