How to find $int_0^1x^3$ using sums and partitions?












0












$begingroup$


The problem statement is to. Calculate $int_0^1x^3dx$ by partitioning $[0,1]$ into subintervals of equal length.



This is my attempt:
Let $p=3.$ Let $delta x = 0.5$ so that the partition is $[0,0.5],[.5,1]$ Then $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$ Which then becomes $$Bigg[lim_{ntoinfty}dfrac{0.5-0}{n}sum_{k=0}^{n-1}Bigg(0+dfrac{k}{n}(0.5-0)Bigg)cdot0.5Bigg]+Bigg[lim_{ntoinfty}dfrac{1-0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(1-0.5)Bigg)cdot0.5Bigg]$$
which reduces to
$$0.5cdotBigg[Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(dfrac{k}{n}(0.5)Bigg)Bigg]+Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or even better, it reduces to
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}(0.5)Bigg)+Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}Bigg)+0.5Bigg]Bigg]$$
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n^2}Bigg[sum_{k=0}^{n-1}k+sum_{k=0}^{n-1}0.5Bigg]Bigg]$$
which is
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{n^2}Bigg(dfrac{(n-1)(n)}{2}+dfrac{n}{2}Bigg)$$
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{2} = 0.25cdot0.25 = 0.0625$$



But I know the integral $int_0^1x^3=0.25$... What am I doing wrong here?






limits definite-integrals summation riemann-integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
    $endgroup$
    – Damien
    Dec 5 '18 at 10:30












  • $begingroup$
    @Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:34






  • 2




    $begingroup$
    $dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
    $endgroup$
    – glowstonetrees
    Dec 5 '18 at 10:36










  • $begingroup$
    You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
    $endgroup$
    – Damien
    Dec 5 '18 at 10:41
















0












$begingroup$


The problem statement is to. Calculate $int_0^1x^3dx$ by partitioning $[0,1]$ into subintervals of equal length.



This is my attempt:
Let $p=3.$ Let $delta x = 0.5$ so that the partition is $[0,0.5],[.5,1]$ Then $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$ Which then becomes $$Bigg[lim_{ntoinfty}dfrac{0.5-0}{n}sum_{k=0}^{n-1}Bigg(0+dfrac{k}{n}(0.5-0)Bigg)cdot0.5Bigg]+Bigg[lim_{ntoinfty}dfrac{1-0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(1-0.5)Bigg)cdot0.5Bigg]$$
which reduces to
$$0.5cdotBigg[Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(dfrac{k}{n}(0.5)Bigg)Bigg]+Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or even better, it reduces to
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}(0.5)Bigg)+Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}Bigg)+0.5Bigg]Bigg]$$
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n^2}Bigg[sum_{k=0}^{n-1}k+sum_{k=0}^{n-1}0.5Bigg]Bigg]$$
which is
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{n^2}Bigg(dfrac{(n-1)(n)}{2}+dfrac{n}{2}Bigg)$$
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{2} = 0.25cdot0.25 = 0.0625$$



But I know the integral $int_0^1x^3=0.25$... What am I doing wrong here?






limits definite-integrals summation riemann-integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
    $endgroup$
    – Damien
    Dec 5 '18 at 10:30












  • $begingroup$
    @Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:34






  • 2




    $begingroup$
    $dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
    $endgroup$
    – glowstonetrees
    Dec 5 '18 at 10:36










  • $begingroup$
    You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
    $endgroup$
    – Damien
    Dec 5 '18 at 10:41














0












0








0





$begingroup$


The problem statement is to. Calculate $int_0^1x^3dx$ by partitioning $[0,1]$ into subintervals of equal length.



This is my attempt:
Let $p=3.$ Let $delta x = 0.5$ so that the partition is $[0,0.5],[.5,1]$ Then $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$ Which then becomes $$Bigg[lim_{ntoinfty}dfrac{0.5-0}{n}sum_{k=0}^{n-1}Bigg(0+dfrac{k}{n}(0.5-0)Bigg)cdot0.5Bigg]+Bigg[lim_{ntoinfty}dfrac{1-0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(1-0.5)Bigg)cdot0.5Bigg]$$
which reduces to
$$0.5cdotBigg[Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(dfrac{k}{n}(0.5)Bigg)Bigg]+Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or even better, it reduces to
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}(0.5)Bigg)+Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}Bigg)+0.5Bigg]Bigg]$$
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n^2}Bigg[sum_{k=0}^{n-1}k+sum_{k=0}^{n-1}0.5Bigg]Bigg]$$
which is
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{n^2}Bigg(dfrac{(n-1)(n)}{2}+dfrac{n}{2}Bigg)$$
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{2} = 0.25cdot0.25 = 0.0625$$



But I know the integral $int_0^1x^3=0.25$... What am I doing wrong here?






limits definite-integrals summation riemann-integration







share|cite|improve this question











$endgroup$




The problem statement is to. Calculate $int_0^1x^3dx$ by partitioning $[0,1]$ into subintervals of equal length.



