Question regarding the formulation of a proof using the mean value theorem for integrals

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Let $f:[a,b]rightarrowmathbbR$ be continuous, and suppose that $int_c^df=0$ for every interval $[c,d]subset[a,b]$. Show that $f(x)=0$ for every $xin[a,b]$.




I just need assistance writing a formal proof of this. I believe I have the general idea:



By the mean value theorem, there exists an $zin[c,d]$ such that $f(z)(d-c)=0$. Because $int_c^df=0$ for $textitevery$ subinterval $[c,d]subset[a,b]$, it follows that $f(x)=0$ for every $xin[a,b]$.



How do I make this rigorous?




calculus real-analysis proof-verification definite-integrals




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    Let $f:[a,b]rightarrowmathbbR$ be continuous, and suppose that $int_c^df=0$ for every interval $[c,d]subset[a,b]$. Show that $f(x)=0$ for every $xin[a,b]$.




    I just need assistance writing a formal proof of this. I believe I have the general idea:



    By the mean value theorem, there exists an $zin[c,d]$ such that $f(z)(d-c)=0$. Because $int_c^df=0$ for $textitevery$ subinterval $[c,d]subset[a,b]$, it follows that $f(x)=0$ for every $xin[a,b]$.



    How do I make this rigorous?




    calculus real-analysis proof-verification definite-integrals




    share|cite|improve this question
























      up vote
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      up vote
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      Let $f:[a,b]rightarrowmathbbR$ be continuous, and suppose that $int_c^df=0$ for every interval $[c,d]subset[a,b]$. Show that $f(x)=0$ for every $xin[a,b]$.




      I just need assistance writing a formal proof of this. I believe I have the general idea:



      By the mean value theorem, there exists an $zin[c,d]$ such that $f(z)(d-c)=0$. Because $int_c^df=0$ for $textitevery$ subinterval $[c,d]subset[a,b]$, it follows that $f(x)=0$ for every $xin[a,b]$.



      How do I make this rigorous?




      calculus real-analysis proof-verification definite-integrals




      share|cite|improve this question















      Let $f:[a,b]rightarrowmathbbR$ be continuous, and suppose that $int_c^df=0$ for every interval $[c,d]subset[a,b]$. Show that $f(x)=0$ for every $xin[a,b]$.




      I just need assistance writing a formal proof of this. I believe I have the general idea:



      By the mean value theorem, there exists an $zin[c,d]$ such that $f(z)(d-c)=0$. Because $int_c^df=0$ for $textitevery$ subinterval $[c,d]subset[a,b]$, it follows that $f(x)=0$ for every $xin[a,b]$.



      How do I make this rigorous?





      calculus real-analysis proof-verification definite-integrals





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 28 at 14:53









      José Carlos Santos

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      120k16101182










      asked Aug 28 at 14:49









      Atsina

      711115




      711115




















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          You proved correctly that every interval $[c,d]subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $xin[a,b],$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.






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            You proved correctly that every interval $[c,d]subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $xin[a,b],$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.






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              accepted










              You proved correctly that every interval $[c,d]subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $xin[a,b],$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.






              share|cite|improve this answer






















                up vote
                0
                down vote



                accepted







                up vote
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                down vote



                accepted






                You proved correctly that every interval $[c,d]subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $xin[a,b],$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.






                share|cite|improve this answer












                You proved correctly that every interval $[c,d]subset[a,b]$ (with more than one point) has a zero of $f$. Therefore, $xin[a,b],$ is dense in $[a,b]$. Since $f$ is countinuous, it follows from this that $f$ is the null function.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 28 at 14:53









                José Carlos Santos

                120k16101182




                120k16101182



























                     

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