Answer
$\dfrac{-5}{6}$
Work Step by Step
Apply RIEMANN SUM:
$\Sigma_{k=1}^n f(c_k) \triangle x= \dfrac{1}{n}\Sigma_{k=1}^n (\dfrac{-k}{n})-(\dfrac{-k}{n})^2$
where, Here, we have $\triangle x=\dfrac{b-a}{n}$ and $c_k=a+\dfrac{k(b-a)}{n}$
Now, $\lim\limits_{n \to \infty}(\dfrac{-1}{n^3}) \Sigma_{k=1}^nk^2-(\dfrac{1}{n^2}) \Sigma_{k=1}^n k=\lim\limits_{n \to \infty} \dfrac{-(n)(n+1)(2n+1)}{6n^2}-\dfrac{(n+1)}{2n}$
Thus, $\lim\limits_{n \to \infty}[(-\dfrac{1}{6})(1+\dfrac{1}{n})(2+\dfrac{1}{n})((\dfrac{1}{2})(1+\dfrac{1}{n})]=(\dfrac{-1}{3})-(\dfrac{1}{2})=\dfrac{-5}{6}$
