what is the order of integration for : integral of x dx * integral of y dy
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I'm still trying to get my head around he basics of this stuff so please use simple language in your answer !
$$ int dx int f(x,y) dy$$
the first integral limits are from 0 to 1 for dx and the second integral limits are from 0 to y = x + 1 for dy
is this just the same as writing :
$$ int int f(x,y) dy dx $$ ?
The order of integration would be the dy first because it specifies that the inner most integral limit is y = x + 1
So in that example x can go from a to b and y can go from 0 to y = x + 1
To reverse integration you just reverse the signs and use x = 1 - y for the limit of dx and the limit of dy becomes from a to b. Is this correct ?
I've drawn a cub with a few function through it and I am thinking about it like this
$$ int y=f(x) dx $$ gives you a point estimate of y and then you multiply it by the change in x, you use integration and you get the area under the curve.
$$ int z=f(x,y) dx $$
means you get an infinitesimally small slice of z so I picture it like a pane of glass going through a shape when the limits are a function it means that you look at the d variable (i.e dx or dy) and see what possible values it can take, i.e the pane of glass gets cut off at the end, for the other variable it is just a constant and depends on the given limit of integration and the function z=f(x,y)
integration
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
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add a comment |
$begingroup$
I'm still trying to get my head around he basics of this stuff so please use simple language in your answer !
$$ int dx int f(x,y) dy$$
the first integral limits are from 0 to 1 for dx and the second integral limits are from 0 to y = x + 1 for dy
is this just the same as writing :
$$ int int f(x,y) dy dx $$ ?
The order of integration would be the dy first because it specifies that the inner most integral limit is y = x + 1
So in that example x can go from a to b and y can go from 0 to y = x + 1
To reverse integration you just reverse the signs and use x = 1 - y for the limit of dx and the limit of dy becomes from a to b. Is this correct ?
I've drawn a cub with a few function through it and I am thinking about it like this
$$ int y=f(x) dx $$ gives you a point estimate of y and then you multiply it by the change in x, you use integration and you get the area under the curve.
$$ int z=f(x,y) dx $$
means you get an infinitesimally small slice of z so I picture it like a pane of glass going through a shape when the limits are a function it means that you look at the d variable (i.e dx or dy) and see what possible values it can take, i.e the pane of glass gets cut off at the end, for the other variable it is just a constant and depends on the given limit of integration and the function z=f(x,y)
integration
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
$endgroup$
add a comment |
$begingroup$
I'm still trying to get my head around he basics of this stuff so please use simple language in your answer !
$$ int dx int f(x,y) dy$$
the first integral limits are from 0 to 1 for dx and the second integral limits are from 0 to y = x + 1 for dy
is this just the same as writing :
$$ int int f(x,y) dy dx $$ ?
The order of integration would be the dy first because it specifies that the inner most integral limit is y = x + 1
So in that example x can go from a to b and y can go from 0 to y = x + 1
To reverse integration you just reverse the signs and use x = 1 - y for the limit of dx and the limit of dy becomes from a to b. Is this correct ?
I've drawn a cub with a few function through it and I am thinking about it like this
$$ int y=f(x) dx $$ gives you a point estimate of y and then you multiply it by the change in x, you use integration and you get the area under the curve.
$$ int z=f(x,y) dx $$
means you get an infinitesimally small slice of z so I picture it like a pane of glass going through a shape when the limits are a function it means that you look at the d variable (i.e dx or dy) and see what possible values it can take, i.e the pane of glass gets cut off at the end, for the other variable it is just a constant and depends on the given limit of integration and the function z=f(x,y)
integration
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
$endgroup$
I'm still trying to get my head around he basics of this stuff so please use simple language in your answer !
$$ int dx int f(x,y) dy$$
the first integral limits are from 0 to 1 for dx and the second integral limits are from 0 to y = x + 1 for dy
is this just the same as writing :
$$ int int f(x,y) dy dx $$ ?
The order of integration would be the dy first because it specifies that the inner most integral limit is y = x + 1
So in that example x can go from a to b and y can go from 0 to y = x + 1
To reverse integration you just reverse the signs and use x = 1 - y for the limit of dx and the limit of dy becomes from a to b. Is this correct ?
I've drawn a cub with a few function through it and I am thinking about it like this
$$ int y=f(x) dx $$ gives you a point estimate of y and then you multiply it by the change in x, you use integration and you get the area under the curve.
$$ int z=f(x,y) dx $$
means you get an infinitesimally small slice of z so I picture it like a pane of glass going through a shape when the limits are a function it means that you look at the d variable (i.e dx or dy) and see what possible values it can take, i.e the pane of glass gets cut off at the end, for the other variable it is just a constant and depends on the given limit of integration and the function z=f(x,y)
integration
integration
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
edited Oct 24 '14 at 13:56
user3528592
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
asked Oct 24 '14 at 13:48
user3528592user3528592
8910
8910
add a comment |
add a comment |
1 Answer
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Integration in a prefix operator notation (beware: this is no official term) I have encountered often in theoretical physics lectures.
In your case it would be used as
$$
intlimits!!dx int!!dy ,, f(x,y)
$$
So the variant you gave is not using this consistently but a rather unusual mix of notations.
Your post beyond the question mark I do not understand.
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
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1 Answer
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1 Answer
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$begingroup$
Integration in a prefix operator notation (beware: this is no official term) I have encountered often in theoretical physics lectures.
In your case it would be used as
$$
intlimits!!dx int!!dy ,, f(x,y)
$$
So the variant you gave is not using this consistently but a rather unusual mix of notations.
Your post beyond the question mark I do not understand.
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
$endgroup$
add a comment |
$begingroup$
Integration in a prefix operator notation (beware: this is no official term) I have encountered often in theoretical physics lectures.
In your case it would be used as
$$
intlimits!!dx int!!dy ,, f(x,y)
$$
So the variant you gave is not using this consistently but a rather unusual mix of notations.
Your post beyond the question mark I do not understand.
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
$endgroup$
add a comment |
$begingroup$
Integration in a prefix operator notation (beware: this is no official term) I have encountered often in theoretical physics lectures.
In your case it would be used as
$$
intlimits!!dx int!!dy ,, f(x,y)
$$
So the variant you gave is not using this consistently but a rather unusual mix of notations.
Your post beyond the question mark I do not understand.
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
$endgroup$
Integration in a prefix operator notation (beware: this is no official term) I have encountered often in theoretical physics lectures.
In your case it would be used as
$$
intlimits!!dx int!!dy ,, f(x,y)
$$
So the variant you gave is not using this consistently but a rather unusual mix of notations.
Your post beyond the question mark I do not understand.
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
edited Oct 24 '14 at 13:58
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
answered Oct 24 '14 at 13:53
mvwmvw
31.5k22252
31.5k22252
add a comment |
add a comment |
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