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What is the rigor behind U subsitution?

U-Substitution IntuitionAttempted proof of the first part of the Fundamental Theorem of CalculusEvaluating definite integrals whose limits are absolutely equalProve one limit$limlimits_mtoinftyfrac1msum_n=1^mcosleft(frac2pi n xbright)$ to a definite integralRiemann sums to find limitsCan we change $triangle$ratio of Riemann integral? $int_a^b f(x)(a(dx))^b(dx)quad a,bin mathbb R^neq0$Riemann Sum of Ratio of Equation of Lines$int _0^axleft(1-fracxaright)dx:$ using Riemann SumsA question on the limit $lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$Limit of the sum using integral

0

1

$begingroup$

$ int f(g(x)) dx = int frac f(u)u' du$

requires that

$ int f(x) dx = int f(x) cdot dx$

but dx just represents the variable that F(x) +c is a function of. So why is it legal for dx to be treated algebriacally?

I tried investigating this property by using riemann summation:

$limlimits_n to infty(sum_k=1^n f(frackxn )fracxn) = int_0^x f(t)dt= (F(x)+c)-(F(0)+c)$

and so you can define

$ limlimits_n to infty (sum_k=1^n f(frackxn )fracxn)+F(0)+c) = int f(x) dx$

you can write $fracxn = dx$ and $kfracxn=kdx= x_k$

then you have

$ limlimits_n to infty (sum_k=1^n f(x_k)dx)+F(0)+c) = int f(x) dx$

since dx is being written to be multiplied by the series, then you can define dx in different terms to aquire an integral in terms of other variables.

but that exists only in abstraction. I can't quite sufficiently complete the task of doing so.

calculus integration riemann-sum

share|cite|improve this question

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glados.Perhaps of interest: en.m.wikipedia.org/wiki/Integration_by_substitution
  $endgroup$
  – Peter Szilas
  Mar 29 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might find this post interesting: math.stackexchange.com/questions/3114746
  $endgroup$
  – Michael Rybkin
  Mar 29 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you talking about definite or indefinite integration? For indefinite integrals (antiderivatives), it nothing but the chain rule.
  $endgroup$
  – Hans Lundmark
  Mar 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate it but I am more looking in to rigor rather than intuition.
  $endgroup$
  – GLaDOS
  Mar 29 at 10:54
  

  
  
  
  

add a comment | 

0

1

$begingroup$

$ int f(g(x)) dx = int frac f(u)u' du$

requires that

$ int f(x) dx = int f(x) cdot dx$

but dx just represents the variable that F(x) +c is a function of. So why is it legal for dx to be treated algebriacally?

I tried investigating this property by using riemann summation:

$limlimits_n to infty(sum_k=1^n f(frackxn )fracxn) = int_0^x f(t)dt= (F(x)+c)-(F(0)+c)$

and so you can define

$ limlimits_n to infty (sum_k=1^n f(frackxn )fracxn)+F(0)+c) = int f(x) dx$

you can write $fracxn = dx$ and $kfracxn=kdx= x_k$

then you have

$ limlimits_n to infty (sum_k=1^n f(x_k)dx)+F(0)+c) = int f(x) dx$

since dx is being written to be multiplied by the series, then you can define dx in different terms to aquire an integral in terms of other variables.

but that exists only in abstraction. I can't quite sufficiently complete the task of doing so.

calculus integration riemann-sum

share|cite|improve this question

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glados.Perhaps of interest: en.m.wikipedia.org/wiki/Integration_by_substitution
  $endgroup$
  – Peter Szilas
  Mar 29 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might find this post interesting: math.stackexchange.com/questions/3114746
  $endgroup$
  – Michael Rybkin
  Mar 29 at 10:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you talking about definite or indefinite integration? For indefinite integrals (antiderivatives), it nothing but the chain rule.
  $endgroup$
  – Hans Lundmark
  Mar 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate it but I am more looking in to rigor rather than intuition.
  $endgroup$
  – GLaDOS
  Mar 29 at 10:54
  

  
  
  
  

add a comment | 

$begingroup$

$ int f(g(x)) dx = int frac f(u)u' du$

requires that

$ int f(x) dx = int f(x) cdot dx$

but dx just represents the variable that F(x) +c is a function of. So why is it legal for dx to be treated algebriacally?