This is my attempt:
Let $p=3.$ Let $delta x = 0.5$ so that the partition is $[0,0.5],[.5,1]$ Then $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$ Which then becomes $$Bigg[lim_{ntoinfty}dfrac{0.5-0}{n}sum_{k=0}^{n-1}Bigg(0+dfrac{k}{n}(0.5-0)Bigg)cdot0.5Bigg]+Bigg[lim_{ntoinfty}dfrac{1-0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(1-0.5)Bigg)cdot0.5Bigg]$$
which reduces to
$$0.5cdotBigg[Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(dfrac{k}{n}(0.5)Bigg)Bigg]+Bigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or even better, it reduces to
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}(0.5)Bigg)+Bigg(0.5+dfrac{k}{n}(0.5)Bigg)Bigg]Bigg]$$
or
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n}sum_{k=0}^{n-1}Bigg[Bigg(dfrac{k}{n}Bigg)+0.5Bigg]Bigg]$$
$$0.5cdotBigg[lim_{ntoinfty}dfrac{0.5}{n^2}Bigg[sum_{k=0}^{n-1}k+sum_{k=0}^{n-1}0.5Bigg]Bigg]$$
which is
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{n^2}Bigg(dfrac{(n-1)(n)}{2}+dfrac{n}{2}Bigg)$$
$$0.5^2cdotlim_{ntoinfty}dfrac{1}{2} = 0.25cdot0.25 = 0.0625$$



But I know the integral $int_0^1x^3=0.25$... What am I doing wrong here?






limits definite-integrals summation riemann-integration




limits definite-integrals summation riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 10:45









José Carlos Santos

157k22126227




157k22126227










asked Dec 5 '18 at 10:26









kaisakaisa

2019




2019








  • 2




    $begingroup$
    For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
    $endgroup$
    – Damien
    Dec 5 '18 at 10:30












  • $begingroup$
    @Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:34






  • 2




    $begingroup$
    $dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
    $endgroup$
    – glowstonetrees
    Dec 5 '18 at 10:36










  • $begingroup$
    You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
    $endgroup$
    – Damien
    Dec 5 '18 at 10:41














  • 2




    $begingroup$
    For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
    $endgroup$
    – Damien
    Dec 5 '18 at 10:30












  • $begingroup$
    @Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:34






  • 2




    $begingroup$
    $dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
    $endgroup$
    – glowstonetrees
    Dec 5 '18 at 10:36










  • $begingroup$
    You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
    $endgroup$
    – Damien
    Dec 5 '18 at 10:41








2




2




$begingroup$
For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
$endgroup$
– Damien
Dec 5 '18 at 10:30






$begingroup$
For example here $int_0^1x^3dx=int_0^{0.5}x^3cdot0.5+int_{0.5}^1x^3cdot0.5$: why did you multiply by 0.5?. Same remark in the following with some divisions by $n$
$endgroup$
– Damien
Dec 5 '18 at 10:30














$begingroup$
@Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
$endgroup$
– kaisa
Dec 5 '18 at 10:34




$begingroup$
@Damien Because the integrals had a $dx$ in them. For the sub integrals, I assumed $dx = 0.5$ as it went from $0$ to $0.5$ for the first one and $0.5$ to $1$ for the second one.
$endgroup$
– kaisa
Dec 5 '18 at 10:34




2




2




$begingroup$
$dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
$endgroup$
– glowstonetrees
Dec 5 '18 at 10:36




$begingroup$
$dx$ is not an actual value. It just represents a differential length. You can think of it as "summing over $x$"
$endgroup$
– glowstonetrees
Dec 5 '18 at 10:36












$begingroup$
You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
$endgroup$
– Damien
Dec 5 '18 at 10:41




$begingroup$
You don't need to first divide the segment by 2. Just directly divide it by $n$ ...
$endgroup$
– Damien
Dec 5 '18 at 10:41










2 Answers
2






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4












$begingroup$

That's not what “using partitions” mean. It means to use the definition of the Riemann integral.



