Answer
$${\text{Area}} = \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& y = {x^2} - {x^3},{\text{ }}\left[ { - 1,1} \right] \cr
& f\left( x \right) = {x^2} - {x^3} \cr
& f\left( { - 1} \right) = 2{\text{ and }}f\left( 1 \right) = 0,{\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{{1 - \left( { - 1} \right)}}{n} = \frac{2}{n}, \cr
& {c_i} = a + i\Delta x \to - 1 + \frac{{2i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {{{\left( { - 1 + \frac{{2x}}{n}} \right)}^2} - {{\left( { - 1 + \frac{{2x}}{n}} \right)}^3}} \right]} \left( {\frac{2}{n}} \right){\text{ }}\left( {\bf{1}} \right) \cr
& {\text{*Simplifying }}{\left( { - 1 + \frac{{2i}}{n}} \right)^2} - {\left( { - 1 + \frac{{2i}}{n}} \right)^3} \cr
& = \left( {1 - \frac{{4i}}{n} + \frac{{4{i^2}}}{{{n^2}}}} \right) - \left( {\frac{{8{i^3}}}{{{n^3}}} - \frac{{12{i^2}}}{{{n^2}}} + \frac{{6i}}{n} - 1} \right) \cr
& = 1 - \frac{{4i}}{n} + \frac{{4{i^2}}}{{{n^2}}} - \frac{{8{i^3}}}{{{n^3}}} + \frac{{12{i^2}}}{{{n^2}}} - \frac{{6i}}{n} + 1 \cr
& = 2 - \frac{{10i}}{n} + \frac{{16{i^2}}}{{{n^2}}} - \frac{{8{i^3}}}{{{n^3}}} \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {2 - \frac{{10i}}{n} + \frac{{16{i^2}}}{{{n^2}}} - \frac{{8{i^3}}}{{{n^3}}}} \right]} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{4}{n} - \frac{{20i}}{{{n^2}}} + \frac{{32{i^2}}}{{{n^3}}} - \frac{{16{i^3}}}{{{n^4}}}} \right)} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{i = 1}^n {\frac{4}{n}} - \sum\limits_{i = 1}^n {\frac{{20i}}{{{n^2}}}} + \sum\limits_{i = 1}^n {\frac{{32{i^2}}}{{{n^3}}}} - \sum\limits_{i = 1}^n {\frac{{16{i^3}}}{{{n^4}}}} } \right) \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {4 - \frac{{20n\left( {n + 1} \right)}}{{2{n^2}}} + \frac{{32n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6{n^3}}}} \right) \cr
& - \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{16{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left[ {4 - 10\left( {1 + \frac{1}{n}} \right) + \frac{{16}}{3}\left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right)} \right] \cr
& - \mathop {\lim }\limits_{n \to \infty } \left[ {4\left( {1 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} \right] \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 4 - 10\left( {1 + 0} \right) + \frac{{16}}{3}\left( {2 + 0 + 0} \right) - 4\left( {1 + 0 + 0} \right) \cr
& {\text{Area}} = 4 - 10 + \frac{{32}}{3} - 4 \cr
& {\text{Area}} = \frac{2}{3} \cr} $$
