Chapter 4 - Integration - Review Exercises - Page 313: 32

$${\text{Area}} = \frac{{26}}{3}$$

$$\eqalign{ & y = {x^2} + 3,{\text{ }}\left[ {0,2} \right] \cr & f\left( x \right) = {x^2} + 3 \cr & f\left( 0 \right) = 3{\text{ and }}f\left( 2 \right) = 7,{\text{ }} \cr & f\left( x \right){\text{ is continuous and not negative on the interval }}\left[ {0,2} \right]. \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{2 - 0}}{n} = \frac{2}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {{{\left( {\frac{{2i}}{n}} \right)}^2} + 3} \right]} \left( {\frac{2}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{8{i^2}}}{{{n^3}}}} \right)} + \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( 6 \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{8}{{{n^3}}}\sum\limits_{i = 1}^n {\left( {{i^2}} \right)} + 6 \cr & {\text{*Where }}\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{8}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + 6 \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{{{n^2}}}} \right) + 6 \cr & {\text{Area}} = \frac{4}{3}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{2{n^2} + 3n + 1}}{{{n^2}}}} \right) + 6 \cr & {\text{Area}} = \frac{4}{3}\mathop {\lim }\limits_{n \to \infty } \left( {2 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right) + 6 \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = \frac{4}{3}\left( {2 + 0 + 0} \right) + 6 \cr & {\text{Area}} = \frac{{26}}{3} \cr} $$