Answer
$${\text{Area}} = 8$$
Work Step by Step
$$\eqalign{
& f\left( y \right) = 4y,{\text{ }}0 \leqslant y \leqslant 2 \cr
& f\left( 0 \right) = 0{\text{ and }}f\left( 2 \right) = 8,{\text{ }} \cr
& f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta y,{\text{ }}\Delta x = \frac{{2 - 0}}{n} = \frac{2}{n} \cr
& {c_i} = a + i\Delta x \to \frac{{2i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {4\left( {\frac{{2i}}{n}} \right)} \left( {\frac{2}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{16}}{{{n^2}}}\sum\limits_{i = 1}^n i \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{16}}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } 8\left( {\frac{{{n^2} + n}}{{{n^2}}}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } 8\left( {1 + \frac{1}{n}} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 8\left( {1 + 0} \right) \cr
& {\text{Area}} = 8 \cr} $$