So, let $P_n=left{0,frac1n,frac2n,ldots,1right}$. The upper sum with respect to this partition isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac knright)^3&=frac1{n^4}sum_{k=1}^nk^3\&=frac{n^4+2n^3+n^2}{4n^4}\&=frac14+frac1{2n}+frac1{4n^2}end{align}whereas the lower sum isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac{k-1}nright)^3&=frac1{n^4}sum_{k=0}^{n-1}k^3\&=frac{n^4-2n^3+n^2}{4n^4}\&=frac14-frac1{2n}+frac1{4n^2}.end{align}Since$$lim_{ntoinfty}frac14+frac1{2n}+frac1{4n^2}=lim_{ntoinfty}frac14-frac1{2n}+frac1{4n^2}=frac14,$$then$$int_0^1x^3,mathrm dx=frac14.$$






share|cite|improve this answer









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    0












    $begingroup$

    Since $n^3=6binom{n+1}{3}+binom{n}{1}$ we have $sum_{n=1}^{N}n^3 = 6binom{N+2}{4}+binom{N+1}{2} = frac{N^2(N+1)^2}{4}$. It follows that



    $$ int_{0}^{1}x^3,dx = lim_{Nto +infty}frac{1}{N}sum_{k=1}^{N}left(frac{k}{N}right)^3 = lim_{Nto +infty}frac{N^2(N+1)^2}{4N^4}=frac{1}{4}.$$
    Since for any $minmathbb{N}$ we have that $sum_{n=1}^{N}n^m$ is a polynomial in the $N$ variable whose leading term equals $frac{N^{m+1}}{m+1}$, the very same argument ensures that $int_{0}^{1}x^m,dx = frac{1}{m+1}$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      That's not what “using partitions” mean. It means to use the definition of the Riemann integral.



      So, let $P_n=left{0,frac1n,frac2n,ldots,1right}$. The upper sum with respect to this partition isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac knright)^3&=frac1{n^4}sum_{k=1}^nk^3\&=frac{n^4+2n^3+n^2}{4n^4}\&=frac14+frac1{2n}+frac1{4n^2}end{align}whereas the lower sum isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac{k-1}nright)^3&=frac1{n^4}sum_{k=0}^{n-1}k^3\&=frac{n^4-2n^3+n^2}{4n^4}\&=frac14-frac1{2n}+frac1{4n^2}.end{align}Since$$lim_{ntoinfty}frac14+frac1{2n}+frac1{4n^2}=lim_{ntoinfty}frac14-frac1{2n}+frac1{4n^2}=frac14,$$then$$int_0^1x^3,mathrm dx=frac14.$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        That's not what “using partitions” mean. It means to use the definition of the Riemann integral.



        So, let $P_n=left{0,frac1n,frac2n,ldots,1right}$. The upper sum with respect to this partition isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac knright)^3&=frac1{n^4}sum_{k=1}^nk^3\&=frac{n^4+2n^3+n^2}{4n^4}\&=frac14+frac1{2n}+frac1{4n^2}end{align}whereas the lower sum isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac{k-1}nright)^3&=frac1{n^4}sum_{k=0}^{n-1}k^3\&=frac{n^4-2n^3+n^2}{4n^4}\&=frac14-frac1{2n}+frac1{4n^2}.end{align}Since$$lim_{ntoinfty}frac14+frac1{2n}+frac1{4n^2}=lim_{ntoinfty}frac14-frac1{2n}+frac1{4n^2}=frac14,$$then$$int_0^1x^3,mathrm dx=frac14.$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          That's not what “using partitions” mean. It means to use the definition of the Riemann integral.



          So, let $P_n=left{0,frac1n,frac2n,ldots,1right}$. The upper sum with respect to this partition isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac knright)^3&=frac1{n^4}sum_{k=1}^nk^3\&=frac{n^4+2n^3+n^2}{4n^4}\&=frac14+frac1{2n}+frac1{4n^2}end{align}whereas the lower sum isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac{k-1}nright)^3&=frac1{n^4}sum_{k=0}^{n-1}k^3\&=frac{n^4-2n^3+n^2}{4n^4}\&=frac14-frac1{2n}+frac1{4n^2}.end{align}Since$$lim_{ntoinfty}frac14+frac1{2n}+frac1{4n^2}=lim_{ntoinfty}frac14-frac1{2n}+frac1{4n^2}=frac14,$$then$$int_0^1x^3,mathrm dx=frac14.$$






          share|cite|improve this answer









          $endgroup$



          That's not what “using partitions” mean. It means to use the definition of the Riemann integral.



