Double Integral of min(x,y): How Do I Solve This Homework Problem?

[Roni1985]

Homework Statement

\int\intmin(x,y)*dx*dy
both limits are from 0 to 1

what is the general solution of this question ?

Homework Equations

The Attempt at a Solution

I attempted to solve by saying that x<y, finding the integral
and then, y<x and finding its integral.
meaning, I divided my double integral into two double integrals that go from 0 to 1.

Actually, I have never done this before and this was just a guess.

Any help would be appreciated.

 

[dickdick]

Homework Statement

\int\intmin(x,y)*dx*dy
both limits are from 0 to 1

what is the general solution of this question ?

Homework Equations

The Attempt at a Solution

I attempted to solve by saying that x<y, finding the integral
and then, y<x and finding its integral.
meaning, I divided my double integral into two double integrals that go from 0 to 1.

Actually, I have never done this before and this was just a guess.

Any help would be appreciated.

That's a really good guess. I can't think of a better way to handle it.

 

[Roni1985]

That's a really good guess. I can't think of a better way to handle it.

Hello,

Well, I think the answer is wrong because I need to get is 1/3 if I remember correctly... but I keep getting 1.


Thanks.

 

[dickdick]

Hello,

Well, I think the answer is wrong because I need to get is 1/3 if I remember correctly... but I keep getting 1.


Thanks.

Sure. Split it into two integrals. I'd be curious to see how you manage to get 1. Can you show your work? 1/3 is correct.

 

[chiro]

Homework Statement

\int\intmin(x,y)*dx*dy
both limits are from 0 to 1

what is the general solution of this question ?

Homework Equations

The Attempt at a Solution

I attempted to solve by saying that x<y, finding the integral
and then, y<x and finding its integral.
meaning, I divided my double integral into two double integrals that go from 0 to 1.

Actually, I have never done this before and this was just a guess.

Any help would be appreciated.

Have you tried using a signum function?

You can use a signum function to calculate either a minimum or a maximum. However I'm not sure if an integral exists for it. If one did I would think that it would be some special function that's non-polynomial.

In case your interested here is how you calculate min(x,y)

min(x,y) = y * (signum(x-y)+1)/2 + x * (signum(y - x)+1)/2. The signum function in this case is defined as:

signum(x) = -1 if x < 0, +1 if x > 0, 0 if x = 0.

This kind of function has a Fourier series so you may want to ask someone who knows about these kind of functions.

 

[dickdick]

Have you tried using a signum function?

You can use a signum function to calculate either a minimum or a maximum. However I'm not sure if an integral exists for it. If one did I would think that it would be some special function that's non-polynomial.

In case your interested here is how you calculate min(x,y)

min(x,y) = y * (signum(x-y)+1)/2 + x * (signum(y - x)+1)/2. The signum function in this case is defined as:

signum(x) = -1 if x < 0, +1 if x > 0, 0 if x = 0.

This kind of function has a Fourier series so you may want to ask someone who knows about these kind of functions.

It is seriously not that complicated. Just split the integration in the unit square along the line x=y.

 

[Roni1985]

Sure. Split it into two integrals. I'd be curious to see how you manage to get 1. Can you show your work? 1/3 is correct.

OMG, it is 1/3... I didn't change the limits of the integrals even after I divided them...


Thanks for your help...