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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "1.  Evaluate the following
, if they exist.  If they are divergent, state clearly why. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "limit((2*x^2+x-3)/(x^2+3*x-4
), x=infinity);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "lim
it((2*x^2+x-3)/(x^2+3*x-4), x=1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "(By l'Hopital's rule)" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "limit((2*x^2+x-3)/(x^2+3*
x-4), x=-4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%*undefinedG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "L'Hopital's rule is NOT valid her
e, because the numerator does not tend to 0.  It is undefined because \+
the denominator tends to 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
37 "sum(2^k * 3^(-k+1), k = 2..infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "2.  Deter
mine whether the following converge.  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "int(((2+x)/(2*x))^(1/3), x=0..2);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#-%$intG6$,$*&\"\"##F(\"\"$*&,&F(\"\"\"%\"xGF-F-F.!\"
\"#F-F*#F-F(/F.;\"\"!F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "e
valf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+d6J]E!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 104 "Well, that's finite.  But why?  Because \+
the integrand is less than (2/x)^(1/3), which can be integrated:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "int((2/x)^(1/3),x=0..2);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 22 "Convergent, all right." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sum(1/(k*(ln(k)^3)),k=2..i
nfinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sumG6$*&%\"kG!\"\"-%#l
nG6#F'!\"$/F';\"\"#%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 
"Maple is telling us she doesn't know the exact answer.  Let's compare
 with an integral:" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 35 "int(1/(k*(ln(k)^3)),k=2..infinity);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%#lnG6#\"\"#!\"##\"\"\"F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Well, that's finite.  Convergent.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 246 "Problems 3 and 4 are concern
ed with estimating the integral of 3/(1+4x^2).   No credit will be giv
en for an accurate estimate of this integral, only for the approximati
on,.\n\n\n3.  (10 points)  In this problem we use Taylor's polynomial \+
and series." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "taylor(3/(1+
4*x^2),x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++%\"xG\"\"$\"\"!!#7\"
\"#\"#[\"\"%-%\"OG6#\"\"\"\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 30 "int(3-12*x^2+48*x^4,x=0..1/2);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6##\"#8\"#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "For what posit
ive values of x is the Taylor series convergent for int(3/(1+4 t^2),t=
0..x);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 
"Answer:  SAME AS FOR 3/(1+4 x^2), except possibly for the ends of the
 interval.  Since this is " }}{PARA 0 "" 0 "" {TEXT -1 93 "the sum of \+
a geometric series in powers  ((x/2)^2)^n, we have convergence for -2 \+
< x < 2.  We" }}{PARA 0 "" 0 "" {TEXT -1 98 "actually do a little bett
er, for it converges when x=2,-2 by the alternating series test.  Answ
er:" }}{PARA 0 "" 0 "" {TEXT -1 11 "-2 ² x ² 2." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 22 "f := x -> 3/(1+4*x^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fG:6#%\"xG6\"6$%)operatorG%&arrowGF(,$*$,&\"\"\"F/*
$9$\"\"#\"\"%!\"\"\"\"$F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
43 "evalf(student[trapezoid](f(x),x=0..1/2,4));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+x6>u6!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 41 "evalf(student[simpson](f(x),x=0..1/2,2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"++++v6!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "J
ust for fun:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(f(x),x=
0..1/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%#PiG#\"\"$\"\")" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"+Xs4y6!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{MARK "10 0 0" 53 }{VIEWOPTS 1 1 0 1 1 1803 }