How to change the order of integration when limit is a function?Systematic method to change the order of integration in multiple integralsChanging order of integration (multiple integral)change order of integrationChanging integration limits to sine functionDoes changing the order of double integration (both integral limits are constants) alter the final answer?How do you change the order of integration without sketching?How to find the integration limits when changing integration order?
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How to change the order of integration when limit is a function?
Systematic method to change the order of integration in multiple integralsChanging order of integration (multiple integral)change order of integrationChanging integration limits to sine functionDoes changing the order of double integration (both integral limits are constants) alter the final answer?How do you change the order of integration without sketching?How to find the integration limits when changing integration order?
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$begingroup$
Consider the double integral:
$A=displaystyleint_0^5 left( int_0^7 x'^2x^3 dx right) dx'$
We can change the order of integration:
$B=displaystyleint_0^7 left( int_0^5 x'^2x^3 dx' right) dx$
Now consider another double integral:
$C=displaystyleint_0^5 left( int_0^x' x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is a function?
EDIT (Generalization)
Now consider another double integral:
$D=displaystyleint_0^5 left( int_0^f(x') x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is an arbitrary function $f(x')$?
calculus integration multivariable-calculus functions definite-integrals
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
$endgroup$
add a comment
|
$begingroup$
Consider the double integral:
$A=displaystyleint_0^5 left( int_0^7 x'^2x^3 dx right) dx'$
We can change the order of integration:
$B=displaystyleint_0^7 left( int_0^5 x'^2x^3 dx' right) dx$
Now consider another double integral:
$C=displaystyleint_0^5 left( int_0^x' x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is a function?
EDIT (Generalization)
Now consider another double integral:
$D=displaystyleint_0^5 left( int_0^f(x') x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is an arbitrary function $f(x')$?
calculus integration multivariable-calculus functions definite-integrals
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
$endgroup$
1
$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago
add a comment
|
$begingroup$
Consider the double integral:
$A=displaystyleint_0^5 left( int_0^7 x'^2x^3 dx right) dx'$
We can change the order of integration:
$B=displaystyleint_0^7 left( int_0^5 x'^2x^3 dx' right) dx$
Now consider another double integral:
$C=displaystyleint_0^5 left( int_0^x' x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is a function?
EDIT (Generalization)
Now consider another double integral:
$D=displaystyleint_0^5 left( int_0^f(x') x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is an arbitrary function $f(x')$?
calculus integration multivariable-calculus functions definite-integrals
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
$endgroup$
Consider the double integral:
$A=displaystyleint_0^5 left( int_0^7 x'^2x^3 dx right) dx'$
We can change the order of integration:
$B=displaystyleint_0^7 left( int_0^5 x'^2x^3 dx' right) dx$
Now consider another double integral:
$C=displaystyleint_0^5 left( int_0^x' x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is a function?
EDIT (Generalization)
Now consider another double integral:
$D=displaystyleint_0^5 left( int_0^f(x') x'^2x^3 dx right) dx'$
How shall one change the order of this integral, i.e. when the limit is an arbitrary function $f(x')$?
calculus integration multivariable-calculus functions definite-integrals
calculus integration multivariable-calculus functions definite-integrals
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
edited 8 hours ago
Joe
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
asked 8 hours ago
JoeJoe
1892 silver badges16 bronze badges
1892 silver badges16 bronze badges
1
$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago
add a comment
|
1
$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago
1
1
$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago
$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
Often, as in your case, a picture is very helpful:
You see immediately $0leq x leq 5, x leq x' leq 5$.
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
$endgroup$
add a comment
|
$begingroup$
Same as what @metamorphy said:
You should find appropriate limit for $x'$ by $x$:
$$C = int_0^5 left( int_x^5 x'^2x^3 dx' right) dx$$
In fact you integrate on a triangle of points $(0,0),(0,5),(5,5)$, by order of $(x,x')$, as @trancelocation showed it in graph.
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
$begingroup$
For its generalization, in fact you need to solve this:
$$x leqslant f(x') Longrightarrow g(x) leqslant x'$$
At first look, it may seems that $g=f^-1$ is the solution! It isn't true always, but we can do it by some prework. The above conclusion is true for $g=f^-1$, if $f$ be an increasing function in which domain we want to solve that inequality. And when it be decreasing function, just reverse the direction to get answer. So in fact we need to break the domain $0 leqslant x' leqslant 5$ to some $a_i leqslant x' leqslant b_i$ for $1 leqslant i leqslant n$ such that $a_1=0, b_i=a_i+1, b_n=5$ such that all $a_i,b_i$ are optimum points of $f$ that it's monotone in every interval. After that you can change order of integral, but you need to break it over intervals that expressed.
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Often, as in your case, a picture is very helpful:
You see immediately $0leq x leq 5, x leq x' leq 5$.
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
$endgroup$
add a comment
|
$begingroup$
Often, as in your case, a picture is very helpful:
You see immediately $0leq x leq 5, x leq x' leq 5$.
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
$endgroup$
add a comment
|
$begingroup$
Often, as in your case, a picture is very helpful:
You see immediately $0leq x leq 5, x leq x' leq 5$.
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
$endgroup$
Often, as in your case, a picture is very helpful:
You see immediately $0leq x leq 5, x leq x' leq 5$.
