Answer
$${\text{Area}} = 15$$
Work Step by Step
$$\eqalign{
& y = 8 - 2x,{\text{ }}\left[ {0,3} \right] \cr
& f\left( x \right) = 8 - 2x \cr
& f\left( 0 \right) = 8{\text{ and }}f\left( 1 \right) = 2,{\text{ }} \cr
& f\left( x \right){\text{ is continuous and not negative on the interval }}\left[ {0,3} \right]. \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{3 - 0}}{n} = \frac{3}{n} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {8 - 2\left( {\frac{{3i}}{n}} \right)} \right]} \left( {\frac{3}{n}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {\frac{{24}}{n} - \frac{{18i}}{{{n^2}}}} \right]} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( 8 \right)} - \mathop {\lim }\limits_{n \to \infty } \frac{{18}}{{{n^2}}}\sum\limits_{i = 1}^n {\left( i \right)} \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {24n} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{{18}}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) \cr
& {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {24} \right) - 9\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) \cr
& {\text{Evaluate the limit when }}n \to \infty \cr
& {\text{Area}} = 24 - 9 \cr
& {\text{Area}} = 15 \cr} $$
