Chapter 4 - Integration - 4.2 Exercises - Page 264: 50

$${\text{Area}} = \frac{{32}}{3}$$

$$\eqalign{ & y = 4 - {x^2},{\text{ }}\left[ { - 2,2} \right] \cr & {\text{By symmetry, we can find the area over the interval }}\left[ {0,2} \right] \cr & {\text{and then multiply the result by 2}}{\text{.}} \cr & f\left( x \right) = 4 - {x^2},{\text{ }}\left[ { - 2,2} \right] \cr & f\left( 0 \right) = 4{\text{ and }}f\left( 2 \right) = 0,{\text{ }} \cr & f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\ & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{2 - 0}}{n} = \frac{2}{n} \cr & {c_i} = a + i\Delta x \to 0 + \frac{{2i}}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {4 - {{\left( {\frac{{2i}}{n}} \right)}^2}} \right]} \left( {\frac{2}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {4 - \frac{{4{i^2}}}{{{n^2}}}} \right]} \left( {\frac{2}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{8}{n}\sum\limits_{i = 1}^n {\left( {1 - \frac{{{i^2}}}{{{n^2}}}} \right)} \cr & {\text{Where }}\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{8}{n}\left( {n - \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6{n^2}}}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {8 - \frac{{16{n^2} + 24n + 8}}{{6{n^2}}}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {8 - \frac{8}{3} - \frac{4}{n} - \frac{4}{{3{n^2}}}} \right) \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = 8 - \frac{8}{3} - 0 - 0 \cr & {\text{Area}} = \frac{{16}}{3} \cr & {\text{The net area is 2Area,}} \cr & {\text{2Area}} = {\text{2}}\left( {\frac{{16}}{3}} \right) \cr & {\text{2Area}} = \frac{{32}}{3} \cr} $$