Answer
$$
12
$$
Work Step by Step
Definition 5.4.3
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x$
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\left(-\left(1+\frac{\frac{(-1+k)4}{n}+\frac{4 k}{n}}{2}+6\right)\right)$
.$\frac{4}{n}$
$=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\left(-\frac{-4+8 k}{2 n}+5\right) \cdot \frac{4}{n}$
$x_{k}^{*}$ is the midpoint, $f(x)=-x+6, \Delta x=\frac{4}{n}$
$A=\lim _{n \rightarrow+\infty} \frac{20}{n} \sum_{k=1}^{n} 1-\frac{32}{2 n^{2}} \sum_{k=1}^{n} \frac{16}{n^{2}}+k \sum_{k=1}^{n} 1$
Rewrite sum in closed form
$=\lim _{n \rightarrow+\infty} \frac{20 n}{n}-\frac{16 n}{n^{2}}+\frac{(1+n)32 n}{4 n^{2}}$
\[
A=-\frac{32}{4}+\frac{80}{4}=12
\]
