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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257 
"" 0 "" {TEXT -1 34 "Lesson 26:  Parametric Arc Length\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 282 "In this worksheet, we will use the proce
ss of integration to compute the lengths of plane parametric curves.  \+
 The same approach will find the lengths of 3-dimensional curves, but \+
we will not consider that extension.  Suppose we are given a parametri
c curve, described by equations " }{XPPEDIT 18 0 "x = f(t);" "6#/%\"xG
-%\"fG6#%\"tG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "y = g(t);" "6#/%\"yG-%
\"gG6#%\"tG" }{TEXT -1 7 ", t in " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG
%\"bG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f \+
:= t-> t^2 ; g := t-> t^3 - 3*t ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%\"fGR6#%\"tG6\"6$%)operatorG%&arrowGF(*$)9$\"\"#\"\"\"F(F(F(" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGR6#%\"tG6\"6$%)operatorG%&arrowG
F(,&*$)9$\"\"$\"\"\"F1*&F0F1F/F1!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 28 "plot([f(t), g(t), t=-2..2]);" }}{PARA 13 "" 1 "" 
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45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "We par
tition the parameter interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"
bG" }{TEXT -1 6 " into " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 35 " \+
equal subintervals with endpoints " }{XPPEDIT 18 0 "t_0 = a;" "6#/%$t_
0G%\"aG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "t_1;" "6#%$t_1G" }{TEXT -1 
9 ", ... , \n" }{XPPEDIT 18 0 "t_n = b;" "6#/%$t_nG%\"bG" }{TEXT -1 
196 " , and draw the piecewise-linear curve composed of line segments \+
joining the corresponding points on our original curve.  It turns out \+
that, with some care, we can do this with the same procedure " }{TEXT 
256 2 "pl" }{TEXT -1 77 " that we used earlier to compute arclengths o
f graphs: we just have to apply " }{TEXT 257 2 "pl" }{TEXT -1 4 " to \+
" }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g;
" "6#%\"gG" }{TEXT -1 12 " separately." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "pl := proc(f,a,b,n)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "        local u,v,k;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "     \+
   k := 1 + floor(n*(x-a)/(b-a));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
31 "        u := a + (k-1)*(b-a)/n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "        v := a + k*(b-a)/n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "
        unapply(f(u) + (f(v) - f(u))/(v-u)*(x-u), x);" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 15 "      end proc:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "plf := pl(f,-2,2,6):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "plg := pl(g,-2,2,6):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 66 "plot(\{[f(t),g(t),t=-2..2], [plf(t),plg(t),t=-2..2]\}
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45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 31 "(We suppressed the output when " }{XPPEDIT 18 0 "plf;" 
"6#%$plfG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "plg;" "6#%$plgG" }
{TEXT -1 145 " were defined; it is not very interesting.)\n\nIt is eas
y to compute the length of the piecewise-linear curve.  A typical segm
ent joins the points " }}{PARA 0 "" 0 "" {TEXT -1 1 "(" }{XPPEDIT 18 
0 "f(t_(j-1));" "6#-%\"fG6#-%#t_G6#,&%\"jG\"\"\"F+!\"\"" }{TEXT -1 2 "
, " }{XPPEDIT 18 0 "g(t_(j-1));" "6#-%\"gG6#-%#t_G6#,&%\"jG\"\"\"F+!\"
\"" }{TEXT -1 7 ") and (" }{XPPEDIT 18 0 "f(t_j);" "6#-%\"fG6#%$t_jG" 
}{TEXT -1 2 ", " }{XPPEDIT 18 0 "g(t_j);" "6#-%\"gG6#%$t_jG" }{TEXT 
-1 127 "), so its length can be computed from the distance formula.\nA
dding these lengths---one for each subinterval---gives the formula" }}
}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Sum(sqrt((f(t_j)-f(t_(j-1)))^
2+(g(t_j)-g(t_(j-1)))^2),j = 1 .. n);" "6#-%$SumG6$-%%sqrtG6#,&*$,&-%
\"fG6#%$t_jG\"\"\"-F-6#-%#t_G6#,&%\"jGF0F0!\"\"F8\"\"#F0*$,&-%\"gG6#F/
F0-F=6#-F46#,&F7F0F0F8F8F9F0/F7;F0%\"nG" }{TEXT -1 2 " ." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 67 "The following procedure computes this for
mula, given the functions " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 5 
" and " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 25 ", the parameter in
terval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 33 ", and
 the number of subintervals." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 217 "approxlen := proc(f,g,a,b,n)\n               local l;\n        \+
