\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{ulem} % for strikethrough \sout{...} command
\usepackage{color}

\usepackage{wasysym}

\usepackage{pgfplots}
\pgfplotsset{compat=1.14}
\usepackage{tikz}
\usepgfplotslibrary{fillbetween}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.25in}
\setlength{\evensidemargin}{-0.25in}
\setlength{\topmargin}{-0.25in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.75in}
\setlength{\textwidth}{7in}

\newcommand{\bdot}{\, {\scriptstyle{\bullet}} \,}
\newcommand{\B}[1]{{\bf {#1}}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\myspace}{\parbox[c][0.25in]{0.25in}{\hfill}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\force}{{\noindent $\mbox{\;}$}}

\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}

\begin{document}

\pagestyle{empty}

\noindent
\parbox{2.25in}{\bf Math 1110-102}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{2.25in}{\bf \hfill Due: Tues., August $2^\mathrm{nd}$, 2022}

%\vspace{0.2in}
%
%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

%\vspace{0.05in}

%\noindent
%Please remember when submitting any work via email or in person to\dots\\
%\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force


\vspace{0.2in}

\begin{description}
\item[\#1] Differential Equations

\begin{enumerate}[(a)]
\item Verify that $y=te^{-3t}$ is a solution of $y''+6y'+9y=0$. 

\item Solve the initial value problem: $y' = \dfrac{x^2+1}{x^3}+2$ and $y(1)=5$.

\force \hfill {\it Hint:} Do some algebra to simplify your fraction first.
\end{enumerate}


\item[\#2] Approximate! 

\vspace{0.05in}

Recall that $L_n$, $R_n$, and $M_n$ are the left, right, and midpoint rule approximations of an integral when using a regular (=equally spaced) partition of $n$ subintervals. In addition, $T_n = \frac{1}{2}L_n+\frac{1}{2}R_n$ approximates our integral using $n$ trapezoids. Finally, $S_{2n} = \frac{2}{3}M_n+\frac{1}{3}T_n$ approximates using parabolic arcs.

\begin{enumerate}[(a)]
\item Consider the definite integral $\displaystyle \int_{-3}^9 \sqrt[3]{x^2}\,dx$. Compute $L_3$, $R_3$, $M_3$, $T_3$, and $S_6$ (=$S_{2\cdot 3}$).
\item Suppose $I=\displaystyle \int_1^{5} \ln(x)\,dx$. Sketch  a picture of the approximations $R_2$ and $T_2$ for this integral.

\begin{center}
\begin{tikzpicture}
\begin{axis}[
  axis x line=middle, axis y line=middle,
  width=3.5in, height=2.5in,     % size of the image = rigged to keep scaling=constrained
  ymin=-0.5, ymax=2.5, % ytick={0}, ylabel=$y$,
  xmin=0, xmax=6.1, % xtick={0}, xlabel=$x$,
  xtick={1,...,5},
  ytick={1,2},
]

\addplot[domain=0.5:5.5,ultra thick, blue] ({x},{ln(x)});

%\draw[fill=red,red,ultra thick] (axis cs: -2,-1) circle [radius=2pt];
%\draw[fill=red,red,ultra thick] (axis cs: -1,6) circle [radius=2pt];
%\draw[fill=red,red,ultra thick] (axis cs: 1,-10) circle [radius=2pt];

%\node at (axis cs: -2,-2) {$(-2,-1)$};
%\node at (axis cs: -1,7) {$(-1,6)$};
%\node at (axis cs: 1,-11) {$(1,-10)$};

\node at (axis cs: 5.5,1.9) {$y=\ln(x)$};
\end{axis}
\end{tikzpicture}
\end{center}

\item Still considering $I=\displaystyle \int_1^{5} \ln(x)\,dx$, rank $L_n$, $R_n$, $M_n$, $T_n$, $I$.\\
{}\force \hfill [For example, $L_n \leq R_n \leq M_n \leq T_n \leq I$ is definitely not the correct answer!]
\end{enumerate}

\item[\#3] Exact area. 

\begin{enumerate}[(a)]
\item First, carefully sketch $y=2x+2$ on the interval $[-2,3]$. 

Then, compute $\displaystyle \int_{-2}^3 (2x+2)\,dx$ using basic geometry (i.e., interpret the integral as net area).

\item First, carefully sketch $y=\sqrt{16-x^2}$ \quad ({\it Hint:} this is the upper-half of a circle). 

Then, compute $\displaystyle \int_{-4}^4 \sqrt{16-x^2}\,dx$ using basic geometry.
\end{enumerate}
\end{description}



\end{document}


