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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}
\[
\int _{0}^{1}x^{2}dx=\frac{1}{3}.\]
 
\end{example}

\subsubsection*{Solution}

Proceed as with the previous example, presuming that the limit exists. Then,
we can just look at uniform subdivisions and circumscribed rectangles, as before.
With \( n \) subintervals, then \( x_{i}=0+i/n \), and since we are using
circumscribed rectangles and \( f(x) \) is increasing, then \( x_{i}^{*}=x_{i} \)
in each case (the right-hand edge of the \( i^{th} \) subinterval gives \( f \)
the largest value in that subinterval). So the Riemann sum is:\begin{eqnarray*}
\sum _{i=1}^{n}f(x_{i})\Delta x_{i} & = & \sum _{i=1}^{n}\left( x_{i}\right) ^{2}\left( \frac{1}{n}\right) \\
 & = & \sum _{i=1}^{n}\left( \frac{i}{n}\right) ^{2}\left( \frac{1}{n}\right) \\
 & = & \frac{1}{n^{2}}\sum _{i=1}^{n}i^{2}\\
 & = & \frac{1}{n^{2}}\frac{n(n+1)(2n+1)}{6}.
\end{eqnarray*}
Taking the limit as \( n\rightarrow \infty  \), which also implies that the
mesh of these subintervals goes to 0, we have \begin{eqnarray*}
\int _{0}^{1}x^{2}dx & = & \lim _{n\rightarrow \infty }\sum _{i=1}^{n}f(x_{i})\Delta x_{i}\\
 & = & \lim _{n\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^{3}}\\
 & = & \frac{1}{3}.
\end{eqnarray*}

\end{document}
