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