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Given: $int_a^bf=int_a^bg$, Can we always find some point $theta$ in $[a,b]$ such that $f(theta)=g(theta)$

Lebesgue Integration and Convergence TheoremsLine integral on unit circle over non-holomorphic functionAny arbitrary closed smooth curve bounds a orientable surface?Basic facts related to Haar measureHow can I prove that there exists $cin[0,1]$ such that $intlimits_0^1intlimits_0^tf''(x)dxdt=frac12f''(c)$Is $int f$ on [0,1] always equal to $int f$ on [x,1] when take the limit of x to 0Find $f(x)$ such that $int_-infty^infty int_-a^a fracy+x1+y^2 f(x) dx dy<infty$Prove $int_R^+ frac11+x a(x) dx > 0$ given $int_R^+ frac1x a(x) dx > 0$ and $int_R^+ a(x) dx = 0$?Under what conditions are expected value and integral commutable?Can some transformation of a function always be found such that it has a specified integral?

2

$begingroup$

I noted something today, I don't know whether the result holds in general setting or not.

Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_a^bf=int_a^bg$.

Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$

Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_[0,1] g=1/3$

Now we can easily find some points which satisfy $x^2=1/3$.

There are lot of examples that I checked and every time I found that resullt holds good.

Can we write a proof?

Thanks for reading and helping out.

real-analysis integration

share|cite|improve this question

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It suffices to prove that if $int_a^bf=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
  $endgroup$
  – dmtri
  Mar 21 at 6:24
  

  
  
  
  

add a comment | 

2

$begingroup$

I noted something today, I don't know whether the result holds in general setting or not.

Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_a^bf=int_a^bg$.

Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$

Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_[0,1] g=1/3$

Now we can easily find some points which satisfy $x^2=1/3$.

There are lot of examples that I checked and every time I found that resullt holds good.

Can we write a proof?

Thanks for reading and helping out.

real-analysis integration

share|cite|improve this question

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write the integral equality as $int_a^b (f(x)-g(x)), dx = 0$, and $f-g$ is continuous. Also, $f(theta)=g(theta)$ if and only if $f(theta) - g(theta)=0$. Therefore, it is enough for you to consider whether if $h$ is a continuous function such that $int_a^b h(x), dx = 0$, then there exists $thetain[a,b]$ such that $h(theta)=0$. Can you show this?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 6:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It suffices to prove that if $int_a^bf=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
  $endgroup$
  – dmtri
  Mar 21 at 6:24
  

  
  
  
  

add a comment | 

$begingroup$

I noted something today, I don't know whether the result holds in general setting or not.

Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
$int_a^bf=int_a^bg$.

Can we always find some point $theta$ in $[a,b]$ such that
$f(theta)=g(theta)$

Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $int f=1/3$. Take $g=1/3$ then $int_[0,1] g=1/3$

Now we can easily find some points which satisfy $x^2=1/3$.

There are lot of examples that I checked and every time I found that resullt holds good.

Can we write a proof?

Thanks for reading and helping out.

real-analysis integration

share|cite|improve this question

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

$endgroup$

share|cite|improve this question

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

share|cite|improve this question

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

asked Mar 21 at 6:17

StammeringMathematicianStammeringMathematician

2,7211324

4 Answers
4

active

oldest

votes

5

$begingroup$

This is just Rolle's theorem disguised.

Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.

share|cite|improve this answer

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Olof.What a nice solution!
  $endgroup$
  – Peter Szilas
  Mar 21 at 6:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Peter, thank you for the compliment
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 7:41
  

  
  
  
  

add a comment | 

2

$begingroup$

An equivalent formulation (via $h=f-g$) is

Suppose $h$ is a continuous function on $[a,b]$ such that
$int_a^bh(x) , dx=0$.

Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?

and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_a^bh(x) , dx ne 0$.

(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)

share|cite|improve this answer

edited Mar 21 at 7:37

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

$endgroup$

add a comment | 

1

$begingroup$

Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:

$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$

share|cite|improve this answer

answered Mar 21 at 6:24

FredFred

48.7k11849

$endgroup$

add a comment | 

1

$begingroup$

As $int_a^bf(x)dx = int_a^bg(x)dx$, we have
$$int_a^b(f(x)-g(x))dx = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.

share|cite|improve this answer

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

$endgroup$

add a comment | 

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5

$begingroup$

This is just Rolle's theorem disguised.

Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.

share|cite|improve this answer

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Olof.What a nice solution!
  $endgroup$
  – Peter Szilas
  Mar 21 at 6:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Peter, thank you for the compliment
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 7:41
  

  
  
  
  

add a comment | 

5

$begingroup$

This is just Rolle's theorem disguised.

Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.

share|cite|improve this answer

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Olof.What a nice solution!
  $endgroup$
  – Peter Szilas
  Mar 21 at 6:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Dear Peter, thank you for the compliment
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 7:41
  

  
  
  
  

add a comment | 

$begingroup$

This is just Rolle's theorem disguised.

Let $h : [a,b] to mathbb R$ be defined as $h(x) = int_a^x (f(t)-g(t))dt$, then $h$ is differentiable on $(a,b)$ with $h'(x) = f(x) - g(x)$ and $h(a) = h(b) = 0$. By Rolle's theorem you can find a $theta in [a,b]$ such that $h'(theta) = 0$, the same as saying $f(theta) = g(theta)$.

share|cite|improve this answer

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

share|cite|improve this answer

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

answered Mar 21 at 6:25

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

2

$begingroup$

An equivalent formulation (via $h=f-g$) is

Suppose $h$ is a continuous function on $[a,b]$ such that
$int_a^bh(x) , dx=0$.

Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?

and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_a^bh(x) , dx ne 0$.

(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)

share|cite|improve this answer

edited Mar 21 at 7:37

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

$endgroup$

add a comment | 

2

$begingroup$

An equivalent formulation (via $h=f-g$) is

Suppose $h$ is a continuous function on $[a,b]$ such that
$int_a^bh(x) , dx=0$.

Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?

and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_a^bh(x) , dx ne 0$.

(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)

share|cite|improve this answer

edited Mar 21 at 7:37

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

$endgroup$

add a comment | 

$begingroup$

An equivalent formulation (via $h=f-g$) is

Suppose $h$ is a continuous function on $[a,b]$ such that
$int_a^bh(x) , dx=0$.

Can we always find some point $theta$ in $[a,b]$ such that
$h(theta)=0$?

and that is true. If $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently
$int_a^bh(x) , dx ne 0$.

(Strictly speaking, we must assume that $a ne b$, because otherwise $int_a^b h(x) , dx = 0$ for arbitrary functions $h$.)

share|cite|improve this answer

edited Mar 21 at 7:37

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

$endgroup$

share|cite|improve this answer

edited Mar 21 at 7:37

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

answered Mar 21 at 6:23

Martin RMartin R

30.7k33560

1

$begingroup$

Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:

$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$

share|cite|improve this answer

answered Mar 21 at 6:24

FredFred

48.7k11849

$endgroup$

add a comment | 

1

$begingroup$

Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:

$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$

share|cite|improve this answer

answered Mar 21 at 6:24

FredFred

48.7k11849

$endgroup$

add a comment | 

$begingroup$

Let $F(x):= int_a^x f(t) dt, G(x):= int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:

$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$ By Rolle, there is $theta in [a,b]$ such that $H'( theta)=0.$

share|cite|improve this answer

answered Mar 21 at 6:24

FredFred

48.7k11849

$endgroup$

share|cite|improve this answer

answered Mar 21 at 6:24

FredFred

48.7k11849

answered Mar 21 at 6:24

FredFred

48.7k11849

answered Mar 21 at 6:24

FredFred

48.7k11849

1

$begingroup$

As $int_a^bf(x)dx = int_a^bg(x)dx$, we have
$$int_a^b(f(x)-g(x))dx = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.

share|cite|improve this answer

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

$endgroup$

add a comment | 

1

$begingroup$

As $int_a^bf(x)dx = int_a^bg(x)dx$, we have
$$int_a^b(f(x)-g(x))dx = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.

share|cite|improve this answer

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

$endgroup$

add a comment | 

$begingroup$

As $int_a^bf(x)dx = int_a^bg(x)dx$, we have
$$int_a^b(f(x)-g(x))dx = 0$$
The Mean Value Theorem for Integrals shows that $f(c)-g(c)=0$ for some $c in (a,b)$, hence $f(c)=g(c)$.

share|cite|improve this answer

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

$endgroup$

share|cite|improve this answer

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

answered Mar 21 at 6:24

HaydenHayden

13.8k12448

answered Mar 21 at 6:24

HaydenHayden

13.8k12448