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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{The definition of the Riemann Integral}

\begin{example}
If \( f(x)=3x \), then \[
\int _{1}^{2}f(x)dx=9/2.\]
 
\end{example}

\subsubsection*{Solution}

This is not trivial to see. If you trust that the limit \textit{exists}, which
is not really obvious, but believable, then you can always use the circumscribed
rectangles, that is, take \( c_{i}=x_{i} \) in the sum. Also, take a regular
partition, with each subinterval of the same width (which would be \( (2-1)/n \)),
so that \( x_{i}=1+i/n \), then the Riemann sum of that partition, with those
sampling points, is\begin{eqnarray*}
\sum _{i=1}^{n}f(x_{i})\Delta x_{i} & = & \sum _{i=1}^{n}3(1+i/n)(1/n)\\
 & = & \sum _{i=1}^{n}3(1+i/n)(1/n)\\
 & = & (3/n)\sum _{i=1}^{n}1+(3/(n^{2}))\sum _{i=1}^{n}i\\
 & = & (3/n)\times n+(3/(n^{2}))\frac{n(n+1)}{2},
\end{eqnarray*}
so that the limit, when the mesh goes to 0 (here meaning that \( n\rightarrow \infty  \)),
goes to\[
3+\frac{3}{2}=\frac{9}{2},\]
as advertised. 

One approach to the more serious question of whether or not the limit exists
is to change to the inscribed rectangles for the same partition above. Since
the height of the inscribed rectangle over the \( t^{th} \) subinterval is
\begin{eqnarray*}
f(x_{i-1}) & = & f(1+(i-1)/n)\\
 & = & 3(1+(i-1)/n).
\end{eqnarray*}
Then, the Riemann sum, using inscribed rectangles at each stage, is \begin{eqnarray*}
\sum _{i=1}^{n}f(x_{i-1})\Delta x_{i} & = & \sum _{i=1}^{n}3(1+(i-1)/n)(1/n)\\
 & = & \sum _{i=1}^{n}3(1+(i-1)/n)(1/n)\\
 & = & (3/n)\sum _{i=1}^{n}1+(3/(n^{2}))\sum _{i=1}^{n}(i-1)\\
 & = & (3/n)\sum _{i=1}^{n}1+(3/(n^{2}))\sum _{j=1}^{n-1}j,\, \textrm{substituting}\, j=i-1\\
 & = & (3/n)\times n+(3/(n^{2}))\frac{(n-1)(n)}{2},
\end{eqnarray*}
which has the \emph{same} limit as \( n\rightarrow \infty  \).

Then, since \emph{any} Riemann sum with this partition, for \emph{any} choice
of sampling points, lies in between these two, any such sum would tend to the
same limit. This does not quite cover all possibilities, though, since the partitions
could be quite different from these. We'll take care of that last detail in
the next section.
\end{document}