I tried investigating this property by using riemann summation:

$limlimits_n to infty(sum_k=1^n f(frackxn )fracxn) = int_0^x f(t)dt= (F(x)+c)-(F(0)+c)$

and so you can define

$ limlimits_n to infty (sum_k=1^n f(frackxn )fracxn)+F(0)+c) = int f(x) dx$

you can write $fracxn = dx$ and $kfracxn=kdx= x_k$

then you have

$ limlimits_n to infty (sum_k=1^n f(x_k)dx)+F(0)+c) = int f(x) dx$

since dx is being written to be multiplied by the series, then you can define dx in different terms to aquire an integral in terms of other variables.

but that exists only in abstraction. I can't quite sufficiently complete the task of doing so.

calculus integration riemann-sum

share|cite|improve this question

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

$endgroup$

share|cite|improve this question

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

share|cite|improve this question

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

asked Mar 29 at 10:34

GLaDOSGLaDOS

133

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's prove that $int_a^b h(g(x)) g^prime(x) dx=int_g(a)^g(b) h(u) du$ for $g$ monotonic on $[a,,b]$ with $a<b$. Let $H$ denote an antiderivative of $h$, without loss of generality satisfying $H(a)=0$. Then the right-hand side of the putative result is $H(g(b))-H(g(a))$. Differentiating this with respect to $b$ gives $h(g(b)) g^prime(b)$, so $H(g(b))-H(g(a))=int_a^b h(g(x)) g^prime(x) dx$ as required.

share|cite|improve this answer

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I really like this one because it doesn't require the multiplication property of dx.
  $endgroup$
  – GLaDOS
  Mar 29 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you prove it for indefinite integral?
  $endgroup$
  – GLaDOS
  Mar 29 at 11:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GLaDOS An indefinite integral just means "your favourite antiderivative $+C$", so if substitution works for definite integrals it also does for indefinite ones. For example, $$int_a^b 3x^2sin x^3 dx=int_a^3^b^3sin uduimpliesint 3x^2sin x^3 dx=intsin udu.$$
  $endgroup$
  – J.G.
  Mar 29 at 12:02
  
  

  
  
  
  

add a comment | 

Your Answer

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0

$begingroup$

Let's prove that $int_a^b h(g(x)) g^prime(x) dx=int_g(a)^g(b) h(u) du$ for $g$ monotonic on $[a,,b]$ with $a<b$. Let $H$ denote an antiderivative of $h$, without loss of generality satisfying $H(a)=0$. Then the right-hand side of the putative result is $H(g(b))-H(g(a))$. Differentiating this with respect to $b$ gives $h(g(b)) g^prime(b)$, so $H(g(b))-H(g(a))=int_a^b h(g(x)) g^prime(x) dx$ as required.

share|cite|improve this answer

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I really like this one because it doesn't require the multiplication property of dx.
  $endgroup$
  – GLaDOS
  Mar 29 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you prove it for indefinite integral?
  $endgroup$
  – GLaDOS
  Mar 29 at 11:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GLaDOS An indefinite integral just means "your favourite antiderivative $+C$", so if substitution works for definite integrals it also does for indefinite ones. For example, $$int_a^b 3x^2sin x^3 dx=int_a^3^b^3sin uduimpliesint 3x^2sin x^3 dx=intsin udu.$$
  $endgroup$
  – J.G.
  Mar 29 at 12:02
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let's prove that $int_a^b h(g(x)) g^prime(x) dx=int_g(a)^g(b) h(u) du$ for $g$ monotonic on $[a,,b]$ with $a<b$. Let $H$ denote an antiderivative of $h$, without loss of generality satisfying $H(a)=0$. Then the right-hand side of the putative result is $H(g(b))-H(g(a))$. Differentiating this with respect to $b$ gives $h(g(b)) g^prime(b)$, so $H(g(b))-H(g(a))=int_a^b h(g(x)) g^prime(x) dx$ as required.

share|cite|improve this answer

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I really like this one because it doesn't require the multiplication property of dx.
  $endgroup$
  – GLaDOS
  Mar 29 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you prove it for indefinite integral?
  $endgroup$
  – GLaDOS
  Mar 29 at 11:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GLaDOS An indefinite integral just means "your favourite antiderivative $+C$", so if substitution works for definite integrals it also does for indefinite ones. For example, $$int_a^b 3x^2sin x^3 dx=int_a^3^b^3sin uduimpliesint 3x^2sin x^3 dx=intsin udu.$$
  $endgroup$
  – J.G.
  Mar 29 at 12:02
  
  

  
  
  
  

add a comment | 

$begingroup$

Let's prove that $int_a^b h(g(x)) g^prime(x) dx=int_g(a)^g(b) h(u) du$ for $g$ monotonic on $[a,,b]$ with $a<b$. Let $H$ denote an antiderivative of $h$, without loss of generality satisfying $H(a)=0$. Then the right-hand side of the putative result is $H(g(b))-H(g(a))$. Differentiating this with respect to $b$ gives $h(g(b)) g^prime(b)$, so $H(g(b))-H(g(a))=int_a^b h(g(x)) g^prime(x) dx$ as required.

share|cite|improve this answer

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

$endgroup$

share|cite|improve this answer

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250

answered Mar 29 at 10:48

J.G.J.G.

32.6k23250