Evaluate integral of min function [on hold]

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?

asked Jan 5 at 22:25

James 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?

asked Jan 5 at 22:25

James 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?

asked Jan 5 at 22:25

James 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

asked Jan 5 at 22:25

James Arten

12911

asked Jan 5 at 22:25

James Arten

12911

asked Jan 5 at 22:25

James Arten

12911

asked Jan 5 at 22:25

James Arten

12911

asked Jan 5 at 22:25

James 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.

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

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

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

pendermath

18510

New contributor

add a comment |

1 Answer
1

active

oldest

votes

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

pendermath

18510

New contributor

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

pendermath

18510

New contributor

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

pendermath

18510

New contributor

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

pendermath

18510

New contributor

answered Jan 6 at 0:05

pendermath

18510

New contributor

answered Jan 6 at 0:05

pendermath

18510

answered Jan 6 at 0:05

pendermath

18510

New contributor

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