Answer
$\dfrac{13}{6}$
Work Step by Step
Consider $f(x)=3x+2x^2$
This implies that
$\Sigma_{i=1}^n (\dfrac{1}{n}) (3k_i+2k_i^2)=(\dfrac{1}{n}) \Sigma_{i=1}^n (\dfrac{3i}{n}+\dfrac{2i^2}{n^2})$
$(\dfrac{3}{n^2})\Sigma_{i=1}^n i+(\dfrac{2}{n^3})\Sigma_{i=1}^n i^2=\dfrac{(3n^2+3n)}{2n^2}+\dfrac{(2n^2+3n+1)}{3n^2}$
Now, $\lim\limits_{n \to \infty}[\dfrac{3+3/n}{2}+\dfrac{2+3/n+1/n^2}{3}]=\dfrac{3}{2}+\dfrac{2}{3}$
or, $\dfrac{9+4}{6}=\dfrac{13}{6}$
