Integral of $int_0^infty {x^2}{e^{-3x}},dx$ [closed]












-1












$begingroup$


$$int_0^infty {x^2}{ e^{-3x}},dx$$



What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.






improper-integrals







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closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    That does tend to zero: exponentials beat powers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 12 '18 at 5:07










  • $begingroup$
    See math.stackexchange.com/questions/3015570/…
    $endgroup$
    – lab bhattacharjee
    Dec 12 '18 at 5:17
















-1












$begingroup$


$$int_0^infty {x^2}{ e^{-3x}},dx$$



What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.






improper-integrals







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    That does tend to zero: exponentials beat powers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 12 '18 at 5:07










  • $begingroup$
    See math.stackexchange.com/questions/3015570/…
    $endgroup$
    – lab bhattacharjee
    Dec 12 '18 at 5:17














-1












-1








-1





$begingroup$


$$int_0^infty {x^2}{ e^{-3x}},dx$$



What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.






improper-integrals







share|cite|improve this question











$endgroup$




$$int_0^infty {x^2}{ e^{-3x}},dx$$



What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.






improper-integrals




improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 5:10









Key Flex

8,28261233




8,28261233










asked Dec 12 '18 at 5:04









user623028user623028

235




235




closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    That does tend to zero: exponentials beat powers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 12 '18 at 5:07










  • $begingroup$
    See math.stackexchange.com/questions/3015570/…
    $endgroup$
    – lab bhattacharjee
    Dec 12 '18 at 5:17














  • 1




    $begingroup$
    That does tend to zero: exponentials beat powers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 12 '18 at 5:07










  • $begingroup$
    See math.stackexchange.com/questions/3015570/…
    $endgroup$
    – lab bhattacharjee
    Dec 12 '18 at 5:17








1




1




$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07




$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07












$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17




$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17










3 Answers
3






active

oldest

votes


















5












$begingroup$

Given $$int_0^infty x^2e^{-3x} dx$$



First compute without boundaries.



By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
    $endgroup$
    – user623028
    Dec 12 '18 at 5:16










  • $begingroup$
    @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
    $endgroup$
    – Yadati Kiran
    Dec 12 '18 at 5:20










  • $begingroup$
    Oh did you make a typo, because that's not written? It's fine though thanks.
    $endgroup$
    – user623028
    Dec 12 '18 at 5:45



















1












$begingroup$

The answer is $boxed{frac{2}{27}}$



More generally, it can be shown that



$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$



Here, we have $n = 2$ and $a = 3$, so the result follows.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    using the substitution $u = 3x$ the integral becomes:
    $$
    I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
    = frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
    = frac{Gamma(3)}{27} \
    = frac2{27}
    $$






    share|cite|improve this answer









    $endgroup$




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Given $$int_0^infty x^2e^{-3x} dx$$



      First compute without boundaries.



      By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
      By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
      $$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
      Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
      By L hospitals rule
      $lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
        $endgroup$
        – user623028
        Dec 12 '18 at 5:16










      • $begingroup$
        @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
        $endgroup$
        – Yadati Kiran
        Dec 12 '18 at 5:20










      • $begingroup$
        Oh did you make a typo, because that's not written? It's fine though thanks.
        $endgroup$
        – user623028
        Dec 12 '18 at 5:45
















      5












      $begingroup$

      Given $$int_0^infty x^2e^{-3x} dx$$



      First compute without boundaries.



      By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
      By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
      $$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
      Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
      By L hospitals rule
      $lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
        $endgroup$
        – user623028
        Dec 12 '18 at 5:16










      • $begingroup$
        @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
        $endgroup$
        – Yadati Kiran
        Dec 12 '18 at 5:20










      • $begingroup$
        Oh did you make a typo, because that's not written? It's fine though thanks.
        $endgroup$
        – user623028
        Dec 12 '18 at 5:45














      5












      5








      5





      $begingroup$

      Given $$int_0^infty x^2e^{-3x} dx$$



      First compute without boundaries.



      By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
      By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
      $$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
      Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
      By L hospitals rule
      $lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$






      share|cite|improve this answer











      $endgroup$



      Given $$int_0^infty x^2e^{-3x} dx$$



      First compute without boundaries.



      By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
      By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
      $$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
      Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
      By L hospitals rule
      $lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 12 '18 at 13:48

























      answered Dec 12 '18 at 5:10









      Key FlexKey Flex

      8,28261233




      8,28261233












      • $begingroup$
        When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
        $endgroup$
        – user623028
        Dec 12 '18 at 5:16










      • $begingroup$
        @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
        $endgroup$
        – Yadati Kiran
        Dec 12 '18 at 5:20










      • $begingroup$
        Oh did you make a typo, because that's not written? It's fine though thanks.
        $endgroup$
        – user623028
        Dec 12 '18 at 5:45


















      • $begingroup$
        When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
        $endgroup$
        – user623028
        Dec 12 '18 at 5:16










      • $begingroup$
        @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
        $endgroup$
        – Yadati Kiran
        Dec 12 '18 at 5:20










      • $begingroup$
        Oh did you make a typo, because that's not written? It's fine though thanks.
        $endgroup$
        – user623028
        Dec 12 '18 at 5:45
















      $begingroup$
      When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
      $endgroup$
      – user623028
      Dec 12 '18 at 5:16




      $begingroup$
      When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
      $endgroup$
      – user623028
      Dec 12 '18 at 5:16












      $begingroup$
      @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
      $endgroup$
      – Yadati Kiran
      Dec 12 '18 at 5:20




      $begingroup$
      @BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
      $endgroup$
      – Yadati Kiran
      Dec 12 '18 at 5:20












      $begingroup$
      Oh did you make a typo, because that's not written? It's fine though thanks.
      $endgroup$
      – user623028
      Dec 12 '18 at 5:45




      $begingroup$
      Oh did you make a typo, because that's not written? It's fine though thanks.
      $endgroup$
      – user623028
      Dec 12 '18 at 5:45











      1












      $begingroup$

      The answer is $boxed{frac{2}{27}}$



      More generally, it can be shown that



      $$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$



      Here, we have $n = 2$ and $a = 3$, so the result follows.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The answer is $boxed{frac{2}{27}}$



        More generally, it can be shown that



        $$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$



        Here, we have $n = 2$ and $a = 3$, so the result follows.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The answer is $boxed{frac{2}{27}}$



          More generally, it can be shown that



          $$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$



          Here, we have $n = 2$ and $a = 3$, so the result follows.






          share|cite|improve this answer









          $endgroup$



          The answer is $boxed{frac{2}{27}}$



          More generally, it can be shown that



          $$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$



          Here, we have $n = 2$ and $a = 3$, so the result follows.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 5:46









          Ekesh KumarEkesh Kumar

          1,00428




          1,00428























              1












              $begingroup$

              using the substitution $u = 3x$ the integral becomes:
              $$
              I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
              = frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
              = frac{Gamma(3)}{27} \
              = frac2{27}
              $$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                using the substitution $u = 3x$ the integral becomes:
                $$
                I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
                = frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
                = frac{Gamma(3)}{27} \
                = frac2{27}
                $$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  using the substitution $u = 3x$ the integral becomes:
                  $$
                  I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
                  = frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
                  = frac{Gamma(3)}{27} \
                  = frac2{27}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  using the substitution $u = 3x$ the integral becomes:
                  $$
                  I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
                  = frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
                  = frac{Gamma(3)}{27} \
                  = frac2{27}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 5:48









                  David HoldenDavid Holden

                  14.9k21224




                  14.9k21224















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