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\begin{document}

\flushleft{\textbf{6.1}}

\begin{itemize}

\item Graph $y=x$ and $y=x^3$

\begin{enumerate}
\item Find the area between $y=x$ and $y=x^3$ from $x=0$ to $x=1$
\begin{enumerate}
\item Shade in the region between the two graphs from $x=0$ to $x=1$.
\item Which function is ``on top''?
\item Write down the integral representing the area between the two graphs.
\item Now calculate the area by computing the integral.
\end{enumerate}
\item Find the area between $y=x$ and $y=x^3$ from $x=0$ to $x=2$.
\begin{enumerate}
\item Shade in the region between the two graphs from $x=0$ to $x=2$.
\item Which function is ``on top''? When does it change?
\item Write down the integral representing the area between the two graphs using the absolute value formula.
\item Now rewrite that as a sums of integrals based on your answer to part b). 
\item Now calculate the area by computing the integrals.
\end{enumerate}
\item Find the area between $y=x$ and $y=x^3$.
\begin{enumerate}
\item We have not been given bounds on $x$. What could the question be asking for?
\item Shade in the regions between the two graphs that are bounded. What are the $x$ values for the bounded regions between the two graphs?
\item Which function is ``on top''? When does it change?
\item Write down the integral representing the area between the two graphs using the absolute value formula.
\item Now rewrite that as a sums of integrals based on your answer to parts b) and c). 
\item Now calculate the area by computing the integrals.
\end{enumerate}
\end{enumerate}

\item Find the area between $\sin x$ and $\cos x$ from $x=0$ to $x=\pi$.
\begin{enumerate}
\item Graph both functions on the interval $[ 0, \pi ]$.
\item Shade in the region between the two graphs from $x=0$ to $x=\pi$.
\item Which function is ``on top''?When does it change?
\item Write down the integral representing the area between the two graphs using the absolute value formula.
\item Now rewrite that as a sums of integrals based on your answer to part b). 
\item Now calculate the area by computing the integrals.
\end{enumerate}

\item Find the area between $y = \tan^{-1} x$ and $y = \frac{\pi}{4}x^2$. 
\begin{enumerate}
\item Graph both functions. Note that these two graphs intersect at $(0,0)$ and $(1,\pi/4)$.
\item Shade in the bounded region. Which function is on top?
\item Write an integral expressing the area between these two graphs. \\
\textbf{Instructor:} Students should get the integral
\[
\int_0^1 \tan^{-1}x - \frac{\pi}{4} x^2 \, dx
\]
After some time, they will notice that they can not calculate $\int \tan^{-1}x \, dx$. Ask for suggestions, and if no one suggests it, then explain that the graphs should be expressed as functions of $y$. This will lead to the following question:
\item Find the area between $x = \tan y$ and $x = \frac{2}{\sqrt{\pi}} \sqrt{y}$. \\
\textbf{Instructor:} Now is a good time to discuss why we choose the positive square root when solving for $x$ in the equation $y = \frac{\pi}{4} x^2$. 
\item Graph both functions. Note that these two graphs intersect at $(0,0)$ and $(\pi/4,1)$.
\item Shade in the bounded region. Which function is on top? \\
\textbf{Instructor:} This is a good time to point out that ``on top'' for functions of $y$ means to the right.
\item Write an integral expressing the area between these two graphs.
\item Now calculate the area by computing the integral.
\end{enumerate}


\end{itemize}

\newpage

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\flushleft{\textbf{5.5}}

\begin{itemize}

\item $\int \tan x \, dx$

\textbf{Instructor:} Ask for suggestions for this integral and eventually students will ask to rewrite it.

\[
\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx
\]

Now the instructor should attempt the $u$ substitution $u = \sin x$, and fail. Most likely, a student will suggest the substitution $u = \cos x$. Ask the students to try this substitution on their own. Once they have
\[
\int \tan x \, dx = - \ln | \cos x \, | + C \textrm{,}
\]
ask for another way this can be written. I usually ask ``what will Quest do?''
\[
\int \tan x \, dx = - \ln | \cos x \, | + C = \ln | \left( \cos x \right)^{-1} \, | + C = \ln | \sec x \, | + C
\]

