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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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2

$begingroup$

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

share|cite|improve this question

edited 7 hours ago

Joe

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$begin{cases}0leqslant x'leqslant 5\0leqslant xleqslant x'end{cases}iffbegin{cases}0leqslant xleqslant 5\xleqslant x'leqslant 5end{cases}.$$
  $endgroup$
  – metamorphy
  8 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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

share|cite|improve this question

edited 7 hours ago

Joe

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$begin{cases}0leqslant x'leqslant 5\0leqslant xleqslant x'end{cases}iffbegin{cases}0leqslant xleqslant 5\xleqslant x'leqslant 5end{cases}.$$
  $endgroup$
  – 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

share|cite|improve this question

edited 7 hours ago

Joe

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

Joe

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

share|cite|improve this question

edited 7 hours ago

Joe

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

asked 8 hours ago

JoeJoe

18922 silver badges1616 bronze badges

3 Answers
3

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3

$begingroup$

Often, as in your case, a picture is very helpful:

You see immediately $0leq x leq 5, x leq x' leq 5$.

share|cite|improve this answer

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

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.

share|cite|improve this answer

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

add a comment
 | 

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3

$begingroup$

Often, as in your case, a picture is very helpful:

You see immediately $0leq x leq 5, x leq x' leq 5$.

share|cite|improve this answer

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Often, as in your case, a picture is very helpful:

You see immediately $0leq x leq 5, x leq x' leq 5$.

share|cite|improve this answer

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Often, as in your case, a picture is very helpful:

You see immediately $0leq x leq 5, x leq x' leq 5$.

share|cite|improve this answer

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

answered 7 hours ago

trancelocationtrancelocation

17.1k11 gold badge1111 silver badges3030 bronze badges

3

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 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.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

answered 8 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

0

$begingroup$

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.

share|cite|improve this answer

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

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.

share|cite|improve this answer

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 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.

share|cite|improve this answer

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

$endgroup$

share|cite|improve this answer

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges

answered 5 hours ago

Ali Ashja'Ali Ashja'

86022 silver badges88 bronze badges