Answer
$$
16
$$
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( \frac{\frac{4 k}{n}+\frac{4(-1+k)}{n}}{2}\cdot2\right) \cdot \frac{4}{n} $
$=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\frac{4}{n} \cdot\left(\frac{-4+8 k}{n}\right) $
$x_{k}^{*}$ is the midpoint, $f(x)=2 x, \Delta x=\frac{4}{n}$
$A=\lim _{n \rightarrow+\infty} \frac{32}{n^{2}} \sum_{k=1}^{n} -\frac{16}{n^{2}} +k \sum_{k=1}^{n} 1$
Rewrite sum in closed form:
$=\lim _{n \rightarrow+\infty}-\frac{16 n}{n^{2}}+ \frac{(1+n)32 n}{2 n^{2}}$
\[
A=\frac{32}{2}=16
\]
