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JJChicken

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Apr 9, 2007
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Don't need to give me the answer, but perhaps point me in the right direction. How to integrate f(x,y) over its domain (all real x, y) such that x + y < 0.5?

Kinda stumped cause the values of y depend on x, so dunno how to integrate y out first.

Thanks resident ATOT geniuses!
 

Born2bwire

Diamond Member
Oct 28, 2005
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Ummmm...

x+y < 0.5
x < 0.5 - y
....

Wait...

What do you mean you do not know how to integrate y first?
 

JJChicken

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Apr 9, 2007
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On my iPhone, so bit hard to write this.

Say I want to integrate f(x,y)=xy where 0<x<1, 0<y<1, x + y < 0.5

So straight away, I know x < 0.5 and y < 0.5. What I did next was integrate y, holding x constant and over 0 to 0.5-x. Since I'm taking x constant, I thought its okay to do this. This leaves me the expression x(0.5-x)^2/2. I then integrate x over 0 to 0.5. Is this correct?

Thanks :)
 

Farmer

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Dec 23, 2003
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Your question is, "what are the limits of integration for the double integral"

Your previous post might be right, but here it is anyway.

Your boundary is a straight line in the x-y space (y = -x + 0.5 is the equation of that line). There are lots of ways to do this. Most intuitive way (may not be easiest depending on what f(x,y) is):

Let the x integral be the outer integral, and the y integral the inner integral.

gif.latex


If it's all positive (x,y) instead of all (x,y), then the lower limit in integration in both cases is 0, not -Inf.
 
Last edited:

JJChicken

Diamond Member
Apr 9, 2007
6,165
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81
Your question is, "what are the limits of integration for the double integral"

Your previous post might be right, but here it is anyway.

Your boundary is a straight line in the x-y space (y = -x + 0.5 is the equation of that line). There are lots of ways to do this. Most intuitive way (may not be easiest depending on what f(x,y) is):

Let the x integral be the outer integral, and the y integral the inner integral.

gif.latex


If it's all positive (x,y) instead of all (x,y), then the lower limit in integration in both cases is 0, not -Inf.

Dude, thanks so much.
 
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