Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 300: 45

$\dfrac{1}{2}$

Here, we have $f(x)=2x^3$ Then, $\Sigma_{i=1}^n (\dfrac{2}{n}) (c_i^3)=(\dfrac{2}{n}) \Sigma_{i=1}^n (\dfrac{i^3}{n^3})$ This implies that $(\dfrac{2}{n^4})\Sigma_{i=1}^n i^3=\dfrac{n^2+2n+1}{2n^2}$ Thus, $\Sigma_{i=1}^n (\dfrac{2}{n}) (c_i^3)=\lim\limits_{n \to \infty}\dfrac{n^2+2n+1}{2n^2}=\dfrac{1}{2}$