Given: $int_{a}^{b}f=int_{a}^{b}g$, Can we always find some point $theta$ in $[a,b]$ such that...

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



The 2019 Stack Overflow Developer Survey Results Are InLebesgue 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 $intlimits_0^1intlimits_0^tf''(x)dxdt=frac{1}{2}f''(c)$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 $int_{-infty}^infty int_{-a}^a frac{|y+x|^p}{1+y^2} f(x) dx dy<infty$Prove $int_{R^{+}} frac{1}{1+x} a(x) dx > 0$ given $int_{R^{+}} frac{1}{x} a(x) dx > 0$ 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
$int_{a}^{b}f=int_{a}^{b}g$.



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









$endgroup$








  • 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 $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
    $endgroup$
    – dmtri
    Mar 21 at 6:24
















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
$int_{a}^{b}f=int_{a}^{b}g$.



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









$endgroup$








  • 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 $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
    $endgroup$
    – dmtri
    Mar 21 at 6:24














2












2








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
$int_{a}^{b}f=int_{a}^{b}g$.



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









$endgroup$




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
$int_{a}^{b}f=int_{a}^{b}g$.



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




real-analysis integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 21 at 6:17









StammeringMathematicianStammeringMathematician

2,7951324




2,7951324








  • 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 $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
    $endgroup$
    – dmtri
    Mar 21 at 6:24














  • 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 $int_{a}^{b}f=0$ then there is $c$ in $[a, b]$ such that $f(c)=0$ .
    $endgroup$
    – dmtri
    Mar 21 at 6:24








1




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






$begingroup$
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




1




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




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










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









$endgroup$









  • 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



















2












$begingroup$

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




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



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
$int_{a}^{b}h(x) , dx ne 0$.



(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











$endgroup$





















    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









    $endgroup$





















      1












      $begingroup$

      As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, 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









      $endgroup$














        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        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









        $endgroup$









        • 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
















        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









        $endgroup$









        • 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














        5












        5








        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









        $endgroup$



        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 21 at 6:25









        астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

        40.3k33678




        40.3k33678








        • 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














        • 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








        2




        2




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




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




        1




        1




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




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











        2












        $begingroup$

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




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



        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
        $int_{a}^{b}h(x) , dx ne 0$.



        (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











        $endgroup$


















          2












          $begingroup$

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




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



          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
          $int_{a}^{b}h(x) , dx ne 0$.



          (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











          $endgroup$
















            2












            2








            2





            $begingroup$

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




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



            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
            $int_{a}^{b}h(x) , dx ne 0$.



            (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











            $endgroup$



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




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



            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
            $int_{a}^{b}h(x) , dx ne 0$.



            (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














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 21 at 7:37

























            answered Mar 21 at 6:23









            Martin RMartin R

            30.8k33561




            30.8k33561























                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









                $endgroup$


















                  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









                  $endgroup$
















                    1












                    1








                    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









                    $endgroup$



                    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












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 21 at 6:24









                    FredFred

                    48.5k11849




                    48.5k11849























                        1












                        $begingroup$

                        As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, 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









                        $endgroup$


















                          1












                          $begingroup$

                          As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, 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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, 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









                            $endgroup$



                            As $int_a^b{f(x)dx} = int_a^b{g(x)dx}$, 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












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                            answered Mar 21 at 6:24









                            HaydenHayden

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