Answer
$${\text{Area}} = 34$$
Work Step by Step
$$\eqalign{
& y = 27 - {x^3},{\text{ }}\left[ {1,3} \right] \cr
& f\left( x \right) = 27 - {x^3} \cr
& f\left( 1 \right) = 26{\text{ and }}f\left( 3 \right) = 0,{\text{ }} \cr
& f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{3 - 1}}{n} = \frac{2}{n} \cr
& {c_i} = a + i\Delta x \to 1 + \frac{{2i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {27 - {{\left( {1 + \frac{{2i}}{n}} \right)}^3}} \right]} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {27 - \left( {1 + \frac{{6i}}{n} + \frac{{12{i^2}}}{{{n^2}}} + \frac{{8{i^3}}}{{{n^3}}}} \right)} \right]} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\sum\limits_{i = 1}^n {\left( {26 - \frac{{6i}}{n} - \frac{{12{i^2}}}{{{n^2}}} - \frac{{8{i^3}}}{{{n^3}}}} \right)} \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {23n - 3 - \frac{{2\left( {2{n^2} + 3n + 1} \right)}}{n} - \frac{{2{{\left( {n + 1} \right)}^2}}}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {46 - \frac{6}{n} - \frac{{4\left( {2{n^2} + 3n + 1} \right)}}{{{n^2}}} - \frac{{4{{\left( {n + 1} \right)}^2}}}{{{n^2}}}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {46 - \frac{6}{n} - 4\left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right) - 4\left( {1 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 46 - 0 - 4\left( {2 + 0 + 0} \right) - 4\left( {1 + 0 + 0} \right) \cr
& {\text{Area}} = 46 - 8 - 4 \cr
& {\text{Area}} = 34 \cr} $$
