Integral of infinitely high curve












1












$begingroup$


I need to find the integral of the curve $frac{1}{x^3}$ bounded by $y = 0, x = 3$. I am unsure of how to take the integral when I have no real lower limit. Is there a way for me to set the lower limit to the limit of $f(x)$ as it approaches 0?






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
    $endgroup$
    – Neeyanth Kopparapu
    Dec 21 '18 at 23:49






  • 1




    $begingroup$
    There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
    $endgroup$
    – jmerry
    Dec 22 '18 at 0:00










  • $begingroup$
    @jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
    $endgroup$
    – KKZiomek
    Dec 22 '18 at 0:05












  • $begingroup$
    @KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
    $endgroup$
    – Henry
    Dec 22 '18 at 0:15










  • $begingroup$
    @Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
    $endgroup$
    – user3482749
    Dec 22 '18 at 0:31
















1












$begingroup$


I need to find the integral of the curve $frac{1}{x^3}$ bounded by $y = 0, x = 3$. I am unsure of how to take the integral when I have no real lower limit. Is there a way for me to set the lower limit to the limit of $f(x)$ as it approaches 0?






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
    $endgroup$
    – Neeyanth Kopparapu
    Dec 21 '18 at 23:49






  • 1




    $begingroup$
    There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
    $endgroup$
    – jmerry
    Dec 22 '18 at 0:00










  • $begingroup$
    @jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
    $endgroup$
    – KKZiomek
    Dec 22 '18 at 0:05












  • $begingroup$
    @KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
    $endgroup$
    – Henry
    Dec 22 '18 at 0:15










  • $begingroup$
    @Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
    $endgroup$
    – user3482749
    Dec 22 '18 at 0:31














1












1








1





$begingroup$


I need to find the integral of the curve $frac{1}{x^3}$ bounded by $y = 0, x = 3$. I am unsure of how to take the integral when I have no real lower limit. Is there a way for me to set the lower limit to the limit of $f(x)$ as it approaches 0?






calculus integration







share|cite|improve this question











$endgroup$




I need to find the integral of the curve $frac{1}{x^3}$ bounded by $y = 0, x = 3$. I am unsure of how to take the integral when I have no real lower limit. Is there a way for me to set the lower limit to the limit of $f(x)$ as it approaches 0?






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 23:58









Foobaz John

22.3k41452




22.3k41452










asked Dec 21 '18 at 23:44









ElijahElijah

626




626












  • $begingroup$
    Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
    $endgroup$
    – Neeyanth Kopparapu
    Dec 21 '18 at 23:49






  • 1




    $begingroup$
    There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
    $endgroup$
    – jmerry
    Dec 22 '18 at 0:00










  • $begingroup$
    @jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
    $endgroup$
    – KKZiomek
    Dec 22 '18 at 0:05












  • $begingroup$
    @KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
    $endgroup$
    – Henry
    Dec 22 '18 at 0:15










  • $begingroup$
    @Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
    $endgroup$
    – user3482749
    Dec 22 '18 at 0:31


















  • $begingroup$
    Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
    $endgroup$
    – Neeyanth Kopparapu
    Dec 21 '18 at 23:49






  • 1




    $begingroup$
    There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
    $endgroup$
    – jmerry
    Dec 22 '18 at 0:00










  • $begingroup$
    @jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
    $endgroup$
    – KKZiomek
    Dec 22 '18 at 0:05












  • $begingroup$
    @KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
    $endgroup$
    – Henry
    Dec 22 '18 at 0:15










  • $begingroup$
    @Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
    $endgroup$
    – user3482749
    Dec 22 '18 at 0:31
















$begingroup$
Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
$endgroup$
– Neeyanth Kopparapu
Dec 21 '18 at 23:49




$begingroup$
Try setting the lower bound of the integral as k, and taking the limit as k approaches 0 of the integral. It turns out this limit does not converge, so the integral doesn't either.
$endgroup$
– Neeyanth Kopparapu
Dec 21 '18 at 23:49




1




1




$begingroup$
There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
$endgroup$
– jmerry
Dec 22 '18 at 0:00




$begingroup$
There are two possible regions here - either $x<3$ or $x>3$. Which one is intended? You haven't provided enough information to determine.
$endgroup$
– jmerry
Dec 22 '18 at 0:00












$begingroup$
@jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
$endgroup$
– KKZiomek
Dec 22 '18 at 0:05






$begingroup$
@jmerry he wants $x < 3$ from what I can infer, because otherwise the title wouldn't be named "Integral of infinitely high curve" ;)
$endgroup$
– KKZiomek
Dec 22 '18 at 0:05














$begingroup$
@KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
$endgroup$
– Henry
Dec 22 '18 at 0:15




