Definite integral giving negative value as a result?












4












$begingroup$


I want to calculate definite integral



$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$



$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$



so:



$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$



which is a negative value. I believe it should be positive.



What went wrong in the process?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    yesterday






  • 2




    $begingroup$
    Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
    $endgroup$
    – T. Bongers
    yesterday












  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
    $endgroup$
    – weno
    yesterday








  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    yesterday










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    yesterday
















4












$begingroup$


I want to calculate definite integral



$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$



$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$



so:



$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$



which is a negative value. I believe it should be positive.



What went wrong in the process?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    yesterday






  • 2




    $begingroup$
    Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
    $endgroup$
    – T. Bongers
    yesterday












  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
    $endgroup$
    – weno
    yesterday








  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    yesterday










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    yesterday














4












4








4





$begingroup$


I want to calculate definite integral



$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$



$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$



so:



$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$



which is a negative value. I believe it should be positive.



What went wrong in the process?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




I want to calculate definite integral



$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$



$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$



so:



$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$



which is a negative value. I believe it should be positive.



What went wrong in the process?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Eevee Trainer

10k31740




10k31740










asked yesterday









wenoweno

39611




39611








  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    yesterday






  • 2




    $begingroup$
    Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
    $endgroup$
    – T. Bongers
    yesterday












  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
    $endgroup$
    – weno
    yesterday








  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    yesterday










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    yesterday














  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    yesterday






  • 2




    $begingroup$
    Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
    $endgroup$
    – T. Bongers
    yesterday












  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
    $endgroup$
    – weno
    yesterday








  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    yesterday










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    yesterday








2




2




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
yesterday




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
yesterday




2




2




$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
yesterday






$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^{1/x^2}$ is $e^{1/x^2} / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
yesterday














$begingroup$
Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
$endgroup$
– weno
yesterday






$begingroup$
Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
$endgroup$
– weno
yesterday






5




5




$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
yesterday




$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
yesterday












$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
yesterday




$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
yesterday










1 Answer
1






active

oldest

votes


















4












$begingroup$

What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:



$$int_a^b f(x)dx = F(b) - F(a)$$



when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.






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%2f3176540%2fdefinite-integral-giving-negative-value-as-a-result%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:



    $$int_a^b f(x)dx = F(b) - F(a)$$



    when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:



      $$int_a^b f(x)dx = F(b) - F(a)$$



      when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:



        $$int_a^b f(x)dx = F(b) - F(a)$$



        when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.






        share|cite|improve this answer









        $endgroup$



        What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:



        $$int_a^b f(x)dx = F(b) - F(a)$$



        when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Eevee TrainerEevee Trainer

        10k31740




        10k31740






























            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%2f3176540%2fdefinite-integral-giving-negative-value-as-a-result%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

            What other Star Trek series did the main TNG cast show up in?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
            Read more

            Berlina muro

            Image
            La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
            Read more

            Berlina aerponto

            Image
            Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
            Read more