Answer
(b) $\max x_{k}=\frac{3 \pi}{4}$
$(\mathrm{a}) \sum_{k=1}^{4} f\left(x_{k}^{*}\right) \Delta x_{k}^{*}=\frac{(-2+\sqrt{2}) 3\pi}{8}$
Work Step by Step
We find:
$\begin{aligned} \sum_{k=1}^{4} \Delta x_{k}^{*}
f\left(x_{k}^{*}\right) &=f\left(x_{1}^{*}\right) \Delta x_{1}^{*}+f\left(x_{2}^{*}\right) \Delta x_{2}^{*}+f\left(x_{3}^{*}\right) \Delta x_{3}^{*}+f\left(x_{4}^{*}\right) \Delta x_{4}^{*} \\ &= \frac{\pi}{2} \cdot f\left(\frac{\pi}{4}\right)+\frac{3 \pi} \cdot {4}f(\pi) + \frac{\pi}{2} \cdot f\left(\frac{3 \pi}{2}\right) + \frac{\pi}{4}\cdot f\left(\frac{7 \pi}{4}\right) \\ &= \frac{\pi}{2}\cdot \frac{1}{\sqrt{2}} +\frac{3 \pi}{4}\cdot (-1) +0 \cdot \frac{\pi}{2}+ \frac{\pi}{4}\frac{1} \cdot {\sqrt{2}} \\ &=-\frac{3 \pi}{4}+0+\frac{\pi}{4 \sqrt{2}} +\frac{\pi}{2 \sqrt{2}}\\ &=\pi \frac{-3 \sqrt{2}+3}{4 \sqrt{2}} \\ &=\frac{(-2+\sqrt{2}) 3\pi}{8} \end{aligned}$
(b) $\max x_{k}=\left\{ \frac{3 \pi}{4}, \frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{2}\right\}=\frac{3 \pi}{4}$
