Answer
$\dfrac{7}{12}$
Work Step by Step
Here, we have $f(x)=x^2-x^3$
Then, $\Sigma_{i=1}^n (\dfrac{1}{n}) (c_i^2-c_i^3)=(\dfrac{1}{n}) \Sigma_{i=1}^n -(-1+\dfrac{i}{n})^3$
This implies that
$(\dfrac{1}{n})\Sigma_{i=1}^n 2-(\dfrac{5}{n^2})\Sigma_{i=1}^n i+(\dfrac{4}{n^3})\Sigma_{i=1}^n i^2-(\dfrac{1}{n^4})\Sigma_{i=1}^n i^3=2-\dfrac{5n+5}{2n}+\dfrac{4n^2+6n+2}{3n^2}-\dfrac{n^2+2n+1}{4n^2}$
Thus, $\Sigma_{i=1}^n (\dfrac{1}{n}) (c_i^2-c_i^3)=\lim\limits_{n \to \infty}2-\dfrac{5n+5}{2n}+\dfrac{4n^2+6n+2}{3n^2}-\dfrac{n^2+2n+1}{4n^2}=\dfrac{7}{12}$
