Answer
$$\frac{25}{2}$$
Work Step by Step
Definition 5.4.3
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x$
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\frac{5}{n} \cdot \left(-\frac{5 k}{n}+5\right) $
$x_{k}^{*}$ is the right endpoint, $f(x)=-x+5, \Delta x=\frac{5}{n}$
$A=\lim _{n \rightarrow+\infty} \frac{25}{n} \sum_{k=1}^{n} -\frac{25}{n^{2}+1} \sum_{k=1}^{n} k$
$=\lim _{n \rightarrow+\infty}\left(\frac{25 n}{n}-\frac{(1+n)25 n}{2 n^{2}}\right)$
$=\lim _{n \rightarrow+\infty}\left(\frac{25 n+25 n^{2}}{2 n^{2}}\right)$
Rewrite sum in closed form
\[
\frac{25}{2}=A
\]