\documentclass{amsart}
\newcommand{\solution}{\noindent\textbf{Solution: }}

\begin{document}

\section{Additional Review Problems!}

\begin{enumerate}
\item Using $u$ substitution, find the integral of $\tan x$\\

\solution
\begin{align*}
\int \tan x dx =& \int \frac{\sin x }{\cos x} dx\\
\intertext{ Set $u= \cos x$, so that $du=-\sin x dx$}
=&\int \frac{-du}{u}\\
=&- \ln{|u|}+C\\
=&-\ln{|\cos x|}+C
\end{align*}

\item  \[\lim_{n\to\infty}\left( \frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\ldots\frac{1}{\sqrt{n}\sqrt{n+n}}\right)\]
Think the fundamental theorem of calculus!

\solution We are going to massage the limit to make it a little better to look at. I'm going to use summation sign here.
\begin{align*}
\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{\sqrt{n}\sqrt{n+k}}
=&\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{\sqrt{1+k/n}}\\
\intertext{The right side is a Riemann sum--- this is easier to see by setting $f(x)=\frac{1}{\sqrt{x}}$}
=&\lim_{n\to\infty} \sum_{k=1}^n  \frac{ f(1+k/n)}{n}
\intertext{Using the definition of the integral}
=&\int_1^2 f(x) dx\\
=&\int_1^2 \frac{1}{\sqrt x} dx\\
=&2x^{1/2}\bigg|_{x=1}^2\\
=&2(\sqrt 2- 1)
\end{align*}


\item Compute the following limit \[\lim_{x\to 4} \frac{1}{x-4}\int_{4}^x \frac{e^t}{t} dt\]

\solution There were several good solutions to this one. (Thanks, Vincent and Taylor.) Here are a few:

\begin{enumerate}
\item One way to think of $ \frac{1}{x-4}\int_{4}^x\frac{e^t}{t} dt$ is the average value of $\frac{e^t}{t}$ between $4$ and $x$. As the function $\frac{e^t}{t}$ is continuous at $4$, the average value of the function on the interval $[4,x]$ should tend to the value of the function at $4$ as $x $ goes to $4$. So the answer is $e^4/4$
\item You can also use L'Hopital rule for this one. Rewriting at 
 \[\lim_{x\to 4}\frac{\int_{4}^x \frac{e^t}{t} dt}{x-4}\]
The top of this fraction and bottom of this fraction are both differentiable, and go to $0$ as $x$ tends to $4$. We may use L'Hopital and get 
\[\lim_{x\to 4}\frac{\frac{d}{dx}\left(\int_{4}^x \frac{e^t}{t} dt\right)}{1}\]
By the fundamental theorem of calculus, this is $e^4/4$.
\item Suppose that $F(t)$ is the anti-derivative of $e^t/t$. Then we can use the fundamental theorem of calculus and get
\begin{align*}
 \lim_{x\to 4} \frac{1}{x-4}\int_{4}^x \frac{e^t}{t} dt=&  \lim_{x\to 4} \frac{F(x)-F(4)}{x-4}
\intertext{The right hand side is the definition of the derivative of $F$ at $4$}
=&F'(4)\\
=&e^4/4
\end{align*}

\end{enumerate}•

\item Japanese Napkin Ring Theorem. \\

\solution Wikipedia has a good write up of this problem, so I will leave it to them. Google : ``Japanese Napkin Ring Theorem'' and look it up!\\

\noindent If you want additional problems to review, try these two:\\

\item Assume a snowball melts so that its volume decreases at a rate proportional to its surface area.  If it takes 3 hours for the snowball to decrease to half its original volume, how much longer will it take for the snowball to melt completely\\


\item Dart Board Probability Problem: Suppose that I am throwing darts at a dart board of radius 1. The score that I get when I hit the board is equal to $1-x$, where $x$ is the distance from the center of the board to where my dart landed. So if I make a bullseye, I get 1 point. If I hit the rim, I get zero points. What is my average score?
\end{enumerate}

\end{document}
