My question refers to questions 77 and 78 of the P1 practice problems. When I study the two questions, they appear to be almost identical, other than a slightly different density function...except there is a contradiction between how the two are solved. My problem is with the integration. I understand what is being done in both problems to integrate, but I'm confused as to why they are integrated differently. Problem 77 keeps (x+y) together when integrating with respect to x, yet problem 78 doesn't. In other words, if one was to follow the same logic of integration as in 78, problem 77 would integrate as (x^2/2 + xy) rather than ((x+y)^2)/2. The two results yield different answers when worked out. Am I missing something here? If not, can someone please explain why the two are integrated differently with respect to x? Thanks!
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Joint Density Function Integration Contradiction
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This is just a calculus trick. I didn't like this problem at first but I understand it. In it X+Y is the unknown rather than "X"+"Y" when you integrate. The Assumption is that when integrating with respect to X the Y is significantly small so its like X=X+Y....OR held constant at the time it fails
Originally posted by sulfysulf View PostMy question refers to questions 77 and 78 of the P1 practice problems. When I study the two questions, they appear to be almost identical, other than a slightly different density function...except there is a contradiction between how the two are solved. My problem is with the integration. I understand what is being done in both problems to integrate, but I'm confused as to why they are integrated differently. Problem 77 keeps (x+y) together when integrating with respect to x, yet problem 78 doesn't. In other words, if one was to follow the same logic of integration as in 78, problem 77 would integrate as (x^2/2 + xy) rather than ((x+y)^2)/2. The two results yield different answers when worked out. Am I missing something here? If not, can someone please explain why the two are integrated differently with respect to x? Thanks!Last edited by djerry81; November 6 2007, 05:59 PM.

No contradiction. Remember in Calc I when your instructor had to threaten physical violence to get people to put the '+C' after integrals? Well, when integrating a function of two variables with respect to x, that +C is really +g(y). Now, multiply out your two answers and look at the difference.
The extra term cancels once you evaluate the limits of integration.
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Resurrecting an old post mainly because I have the exact same question and don't quite understand the answer.
Why can't I just integrate with respect to x, and then integrate with respect to y?why is this case different? (referring to question 77).
(never mind, I figured it out).Last edited by Juiom; May 17 2009, 09:34 PM.
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I'm assuming you get why the integral is set up the way it is and are just curious as to the actual integration process. OK let's do it the way you suggested,
first let's integrate w.r.t. x, we get x^2/2 + xy, integrated from 1 to 2 yields (2 + 2y)  (1/2 + y) = y + 3/2.
Now let's integrate w.r.t. y, we get y^2/2 + 3y/2, once again from 1 to 2 yields
(2 + 3)  (1/2 + 3/2) = 5  2 = 3
Now remember we have a 1/8 term out from so 3*1/8 = 3/8 and subtracting that from 1 gives us 5/8 which looks a lot like .625 which is the answer they give.
Questions?
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I solved it a different way, not like the solution given.
(#77)
The integration; try splitting it to (x/8) + (y/8); that works better for me. Might seem easier to you.
Integral of x/8 with respect to x = (x^2)/16 and integral of (y/8) with respect to x = xy/8.P FM MFE C
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