Answer
$\dfrac{13}{6}$
Work Step by Step
Here, we have $f(x)=3x+2x^2$
Then, $\Sigma_{i=1}^n (\dfrac{1}{n}) (3c_i+2c_i^2)=(\dfrac{1}{n}) \Sigma_{i=1}^n (\dfrac{3i}{n}+\dfrac{2i^2}{n^2})$
or,
$(\dfrac{3}{n^2})\Sigma_{i=1}^n i+(\dfrac{2}{n^3})\Sigma_{i=1}^n i^2=\dfrac{3n^2+3n}{2n^2}+\dfrac{2n^2+3n+1}{3n^2}$
Thus, $\Sigma_{i=1}^n (\dfrac{1}{n}) (3c_i+2c_i^2)=\lim\limits_{n \to \infty}[\dfrac{3n^2+3n}{2n^2}+\dfrac{2n^2+3n+1}{3n^2}]=\dfrac{3}{2}+\dfrac{2}{3}=\dfrac{13}{6}$