$begingroup$
@KKZiomek but he does not say bounded by $x=0$ despite saying bounded by $y=0$
$endgroup$
– Henry
Dec 22 '18 at 0:15












$begingroup$
@Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
$endgroup$
– user3482749
Dec 22 '18 at 0:31




$begingroup$
@Henry I'm reasonably sure it's a typo, and he means $x = 0$ instead of $y = 0$.
$endgroup$
– user3482749
Dec 22 '18 at 0:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

Note that (by definition)
$$
int_{0}^3frac{1}{x^3}, dx=lim_{adownarrow 0}int_{a}^3frac{1}{x^3}=lim_{adownarrow 0}left[-frac{1}{2}x^{-2}right]_{a}^3
$$
which you can compute.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If that is the integral the OP meant then it doesn't converge
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 0:48



















1












$begingroup$

Assuming you mean the tiny sliver below the black curve and above and to the right of the red lines in this picture



enter image description here



then you want $$int_{x=3}^infty left(frac1{x^3}-0right) , dx = left[-frac{1}{2x^2}right]^infty_{3} =frac{1}{18}$$






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%2f3049007%2fintegral-of-infinitely-high-curve%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









    1












    $begingroup$

    Note that (by definition)
    $$
    int_{0}^3frac{1}{x^3}, dx=lim_{adownarrow 0}int_{a}^3frac{1}{x^3}=lim_{adownarrow 0}left[-frac{1}{2}x^{-2}right]_{a}^3
    $$
    which you can compute.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If that is the integral the OP meant then it doesn't converge
      $endgroup$
      – DonAntonio
      Dec 22 '18 at 0:48
















    1












    $begingroup$

    Note that (by definition)
    $$
    int_{0}^3frac{1}{x^3}, dx=lim_{adownarrow 0}int_{a}^3frac{1}{x^3}=lim_{adownarrow 0}left[-frac{1}{2}x^{-2}right]_{a}^3
    $$
    which you can compute.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If that is the integral the OP meant then it doesn't converge
      $endgroup$
      – DonAntonio
      Dec 22 '18 at 0:48














    1












    1








    1





    $begingroup$

    Note that (by definition)
    $$
    int_{0}^3frac{1}{x^3}, dx=lim_{adownarrow 0}int_{a}^3frac{1}{x^3}=lim_{adownarrow 0}left[-frac{1}{2}x^{-2}right]_{a}^3
    $$
    which you can compute.






    share|cite|improve this answer









    $endgroup$



    Note that (by definition)
    $$
    int_{0}^3frac{1}{x^3}, dx=lim_{adownarrow 0}int_{a}^3frac{1}{x^3}=lim_{adownarrow 0}left[-frac{1}{2}x^{-2}right]_{a}^3
    $$
    which you can compute.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 23:58









    Foobaz JohnFoobaz John

    22.3k41452




    22.3k41452












    • $begingroup$
      If that is the integral the OP meant then it doesn't converge
      $endgroup$
      – DonAntonio
      Dec 22 '18 at 0:48


















    • $begingroup$
      If that is the integral the OP meant then it doesn't converge
      $endgroup$
      – DonAntonio
      Dec 22 '18 at 0:48
















    $begingroup$
    If that is the integral the OP meant then it doesn't converge
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 0:48




    $begingroup$
    If that is the integral the OP meant then it doesn't converge
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 0:48











    1












    $begingroup$

    Assuming you mean the tiny sliver below the black curve and above and to the right of the red lines in this picture



    enter image description here



    then you want $$int_{x=3}^infty left(frac1{x^3}-0right) , dx = left[-frac{1}{2x^2}right]^infty_{3} =frac{1}{18}$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Assuming you mean the tiny sliver below the black curve and above and to the right of the red lines in this picture



      enter image description here



      then you want $$int_{x=3}^infty left(frac1{x^3}-0right) , dx = left[-frac{1}{2x^2}right]^infty_{3} =frac{1}{18}$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Assuming you mean the tiny sliver below the black curve and above and to the right of the red lines in this picture



        enter image description here



        then you want $$int_{x=3}^infty left(frac1{x^3}-0right) , dx = left[-frac{1}{2x^2}right]^infty_{3} =frac{1}{18}$$






        share|cite|improve this answer











        $endgroup$



        Assuming you mean the tiny sliver below the black curve and above and to the right of the red lines in this picture



        enter image description here



        then you want $$int_{x=3}^infty left(frac1{x^3}-0right) , dx = left[-frac{1}{2x^2}right]^infty_{3} =frac{1}{18}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 22 '18 at 10:42

























        answered Dec 22 '18 at 0:13









        HenryHenry

        100k481168




        100k481168






























            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%2f3049007%2fintegral-of-infinitely-high-curve%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