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\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

March 21, 2003}
\end{center}

\begin{enumerate}

\item  As you come in, please tell me your height in inches.

\item  Consider the function $f(t)=A e^{-ct}$ for $t \geq 0$ where $A$ and $c$ are constants.
	\begin{enumerate}
	\item  Suppose you want to ensure $\int_{0}^{\infty} f(t)\; dt =1$.  Find conditions on $A$ and $c$ to ensure this.
	\\\sol We just compute
	\begin{eqnarray*}
	\int_0^\infty Ae^{-ct} \; dt & = & \lim_{u \ra \infty} \int_0^u Ae^{-ct} \; dt
	= \lim_{u \ra \infty} \left. -\frac{A}{c}e^{-ct} \right|_0^u
	= \lim_{u \ra \infty} -\frac{A}{c}( e^{-cu} - e^{-0c}) \\
	& = & \lim_{u \ra \infty} -\frac{A}{c}(e^{cu} -1) = \frac{A}{c}
	\end{eqnarray*}
	So, we need $\frac{A}{c}=1$ or $A=c$.

	\item  Suppose, in addition, you know that $A=0.01$, what is $c$?
	\\\sol According to the previous work, $c=0.01$.

	\item  Using your $A=0.01$ and the $c$ that you just found, find a nice expression for the function:
		\[
		F(t) = \int_0^{t} f(u) \; du
		\]
	\\\sol Just do the integration:
	\[
	F(t) = \int_0^t (0.01)e^{-(0.01)u} \; du = \left. -e^{-(0.01)u} \right|_0^t = 1-e^{-(0.01)t}
	\]

	\item  What is $\lim_{t \ra \infty} F(t)$?
	\\\sol Just take the limit:
	\[
	\lim_{t \ra \infty} F(t) = 1
	\]

	\item  Compute: $\int_0^\infty tf(t)\; dt$.
	\\\sol  This is integration by parts:
	\[
	\int_0^\infty t(0.01)e^{-0.01t} \; dt
	\]
	Let $u=0.01t$ and $dv=e^{-0.01t}\;dt$.  This gives $du=0.01\; dt$ and $v=-100e^{-0.01t}$.
	\begin{eqnarray*}
	uv-\int v\; du & = & -te^{-0.01t} + \int e^{-0.01t} \; dt \\
	& = & -te^{-0.01t} - 100e^{-0.01t} +C 
	\end{eqnarray*}
	We can take limits and plug in our endpoints:
	\begin{eqnarray*}
	\int_0^\infty t(0.01)e^{-0.01t} \; dt & = & 
	\lim_{u \ra \infty} \int_0^u t(0.01)e^{-0.01t} \; dt 
	= \lim_{u \ra \infty} \left. -te^{-0.01t} - 100e^{-0.01t} \right|_0^u \\
	& = & \lim_{u \ra \infty} (-ue^{-0.01u} - 100e^{-0.01u}) - (-100) = 100
	\end{eqnarray*}


	\end{enumerate}

	
\end{enumerate}

\end{document}