Evaluate integral of min function [on hold]
Multi tool use
Probably/surely a trivial thing, but I wanted to know how to evaluate in general a thing like this: If I know thaat $x,y in [0,1]$ I want to evaluate
begin{equation}
int_0^1 min{x,y},dy
end{equation}
Can you help me?
calculus integration definite-integrals
asked Jan 5 at 22:25
James ArtenJames Arten
12911
put on hold as off-topic by Nosrati, RRL, StubbornAtom, user91500, Paul Frost 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, StubbornAtom, user91500, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Probably/surely a trivial thing, but I wanted to know how to evaluate in general a thing like this: If I know thaat $x,y in [0,1]$ I want to evaluate
begin{equation}
int_0^1 min{x,y},dy
end{equation}
Can you help me?
calculus integration definite-integrals
asked Jan 5 at 22:25
James ArtenJames Arten
12911
put on hold as off-topic by Nosrati, RRL, StubbornAtom, user91500, Paul Frost 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, StubbornAtom, user91500, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
1
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31
add a comment |
Probably/surely a trivial thing, but I wanted to know how to evaluate in general a thing like this: If I know thaat $x,y in [0,1]$ I want to evaluate
begin{equation}
int_0^1 min{x,y},dy
end{equation}
Can you help me?
calculus integration definite-integrals
asked Jan 5 at 22:25
James ArtenJames Arten
12911
Probably/surely a trivial thing, but I wanted to know how to evaluate in general a thing like this: If I know thaat $x,y in [0,1]$ I want to evaluate
begin{equation}
int_0^1 min{x,y},dy
end{equation}
Can you help me?
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 5 at 22:25
James ArtenJames Arten
12911
asked Jan 5 at 22:25
James ArtenJames Arten
12911
asked Jan 5 at 22:25
James ArtenJames Arten
12911
asked Jan 5 at 22:25
James ArtenJames Arten
12911
asked Jan 5 at 22:25
James ArtenJames Arten
12911
12911
put on hold as off-topic by Nosrati, RRL, StubbornAtom, user91500, Paul Frost 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, StubbornAtom, user91500, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Nosrati, RRL, StubbornAtom, user91500, Paul Frost 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, StubbornAtom, user91500, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
1
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31
add a comment |
1
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
1
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31
1
1
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
1
1
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31
add a comment |
1 Answer
1
active
oldest
votes
As $x in [0,1]$, $int_0^1 min(x,y) dy$ can be split into two parts:
$int_0^1 min(x,y) dy= int_0^x min(x,y) dy + int_x^1 min(x,y) dy$
In the first integral, the integration variable $y$ varies between 0 and $x$. Therefore, as @Kenny Wong correctly points out, $y$ is always less than or equal to $x$, so $min(x,y)=y$.
Likewise, in the second integral, the integration variable $y$ varies between x and $1$. Therefore, $y$ is always greater than or equal to $x$, so $min(x,y)=x$.
Therefore we get:
$int_0^1 min(x,y) dy= int_0^x y dy + int_x^1 x dy$ =
$frac{1}{2}x^2 + x(1-x)$ = $frac{2x-x^2}{2}$
answered Jan 6 at 0:05
pendermathpendermath
18510
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pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
As $x in [0,1]$, $int_0^1 min(x,y) dy$ can be split into two parts:
$int_0^1 min(x,y) dy= int_0^x min(x,y) dy + int_x^1 min(x,y) dy$
In the first integral, the integration variable $y$ varies between 0 and $x$. Therefore, as @Kenny Wong correctly points out, $y$ is always less than or equal to $x$, so $min(x,y)=y$.
Likewise, in the second integral, the integration variable $y$ varies between x and $1$. Therefore, $y$ is always greater than or equal to $x$, so $min(x,y)=x$.
Therefore we get:
$int_0^1 min(x,y) dy= int_0^x y dy + int_x^1 x dy$ =
$frac{1}{2}x^2 + x(1-x)$ = $frac{2x-x^2}{2}$
answered Jan 6 at 0:05
pendermathpendermath
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
As $x in [0,1]$, $int_0^1 min(x,y) dy$ can be split into two parts:
$int_0^1 min(x,y) dy= int_0^x min(x,y) dy + int_x^1 min(x,y) dy$
In the first integral, the integration variable $y$ varies between 0 and $x$. Therefore, as @Kenny Wong correctly points out, $y$ is always less than or equal to $x$, so $min(x,y)=y$.
Likewise, in the second integral, the integration variable $y$ varies between x and $1$. Therefore, $y$ is always greater than or equal to $x$, so $min(x,y)=x$.
Therefore we get:
$int_0^1 min(x,y) dy= int_0^x y dy + int_x^1 x dy$ =
$frac{1}{2}x^2 + x(1-x)$ = $frac{2x-x^2}{2}$
answered Jan 6 at 0:05
pendermathpendermath
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
As $x in [0,1]$, $int_0^1 min(x,y) dy$ can be split into two parts:
$int_0^1 min(x,y) dy= int_0^x min(x,y) dy + int_x^1 min(x,y) dy$
In the first integral, the integration variable $y$ varies between 0 and $x$. Therefore, as @Kenny Wong correctly points out, $y$ is always less than or equal to $x$, so $min(x,y)=y$.
Likewise, in the second integral, the integration variable $y$ varies between x and $1$. Therefore, $y$ is always greater than or equal to $x$, so $min(x,y)=x$.
Therefore we get:
$int_0^1 min(x,y) dy= int_0^x y dy + int_x^1 x dy$ =
$frac{1}{2}x^2 + x(1-x)$ = $frac{2x-x^2}{2}$
answered Jan 6 at 0:05
pendermathpendermath
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
As $x in [0,1]$, $int_0^1 min(x,y) dy$ can be split into two parts:
$int_0^1 min(x,y) dy= int_0^x min(x,y) dy + int_x^1 min(x,y) dy$
In the first integral, the integration variable $y$ varies between 0 and $x$. Therefore, as @Kenny Wong correctly points out, $y$ is always less than or equal to $x$, so $min(x,y)=y$.
Likewise, in the second integral, the integration variable $y$ varies between x and $1$. Therefore, $y$ is always greater than or equal to $x$, so $min(x,y)=x$.
Therefore we get:
$int_0^1 min(x,y) dy= int_0^x y dy + int_x^1 x dy$ =
$frac{1}{2}x^2 + x(1-x)$ = $frac{2x-x^2}{2}$
answered Jan 6 at 0:05
pendermathpendermath
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 6 at 0:05
pendermathpendermath
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 6 at 0:05
pendermathpendermath
18510
answered Jan 6 at 0:05
pendermathpendermath
18510
18510
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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1
So you're evaluating $int_0^x y dy + int_x^1 x dy$...
– Kenny Wong
Jan 5 at 22:29
why does it become like this? Thank you
– James Arten
Jan 5 at 22:30
1
If $y in [0,x]$, then $min(x,y) = y$. If $y in [x, 1]$, then $min(x,y) = x$.
– Kenny Wong
Jan 5 at 22:31