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Can we allow $f$ to be undefined at finitely many points in $(a,b)$ when formulating $int_a^b f(x) dx=F(b)-F(a)$, ($F$ is the antiderivative of $f$)

4

$begingroup$

Let $f$ be a real-valued function on a closed interval $[a,b]$ undefined only at finite points in $(a,b)$. Let $F$ be antiderivative of $f$. Then:

$$int_a^b f(x) dx=F(b)-F(a)$$

Is the theorem true? How shall we prove it?

I ask this because I have read that function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

calculus integration limits functions definite-integrals

share|cite|improve this question

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Asaf Karagila: Thanks for the edit.
  $endgroup$
  – Joe
  Mar 18 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Titles have 150 characters, and you should use them whenever possible to give a clear description of your question.
  $endgroup$
  – Asaf Karagila♦
  Mar 18 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Understood.......................
  $endgroup$
  – Joe
  Mar 18 at 11:29
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $f$ be a real-valued function on a closed interval $[a,b]$ undefined only at finite points in $(a,b)$. Let $F$ be antiderivative of $f$. Then:

$$int_a^b f(x) dx=F(b)-F(a)$$

Is the theorem true? How shall we prove it?

I ask this because I have read that function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

calculus integration limits functions definite-integrals

share|cite|improve this question

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Asaf Karagila: Thanks for the edit.
  $endgroup$
  – Joe
  Mar 18 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Titles have 150 characters, and you should use them whenever possible to give a clear description of your question.
  $endgroup$
  – Asaf Karagila♦
  Mar 18 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Understood.......................
  $endgroup$
  – Joe
  Mar 18 at 11:29
  

  
  
  
  

add a comment | 

$begingroup$

Let $f$ be a real-valued function on a closed interval $[a,b]$ undefined only at finite points in $(a,b)$. Let $F$ be antiderivative of $f$. Then:

$$int_a^b f(x) dx=F(b)-F(a)$$

Is the theorem true? How shall we prove it?

I ask this because I have read that function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

calculus integration limits functions definite-integrals

share|cite|improve this question

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

$endgroup$

share|cite|improve this question

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

share|cite|improve this question

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

edited Mar 18 at 11:21

Asaf Karagila♦

307k33439771

asked Mar 18 at 7:06

JoeJoe

313214

asked Mar 18 at 7:06

JoeJoe

313214

1 Answer
1

active

oldest

votes

6

$begingroup$

That statement doesn't make sense, since $f$ may well not be integrable on $[a,b]$ then. And if it is, the statment is not necessarily true. Take, for instance, the null function $f$ on $[0,2]setminus1$. Then$$beginarrayrcccFcolon&[0,2]setminus1&longrightarrow&mathbb R\&x&mapsto&begincases0&text if xin[0,1)\1&text if xin(1,2]endcasesendarray$$is an antiderivative of $f$, but$$int_0^2f(x),mathrm dx=0neq1=F(2)-F(0).$$

share|cite|improve this answer

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... But what about the statement "function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral."
  $endgroup$
  – Joe
  Mar 18 at 7:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's rather vague. I suppose that the idea was that if the domain of $f$ is something like $(a,b)setminusc,d$, with $c<d$, we could use the standard method of antiderivatives to compute $int_a^cf(x),mathrm dx$, $int_c^df(x),mathrm dx$, and $int_d^bf(x),mathrm dx$.
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes exactly that... Why is it so?
  $endgroup$
  – Joe
  Mar 18 at 7:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because, if $f$ is integrable,$$int_a^bf(x),mathrm dx=int_a^cf(x),mathrm dx+int_c^df(x),mathrm dx+int_d^af(x),mathrm dx.$$
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe To begin answering that question, you can't integrate a function that isn't even well-defined! A better statement would be, if $f$ and $g$ are integrable functions so that their values differ at finitely many points, then they have the same integrals. Now if you have learned about Riemann sum, the main idea is that you can always take a small enough partition so that the intervals containing those finitely many points have lengths as small as you wish.
  $endgroup$
  – lEm
  Mar 18 at 7:50
  

  
  
