Answer
$${\text{Area}} = \frac{{125}}{3}$$
Work Step by Step
$$\eqalign{
& f\left( y \right) = {y^2},{\text{ }}0 \leqslant y \leqslant 5 \cr
& f\left( 0 \right) = 0{\text{ and }}f\left( 5 \right) = 25,{\text{ }} \cr
& f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta y,{\text{ }}\Delta x = \frac{{5 - 0}}{n} = \frac{5}{n} \cr
& {c_i} = a + i\Delta x \to \frac{{5i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{{\left( {\frac{{5i}}{n}} \right)}^2}} \left( {\frac{5}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{125}}{{{n^3}}}\sum\limits_{i = 1}^n i \cr
& {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{125}}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) \cr
& {\text{Area}} = \frac{{125}}{6}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{2{n^2} + 3n + 1}}{{{n^2}}}} \right) \cr
& {\text{Area}} = \frac{{125}}{6}\mathop {\lim }\limits_{n \to \infty } \left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = \frac{{125}}{6}\left( {2 + 0 + 0} \right) \cr
& {\text{Area}} = \frac{{125}}{3} \cr} $$