          So, let $P_n=left{0,frac1n,frac2n,ldots,1right}$. The upper sum with respect to this partition isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac knright)^3&=frac1{n^4}sum_{k=1}^nk^3\&=frac{n^4+2n^3+n^2}{4n^4}\&=frac14+frac1{2n}+frac1{4n^2}end{align}whereas the lower sum isbegin{align}sum_{k=1}^nfrac1ntimesleft(frac{k-1}nright)^3&=frac1{n^4}sum_{k=0}^{n-1}k^3\&=frac{n^4-2n^3+n^2}{4n^4}\&=frac14-frac1{2n}+frac1{4n^2}.end{align}Since$$lim_{ntoinfty}frac14+frac1{2n}+frac1{4n^2}=lim_{ntoinfty}frac14-frac1{2n}+frac1{4n^2}=frac14,$$then$$int_0^1x^3,mathrm dx=frac14.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 10:42









          José Carlos SantosJosé Carlos Santos

          157k22126227




          157k22126227























              0












              $begingroup$

              Since $n^3=6binom{n+1}{3}+binom{n}{1}$ we have $sum_{n=1}^{N}n^3 = 6binom{N+2}{4}+binom{N+1}{2} = frac{N^2(N+1)^2}{4}$. It follows that



              $$ int_{0}^{1}x^3,dx = lim_{Nto +infty}frac{1}{N}sum_{k=1}^{N}left(frac{k}{N}right)^3 = lim_{Nto +infty}frac{N^2(N+1)^2}{4N^4}=frac{1}{4}.$$
              Since for any $minmathbb{N}$ we have that $sum_{n=1}^{N}n^m$ is a polynomial in the $N$ variable whose leading term equals $frac{N^{m+1}}{m+1}$, the very same argument ensures that $int_{0}^{1}x^m,dx = frac{1}{m+1}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $n^3=6binom{n+1}{3}+binom{n}{1}$ we have $sum_{n=1}^{N}n^3 = 6binom{N+2}{4}+binom{N+1}{2} = frac{N^2(N+1)^2}{4}$. It follows that



                $$ int_{0}^{1}x^3,dx = lim_{Nto +infty}frac{1}{N}sum_{k=1}^{N}left(frac{k}{N}right)^3 = lim_{Nto +infty}frac{N^2(N+1)^2}{4N^4}=frac{1}{4}.$$
                Since for any $minmathbb{N}$ we have that $sum_{n=1}^{N}n^m$ is a polynomial in the $N$ variable whose leading term equals $frac{N^{m+1}}{m+1}$, the very same argument ensures that $int_{0}^{1}x^m,dx = frac{1}{m+1}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $n^3=6binom{n+1}{3}+binom{n}{1}$ we have $sum_{n=1}^{N}n^3 = 6binom{N+2}{4}+binom{N+1}{2} = frac{N^2(N+1)^2}{4}$. It follows that



                  $$ int_{0}^{1}x^3,dx = lim_{Nto +infty}frac{1}{N}sum_{k=1}^{N}left(frac{k}{N}right)^3 = lim_{Nto +infty}frac{N^2(N+1)^2}{4N^4}=frac{1}{4}.$$
                  Since for any $minmathbb{N}$ we have that $sum_{n=1}^{N}n^m$ is a polynomial in the $N$ variable whose leading term equals $frac{N^{m+1}}{m+1}$, the very same argument ensures that $int_{0}^{1}x^m,dx = frac{1}{m+1}$.






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                  $endgroup$



                  Since $n^3=6binom{n+1}{3}+binom{n}{1}$ we have $sum_{n=1}^{N}n^3 = 6binom{N+2}{4}+binom{N+1}{2} = frac{N^2(N+1)^2}{4}$. It follows that



                  $$ int_{0}^{1}x^3,dx = lim_{Nto +infty}frac{1}{N}sum_{k=1}^{N}left(frac{k}{N}right)^3 = lim_{Nto +infty}frac{N^2(N+1)^2}{4N^4}=frac{1}{4}.$$
                  Since for any $minmathbb{N}$ we have that $sum_{n=1}^{N}n^m$ is a polynomial in the $N$ variable whose leading term equals $frac{N^{m+1}}{m+1}$, the very same argument ensures that $int_{0}^{1}x^m,dx = frac{1}{m+1}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 22:22









                  Jack D'AurizioJack D'Aurizio

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