Split up the interval [0, 2] into n equally spaced subintervals:
[tex]\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right][/tex]
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
[tex]r_i=\dfrac{2i}n[/tex]
where [tex]1\le i\le n[/tex]. Each interval has length [tex]\Delta x_i=\frac{2-0}n=\frac2n[/tex].
At these sampling points, the function takes on values of
[tex]f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}[/tex]
We approximate the integral with the Riemann sum:
[tex]\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3[/tex]
Recall that
[tex]\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4[/tex]
so that the sum reduces to
[tex]\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}[/tex]
Take the limit as n approaches infinity, and the Riemann sum converges to the value of the integral:
[tex]\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}[/tex]
Just to check:
[tex]\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28[/tex]
