Proof that $F$ is continuous in fundamental theorem of calculus












1












$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40


















1












$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40
















1












1








1





$begingroup$


I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus







share|cite|improve this question











$endgroup$




I am studying calculus and have arrived at the first FTC. In our book it states that:



let $f: [a,b]tomathbb{R}$ be continuous on the interval $[a,b]$ then we define the function
$F_f : [a,b]tomathbb{R} : x mapsto F_f(x) := int_a^x f(t),dt$



Then these properties hold:





  • $F_f$ is continuous on $[a,b]$


  • $F_f$ is differentiable on $]a,b[$


  • $F_f$ is a primitive for $f$


The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:



The continuity of $F_f$ follows immediately from the fact that for every $x,y in [a,b]$ we have:



$|F_f(x) - F_f(y)| = int_a^x f(t),dt - int_a^y f(t),dt| = |int_y^x f(t),dt| leq max_{x in [a,b]} |f(x)|cdot |x-y|$



Now here is what I am not sure of:




  • Where it says $max_{x in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $max_{t in [a,b]} |f(t)|$, right?


  • This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?







calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 0:26







Chai

















asked Jan 4 at 22:03









ChaiChai

1276




1276








  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40
















  • 2




    $begingroup$
    The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
    $endgroup$
    – Math1000
    Jan 4 at 22:16






  • 1




    $begingroup$
    The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
    $endgroup$
    – Math1000
    Jan 4 at 22:17










  • $begingroup$
    Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
    $endgroup$
    – Chai
    Jan 5 at 0:40










2




2




$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16




$begingroup$
The $max_{xin[a,b]}|f(x)|$ refers to the maximum value that $f$ attains on $[a,b]$ (which exists since $f$ is continuous and $[a,b]$ is a closed interval) - indeed using a different "dummy variable" such as $t$ would be less confusing there.
$endgroup$
– Math1000
Jan 4 at 22:16




1




1




$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17




$begingroup$
The given inequality actually implies that $F_f$ is Lipschitz continuous, which is a stronger condition than regular continuity. It would be a good exercise to look up this definition and show that it indeed implies continuity.
$endgroup$
– Math1000
Jan 4 at 22:17












$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40






$begingroup$
Thanks for your comments. It would seem that this definition (from Wiki): In particular, a real-valued function f : R → R is called Lipschitz continuous if there exists a positive real constant K such that, for all real $x_1$ and $x_2$, $| f(x_1) − f(x_2) | leq K | x_1 − x_2 |$ is trivially satisfied taking $| F_f(x) − F_f(y) | leq K = max_{x in [a,b]} |f(x)| cdot |x-y|$
$endgroup$
– Chai
Jan 5 at 0:40












2 Answers
2






active

oldest

votes


















4












$begingroup$


  • Right.

  • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56






  • 2




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17



















1












$begingroup$

$F$ is continuous if



$lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



$forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



$F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



$f(x)$ is bounded over the interval, and has a least upper bound.
Let $M$ be the least upper bound of $|f(x)|$



$M = sup {|f(x)|: xin[x,y]}$



$|int_y^x f(x) dx| le |x-y|M$



$delta = frac {epsilon}{M}$



$F(x)$ is continuous.






share|cite|improve this answer









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062167%2fproof-that-f-is-continuous-in-fundamental-theorem-of-calculus%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17
















    4












    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17














    4












    4








    4





    $begingroup$


    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$






    share|cite|improve this answer











    $endgroup$




    • Right.

    • Call it $M$. That is, let $M=sup_{xin[a,b]}bigllvert f(x)bigrrvert$. Then$$(forall x,yin[a,b]):bigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert.$$So, for any $varepsilon>0$, if you take $delta=fracvarepsilon M$, then$$lvert x-yrvert<delta=fracvarepsilon Mimpliesbigllvert F_f(x)-F_f(y)bigrrvertleqslant Mlvert x-yrvert<Mfracvarepsilon M=varepsilon.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 9:13

























    answered Jan 4 at 22:18









    José Carlos SantosJosé Carlos Santos

    173k23133241




    173k23133241












    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17


















    • $begingroup$
      Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
      $endgroup$
      – Chai
      Jan 5 at 0:56






    • 2




      $begingroup$
      Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
      $endgroup$
      – José Carlos Santos
      Jan 5 at 9:17
















    $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56




    $begingroup$
    Thanks for a clear answer. I assume the m in the last part should be capital M. Also, we could use the extreme value theorem and then write $ max |f(x)| $ instead of $sup f(x)$ although for continuity of $F_f$ I don't think it matters.
    $endgroup$
    – Chai
    Jan 5 at 0:56




    2




    2




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17




    $begingroup$
    Yes, it was a typo and I've edited my answer. Using $sup$ instead of $max$, we get a proof that holds for every Riemann-integrable functions, not just for the continuous ones.
    $endgroup$
    – José Carlos Santos
    Jan 5 at 9:17











    1












    $begingroup$

    $F$ is continuous if



    $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



    $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



    $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



    $f(x)$ is bounded over the interval, and has a least upper bound.
    Let $M$ be the least upper bound of $|f(x)|$



    $M = sup {|f(x)|: xin[x,y]}$



    $|int_y^x f(x) dx| le |x-y|M$



    $delta = frac {epsilon}{M}$



    $F(x)$ is continuous.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $F$ is continuous if



      $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



      $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



      $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



      $f(x)$ is bounded over the interval, and has a least upper bound.
      Let $M$ be the least upper bound of $|f(x)|$



      $M = sup {|f(x)|: xin[x,y]}$



      $|int_y^x f(x) dx| le |x-y|M$



      $delta = frac {epsilon}{M}$



      $F(x)$ is continuous.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $F$ is continuous if



        $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



        $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



        $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



        $f(x)$ is bounded over the interval, and has a least upper bound.
        Let $M$ be the least upper bound of $|f(x)|$



        $M = sup {|f(x)|: xin[x,y]}$



        $|int_y^x f(x) dx| le |x-y|M$



        $delta = frac {epsilon}{M}$



        $F(x)$ is continuous.






        share|cite|improve this answer









        $endgroup$



        $F$ is continuous if



        $lim_limits{xto y} F(x) = F(y)$ for all $y$ in $[a,b]$



        $forallepsilon>0,exists delta >0,forall x,y in [a,b]: |x-y|<delta implies |F(x) - F(y)|<epsilon$



        $F(x) - F(y) = int_a^x f(x) dx - int_a^y f(x) dx = int_y^x f(x) dx$



        $f(x)$ is bounded over the interval, and has a least upper bound.
        Let $M$ be the least upper bound of $|f(x)|$



        $M = sup {|f(x)|: xin[x,y]}$



        $|int_y^x f(x) dx| le |x-y|M$



        $delta = frac {epsilon}{M}$



        $F(x)$ is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:37









        Doug MDoug M

        45.3k31954




        45.3k31954






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062167%2fproof-that-f-is-continuous-in-fundamental-theorem-of-calculus%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more