Answer
$$\frac{25}{2}$$
Work Step by Step
Definition 5.4 \.3
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x$
$x_{k}^{*}$ is the left endpoint,
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\left(-\frac{(-1+k)5}{n}+5\right) \cdot \frac{5}{n}$
$f(x)=-x+5, \Delta x=\frac{5}{n}$
$A=\lim _{n \rightarrow+\infty} \frac{25}{n} \sum_{k=1}^{n}1 -\frac{25}{n^{2}} \sum_{k=1}^{n} k+\frac{25}{n^{2}} \sum_{k=1}^{n} 1$
Rewrite sum in closed form
$=\lim _{n \rightarrow+\infty}\left(\frac{25 n}{n^{2}+\frac{25 n}{n}-\frac{(1+n)25 n}{2 n^{2}}}\right)$
$=\lim _{n \rightarrow+\infty}\left(\frac{25 n^{2}}{2 n^{2}}\right)$
$A=\frac{25}{2}$
