Answer
(b) $2=\max \Delta x_{k}$
$(\mathrm{a}) \sum_{k=1}^{3} f\left(x_{k}^{*}\right) \Delta x_{k}^{*}=\frac{71}{6}$
Work Step by Step
We find:
$\begin{aligned} \sum_{k=1}^{3} \Delta x_{k}^{*}
f\left(x_{k}^{*}\right) &=\Delta x_{1}^{*} f\left(x_{1}^{*}\right) + \Delta x_{2}^{*} f\left(x_{2}^{*}\right)+ \Delta x_{3}^{*}
f\left(x_{3}^{*}\right)\\ &=1\cdot f\left(\frac{1}{3}\right) +1\cdot f\left(\frac{3}{2}\right) +2\cdot f(3) \\ &=1\cdot \frac{4}{3} +1\cdot \frac{5}{2} +2 \cdot 4 \\ &=8+\frac{5}{2}+\frac{4}{3}\end{aligned}$
$=\frac{71}{6}$
(b) $\max \Delta x_{k}=\max \{1,2,1\}=2$
