Answer
$$\frac{b^{4}}{4}$$
Work Step by Step
Since
\begin{aligned}
R_{N} &=\Delta x \sum_{k=1}^{N} f\left(x_{k}\right)\\
&=\frac{b}{N} \sum_{k=1}^{N}\left(k^{3} \cdot \frac{b^{3}}{N^{3}}\right) \\
&=\frac{b^{4}}{N^{4}} \sum_{k=1}^{N} k^{3}\\
&=\frac{b^{4}}{N^{4}}\left(\frac{N^{4}}{4}+\frac{N^{3}}{2}+\frac{N^{2}}{4}\right) \\
&=\frac{b^{4}}{4}+\frac{b^{4}}{2 N}+\frac{b^{4}}{4 N^{2}}
\end{aligned}
Then
\begin{aligned}
\int_{0}^{b} x^{3} d x &=\lim _{N \rightarrow \infty} R_{N}\\
&=\lim _{N \rightarrow \infty}\left(\frac{b^{4}}{4}+\frac{b^{4}}{2 N}+\frac{b^{4}}{4 N^{2}}\right) \\
&=\frac{b^{4}}{4}
\end{aligned}
