Answer
$${\text{Area}} = 12$$
Work Step by Step
$$\eqalign{
& y = 5 - {x^2},{\text{ }}\left[ { - 2,1} \right] \cr
& f\left( x \right) = 5 - {x^2} \cr
& f\left( { - 2} \right) = 1{\text{ and }}f\left( 1 \right) = 4, \cr
& f\left( x \right){\text{ is continuous and not negative on the interval }}\left[ { - 2,1} \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{{1 - \left( { - 2} \right)}}{n} = \frac{3}{n} \cr
& {c_i} = a + i\Delta x \to - 2 + \frac{{3i}}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {5 - {{\left( { - 2 + \frac{{3i}}{n}} \right)}^2}} \right]} \left( {\frac{3}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {5 - 4 + \frac{{12i}}{n} - \frac{{9{i^2}}}{{{n^2}}}} \right]} \left( {\frac{3}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\sum\limits_{i = 1}^n {\left( {1 + \frac{{12i}}{n} - \frac{{9{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},{\text{ and }}\sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}{2} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\left( {n + \frac{{12}}{n} \cdot \frac{{n\left( {n + 1} \right)}}{2} - \frac{9}{{{n^2}}} \cdot \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\left( {n + 6\left( {n + 1} \right) - \frac{{3n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2n}}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {3 + 18\left( {1 + \frac{1}{n}} \right) - \frac{{9\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2{n^2}}}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {3 + 18\left( {1 + \frac{1}{n}} \right) - 9\left( {1 + \frac{3}{n} + \frac{1}{{2n}}} \right)} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 3 + 18\left( {1 + 0} \right) - 9\left( {1 + 0 + 0} \right) \cr
& {\text{Area}} = 3 + 18 - 9 \cr
& {\text{Area}} = 12 \cr} $$
