Answer
$${\text{Area}} = 10$$
Work Step by Step
$$\eqalign{
& y = 3{x^2} + 1,{\text{ }}\left[ {0,2} \right] \cr
& f\left( x \right) = 3{x^2} + 1 \cr
& f\left( 0 \right) = 1{\text{ and }}f\left( 2 \right) = 13,{\text{ }} \cr
& f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{2 - 0}}{n} = \frac{2}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {3{{\left( {\frac{{2i}}{n}} \right)}^2} + 1} \right]} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {\frac{{24{i^2}}}{{{n^3}}} + \frac{2}{n}} \right]} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{24}}{{{n^3}}}\sum\limits_{i = 1}^n {\left( {{i^2}} \right)} + 2 \cr
& {\text{Where }}\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{24}}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + 2 \cr
& {\text{Area}} = 4\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{{{n^2}}}} \right) + 2 \cr
& {\text{Area}} = 4\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{2{n^2} + 3n + 1}}{{{n^2}}}} \right) + 2 \cr
& {\text{Area}} = 4\mathop {\lim }\limits_{n \to \infty } \left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right) + 2 \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 4\left( {2 + 0 + 0} \right) + 2 \cr
& {\text{Area}} = 10 \cr} $$
