Riemann Sums for computing $\int_0^3 x^3 dx$

First compute $\Delta x=\frac{b-a}{n}=\frac{3-0}{n}=\frac{3}{n}$, where $n$ is the number of subintervals. Also 
$$x_i=a+i\Delta x=0+i\times \frac{3}{n}=\frac{3 i}{n}.$$
Since $f(x)=x^3$, using the righ hand rule we have
\[S_R(n)=\sum_{i=1}^{n}f(x_i)\Delta x=\sum_{i=1}^{n}(x_i)^3 \Delta x=\sum_{i=1}^{n}(\frac{3 i}{n})^3 \times \frac{3}{n} = \sum_{i=1}^{n} \frac{3^4}{n^4} i^3\]
\[=\frac{81}{n^4}\sum_{i=1}^{n} i^3=\frac{81}{n^4} (\frac{n(n+1)}{2})^2=\frac{81}{4}\frac{(n+1)^2}{n^2}.\]
Hence
\[\int_0^3 x^3 dx=\lim_{n \rightarrow \infty} \frac{81}{4}\frac{(n+1)^2}{n^2}=\lim_{n \rightarrow \infty} \frac{81}{4}\frac{n^2}{n^2}=\frac{81}{4}.\]