How to change the limits of integration The 2019 Stack Overflow Developer Survey Results Are...

Evaluating number of iteration with a certain map with While

Which Sci-Fi work first showed weapon of galactic-scale mass destruction?

Any good smartcontract for "business calendar" oracles?

Why isn't airport relocation done gradually?

If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?

Can I write a for loop that iterates over both collections and arrays?

The difference between dialogue marks

What do hard-Brexiteers want with respect to the Irish border?

Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?

How to answer pointed "are you quitting" questioning when I don't want them to suspect

If a poisoned arrow's piercing damage is reduced to 0, do you still get poisoned?

Carnot-Caratheodory metric

How was Skylab's orbit inclination chosen?

Where to refill my bottle in India?

Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints

Where does the "burst of radiance" from Holy Weapon originate?

Is domain driven design an anti-SQL pattern?

How come people say “Would of”?

It's possible to achieve negative score?

Why could you hear an Amstrad CPC working?

Idiomatic way to prevent slicing?

What tool would a Roman-age civilization have to grind silver and other metals into dust?

Should I use my personal or workplace e-mail when registering to external websites for work purpose?

How to deal with fear of taking dependencies



How to change the limits of integration



The 2019 Stack Overflow Developer Survey Results Are InIntegration limits of the double integral after conversion to the polar coordinatesConvergent or Divergent using LimitsHow do I solve a double integral with an absolute value?How are limits of integration changed?Converting limits of integrationHow to set the limits for Jacobian IntegrationLimits of integration in multivariable integrals during change of variablesChange of limits of integrationWhat theorem(s) is(are) used to change between the various improper integrals.how does an integral becoming negative effect limits of integration?












4












$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    57 mins ago


















4












$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    57 mins ago
















4












4








4





$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$




I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits




calculus integration limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









BolboaBolboa

391516




391516








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    57 mins ago
















  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    57 mins ago










1




1




$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago






$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago












2 Answers
2






active

oldest

votes


















6












$begingroup$

$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    55 mins ago






  • 1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    52 mins ago



















0












$begingroup$

Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






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%2f3181696%2fhow-to-change-the-limits-of-integration%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









    6












    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      55 mins ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      52 mins ago
















    6












    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      55 mins ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      52 mins ago














    6












    6








    6





    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$



    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 57 mins ago









    Kavi Rama MurthyKavi Rama Murthy

    73.6k53170




    73.6k53170








    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      55 mins ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      52 mins ago














    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      55 mins ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      52 mins ago








    1




    1




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    55 mins ago




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    55 mins ago




    1




    1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    52 mins ago




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    52 mins ago











    0












    $begingroup$

    Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






        share|cite|improve this answer











        $endgroup$



        Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 54 mins ago

























        answered 56 mins ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,836212




        1,836212






























            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%2f3181696%2fhow-to-change-the-limits-of-integration%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

            El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

            Image
            Feminismo en EspañaAborto en EspañaManifestaciones y protestas en EspañaManifestaciones y protestas en 2014España en 2014Historia de las mujeres en España movimiento socialfeministaderechos sexuales y reproductivosmujeresLey del aborto en España1 de febrero2014MadridAlberto Ruíz Gallardónla ley del aborto vigente del 2010AsturiasTertulia Feminista Les ComadresCongreso de DiputadosAlicia MiyaresAna MatoAlberto Ruiz GallardónEdimburgoRomaParísLuxemburgoBuenos AiresGracia QuerejetaÁngeles González-SindeMabel LozanoIcíar BollaínIsabel CoixetGeorgina CisquellaTeresa FontVirginia Yagüe2010José Luis Rodríguez ZapateroCongreso de Diputados19 de diciembre20095 de julio2010Partido PopularTribunal Constitucional201120 de diciembre2013Partido PopularMariano RajoyGijónTertulia Feminista Les ComadresBegoña PiñeroRenfeAlicia MiyaresTertulia Feminista Les Comadres1 de febrero2014Congreso de DiputadosMariano RajoyAlberto Ruiz GallardónAna MatoAsturiasAndalucíaCanariasCataluñaComunidad ValencianaCasti...
            Read more

            Castillo d'Acher Características Menú de navegación

            Image
            Montañas de los Pirineos de AragónMontañas de la provincia de Huesca msnmSelva de OzaPirineo aragonésEspañabosque Este artículo o sección necesita referencias que aparezcan en una publicación acreditada. Este aviso fue puesto el 11 de agosto de 2011. El pico desde Selva de Oza. El Castillo d' Acher es una cima de 2.384msnm en la zona de la Selva de Oza en la parte occidental del Pirineo aragonés, en España. Debe su nombre a su aspecto de fortaleza que presenta hacia todas sus vertientes. Características En orden ascendente, la cima pasa de un bosque frondoso de tonalidades verde-amarillenta a laderas rojizas. Su cima, que suele presentarse mucha nieve, es un pequeño valle rectangular rodeado en sus cuatro caras por una muralla que le da un aspecto de fortaleza. This page is only for reference, If you need detailed information, please check here
            Read more

            Connecting two nodes from the same mother node horizontallyTikZ: What EXACTLY does the the |- notation for...

            Image
            Pigeonhole Principle Problem Was Hulk present at this event? Attending a conference where my ex-supervisor and his collaborator are present, should I attend? How did Arya manage to disguise herself? What is the word which sounds like "shtrass"? Does higher resolution in an image imply more bits per pixel? What was the state of the German rail system in 1944? Packet sniffer for MacOS Mojave and above Pressure to defend the relevance of one's area of mathematics Why are there synthetic chemicals in our bodies? Where do they come from? Does the Darkness spell dispel the Color Spray ...
            Read more