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
answered 7 hours ago
trancelocationtrancelocation
17.1k1 gold badge11 silver badges30 bronze badges
17.1k1 gold badge11 silver badges30 bronze badges
add a comment
|
add a comment
|
$begingroup$
Same as what @metamorphy said:
You should find appropriate limit for $x'$ by $x$:
$$C = int_0^5 left( int_x^5 x'^2x^3 dx' right) dx$$
In fact you integrate on a triangle of points $(0,0),(0,5),(5,5)$, by order of $(x,x')$, as @trancelocation showed it in graph.
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
$begingroup$
Same as what @metamorphy said:
You should find appropriate limit for $x'$ by $x$:
$$C = int_0^5 left( int_x^5 x'^2x^3 dx' right) dx$$
In fact you integrate on a triangle of points $(0,0),(0,5),(5,5)$, by order of $(x,x')$, as @trancelocation showed it in graph.
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
$begingroup$
Same as what @metamorphy said:
You should find appropriate limit for $x'$ by $x$:
$$C = int_0^5 left( int_x^5 x'^2x^3 dx' right) dx$$
In fact you integrate on a triangle of points $(0,0),(0,5),(5,5)$, by order of $(x,x')$, as @trancelocation showed it in graph.
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
Same as what @metamorphy said:
You should find appropriate limit for $x'$ by $x$:
$$C = int_0^5 left( int_x^5 x'^2x^3 dx' right) dx$$
In fact you integrate on a triangle of points $(0,0),(0,5),(5,5)$, by order of $(x,x')$, as @trancelocation showed it in graph.
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
edited 6 hours ago
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
answered 8 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
8602 silver badges8 bronze badges
add a comment
|
add a comment
|
$begingroup$
For its generalization, in fact you need to solve this:
$$x leqslant f(x') Longrightarrow g(x) leqslant x'$$
At first look, it may seems that $g=f^-1$ is the solution! It isn't true always, but we can do it by some prework. The above conclusion is true for $g=f^-1$, if $f$ be an increasing function in which domain we want to solve that inequality. And when it be decreasing function, just reverse the direction to get answer. So in fact we need to break the domain $0 leqslant x' leqslant 5$ to some $a_i leqslant x' leqslant b_i$ for $1 leqslant i leqslant n$ such that $a_1=0, b_i=a_i+1, b_n=5$ such that all $a_i,b_i$ are optimum points of $f$ that it's monotone in every interval. After that you can change order of integral, but you need to break it over intervals that expressed.
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
$begingroup$
For its generalization, in fact you need to solve this:
$$x leqslant f(x') Longrightarrow g(x) leqslant x'$$
At first look, it may seems that $g=f^-1$ is the solution! It isn't true always, but we can do it by some prework. The above conclusion is true for $g=f^-1$, if $f$ be an increasing function in which domain we want to solve that inequality. And when it be decreasing function, just reverse the direction to get answer. So in fact we need to break the domain $0 leqslant x' leqslant 5$ to some $a_i leqslant x' leqslant b_i$ for $1 leqslant i leqslant n$ such that $a_1=0, b_i=a_i+1, b_n=5$ such that all $a_i,b_i$ are optimum points of $f$ that it's monotone in every interval. After that you can change order of integral, but you need to break it over intervals that expressed.
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
add a comment
|
$begingroup$
For its generalization, in fact you need to solve this:
$$x leqslant f(x') Longrightarrow g(x) leqslant x'$$
At first look, it may seems that $g=f^-1$ is the solution! It isn't true always, but we can do it by some prework. The above conclusion is true for $g=f^-1$, if $f$ be an increasing function in which domain we want to solve that inequality. And when it be decreasing function, just reverse the direction to get answer. So in fact we need to break the domain $0 leqslant x' leqslant 5$ to some $a_i leqslant x' leqslant b_i$ for $1 leqslant i leqslant n$ such that $a_1=0, b_i=a_i+1, b_n=5$ such that all $a_i,b_i$ are optimum points of $f$ that it's monotone in every interval. After that you can change order of integral, but you need to break it over intervals that expressed.
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
$endgroup$
For its generalization, in fact you need to solve this:
$$x leqslant f(x') Longrightarrow g(x) leqslant x'$$
At first look, it may seems that $g=f^-1$ is the solution! It isn't true always, but we can do it by some prework. The above conclusion is true for $g=f^-1$, if $f$ be an increasing function in which domain we want to solve that inequality. And when it be decreasing function, just reverse the direction to get answer. So in fact we need to break the domain $0 leqslant x' leqslant 5$ to some $a_i leqslant x' leqslant b_i$ for $1 leqslant i leqslant n$ such that $a_1=0, b_i=a_i+1, b_n=5$ such that all $a_i,b_i$ are optimum points of $f$ that it's monotone in every interval. After that you can change order of integral, but you need to break it over intervals that expressed.
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
answered 5 hours ago
Ali Ashja'Ali Ashja'
8602 silver badges8 bronze badges
8602 silver badges8 bronze badges
add a comment
|
add a comment
|
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$begingroup$
$$begincases0leqslant x'leqslant 5\0leqslant xleqslant x'endcasesiffbegincases0leqslant xleqslant 5\xleqslant x'leqslant 5endcases.$$
$endgroup$
– metamorphy
8 hours ago