Answer
$$
18
$$
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, so:
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\left(-\frac{(-1+k)^{2}9}{n^{2}}+9\right) \cdot \frac{3}{n}$
$f(x)=-x^{2}+9, \Delta x=\frac{3}{n}$
$A=\lim _{n \rightarrow+\infty} \frac{27}{n} \sum_{k=1}^{n} -\frac{27}{n^{3}}+1 \sum_{k=1}^{n-1} k^{2}$
$=\lim _{n \rightarrow+\infty}\left(\frac{27 n}{n}-\frac{(-1+n)(-1+2 n)27 n}{6 n^{3}}\right)$
$=\lim _{n \rightarrow+\infty}\left(\frac{-27 n(-1+n)(-1+2 n)+162 n^{3}}{6 n^{3}}\right)$
Rewrite sum in closed form:
\[
A=-\frac{54}{6}+\frac{162}{6}=18
\]
