Answer
$(\mathrm{c}) \sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=-\frac{\pi}{4}$
$(\mathrm{b}) \sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=0$
$(\mathrm{a}) \sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\frac{\pi}{4}$
Work Step by Step
(a) The left end points:
\[
\Delta x=\frac{-a+b}{n}=\frac{-0+\pi}{4}=\frac{\pi}{4}
\]
\[
\begin{array}{c}
x_{k} *=x_{k-1}=(-1+k)+a \Delta x=(-1+k)+0\frac{\pi}{4}=\frac{\pi(k-1)}{4} \\
f\left(x_{k} *\right)=\cos \frac{(-1+k)\pi}{4}
\end{array}
\]
$\therefore \sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\frac{\pi}{4} \left(\cos \frac{\pi}{4}+\cos \frac{\pi}{2}+\cos \frac{3 \pi}{4}+\cos 0\right) =\frac{\pi}{4}\left(\frac{1}{\sqrt{2}}+0-\frac{1}{\sqrt{2}+1}\right) =\left[\frac{\pi}{4}\right]$
(b) The midpoints:
\[
\begin{array}{c}
x_{k^{*}}=a+\left(-\frac{1}{2}+k\right) \Delta x=0+\frac{\pi}{4} \left(-\frac{1}{2}+k\right) =\left(-\frac{1}{2}+k\right) \frac{\pi}{4} \\
f\left(x_{k^{*}}\right)=\cos \left(-\frac{1}{2}+k\right) \frac{\pi}{4} \\
\therefore \sum_{k=1}^{4} f\left(x_{k^{*}}\right) \Delta x=\left(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}+\cos \frac{5 \pi}{8}+\cos \frac{7 \pi}{8}\right) \frac{\pi}{4}
\end{array}
\]
$\left.\Rightarrow\left(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}-\cos \left(\pi-\frac{5 \pi}{8}\right)-\cos \left(\pi-\frac{7 \pi}{8}\right)\right) \frac{\pi}{4}=\cos \frac{3 \pi}{8}+\cos \frac{\pi}{8}-\cos \frac{1 \pi}{8}-\cos \frac{3 \pi}{8}\right) \frac{\pi}{4}=0$
(c) The right end points:
\[
c_{k} *=x_{k}=a+k \Delta x=k+0 \frac{\pi}{1}=\frac{\pi k}{1}
\]
\[
f\left(x_{k} *\right)=\cos \frac{\pi k}{4}
\]
$\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\left(\cos \frac{\pi}{4}+\cos \frac{\pi}{2}+\cos \frac{3 \pi}{4}+\cos \pi\right) \frac{\pi}{4}= -\frac{\pi}{4}$
