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