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?










4












$begingroup$


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
    1 hour ago
















4












$begingroup$


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
    1 hour ago














4












4








4





$begingroup$


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 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
    1 hour 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
    1 hour 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
1 hour 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
1 hour 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
    1 hour ago






  • 1




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











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    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
      1 hour ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      57 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
      1 hour ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      57 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 1 hour 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
      1 hour ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      57 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
      1 hour ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      57 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
    1 hour 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
    1 hour ago




    1




    1




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




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    57 mins 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 59 mins ago

























        answered 1 hour ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,836212




        1,836212



























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