Mixing
Simplification of an expression of a double integral
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Is the following expression able to be simplified?
$$I = int_0^{x/2} int_0^s f(s,r),dr,ds + int_{x/2}^x int_{2s - x}^s f(s,r),dr,ds . $$
Here $f(s,r)=u(r,2s+x-r)$ so that we could also write
$$I = int_0^{x/2} int_0^s u(r,2s+x-r),dr,ds + int_{x/2}^x int_{2s - x}^s u(r,2s+x-r),dr,ds . $$
$u$ is unknown.
I made a lot a variables substitutions but without any success. I feel that I can simplify it much more than that. Any ideas? Thanks.
integration definite-integrals multiple-integral
edited Jan 29 at 15:55
Adrian Keister
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
$endgroup$
add a comment |
$begingroup$
Is the following expression able to be simplified?
$$I = int_0^{x/2} int_0^s f(s,r),dr,ds + int_{x/2}^x int_{2s - x}^s f(s,r),dr,ds . $$
Here $f(s,r)=u(r,2s+x-r)$ so that we could also write
$$I = int_0^{x/2} int_0^s u(r,2s+x-r),dr,ds + int_{x/2}^x int_{2s - x}^s u(r,2s+x-r),dr,ds . $$
$u$ is unknown.
I made a lot a variables substitutions but without any success. I feel that I can simplify it much more than that. Any ideas? Thanks.
integration definite-integrals multiple-integral
edited Jan 29 at 15:55
Adrian Keister
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
$endgroup$
$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47
add a comment |
$begingroup$
Is the following expression able to be simplified?
$$I = int_0^{x/2} int_0^s f(s,r),dr,ds + int_{x/2}^x int_{2s - x}^s f(s,r),dr,ds . $$
Here $f(s,r)=u(r,2s+x-r)$ so that we could also write
$$I = int_0^{x/2} int_0^s u(r,2s+x-r),dr,ds + int_{x/2}^x int_{2s - x}^s u(r,2s+x-r),dr,ds . $$
$u$ is unknown.
I made a lot a variables substitutions but without any success. I feel that I can simplify it much more than that. Any ideas? Thanks.
integration definite-integrals multiple-integral
edited Jan 29 at 15:55
Adrian Keister
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
$endgroup$
Is the following expression able to be simplified?
$$I = int_0^{x/2} int_0^s f(s,r),dr,ds + int_{x/2}^x int_{2s - x}^s f(s,r),dr,ds . $$
Here $f(s,r)=u(r,2s+x-r)$ so that we could also write
$$I = int_0^{x/2} int_0^s u(r,2s+x-r),dr,ds + int_{x/2}^x int_{2s - x}^s u(r,2s+x-r),dr,ds . $$
$u$ is unknown.
I made a lot a variables substitutions but without any success. I feel that I can simplify it much more than that. Any ideas? Thanks.
integration definite-integrals multiple-integral
integration definite-integrals multiple-integral
edited Jan 29 at 15:55
Adrian Keister
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
edited Jan 29 at 15:55
Adrian Keister
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
edited Jan 29 at 15:55
Adrian Keister
5,28171933
edited Jan 29 at 15:55
Adrian Keister
5,28171933
edited Jan 29 at 15:55
Adrian Keister
5,28171933
5,28171933
asked Jan 29 at 15:05
GustaveGustave
734211
asked Jan 29 at 15:05
GustaveGustave
734211
asked Jan 29 at 15:05
GustaveGustave
734211
734211
$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47
add a comment |
$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47
$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47
add a comment |
1 Answer
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$begingroup$
With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:
So reversing the order of integration allows the integral to be written as the single double integral
$$ I=int_0^xint_r^{(r+x)/2}f(s,r),dsdr $$
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
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1 Answer
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$begingroup$
With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:
So reversing the order of integration allows the integral to be written as the single double integral
$$ I=int_0^xint_r^{(r+x)/2}f(s,r),dsdr $$
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:
So reversing the order of integration allows the integral to be written as the single double integral
$$ I=int_0^xint_r^{(r+x)/2}f(s,r),dsdr $$
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:
So reversing the order of integration allows the integral to be written as the single double integral
$$ I=int_0^xint_r^{(r+x)/2}f(s,r),dsdr $$
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:
So reversing the order of integration allows the integral to be written as the single double integral
$$ I=int_0^xint_r^{(r+x)/2}f(s,r),dsdr $$
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 30 at 2:19
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
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$begingroup$
Do you know what $f$ is? If so, can you please write it out: $f(s,r)=dots?$
$endgroup$
– Adrian Keister
Jan 29 at 15:14
$begingroup$
Yes, $f(s,r)=u(r,2s+x-r)$ for certain function $u$.
$endgroup$
– Gustave
Jan 29 at 15:15
$begingroup$
Ok, what's $u?$ Can you write out what $u$ is?
$endgroup$
– Adrian Keister
Jan 29 at 15:24
$begingroup$
No sir, it is uknown.
$endgroup$
– Gustave
Jan 29 at 15:25
$begingroup$
For what it's worth, changing the order of integration results in this: begin{align*}int_0^{x/2}int_0^sdr,ds&=int_0^{x/2}int_r^{x/2}ds,dr \ int_{x/2}^xint_{2s-x}^sdr,ds&=int_0^{x/2}int_{x/2}^{(r+x)/2}ds,dr+int_{x/2}^xint_r^{(r+x)/2}ds,dr.end{align*}
$endgroup$
– Adrian Keister
Jan 29 at 15:47