## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{-5}{6}$
Here, we have $\triangle x=\dfrac{b-a}{n}$ and $c_k=a+\dfrac{k(b-a)}{n}$ Consider the RIEMANN SUM: $\Sigma_{k=1}^n f(c_k) \triangle x= \dfrac{1}{n}\Sigma_{k=1}^n (\dfrac{-k}{n})-(\dfrac{-k}{n})^2$ 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, $\lim\limits_{n \to \infty}(-\dfrac{1}{6})(1+1/n)(2+1/n)((1/2)(1+1/n)=\dfrac{-1}{3}-\dfrac{1}{2}=\dfrac{-5}{6}$