\item Moving a constant over to $du$

\begin{enumerate}
\item $\int x e^{x^2} \, dx$ \\
\textbf{Instructor:} This example is easy enough, just point out that when choosing $u$, we don't need to see the entire $du$ in the integral. Just up to multiplication by a constant. In this case, $u = x^2$, but $du = 2x \, dx$ is not in the integral. However, $\frac{1}{2} \, du = x \, dx$ is there.
\item $\int e^{5x} \, dx$ \\
\textbf{Instructor:} Again, this example is easy, but notice that whenever $u$ is linear, then $du$ is just a number, and since we can change $du$ by a constant, \emph{we never need to see $du$}. We can always do a linear substitution: $u = mx+b$.
\item $\displaystyle \int \frac{x}{5-2x} \, dx$ \\
\textbf{Instructor:} Here the students will do a linear substitution (they've been led to this by the previous problems) in order to simplify the denominator, but this will complicate the numerator. This leads them into more difficult substitutions that require ``solving for $x$'' after having chosen $u$. Or even more difficult substitutions like the next one.
\end{enumerate}

\item
\[
\int \frac{x^3}{\sqrt{x^2+1}} \, dx
\]
\textbf{Instructor:} I would consider letting the students try for a few minutes, then trying to elicit the suggestion of $u = x^2+1$. Note that it is not easy to see $du$ now (up to a constant, we're looking for $x \, dx$), \emph{then} give them the hint to write
\[
\int \frac{x^3}{\sqrt{x^2+1}} \, dx = \int \frac{x^2}{\sqrt{x^2+1}} \cdot x \, dx
\]
Now
\[
u = x^2 + 1 \hspace{10pt} \frac{1}{2} \, du = x \, dx
\]
At this point I say ``let's try to swap out everything we can'' and write
\[
\int \frac{x^2}{\sqrt{x^2+1}} \cdot x \, dx = \int \frac{}{\sqrt{u}} \cdot \frac{1}{2} \, du
\]
and point out that we don't know the numerator still. Invariably, a student will produce the solution that we should replace $x^2$ with $u-1$. 

\item
\[
\int \tan \theta \sec^2 \theta \, d \theta
\]
\textbf{Note:} This example will also be included in section 7.3, Trigonometric Integrals, which is where I prefer to address it. \\
\textbf{Instructor:} I would begin this with the class, and ask for suggestions. Some student will come up with the substitution $u = \tan \theta$. Then have the students work the problem that way to get
\[
\int \tan \theta \sec^2 \theta \, d \theta = \int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2} \tan^2 \theta + C
\]
Then ask for another way to solve this integral. If no one has any suggestions, give them the hint that it's another substitution. If no one has any suggestions, rewrite
\[
\int \tan \theta \sec^2 \theta \, d \theta = \int \sec \theta \tan \theta \sec \theta \, d \theta
\]
and someone will eventually suggest the substitution $u = \sec \theta$, giving
\[
\int \tan \theta \sec^2 \theta \, d \theta = \int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2} \sec^2 \theta + C
\]
At this point I always wait, leaving both solutions on the board. I just wait for as long as it takes until someone notices there might be an issue. At this point, the students really need to be engaged enough to see for themselves that they've gotten two different answers to the same problem, and to be concerned about that. Then I talk about how these answers only differ by a constant, which is wrapped up in the ``$+ C$''. I also bring up the example that $\ln ( 2 x ) + C$ is the same as $\ln ( x ) + C$ for the same reason
\end{itemize}

