Let $f$ be continuous on $[0,1]$ such that $int_0^x f = int_x^1 f$, what is $f$?












4












$begingroup$



Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,



$int_0^x f = int_x^1 f$



Can you determine $f$?




I've argued the following:



Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.



Evaluating both sides of the equation we get



$F(x)-F(0) = F(1)-F(x)$=



$2F(x) = F(1) + F(0)$



$F(x) = frac{F(1) + F(0)}{2}$



Taking the derivative of both sides we get



$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.



I'm not sure if I'm going about this problem correctly.






calculus analysis







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  • 3




    $begingroup$
    Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
    $endgroup$
    – T. Bongers
    Dec 1 '18 at 1:19
















4












$begingroup$



Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,



$int_0^x f = int_x^1 f$



Can you determine $f$?




I've argued the following:



Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.



Evaluating both sides of the equation we get



$F(x)-F(0) = F(1)-F(x)$=



$2F(x) = F(1) + F(0)$



$F(x) = frac{F(1) + F(0)}{2}$



Taking the derivative of both sides we get



$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.



I'm not sure if I'm going about this problem correctly.






calculus analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
    $endgroup$
    – T. Bongers
    Dec 1 '18 at 1:19














4












4








4





$begingroup$



Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,



$int_0^x f = int_x^1 f$



Can you determine $f$?




I've argued the following:



Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.



Evaluating both sides of the equation we get



$F(x)-F(0) = F(1)-F(x)$=



$2F(x) = F(1) + F(0)$



$F(x) = frac{F(1) + F(0)}{2}$



Taking the derivative of both sides we get



$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.



I'm not sure if I'm going about this problem correctly.






calculus analysis







share|cite|improve this question











$endgroup$





Let $f$ be continuous on $[0,1]$, and suppose that for all $x$, $0<x<1$,



$int_0^x f = int_x^1 f$



Can you determine $f$?




I've argued the following:



Since $f$ is continuous on $[0,1]$, it follows that $f$ has an antiderivative $F$ on $I=[0,1]$.



Evaluating both sides of the equation we get



$F(x)-F(0) = F(1)-F(x)$=



$2F(x) = F(1) + F(0)$



$F(x) = frac{F(1) + F(0)}{2}$



Taking the derivative of both sides we get



$F'(x) = f(x) = frac{F'(1) + F'(0)}{2} = 0$ since $F(1) + F(0)$ is a constant.



I'm not sure if I'm going about this problem correctly.






calculus analysis




calculus analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 1:21







K.M

















asked Dec 1 '18 at 1:17









K.MK.M

686412




686412








  • 3




    $begingroup$
    Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
    $endgroup$
    – T. Bongers
    Dec 1 '18 at 1:19














  • 3




    $begingroup$
    Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
    $endgroup$
    – T. Bongers
    Dec 1 '18 at 1:19








3




3




$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19




$begingroup$
Aside from the "since $F(1) + F(0)$ missing a conclusion, yes this is correct. This is a ridiculously strong assumption: It says that no matter how you split the interval $[0, 1]$ into two subintervals, the mass that $f$ carries on each one is the same. That's too strong for any non-zero function.
$endgroup$
– T. Bongers
Dec 1 '18 at 1:19










2 Answers
2






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8












$begingroup$

Your method is generally correct, but I think this problem is easier than you think. We have:



$$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$



Use the fundamental theorem of calculus to take the derivative of both sides:



$$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$



Thus, $f(x)=0$ for all $xin [0, 1]$.






share|cite|improve this answer









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  • 1




    $begingroup$
    Beat me to it by 2 minutes. +1
    $endgroup$
    – triple_sec
    Dec 1 '18 at 1:26



















7












$begingroup$

The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    8












    $begingroup$

    Your method is generally correct, but I think this problem is easier than you think. We have:



    $$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$



    Use the fundamental theorem of calculus to take the derivative of both sides:



    $$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$



    Thus, $f(x)=0$ for all $xin [0, 1]$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Beat me to it by 2 minutes. +1
      $endgroup$
      – triple_sec
      Dec 1 '18 at 1:26
















    8












    $begingroup$

    Your method is generally correct, but I think this problem is easier than you think. We have:



    $$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$



    Use the fundamental theorem of calculus to take the derivative of both sides:



    $$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$



    Thus, $f(x)=0$ for all $xin [0, 1]$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Beat me to it by 2 minutes. +1
      $endgroup$
      – triple_sec
      Dec 1 '18 at 1:26














    8












    8








    8





    $begingroup$

    Your method is generally correct, but I think this problem is easier than you think. We have:



    $$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$



    Use the fundamental theorem of calculus to take the derivative of both sides:



    $$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$



    Thus, $f(x)=0$ for all $xin [0, 1]$.






    share|cite|improve this answer









    $endgroup$



    Your method is generally correct, but I think this problem is easier than you think. We have:



    $$int_0^x f(t)dt=int_x^1 f(t)dt=-int_1^x f(t)dt$$



    Use the fundamental theorem of calculus to take the derivative of both sides:



    $$f(x)=-f(x)rightarrow 2f(x)=0rightarrow f(x)=0$$



    Thus, $f(x)=0$ for all $xin [0, 1]$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 1 '18 at 1:23









    Noble MushtakNoble Mushtak

    15.2k1735




    15.2k1735








    • 1




      $begingroup$
      Beat me to it by 2 minutes. +1
      $endgroup$
      – triple_sec
      Dec 1 '18 at 1:26














    • 1




      $begingroup$
      Beat me to it by 2 minutes. +1
      $endgroup$
      – triple_sec
      Dec 1 '18 at 1:26








    1




    1




    $begingroup$
    Beat me to it by 2 minutes. +1
    $endgroup$
    – triple_sec
    Dec 1 '18 at 1:26




    $begingroup$
    Beat me to it by 2 minutes. +1
    $endgroup$
    – triple_sec
    Dec 1 '18 at 1:26











    7












    $begingroup$

    The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.






        share|cite|improve this answer









        $endgroup$



        The use of the fundamental theorem of calculus makes the solution simpler. Since the two functions $xmapstoint_0^x f$ and $xmapstoint_x^1 f$ are identical on the interval $(0,1)$, it follows that their derivatives are also equal: $$f(x)=-f(x)quadtext{for every $xin(0,1)$.}$$ Hence, $f(x)=0$ for every $xin(0,1)$ and, by continuity, also for the endpoints $x=0$ and $x=1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 1:26









        triple_sectriple_sec

        15.8k21851




        15.8k21851






























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