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Why did we divide the region R into three regions?
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Here I cannot understand why we divided the region R into three while computing the expected value of |x-y|
integration density-function
asked Jul 24 at 20:43
Roy Rizk
887
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Here I cannot understand why we divided the region R into three while computing the expected value of |x-y|
integration density-function
asked Jul 24 at 20:43
Roy Rizk
887
Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54
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up vote
1
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up vote
1
down vote
favorite
Here I cannot understand why we divided the region R into three while computing the expected value of |x-y|
integration density-function
asked Jul 24 at 20:43
Roy Rizk
887
Here I cannot understand why we divided the region R into three while computing the expected value of |x-y|
integration density-function
asked Jul 24 at 20:43
Roy Rizk
887
asked Jul 24 at 20:43
Roy Rizk
887
asked Jul 24 at 20:43
Roy Rizk
887
asked Jul 24 at 20:43
Roy Rizk
887
887
Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54
add a comment |Â
Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54
Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54
add a comment |Â
2 Answers
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When $1<x<2$, $y$ is necessarily less than $x$. That's one region.
when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).
answered Jul 24 at 20:51
Alon Amit
10.2k3765
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up vote
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The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative.
This makes it very awkward to integrate in just one integral; do you see why?
So we are going to have at least two regions of integration.
One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$
and the other is the trapezoid with vertices
$(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$
If you decide the outer integral should be the integral over $dx,$
and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$
But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.
It turns out that two regions actually are sufficient.
You can integrate over the trapezoid this way:
$$
frac35 int_0^1 dy int_y^2 dx , (x - y)x(y+y^2).
$$
You can substitute this integral into the solution instead of the sum of integrals
$$frac35 int_1^2 dx int_0^1 dy , (x - y)x(y+y^2)+
frac35 int_0^1 dx int_0^x dy , (x - y)x(y+y^2).$$
answered Jul 25 at 5:42
David K
48.2k340107
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
When $1<x<2$, $y$ is necessarily less than $x$. That's one region.
when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).
answered Jul 24 at 20:51
Alon Amit
10.2k3765
add a comment |Â
up vote
0
down vote
When $1<x<2$, $y$ is necessarily less than $x$. That's one region.
when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).
answered Jul 24 at 20:51
Alon Amit
10.2k3765
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When $1<x<2$, $y$ is necessarily less than $x$. That's one region.
when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).
answered Jul 24 at 20:51
Alon Amit
10.2k3765
When $1<x<2$, $y$ is necessarily less than $x$. That's one region.
when $0<x<1$, there are two sub-regions: $y<x$ (that's the second integral) and $y>x$ (that's the third).
answered Jul 24 at 20:51
Alon Amit
10.2k3765
answered Jul 24 at 20:51
Alon Amit
10.2k3765
answered Jul 24 at 20:51
Alon Amit
10.2k3765
answered Jul 24 at 20:51
Alon Amit
10.2k3765
10.2k3765
add a comment |Â
add a comment |Â
up vote
0
down vote
The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative.
This makes it very awkward to integrate in just one integral; do you see why?
So we are going to have at least two regions of integration.
One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$
and the other is the trapezoid with vertices
$(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$
If you decide the outer integral should be the integral over $dx,$
and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$
But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.
It turns out that two regions actually are sufficient.
You can integrate over the trapezoid this way:
$$
frac35 int_0^1 dy int_y^2 dx , (x - y)x(y+y^2).
$$
You can substitute this integral into the solution instead of the sum of integrals
$$frac35 int_1^2 dx int_0^1 dy , (x - y)x(y+y^2)+
frac35 int_0^1 dx int_0^x dy , (x - y)x(y+y^2).$$
answered Jul 25 at 5:42
David K
48.2k340107
add a comment |Â
up vote
0
down vote
The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative.
This makes it very awkward to integrate in just one integral; do you see why?
So we are going to have at least two regions of integration.
One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$
and the other is the trapezoid with vertices
$(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$
If you decide the outer integral should be the integral over $dx,$
and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$
But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.
It turns out that two regions actually are sufficient.
You can integrate over the trapezoid this way:
$$
frac35 int_0^1 dy int_y^2 dx , (x - y)x(y+y^2).
$$
You can substitute this integral into the solution instead of the sum of integrals
$$frac35 int_1^2 dx int_0^1 dy , (x - y)x(y+y^2)+
frac35 int_0^1 dx int_0^x dy , (x - y)x(y+y^2).$$
answered Jul 25 at 5:42
David K
48.2k340107
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative.
This makes it very awkward to integrate in just one integral; do you see why?
So we are going to have at least two regions of integration.
One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$
and the other is the trapezoid with vertices
$(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$
If you decide the outer integral should be the integral over $dx,$
and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$
But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.
It turns out that two regions actually are sufficient.
You can integrate over the trapezoid this way:
$$
frac35 int_0^1 dy int_y^2 dx , (x - y)x(y+y^2).
$$
You can substitute this integral into the solution instead of the sum of integrals
$$frac35 int_1^2 dx int_0^1 dy , (x - y)x(y+y^2)+
frac35 int_0^1 dx int_0^x dy , (x - y)x(y+y^2).$$
answered Jul 25 at 5:42
David K
48.2k340107
The entire region of integration includes some points where $x - y$ is positive and some where $x - y$ is negative.
This makes it very awkward to integrate in just one integral; do you see why?
So we are going to have at least two regions of integration.
One region is the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1),$
and the other is the trapezoid with vertices
$(0,0),$ $(1,1),$ $(2,1),$ and $(2,0).$
If you decide the outer integral should be the integral over $dx,$
and the inner integral will integrate over $dy,$ then it is hard to integrate over the trapezoid using just one region: for $0 < x < 1,$ the upper bound on $y$ is given by $y = x,$ but for $1 < x < 2$ it is given by $y = 1.$
But actually it is not necessary to use three regions (one for the triangle and two for the trapezoid); the writer of this solution merely found it convenient.
It turns out that two regions actually are sufficient.
You can integrate over the trapezoid this way:
$$
frac35 int_0^1 dy int_y^2 dx , (x - y)x(y+y^2).
$$
You can substitute this integral into the solution instead of the sum of integrals
$$frac35 int_1^2 dx int_0^1 dy , (x - y)x(y+y^2)+
frac35 int_0^1 dx int_0^x dy , (x - y)x(y+y^2).$$
answered Jul 25 at 5:42
David K
48.2k340107
answered Jul 25 at 5:42
David K
48.2k340107
answered Jul 25 at 5:42
David K
48.2k340107
answered Jul 25 at 5:42
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
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Do you understand that doing so actually leads to the desired result?
– Henning Makholm
Jul 24 at 20:49
Have you drawn a picture? I think than will make it plain.
– saulspatz
Jul 24 at 20:51
There's usually more than one way to solve a problem like this. I think it can be done with just two regions if you make all your integrals be of the form $int dyint dx ldots.$ But you try putting the two $int dxint dy$ integrals in this solution together to make one region, and see how easily you can say what the bounds of the $dy$ integral should be.
– David K
Jul 24 at 20:54