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Math Help - double integral 'prob'

  1. #16
    Super Member 11rdc11's Avatar
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    Thanks, yeKcim. I can't draw to save my life
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  2. #17
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bluesblues View Post
    thanks this was the reasoning i was trying to do in my head without going over old noted it seems to be correct what you have put
    although should not the line be y=x nvm,,,but having done this integration a hundred times i still keep getting ... as an answer

    \int_o^\infty \int_0^x \frac{1}{y}exp(-\frac{x}{y})exp(-y)dy dx =-1

    hehehehe

    line  x=y is the same as line  y=x (you know that but i'm just pointing anyway )

    as for integral ... put some work that you have done... and we'll see where is the problem


    P.S. your result should be \displaystyle  \frac {e-1}{e} = 1 - \frac {1}{e} = 0.63212
    Last edited by yeKciM; August 10th 2010 at 10:53 AM.
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  3. #18
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    really confused actually,,,,what determines the order of integration i would want to start with,,oh one sec ill see this last post.(ignore) order is naturally x first
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  4. #19
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    yeKciM THANKYOU but are you 100% positive this is correct answer,,,i just want like one other peron to confirm it. i fully understand now that my original thinking is correct due to the diagram drawn above everything makes sense i think,,,,although still trying to do the other question i found
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  5. #20
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bluesblues View Post
    really confused actually,,,,what determines the order of integration i would want to start with,,oh one sec ill see this last post.
    let's go like this

     \displaystyle \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx

    and solve first :

     \displaystyle \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx = (e-1)e^{-y-1}

    after that it's much easier
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  6. #21
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    rv's have joind density fn exp(-x-y) for x,y>0


    find p(x>y) answer given in 1/2 : O

    using same method you dont get 1/2 :s
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  7. #22
    Senior Member yeKciM's Avatar
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    yes u'll get the 1/2... go and check it out there (on that thread)

    just pay attention to the how u can write down "-e^(-x-y)" and you will have no problems
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  8. #23
    Member i_zz_y_ill's Avatar
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    btw i lost my password y day,,,,,,,,,,,,,,,,,,,,created new account,,,,ive confused myslef again,,i cant believe it,,why did we choose to put dydx and then do the dx integral first i thought you cant do that? is the reason we put dy first because x>y?????

    furthermore by doing the x integral first do we not have to change the limits in which case it doesnt work youd get x from y to infinity and y from 0 to infinty thanks?


    ]i.e you cannot do dx first when dy came fist  \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
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  9. #24
    Member i_zz_y_ill's Avatar
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    if y>x i can understand y= \infty but this is x>y so it has to be x= \infty can someone please explain
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  10. #25
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    can someone confirm the answer is simply  <br />
\int_0^\infty \int_o^y f(x,y) dx dy with that working and there is no reason to put dy before dx
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