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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#17}

\bigskip\noindent
3.1, \#9.
We get upper and lower estimates for the speed at $t=5$ by summing
upper and lower estimates for the speed on the intervals between
consecutive readings of the acceleration.  The speed
at $t=0$ is 0, so these estimates are
the left and the right sums
$$\eqalign{
   9.81+8.03+6.53+5.30+4.41&=34.16\hbox{ m/sec}\cr
   8.03+6.53+5.30+4.41+3.61&=27.96\hbox{ m/sec}\cr}$$
Their average is 31.06 m/sec.  Since consecutive changes
in acceleration are decreasing, the graph of acceleration against time
is concave-up.  The average of the left and right sums
approximates the area under the graph by trapezoids.
So the new estimate is greater than the true speed.

\bigskip\noindent
3.2, \#1, 2, 5.
Left and right sums are given in the following table:
$$\vbox{
\halign{\hfil\quad#&\hfil\qquad#&\hfil\quad#&\hfil\qquad#&
     \hfil\quad#&\hfil\qquad#&\quad\hfil#\quad\cr
   \phantom{n}&\multispan2 $\displaystyle\phantom{1}\int_0^3x^3\,dx$
     &\multispan2 $\displaystyle\phantom{111}\int_0^{\pi/2}\cos x\,dx$
     &\multispan2 $\displaystyle\phantom{11}\int_2^3\sin(t^2)\,dt$ \cr
  \noalign{\vskip5pt}
  $n$&Left &Right &Left &Right &Left &Right \cr
  2&.0625&.5625&1.3408&.5554&$-.3950$&$.1895$\cr
  10&.2025&.3025&1.0765&.9194&$-.0921$&$.0248$\cr
  50&.2401&.2601&1.0156&.9842&$-.0430$&$-.0196$\cr
  250&.2480&.2520&1.0031&.9969&$-.0336$&$-.0289$\cr}}$$
$x^3$ and $\cos x$ are monotonic on the given intervals,
so their integrals are between the left and right sums.
We take the averages of the left and right
sums for $n=250$, namely .25 and 1.00,
as estimates for the values of the integrals.
In \#5 $\sin(t^2)$ is {\it not\/} monotonic on $[2,3]$,
so its integral may not be between left and right sums.
Thus the average $-.0312$
of the left and right sums for $n=250$ may not be a reaonable
estimate for the integral.

\bigskip\noindent
3.2, \#13.  Since $2^x$ is increasing on $[1,2]$,
its integral is between left and right sums.
For one decimal accurancy,
(right sum) $-$ (left sum) should be less
than $0.1$.  Thus the number $n$ of subintervals
must satisfy the inequality $(2^2-2^1)(2-1)/n<0.1$ and
$n>20$.  For $n=21$ the left and right sums are
respectively $2.8380$ and $2.9333$.  Their average
$2.8857$ is accurate to one decimal place.

\bigskip\noindent
3.2, \#18.  $t^2$ is
increasing on $[1,3]$.  We reason as in \#13.
The number $n$ of subintervals must
satisfy the inequality $(3^2-1^2)(3-1)/n<0.1$.  Thus
$n> 160$.  An approximation for the integral was not asked for.


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