Evaluating definite integrals using Fundamental Theorem of Calculus












9












$begingroup$


Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




The following, however, seems to give a counterexample.*

Can someone resolve this for me?:



Let $f(x) = frac{1}{4 sin (x)+5}$.



$f$ is continuous on $[0, 2 pi]$:



enter image description here



Consider two antiderivatives of $f$, $F_1$ and $F_2$:



$$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



$$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



However,



$F_1(2pi)-F_1(0)=2pi/3$



$F_2(2pi)-F_2(0)=0$



Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



$F_1 =$



enter image description here



$F_2=$



enter image description here



*I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






calculus definite-integrals







share|cite|improve this question









New contributor




theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    9












    $begingroup$


    Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




    If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




    The following, however, seems to give a counterexample.*

    Can someone resolve this for me?:



    Let $f(x) = frac{1}{4 sin (x)+5}$.



    $f$ is continuous on $[0, 2 pi]$:



    enter image description here



    Consider two antiderivatives of $f$, $F_1$ and $F_2$:



    $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



    $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



    Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



    However,



    $F_1(2pi)-F_1(0)=2pi/3$



    $F_2(2pi)-F_2(0)=0$



    Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



    $F_1 =$



    enter image description here



    $F_2=$



    enter image description here



    *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






    calculus definite-integrals







    share|cite|improve this question









    New contributor




    theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      9












      9








      9





      $begingroup$


      Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




      If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




      The following, however, seems to give a counterexample.*

      Can someone resolve this for me?:



      Let $f(x) = frac{1}{4 sin (x)+5}$.



      $f$ is continuous on $[0, 2 pi]$:



      enter image description here



      Consider two antiderivatives of $f$, $F_1$ and $F_2$:



      $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



      $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



      Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



      However,



      $F_1(2pi)-F_1(0)=2pi/3$



      $F_2(2pi)-F_2(0)=0$



      Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



      $F_1 =$



      enter image description here



      $F_2=$



      enter image description here



      *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






      calculus definite-integrals







      share|cite|improve this question









      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




      If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




      The following, however, seems to give a counterexample.*

      Can someone resolve this for me?:



      Let $f(x) = frac{1}{4 sin (x)+5}$.



      $f$ is continuous on $[0, 2 pi]$:



      enter image description here



      Consider two antiderivatives of $f$, $F_1$ and $F_2$:



      $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



      $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



      Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



      However,



      $F_1(2pi)-F_1(0)=2pi/3$



      $F_2(2pi)-F_2(0)=0$



      Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



      $F_1 =$



      enter image description here



      $F_2=$



      enter image description here



      *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






      calculus definite-integrals




      calculus definite-integrals






      share|cite|improve this question









      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago







      theorist













      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 6 hours ago









      theoristtheorist

      1463




      1463




      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          24












          $begingroup$

          A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You have to subtract the same antiderivative evaluated at starting and end point from each other. The F in the theorem you quote can not be two different antiderivatives. It can be any of them, but only one of them.






            share|cite|improve this answer








            New contributor




            Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$









            • 1




              $begingroup$
              That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
              $endgroup$
              – theorist
              1 hour ago













            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
            });


            }
            });






            theorist is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3079520%2fevaluating-definite-integrals-using-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









            24












            $begingroup$

            A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






            share|cite|improve this answer









            $endgroup$


















              24












              $begingroup$

              A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






              share|cite|improve this answer









              $endgroup$
















                24












                24








                24





                $begingroup$

                A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






                share|cite|improve this answer









                $endgroup$



                A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 5 hours ago









                Martin ArgeramiMartin Argerami

                125k1177178




                125k1177178























                    0












                    $begingroup$

                    You have to subtract the same antiderivative evaluated at starting and end point from each other. The F in the theorem you quote can not be two different antiderivatives. It can be any of them, but only one of them.






                    share|cite|improve this answer








                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$









                    • 1




                      $begingroup$
                      That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                      $endgroup$
                      – theorist
                      1 hour ago


















                    0












                    $begingroup$

                    You have to subtract the same antiderivative evaluated at starting and end point from each other. The F in the theorem you quote can not be two different antiderivatives. It can be any of them, but only one of them.






                    share|cite|improve this answer








                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$









                    • 1




                      $begingroup$
                      That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                      $endgroup$
                      – theorist
                      1 hour ago
















                    0












                    0








                    0





                    $begingroup$

                    You have to subtract the same antiderivative evaluated at starting and end point from each other. The F in the theorem you quote can not be two different antiderivatives. It can be any of them, but only one of them.






                    share|cite|improve this answer








                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    You have to subtract the same antiderivative evaluated at starting and end point from each other. The F in the theorem you quote can not be two different antiderivatives. It can be any of them, but only one of them.







                    share|cite|improve this answer








                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 2 hours ago









                    Georg RempferGeorg Rempfer

                    1




                    1




                    New contributor




                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Georg Rempfer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.








                    • 1




                      $begingroup$
                      That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                      $endgroup$
                      – theorist
                      1 hour ago
















                    • 1




                      $begingroup$
                      That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                      $endgroup$
                      – theorist
                      1 hour ago










                    1




                    1




                    $begingroup$
                    That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                    $endgroup$
                    – theorist
                    1 hour ago






                    $begingroup$
                    That's not what's going on here. If you take another look at my question, you'll see I'm not subtracting either F2(2 Pi) - F1(0) or F1(2 Pi) - F2(0). I believe Martin's answer is the correct one: The problem is that F2 is not an antiderivative of f.
                    $endgroup$
                    – theorist
                    1 hour ago












                    theorist is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    theorist is a new contributor. Be nice, and check out our Code of Conduct.













                    theorist is a new contributor. Be nice, and check out our Code of Conduct.












                    theorist is a new contributor. Be nice, and check out our Code of Conduct.
















                    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%2f3079520%2fevaluating-definite-integrals-using-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

                    الفوسفات في المغرب

                    Image
                    قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
                    Read more

                    Four equal circles intersect: What is the area of the small shaded portion and its height

                    Image
                    3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
                    Read more

                    جامعة ليفربول

                    Image
                    جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
                    Read more