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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Indefinite integration}

\begin{example}
Assume that a thingie is moving along a line so that its acceleration is \( 4feet/sec^{2} \).
If at \( t=0 \) the thingie is moving at \( 5feet/sec \) and is \( 7feet \)
from origin, what is its position at any time \( t \)? 
\end{example}

\subsubsection*{Solution}

Those conditions, contained in {}``If at \( t=0 \) the thingie is moving at
\( 5feet/sec \) and is \( 7feet \) from origin{}'', are called \emph{initial
conditions}, for reasons that should be clear. Initial conditions allow you
to solve for and determine the constants of integration, so that the answer
we get will not involve some unknown \( C \). 

Since the acceleration \( a=4 \), and acceleration is the derivative of velocity,
\begin{eqnarray*}
v & = & \int a\, dt\\
 & = & \int 4\, dt\\
 & = & 4t+C.
\end{eqnarray*}
Note that \( t \) is the independent variable, representing time. Now, since
at time \( t=0 \), \( v=5, \) we have a string of equalities:\[
5=v(0)=4\cdot 0+C=C,\]
so that \( C=5, \) or \[
v=4t+5.\]
Furthermore, since the position \( s(t) \) is the integral of the velocity,
\[
s(t)=\int v(t)\, dt,\]
 we have \begin{eqnarray*}
s(t) & = & \int v(t)\, dt\\
 & = & \int (4t+5)dt\\
 & = & 2t^{2}+5t+c.
\end{eqnarray*}
Again, the initial conditions can resolve the constant of integration, since
\( s(0)=7 \):\[
7=s(0)=2\cdot 0^{2}+5\cdot 0+c=c,\]
so this constant is \( c=7 \), and \[
s(t)=2t^{2}+5t+7.\]
 This is the position of the thingie at time \( t \), which is what we needed
to find.
\end{document}
