Answer
$${\text{Area}} = \frac{{31}}{6}$$
Work Step by Step
$$\eqalign{
& y = 2{x^3} - {x^2},{\text{ }}\left[ {1,2} \right] \cr
& f\left( x \right) = 2{x^3} - {x^2} \cr
& f\left( 1 \right) = 1{\text{ and }}f\left( 2 \right) = 12,{\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 - 1}}{n} = \frac{1}{n}, \cr
& {c_i} = a + i\Delta x \to 1 + \frac{i}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {2{{\left( {1 + \frac{x}{n}} \right)}^3} - {{\left( {1 + \frac{x}{n}} \right)}^2}} \right]} \left( {\frac{1}{n}} \right){\text{ }} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{2{i^3}}}{{{n^3}}} + \frac{{5{i^2}}}{{{n^2}}} + \frac{{4i}}{n} + 1} \right)} \left( {\frac{1}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{2{i^3}}}{{{n^4}}} + \frac{{5{i^2}}}{{{n^3}}} + \frac{{4i}}{{{n^2}}} + \frac{1}{n}} \right)} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{2}{{{n^4}}}\sum\limits_{i = 1}^n {{i^3}} + \frac{5}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} + \frac{4}{{{n^2}}}\sum\limits_{i = 1}^n i + \sum\limits_{i = 1}^n {\frac{1}{n}} } \right) \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{2{n^2}{{\left( {n + 1} \right)}^2}}}{{4{n^4}}} + \frac{{5n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6{n^3}}} + \frac{{4n\left( {n + 1} \right)}}{{2{n^2}}} + 1} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{2{n^2}}} + \frac{1}{n} + \frac{1}{2} + \frac{5}{{6{n^2}}} + \frac{5}{{2n}} + \frac{5}{3} + \frac{2}{n} + 2 + 1} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 0 + 0 + \frac{1}{2} + 0 + 0 + \frac{5}{3} + 0 + 2 + 1 \cr
& {\text{Area}} = \frac{{31}}{6} \cr} $$
