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\begin{document}

\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}

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