Answer
$\dfrac{7}{12}$
Work Step by Step
Consider $f(x)=x^2-x^3$
Then, $\Sigma_{i=1}^n (\dfrac{1}{n}) (k_i^2-k_i^3)=(\dfrac{1}{n}) \Sigma_{i=1}^n [-(-1+\dfrac{i}{n})^3]$
This implies that
$(\dfrac{1}{n})\Sigma_{i=1}^n 2-(\dfrac{5}{n^2})\Sigma_{i=1}^n i+(\dfrac{4}{n^3})\Sigma_{i=1}^n i^2-(\dfrac{1}{n^4})\Sigma_{i=1}^n i^3=2-[\dfrac{5n+5}{2n}]+[\dfrac{4n^2+6n+2}{3n^2}]-[\dfrac{n^2+2n+1}{4n^2}]$
Now,
$\lim\limits_{n \to \infty}[2-\dfrac{5+5/n}{2}+\dfrac{4+6/n+2/n^2}{3n^2}-\dfrac{1+2/n+1/n^2}{4}]=\dfrac{7}{12}$