\newpage 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\flushleft{\textbf{5.3}}

\begin{itemize}

\item Exploring 
\[
g(x) = \int_0^x f(t) \, dt
\]

\begin{enumerate}

\item
\[
g(x) = \int_0^x t \, dt, x \geq 0
\]

\begin{enumerate}
\item Let $f(t) = t, t \geq 0$. Graph $f(t)$. Make sure to use a $t$-axis and a $y$-axis.
\item We're going to compute 
\[
g(x) = \int_0^x t \, dt, x \geq 0 \textrm{.}
\]
First, plot an $x > 0$ on the $t$-axis. This is some positive real number, but we're not choosing its value. We're treating the variable $x$ like a constant here, usually reserved for symbols like $a$, $b$, or especially $c$.
\item Now graph $f(x)$, which is just $x$ again in this case, but on the $y$-axis.
\item Let's use the geometric interpretation of $g(x)$. Thus $g(x)$ is the area under the graph of $f(t)$ and above the $t$-axis from $t=0$ to $t=x$. What shape does this give us?
\item Use the geometry formula for the area of a triangle to determine $g(x)$ (i.e. $A = \frac{1}{2} \cdot b\cdot h$.)
\item We see that $g(x) = \frac{1}{2} x^2$, an antiderivative of $f(x)$. (Note that the variables have switched. $g(x)$ is an antiderivative of $f(x)$, not $f(t)$.)
\end{enumerate}

\item
\[
g(x) = \int_0^x \cos t \, dt, 0 \leq x \leq 2 \pi
\]

\begin{enumerate}
\item Let $f(t) = \cos t, 0 \leq t \leq 2 \pi$. Graph $f(t)$. Make sure to use a $t$-axis and a $y$-axis.
\item We're going to try to graph $g(x)$. Let's start by filling in the following table as best as we can: \\ 
\vspace{10pt}

\begin{tabular}{l|l}
$x$ & $g(x)$ \\
\hline
$0$ & \\
$\frac{\pi}{2}$ & \\
$\pi$ & \\
$\frac{3 \pi}{2}$ & \\
$2 \pi$ & \\
\end{tabular} \\
\vspace{10pt}
\textbf{Instructor:} Students should reach the following conclusions:
\begin{enumerate}
\item $g(0) = \int_0^0 \cos t \, dt = 0$ using a basic property of definite integrals.
\item $g(\pi / 2) = \int_0^{\pi /2} \cos t \, dt$: students will need more help here to see that they can not find an exact value, but the class will reach a consensus that it is some positive value $\alpha$, and you want to make sure they identify $\alpha$ with the area of that geometric shape, the ``first hill'' of $f(t)$.
\item $g(\pi) = \int_0^{\pi} \cos t \, dt = \int_0^{\pi /2} \cos t \, dt + \int_{\pi/2}^{\pi} \cos t \, dt = 0$: students should now see that the ``first hill'' of $f(t)$ will cancel with the ``first valley'' of $f(t)$. Point out that this means that $\int_{\pi/2}^{\pi} \cos t \, dt = - \alpha$. Make sure they identify this in the graph of $f(t)$.
\item $g(3 \pi /2) = - \alpha$: using the above logic about the graph of $f(t)$, students should be able to see that the total integral here will be $-\alpha$.
\item $g(2 \pi) = 0$ follows similarly.
\end{enumerate}
The table will look like this now: \\
\vspace{10pt}
\begin{tabular}{l|l}
$x$ & $g(x)$ \\
\hline
$0$ & $0$ \\
$\frac{\pi}{2}$ & $\alpha$ \\
$\pi$ & $0$ \\
$\frac{3 \pi}{2}$ & $-\alpha$ \\
$2 \pi$ & $0$ \\
\end{tabular} \\
\vspace{10pt}

\item Plot the points in the chart. What curve fits these points?
\item What do you guess $g(x)$ is?\\
\textbf{Instructor:} It should be clear now that $g(x) = \sin x$. Point out that this is an antiderivative of $f(x)$. Again, note the change in variables.

