Answer
$$\frac{15}{4}$$
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:
$A=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{\frac{3(-1+k)}{n}+1}{2} \cdot \frac{3}{n}$
$\mathrm{f}(\mathrm{x})=\mathrm{x} / 2, \Delta x=\frac{3}{n}$
Rewrite sum in closed form:
$A=\lim _{n \rightarrow+\infty}\left(-9 n+\frac{15 n^{2}}{4 n^{2}}\right)$
$A=\frac{15}{4}$
