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\begin{document}

\textbf{Solution to Problem 1: } \newline

\textbf{(i)} $f(x) = x$ on $P = \{0, \frac{3}{4}, \frac{3}{2}, \frac{9}{4}, 3\}$: \begin{align*}
U(f, P) &= \frac{3}{4}\bigg(\frac{3}{4} - 0\bigg) + \frac{6}{4}\bigg(\frac{6}{4} - \frac{3}{4}\bigg) + \frac{9}{4}\bigg(\frac{9}{4} - \frac{6}{4}\bigg)+ \frac{12}{4}\bigg(\frac{12}{4} - \frac{9}{4}\bigg) \\
&= \frac{3}{4}\bigg(\frac{3}{4}\bigg) + \frac{6}{4}\bigg(\frac{3}{4}\bigg) + \frac{9}{4}\bigg(\frac{3}{4}\bigg) + \frac{12}{4}\bigg(\frac{3}{4}\bigg) = \frac{9}{16} + \frac{18}{16} + \frac{27}{16} + \frac{36}{16} \\
&= \frac{45}{8} \\
L(f, P) &= 0\bigg(\frac{3}{4}\bigg) + \frac{3}{4}\bigg(\frac{3}{4}\bigg) + \frac{6}{4}\bigg(\frac{3}{4}\bigg) + \frac{9}{4}\bigg(\frac{3}{4}\bigg) = \frac{9}{16} + \frac{18}{16} + \frac{27}{16} \\
&= \frac{27}{8}
\end{align*}

\textbf{(ii)} $f(x) = \sqrt{x}$ on $P = \{0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\}$: \begin{align*}
U(f, P) &= \sqrt{\frac{1}{4}}\bigg(\frac{1}{4}\bigg) + \sqrt{\frac{1}{2}}\bigg(\frac{1}{4}\bigg) + \sqrt{\frac{3}{4}}\bigg(\frac{1}{4}\bigg) + \sqrt{1}\bigg(\frac{1}{4}\bigg) \\
&= \frac{1}{4}\sqrt{\frac{1}{4}} + \frac{1}{4}\sqrt{\frac{1}{2}} + \frac{1}{4}\sqrt{\frac{3}{4}} + \frac{1}{4} \\
L(f, P) &= \sqrt{0}\bigg(\frac{1}{4}\bigg) + \sqrt{\frac{1}{4}}\bigg(\frac{1}{4}\bigg) + \sqrt{\frac{1}{2}}\bigg(\frac{1}{4}\bigg) + \sqrt{\frac{3}{4}}\bigg(\frac{1}{4}\bigg) \\
&= \frac{1}{4}\sqrt{\frac{1}{4}} + \frac{1}{4}\sqrt{\frac{1}{2}} + \frac{1}{4}\sqrt{\frac{3}{4}}
\end{align*}

\textbf{(iii)} $f(x) = \sin(x)$ on $P = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\}$: \begin{align*}
U(f, P) &= \sin\bigg(\frac{\pi}{2}\bigg)\bigg(\frac{\pi}{2}\bigg) + \sin\bigg(\frac{\pi}{2}\bigg)\bigg(\frac{\pi}{2}\bigg) + \sin(\pi)\bigg(\frac{\pi}{2}\bigg) + \sin(2\pi)\bigg(\frac{\pi}{2}\bigg) \\
&= \frac{\pi}{2} + \frac{\pi}{2} \\
&= \pi \\
L(f, P) &= \sin(0)\bigg(\frac{\pi}{2}\bigg) + \sin(\pi)\bigg(\frac{\pi}{2}\bigg) + \sin\bigg(\frac{3\pi}{2}\bigg)\bigg(\frac{\pi}{2}\bigg) + \sin\bigg(\frac{3\pi}{2}\bigg)\bigg(\frac{\pi}{2}\bigg) \\
&= -\frac{\pi}{2} - \frac{\pi}{2} = \\
&= -\pi
\end{align*}

\textbf{(iv)} $f(x) = 1 - 2x$ on $P = \{1, 2, 3\}$: \begin{align*}
U(f, P) &= (1 - 2(1))(1) + (1 - 2(2))(1) = 1 - 2 + 1 - 4 = -4 \\
L(f, P) &= (1 - 2(2))(1) + (1 - 2(3))(1) = 1 - 4 + 1 - 6 = -8
\end{align*}
\end{document}