Answer
$$\frac{2}{3}$$
Work Step by Step
Definition 5.4.3
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x$
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n}\left(\frac{\frac{(-1+k)2}{n}+\frac{2 k}{n}}{2}-1\right)^{2} \cdot \frac{2}{n}$
$x_{k}^{*}$ is the midpoint, $f(x)=x^{2}, \Delta x=\frac{2}{n}$
Rewrite sum in closed form
\[
A=\lim _{n \rightarrow+\infty} \frac{-2+2 n^{2}}{3 n^{2}}
\]
Evaluate limit
$A=\frac{2}{3}$
