Answer
$9$
Work Step by Step
Apply formula:$\int_{A}^{B} f(x)dx=\lim\limits_{n \to \infty} \Sigma_{k=1}^nf(a+k \triangle x)$
$\int_{-1}^{2} 3x^2-2x+1 dx=\lim\limits_{n \to \infty}(\dfrac{3}{n}) \Sigma_{k=1}^n 3(-1+\dfrac{3k}{n})^2-2 (-1+\dfrac{3k}{n})+1$
Thus, we have $\lim\limits_{n \to \infty}(\dfrac{18}{n})\Sigma_{k=1}^n1+(\dfrac{18}{n^3})\Sigma_{k=1}^nk^2-(\dfrac{72}{n^2})\Sigma_{k=1}^n k=(\dfrac{18}{n})(n)+(\dfrac{18}{n^3})(\dfrac{n(n+1)(2n+1)}{6})-(\dfrac{72}{n^2}) (\dfrac{n(n+1)}{2})$
This implies that
$18+(27/2)(1+\dfrac{1}{n})(2+\dfrac{1}{n})-36(1+\dfrac{1}{n})=18+27-36=9$
