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?

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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?










5












$begingroup$


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



$$int_0^2 piint_0^inftye^-r^2rdrTheta $$



So I do the following step...



$$=2 piint_0^inftye^-r^2rdr$$



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



$$=2 piint_- infty^0frac12e^sds$$



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
    2 days ago











  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    yesterday















5












$begingroup$


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



$$int_0^2 piint_0^inftye^-r^2rdrTheta $$



So I do the following step...



$$=2 piint_0^inftye^-r^2rdr$$



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



$$=2 piint_- infty^0frac12e^sds$$



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
    2 days ago











  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    yesterday













5












5








5





$begingroup$


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



$$int_0^2 piint_0^inftye^-r^2rdrTheta $$



So I do the following step...



$$=2 piint_0^inftye^-r^2rdr$$



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



$$=2 piint_- infty^0frac12e^sds$$



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...



$$int_0^2 piint_0^inftye^-r^2rdrTheta $$



So I do the following step...



$$=2 piint_0^inftye^-r^2rdr$$



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



$$=2 piint_- infty^0frac12e^sds$$



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 2 days ago









BolboaBolboa

408616




408616







  • 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
    2 days ago











  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    yesterday












  • 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
    2 days ago











  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    yesterday







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
2 days 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
2 days ago













$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday




$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday










2 Answers
2






active

oldest

votes


















8












$begingroup$

$s=-r^2$ 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
    2 days ago






  • 1




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


















0












$begingroup$

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






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    $s=-r^2$ 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
      2 days ago






    • 1




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















    8












    $begingroup$

    $s=-r^2$ 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
      2 days ago






    • 1




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













    8












    8








    8





    $begingroup$

    $s=-r^2$ 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 2 days ago









    Kavi Rama MurthyKavi Rama Murthy

    74.1k53270




    74.1k53270







    • 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
      2 days ago






    • 1




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






    • 1




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




    1




    1




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




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











    0












    $begingroup$

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






    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 $-e^-r^2over 2$






      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 $-e^-r^2over 2$






        share|cite|improve this answer











        $endgroup$



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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,953212




        1,953212



























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