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