Answer
$${\text{Area}} = \frac{{19}}{4}$$
Work Step by Step
$$\eqalign{
& h\left( y \right) = {y^3} + 1,{\text{ }}1 \leqslant y \leqslant 2 \cr
& h\left( 1 \right) = 2{\text{ and }}h\left( 2 \right) = 9,{\text{ }} \cr
& g\left( y \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 {h\left( {{c_i}} \right)} \Delta y,{\text{ }}\Delta y = \frac{{2 - 1}}{n} = \frac{1}{n} \cr
& {c_i} = a + i\Delta y \to 1 + \frac{i}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {{{\left( {1 + \frac{x}{n}} \right)}^3} + 1} \right]} \left( {\frac{1}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{{i^3}}}{{{n^3}}} + \frac{{3{i^2}}}{{{n^2}}} + \frac{{3i}}{n} + 2} \right)} \left( {\frac{1}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{{i^3}}}{{{n^4}}} + \frac{{3{i^2}}}{{{n^3}}} + \frac{{3i}}{{{n^2}}} + \frac{2}{n}} \right)} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{{i^3}}}{{{n^4}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{3{i^2}}}{{{n^3}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{3i}}{{{n^2}}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{2}{n}} \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{4{n^2}}} + \frac{1}{{2n}} + \frac{1}{4}} \right) + \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{2{n^2}}} + \frac{3}{{2n}} + 1} \right) + 2 \cr
& + \mathop {\lim }\limits_{n \to \infty } \left( {\frac{3}{{2n}} + \frac{3}{2}} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = \left( {0 + 0 + \frac{1}{4}} \right) + \left( {0 + 0 + 1} \right) + 2 + \left( {0 + \frac{3}{2}} \right) \cr
& {\text{Area}} = \frac{{19}}{4} \cr} $$
