%% Math 131, Fall, 2003
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\usepackage{math131}

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\newcommand{\ddx}[1]{\ds\frac{d}{d#1}}

\begin{document}

\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Solutions to Worksheet on Riemann Sums
\end{center}

\be

\item In the figure below, sketch the inscribed and circumscribed
  rectangles for 4~intervals. Then use your calculator to find upper
  and lower bounds for $\ds\int_1^2\frac{1}{t}\,dt$.

  \begin{center}
  \input{hsoln09a.eepic}
  \end{center}

  \soln Label the $t$~values $t_0$,\ldots, $t_4$; we have
  $f(t_0)=1$, $f(t_1)=\frac{4}{5}$, $f(t_2)=\frac{2}{3}$,
  $f(t_3)=\frac{4}{7}$, and $f(t_4)=\frac{1}{2}$. Since the integrand
  is decreasing, the left-hand sum over-estimates and the right-hand
  sum under-estimates. In exact arithmetic (you're welcome to use
  decimals!), 
  \begin{align*}
    \text{Upper} &=
    \frac{1}{4}\Bigl(1+\frac{4}{5}+\frac{2}{3}+\frac{4}{7}\Bigr)
    =\frac{319}{420}\simeq0.76, \\
    & \\
    \text{Lower} &=
    \frac{1}{4}\Bigl(\frac{4}{5}+\frac{2}{3}+\frac{4}{7}+\frac{1}{2}\Bigr)
    =\frac{533}{840}\simeq0.635, \\
  \end{align*}

\vfil\eject

  \bi
  \item By how much do your bounds differ? 

  \item If 12~intervals are used instead of~4, by how much do the
  upper and lower bounds differ?

    \soln With 4~intervals, $|f(b)-f(a)|(b-a)/4=1/8=0.125$. With
    12~intervals, the difference would be~$1/24$.

  \item What is your best guess of the actual value of the integral?

  \item The actual value of the integral is~$\ln 2\simeq0.69315$. How
  far off is your best guess?

    \soln The best guess is the average, $\frac{1171}{1680}\simeq
    0.697$, which differs from the actual value by about~$0.0039$.
    The red curve in the picture is the ``connect-the-dots'' graph,
    which encloses an area equal to the best guess. As you can see,
    the difference between the actual area and the best guess is much
    smaller than the difference between the actual area and either the
    lower or the upper bound using rectangles.

  \ei

\item Let $f(t)=4t^3+1$, and suppose we want to approximate
  $\ds\int_{-1}^1 f(t)\,dt$. (Note~$f$ is increasing.)

  \bd

  \item{(a)} If we use~100 intervals of equal length, by how much will
    the left-hand and right-hand estimates differ? (Do not compute the
    actual estimates!)
  
    \soln Here $|f(b)-f(a)|(b-a)=8\cdot2=16$; with 100~intervals, the
    difference between upper and lower sums is $\frac{16}{100}=0.16$.

  \item{(b)} How many intervals would guarantee an accuracy of
    4~decimals? 6~decimals? 

    \soln Respectively, $160,000$~intervals and
    $16,000,000$~intervals. 

  \vfil

  \ed

\ee

\eject

\end{document}
