Calculating integral using summation notation

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I'm trying to understand $lim_nrightarrowinftyfrac3nsum_k=1^n((frac3kn)^2-(frac3kn))$. I believe we can take $a=0, b=3$ and so this is equivalent to $int_0^3(x^2-x) dx$. Is this correct?






calculus riemann-sum







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    Yes, this is correct.
    – szw1710
    4 hours ago














up vote
3
down vote

favorite












I'm trying to understand $lim_nrightarrowinftyfrac3nsum_k=1^n((frac3kn)^2-(frac3kn))$. I believe we can take $a=0, b=3$ and so this is equivalent to $int_0^3(x^2-x) dx$. Is this correct?






calculus riemann-sum







share|cite|improve this question

















  • 2




    Yes, this is correct.
    – szw1710
    4 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I'm trying to understand $lim_nrightarrowinftyfrac3nsum_k=1^n((frac3kn)^2-(frac3kn))$. I believe we can take $a=0, b=3$ and so this is equivalent to $int_0^3(x^2-x) dx$. Is this correct?






calculus riemann-sum







share|cite|improve this question













I'm trying to understand $lim_nrightarrowinftyfrac3nsum_k=1^n((frac3kn)^2-(frac3kn))$. I believe we can take $a=0, b=3$ and so this is equivalent to $int_0^3(x^2-x) dx$. Is this correct?






calculus riemann-sum




calculus riemann-sum






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share|cite|improve this question











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asked 4 hours ago









confusedmath

1166




1166







  • 2




    Yes, this is correct.
    – szw1710
    4 hours ago












  • 2




    Yes, this is correct.
    – szw1710
    4 hours ago







2




2




Yes, this is correct.
– szw1710
4 hours ago




Yes, this is correct.
– szw1710
4 hours ago










2 Answers
2






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If $f$ is Riemann integrable at $[a,b]$ then
$$lim_nto +inftyfracb-ansum_k=1^nf(a+kfracb-an)=int_a^bf(x)dx$$



in your case
$$a=0, ; b=3,; f(x)=x^2-x$$
and
$$int_0^3f=frac 92$$






share|cite|improve this answer



























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    down vote













    Recall that in general by Riemann sum



    $$lim_nto inftyfracb-ansum_k=0^n fleft(a+kover n(b-a)right)=int_a^b f(x) dx$$



    and in your case by $a=0$, $b=3$ we have



    $$lim_nrightarrowinftyfrac3nsum_k=1^nleft(left(frac3knright)^2-frac3knright)=int_0^3(x^2-x) dx$$






    share|cite|improve this answer




















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      2 Answers
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      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      up vote
      3
      down vote



      accepted










      If $f$ is Riemann integrable at $[a,b]$ then
      $$lim_nto +inftyfracb-ansum_k=1^nf(a+kfracb-an)=int_a^bf(x)dx$$



      in your case
      $$a=0, ; b=3,; f(x)=x^2-x$$
      and
      $$int_0^3f=frac 92$$






      share|cite|improve this answer
























        up vote
        3
        down vote



        accepted










        If $f$ is Riemann integrable at $[a,b]$ then
        $$lim_nto +inftyfracb-ansum_k=1^nf(a+kfracb-an)=int_a^bf(x)dx$$



        in your case
        $$a=0, ; b=3,; f(x)=x^2-x$$
        and
        $$int_0^3f=frac 92$$






        share|cite|improve this answer






















          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          If $f$ is Riemann integrable at $[a,b]$ then
          $$lim_nto +inftyfracb-ansum_k=1^nf(a+kfracb-an)=int_a^bf(x)dx$$



          in your case
          $$a=0, ; b=3,; f(x)=x^2-x$$
          and
          $$int_0^3f=frac 92$$






          share|cite|improve this answer












          If $f$ is Riemann integrable at $[a,b]$ then
          $$lim_nto +inftyfracb-ansum_k=1^nf(a+kfracb-an)=int_a^bf(x)dx$$



          in your case
          $$a=0, ; b=3,; f(x)=x^2-x$$
          and
          $$int_0^3f=frac 92$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          Salahamam_ Fatima

          34.1k21430




          34.1k21430




















              up vote
              3
              down vote













              Recall that in general by Riemann sum



              $$lim_nto inftyfracb-ansum_k=0^n fleft(a+kover n(b-a)right)=int_a^b f(x) dx$$



              and in your case by $a=0$, $b=3$ we have



              $$lim_nrightarrowinftyfrac3nsum_k=1^nleft(left(frac3knright)^2-frac3knright)=int_0^3(x^2-x) dx$$






              share|cite|improve this answer
























                up vote
                3
                down vote













                Recall that in general by Riemann sum



                $$lim_nto inftyfracb-ansum_k=0^n fleft(a+kover n(b-a)right)=int_a^b f(x) dx$$



                and in your case by $a=0$, $b=3$ we have



                $$lim_nrightarrowinftyfrac3nsum_k=1^nleft(left(frac3knright)^2-frac3knright)=int_0^3(x^2-x) dx$$






                share|cite|improve this answer






















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Recall that in general by Riemann sum



                  $$lim_nto inftyfracb-ansum_k=0^n fleft(a+kover n(b-a)right)=int_a^b f(x) dx$$



                  and in your case by $a=0$, $b=3$ we have



                  $$lim_nrightarrowinftyfrac3nsum_k=1^nleft(left(frac3knright)^2-frac3knright)=int_0^3(x^2-x) dx$$






                  share|cite|improve this answer












                  Recall that in general by Riemann sum



                  $$lim_nto inftyfracb-ansum_k=0^n fleft(a+kover n(b-a)right)=int_a^b f(x) dx$$



                  and in your case by $a=0$, $b=3$ we have



                  $$lim_nrightarrowinftyfrac3nsum_k=1^nleft(left(frac3knright)^2-frac3knright)=int_0^3(x^2-x) dx$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  gimusi

                  73.9k73889




                  73.9k73889



























                       

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