Answer
$$
\frac{1}{3}
$$
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} \frac{1}{n} \cdot \left(\frac{\frac{(-1+k)}{n}+\frac{k}{n}}{2}\right)^{2} $
$x_{k}^{*}$ is the midpoint, $f(x)=x^{2}, \Delta x=\frac{1}{n}$
Rewrite sum in closed form
\[
A=\lim _{n \rightarrow+\infty} \frac{-1+4 n^{2}}{12 n^{2}}
\]
Evaluate limit
\[
A=\frac{1}{3}
\]
