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

Question

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 $\int_0^1x^3dx=\int_0^{0.5}x^3\cdot0.5+\int_{0.5}^1x^3\cdot0.5$ Which then becomes $$\Bigg[\lim_{n\to\infty}\dfrac{0.5-0}{n}\sum_{k=0}^{n-1}\Bigg(0+\dfrac{k}{n}(0.5-0)\Bigg)\cdot0.5\Bigg]+\Bigg[\lim_{n\to\infty}\dfrac{1-0.5}{n}\sum_{k=0}^{n-1}\Bigg(0.5+\dfrac{k}{n}(1-0.5)\Bigg)\cdot0.5\Bigg]$$ which reduces to $$0.5\cdot\Bigg[\Bigg[\lim_{n\to\infty}\dfrac{0.5}{n}\sum_{k=0}^{n-1}\Bigg(\dfrac{k}{n}(0.5)\Bigg)\Bigg]+\Bigg[\lim_{n\to\infty}\dfrac{0.5}{n}\sum_{k=0}^{n-1}\Bigg(0.5+\dfrac{k}{n}(0.5)\Bigg)\Bigg]\Bigg]$$ or even better, it reduces to $$0.5\cdot\Bigg[\lim_{n\to\infty}\dfrac{0.5}{n}\sum_{k=0}^{n-1}\Bigg[\Bigg(\dfrac{k}{n}(0.5)\Bigg)+\Bigg(0.5+\dfrac{k}{n}(0.5)\Bigg)\Bigg]\Bigg]$$ or $$0.5\cdot\Bigg[\lim_{n\to\infty}\dfrac{0.5}{n}\sum_{k=0}^{n-1}\Bigg[\Bigg(\dfrac{k}{n}\Bigg)+0.5\Bigg]\Bigg]$$ $$0.5\cdot\Bigg[\lim_{n\to\infty}\dfrac{0.5}{n^2}\Bigg[\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}0.5\Bigg]\Bigg]$$ which is $$0.5^2\cdot\lim_{n\to\infty}\dfrac{1}{n^2}\Bigg(\dfrac{(n-1)(n)}{2}+\dfrac{n}{2}\Bigg)$$ $$0.5^2\cdot\lim_{n\to\infty}\dfrac{1}{2} = 0.25\cdot0.25 = 0.0625$$

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

Answer 1

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

So, let $P_n=\left\{0,\frac1n,\frac2n,\ldots,1\right\}$. The upper sum with respect to this partition is\begin{align}\sum_{k=1}^n\frac1n\times\left(\frac kn\right)^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 is\begin{align}\sum_{k=1}^n\frac1n\times\left(\frac{k-1}n\right)^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$$\lim_{n\to\infty}\frac14+\frac1{2n}+\frac1{4n^2}=\lim_{n\to\infty}\frac14-\frac1{2n}+\frac1{4n^2}=\frac14,$$then$$\int_0^1x^3\,\mathrm dx=\frac14.$$

Answer 2

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

$$ \int_{0}^{1}x^3\,dx = \lim_{N\to +\infty}\frac{1}{N}\sum_{k=1}^{N}\left(\frac{k}{N}\right)^3 = \lim_{N\to +\infty}\frac{N^2(N+1)^2}{4N^4}=\frac{1}{4}.$$ Since for any $m\in\mathbb{N}$ 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 $\int_{0}^{1}x^m\,dx = \frac{1}{m+1}$.