  
  

 | 
show 3 more comments

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6

$begingroup$

That statement doesn't make sense, since $f$ may well not be integrable on $[a,b]$ then. And if it is, the statment is not necessarily true. Take, for instance, the null function $f$ on $[0,2]setminus1$. Then$$beginarrayrcccFcolon&[0,2]setminus1&longrightarrow&mathbb R\&x&mapsto&begincases0&text if xin[0,1)\1&text if xin(1,2]endcasesendarray$$is an antiderivative of $f$, but$$int_0^2f(x),mathrm dx=0neq1=F(2)-F(0).$$

share|cite|improve this answer

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... But what about the statement "function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral."
  $endgroup$
  – Joe
  Mar 18 at 7:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's rather vague. I suppose that the idea was that if the domain of $f$ is something like $(a,b)setminusc,d$, with $c<d$, we could use the standard method of antiderivatives to compute $int_a^cf(x),mathrm dx$, $int_c^df(x),mathrm dx$, and $int_d^bf(x),mathrm dx$.
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes exactly that... Why is it so?
  $endgroup$
  – Joe
  Mar 18 at 7:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because, if $f$ is integrable,$$int_a^bf(x),mathrm dx=int_a^cf(x),mathrm dx+int_c^df(x),mathrm dx+int_d^af(x),mathrm dx.$$
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe To begin answering that question, you can't integrate a function that isn't even well-defined! A better statement would be, if $f$ and $g$ are integrable functions so that their values differ at finitely many points, then they have the same integrals. Now if you have learned about Riemann sum, the main idea is that you can always take a small enough partition so that the intervals containing those finitely many points have lengths as small as you wish.
  $endgroup$
  – lEm
  Mar 18 at 7:50
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

That statement doesn't make sense, since $f$ may well not be integrable on $[a,b]$ then. And if it is, the statment is not necessarily true. Take, for instance, the null function $f$ on $[0,2]setminus1$. Then$$beginarrayrcccFcolon&[0,2]setminus1&longrightarrow&mathbb R\&x&mapsto&begincases0&text if xin[0,1)\1&text if xin(1,2]endcasesendarray$$is an antiderivative of $f$, but$$int_0^2f(x),mathrm dx=0neq1=F(2)-F(0).$$

share|cite|improve this answer

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok... But what about the statement "function undefined at finite points in the interval $(a,b)$ doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral."
  $endgroup$
  – Joe
  Mar 18 at 7:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's rather vague. I suppose that the idea was that if the domain of $f$ is something like $(a,b)setminusc,d$, with $c<d$, we could use the standard method of antiderivatives to compute $int_a^cf(x),mathrm dx$, $int_c^df(x),mathrm dx$, and $int_d^bf(x),mathrm dx$.
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes exactly that... Why is it so?
  $endgroup$
  – Joe
  Mar 18 at 7:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because, if $f$ is integrable,$$int_a^bf(x),mathrm dx=int_a^cf(x),mathrm dx+int_c^df(x),mathrm dx+int_d^af(x),mathrm dx.$$
  $endgroup$
  – José Carlos Santos
  Mar 18 at 7:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Joe To begin answering that question, you can't integrate a function that isn't even well-defined! A better statement would be, if $f$ and $g$ are integrable functions so that their values differ at finitely many points, then they have the same integrals. Now if you have learned about Riemann sum, the main idea is that you can always take a small enough partition so that the intervals containing those finitely many points have lengths as small as you wish.
  $endgroup$
  – lEm
  Mar 18 at 7:50
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

That statement doesn't make sense, since $f$ may well not be integrable on $[a,b]$ then. And if it is, the statment is not necessarily true. Take, for instance, the null function $f$ on $[0,2]setminus1$. Then$$beginarrayrcccFcolon&[0,2]setminus1&longrightarrow&mathbb R\&x&mapsto&begincases0&text if xin[0,1)\1&text if xin(1,2]endcasesendarray$$is an antiderivative of $f$, but$$int_0^2f(x),mathrm dx=0neq1=F(2)-F(0).$$

share|cite|improve this answer

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

share|cite|improve this answer

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240

answered Mar 18 at 7:12

José Carlos SantosJosé Carlos Santos

171k23132240