\input epsf
\nopagenumbers
\magnification=\magstep1

\noindent {\bf Problem statement} The convergent improper integral
${\int_0^{\infty} e^{-x^2} \, dx}$ is important in probability and
statistics. An amazing result of multivariable calculus shows that the
value is exactly ${{\sqrt \pi}\over 2}.$ This problem shows how to get
a numerical approximation of the integral's value.

\medskip

\noindent a) Suppose $a\geq 1$. Use a substitution to calculate
${\int_a^{\infty} x e^{-x^2} \, dx}$. Now use this result to show that
${\int_a^{\infty} e^{-x^2} \, dx \leq {1\over 2}e^{-a^2} }$.

\medskip

\noindent b) Calculate an approximate value for ${\int_0^{\infty}
e^{-x^2} \, dx}$ as follows: find a value of $a$ so that ${1\over
2}e^{-a^2} < 10^{-5}$. Now calculate ${\int_0^a e^{-x^2} \, dx }$
approximately using the numerical integration capability of your
calculator.

\medskip

\noindent c) Discuss all the sources of error in your calculation.
Also discuss whether your answer is consistent with the claimed exact
value of the integral in a).

\medskip

\noindent d) Using your answer to b) calculate a numerical value for
${\int_0^{\infty} x^2 e^{-x^2} \, dx}$.

\medskip

\noindent {\bf Hint} Write the integrand as $x(xe^{-x^2})$ and use
integration by parts.

\vfil\eject\end