\end{enumerate}

\end{enumerate}

\underline{\hspace{450pt}}

\item 

\[
\frac{d}{dx} \left( \int_{\sqrt{x}}^{\ln x} \sin t \, dt \right)
\]

\begin{enumerate}

\item
First, use the property that
\[
\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx
\]
\textbf{Instructor:}
Students should get
\begin{eqnarray*}
\frac{d}{dx} \left( \int_{\sqrt{x}}^{\ln x} \sin t \, dt \right) & = & \frac{d}{dx} \left( \int_{\sqrt{x}}^c \sin t \, dt + \int_c^{\ln x} \sin t \, dt \right) \\
& = & \frac{d}{dx} \left( \int_{\sqrt{x}}^c \sin t \, dt \right) + \frac{d}{dx} \left( \int_c^{\ln x} \sin t \, dt \right) \\
\end{eqnarray*}
\item
Second, use the property that
\[
\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx
\]
\textbf{Instructor:}
Students should get
\begin{eqnarray*}
\frac{d}{dx} \left( \int_{\sqrt{x}}^c \sin t \, dt \right) + \frac{d}{dx} \left( \int_c^{\ln x} \sin t \, dt \right) & = & \frac{d}{dx} \left( - \int_c^{\sqrt{x}} \sin t \, dt \right) + \frac{d}{dx} \left( \int_c^{\ln x} \sin t \, dt \right) \\
& = & - \frac{d}{dx} \left( \int_c^{\sqrt{x}} \sin t \, dt \right) + \frac{d}{dx} \left( \int_c^{\ln x} \sin t \, dt \right) \\
\end{eqnarray*}
\item Last, use the Chain Rule with the First Part of the Fundamental Theorem of Calculus:
\begin{eqnarray*}
\frac{d}{dx} \left( \int_a^u f(t) \, dt \right) & = & \frac{d}{du} \int_a^u f(t) \, dt \cdot \frac{du}{dx} \\
& = & f(u) \cdot \frac{du}{dx} \\
\end{eqnarray*}
\textbf{Instructor:}
Students should get
\begin{eqnarray*}
- \frac{d}{dx} \left( \int_c^{\sqrt{x}} \sin t \, dt \right) + \frac{d}{dx} \left( \int_c^{\ln x} \sin t \, dt \right) & = & - \sin \left( \sqrt{x} \right) \frac{d}{dx} \sqrt{x} + \sin \left( \ln x \right) \frac{d}{dx} \ln x \\
& = & - \sin \left( \sqrt{x} \right) \cdot \frac{1}{2 \sqrt{x}} + \sin \left( \ln x \right) \cdot \frac{1}{x} \\
\end{eqnarray*}
\end{enumerate}

\end{itemize}

\newpage

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\flushleft{
\textbf{NAME:} \underline{\hspace{250pt}} \textbf{EID:} \underline{\hspace{100pt}} \\

\textbf{Final Exam M408S Spring 2015} \\

To receive credit, write your name on every sheet, show all work, give reasons for answers. \\}

\vspace{10pt}

\begin{enumerate}

\item[1.] (10 points)

Estimate $\displaystyle \int_0^1 x \cdot \tan^{-1}(x^2) \, dx$ using the Taylor polynomial $T_3(x)$ for $x \cdot \tan^{-1}(x^2)$ centered at $0$.

\newpage

\item[2.] (10 points)

What is the radius of convergence for
\[
f(x) = \sum_{n=0}^{\infty} \frac{\left(2n\right)!}{\left(n!\right)^2} \, \, x^n \textrm{ ?}
\]

\newpage

\flushleft{ \hspace{-24pt}
\textbf{NAME:} \underline{\hspace{250pt}} \textbf{EID:} \underline{\hspace{100pt}} \\

\hspace{-24pt} \textbf{Final Exam M408S Spring 2015} \\

\hspace{-24pt} To receive credit, write your name on every sheet, show all work, give reasons for answers. \\ }

\vspace{20pt}

\item[3.] (10 points) 

Compute $T_3(x)$ centered at $\pi/3$ for $f(x) = \cos x$.

\newpage

\item[4.] (10 points) 

The power series
\[
f(x) = \sum_{n=1}^{\infty} \frac{n}{n^2+n+1} \, \, x^n
\]
has radius of convergence $R = 1$. What is the interval of convergence?

\end{enumerate}



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