How do I prove that if $f$ is continuous, then $F_{0}'(x) = f(x)$?












1












$begingroup$


If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



Define a function $F_0(x) = int_a^x f$ for $a le x le b$



If $x$ and $x+h$ both lie in $I$, then



$F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



= $int_x^{x+h} f = f(bar{x})h$



where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






calculus real-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



    Define a function $F_0(x) = int_a^x f$ for $a le x le b$



    If $x$ and $x+h$ both lie in $I$, then



    $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



    = $int_x^{x+h} f = f(bar{x})h$



    where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



    Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






    calculus real-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



      Define a function $F_0(x) = int_a^x f$ for $a le x le b$



      If $x$ and $x+h$ both lie in $I$, then



      $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



      = $int_x^{x+h} f = f(bar{x})h$



      where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



      Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






      calculus real-analysis







      share|cite|improve this question









      $endgroup$




      If $f$ is continuous on the interval $I = [a,b]$, then $f$ has an antiderivative on $I$.



      Define a function $F_0(x) = int_a^x f$ for $a le x le b$



      If $x$ and $x+h$ both lie in $I$, then



      $F_{0}(x+h) -F_{0}(x) = int_a^{x+h} f - int_a^x f$



      = $int_x^{x+h} f = f(bar{x})h$



      where we use the mean value theorem for integrals and $bar{x}$ lies between $x$ and $x+h$. divide by $h$, and let $h$ approach $0$; since $bar{x}$ must then approach $x$, and since $f$ is continuous, we find $F'_{0}(x) = f(x)$, and $F_{0}$ is an antiderivative of $f$. (When $x$ is an endpoint of $I$, the argument shows the appropriate one-sided derivative of $F_{0}$ has the correct value.)



      Looking at the part that says that "since $f$ is continuous, we find $F'_{0}(x) = f(x)$." I wanted to know if there's a way to show more precisely that if $f$ is continuous, then $F_{0}'(x) = f(x)$?






      calculus real-analysis




      calculus real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 '18 at 20:43









      K.MK.M

      693412




      693412






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



          So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              where does the $(f -f(x)) + f(x)h$ come from?
              $endgroup$
              – K.M
              Nov 28 '18 at 21:06










            • $begingroup$
              Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
              $endgroup$
              – Will M.
              Nov 28 '18 at 21:09













            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%2f3017685%2fhow-do-i-prove-that-if-f-is-continuous-then-f-0x-fx%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









            2












            $begingroup$

            Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



            So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



              So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



                So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.






                share|cite|improve this answer









                $endgroup$



                Following your argument, you have that $$frac{F_0(x_0+h)-F_0(x_0)}{h}=f(bar x),$$ but the mean value theorem aserts that $bar xin[x_0,x_0+h]$ (or viceversa if $h<0$). So, in fact, $bar x$ depends on $h$.



                So if $hto0$, then the interval $[x_0,x_0+h]$ goes to the singleton ${x_0}$, and $bar xto x_0$ as $hto 0$. So, if you put all together, you'll have $$F_0'(x_0)=lim_{hto0}frac{F_0(x_0+h)-F_0(x_0)}{h}=lim_{hto0}f(bar x)= f(x_0),$$ where the last equality holds because of the continuity of $f$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 20:53









                Tito EliatronTito Eliatron

                1,513622




                1,513622























                    1












                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09


















                    1












                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09
















                    1












                    1








                    1





                    $begingroup$

                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D






                    share|cite|improve this answer











                    $endgroup$



                    If $f$ is continuous at $x$ then, for any $varepsilon > 0$ there is an $1 geq r > 0$ such that the relation $|h| < r$ implies $|f(x+h)-f(x)|<varepsilon.$ By noticing, $F_0(x+h)-F_0(x)=intlimits_x^{x+h}(f-f(x))+f(x)h$ we reach at once $left| dfrac{F_0(x+h)-F_0(x) - f(x)}{h}right| leq varepsilon |h| leq varepsilon.$ (The same proof applies whatever the normed space $F$ and $f$ were defined.) Q.E.D







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 28 '18 at 21:09

























                    answered Nov 28 '18 at 20:58









                    Will M.Will M.

                    2,630315




                    2,630315












                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09




















                    • $begingroup$
                      where does the $(f -f(x)) + f(x)h$ come from?
                      $endgroup$
                      – K.M
                      Nov 28 '18 at 21:06










                    • $begingroup$
                      Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                      $endgroup$
                      – Will M.
                      Nov 28 '18 at 21:09


















                    $begingroup$
                    where does the $(f -f(x)) + f(x)h$ come from?
                    $endgroup$
                    – K.M
                    Nov 28 '18 at 21:06




                    $begingroup$
                    where does the $(f -f(x)) + f(x)h$ come from?
                    $endgroup$
                    – K.M
                    Nov 28 '18 at 21:06












                    $begingroup$
                    Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                    $endgroup$
                    – Will M.
                    Nov 28 '18 at 21:09






                    $begingroup$
                    Addition of numbers: $f(t) = f(t) - f(x) + f(x).$
                    $endgroup$
                    – Will M.
                    Nov 28 '18 at 21:09




















                    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%2f3017685%2fhow-do-i-prove-that-if-f-is-continuous-then-f-0x-fx%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

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                    Read more

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                    Image
                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                    Read more