       l := (b-a)/n;\n               sum( sqrt( (f(a+j*l) - f(a+(j-1)*
l))^2\n                 + (g(a+j*l) - g(a+(j-1)*l))^2 ), j=1..n);\n   \+
          end proc:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "As always,
 " }{TEXT 258 5 "Maple" }{TEXT -1 114 " will evaluate the answer exact
ly, including square roots, so it is convenient to ask for a decimal a
pproximation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(appr
oxlen(f,g,-2,2,5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+**f&3X\"!\"
)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "evalf(approxlen(f,g,-2
,2,10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+91D/:!\")" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "evalf(approxlen(f,g,-2,2,20));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+95!o^\"!\")" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 30 "evalf(approxlen(f,g,-2,2,50));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"+l:E?:!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "evalf(approxlen(f,g,-2,2,100));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+yhv?:!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 200 "
We see a familiar pattern: as the number of subintervals increases, th
e lengths of the piecewise-linear approximations approach a fixed valu
e, which is obviously the true length of the original curve." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" 
{TEXT -1 10 "Question 1" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "Use pi
ecewise-linear approximations to find the length of the loop of the cu
rve above, accurate to within 0.01.  (Hint: what are the two values of
 " }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 35 " at the self-intersecti
on point?)\n\n" }{TEXT 263 9 "Solution." }{TEXT -1 24 "  The curve cro
sses the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 11 "-axis when " }
{XPPEDIT 18 0 "t = -sqrt(3);" "6#/%\"tG,$-%%sqrtG6#\"\"$!\"\"" }{TEXT 
-1 16 " and again when " }{XPPEDIT 18 0 "t = sqrt(3);" "6#/%\"tG-%%sqr
tG6#\"\"$" }{TEXT -1 18 ", both times when " }{XPPEDIT 18 0 "x = 3;" "
6#/%\"xG\"\"$" }{TEXT -1 54 ".\nThe length of the curve between these \+
two values of " }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 27 " can be ap
proximated using " }{TEXT 264 9 "approxlen" }{TEXT -1 1 ":" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "evalf(approxlen(f,g,-sqrt(3),sqrt(3
),5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+n7DI5!\")" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "evalf(approxlen(f,g,-sqrt(3),sqrt(3
),10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\\a[h5!\")" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "evalf(approxlen(f,g,-sqrt(3),sqrt(3
),50));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+N^-t5!\")" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "evalf(approxlen(f,g,-sqrt(3),sqrt(3
),100));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+KeQt5!\")" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 198 "Although we should probably try a larger
 number of subintervals to be sure that the second decimal place will \+
remain stable, we can probably assume that the length is 10.73, accura
te to within 0.01." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" 
{TEXT -1 10 "Question 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Plot th
e polar curve " }{XPPEDIT 18 0 "r = theta;" "6#/%\"rG%&thetaG" }{TEXT 
-1 93 " .  By writing it as a parametric curve, find the length of the
 portion of the curve between " }{XPPEDIT 18 0 "theta = 0;" "6#/%&thet
aG\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "theta = 2*Pi;" "6#/%&thet
aG*&\"\"#\"\"\"%#PiGF'" }{TEXT -1 3 ".\n\n" }{TEXT 265 9 "Solution." }
{TEXT -1 44 "  As a polar curve, our curve has equations " }{XPPEDIT 
18 0 "x = theta*cos(theta);" "6#/%\"xG*&%&thetaG\"\"\"-%$cosG6#F&F'" }
{TEXT -1 2 ", " }{XPPEDIT 18 0 "y = theta*sin(theta);" "6#/%\"yG*&%&th
etaG\"\"\"-%$sinG6#F&F'" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "f1 := theta-> theta*cos(theta) ;" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1GR6#%&theta
G6\"6$%)operatorG%&arrowGF(*&9$\"\"\"-%$cosG6#F-F.F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "g1 := theta-> theta*sin(theta) ;" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#g1GR6#%&thetaG6\"6$%)operatorG%&arrowGF(*&9$\"\"\"-%$sinG6#F-F.F(F
(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot([f1(theta), g1(
theta), theta=0..2*Pi]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
{PLOTDATA 2 "6%-%'CURVESG6$7en7$$\"\"!F)F(7$$\"3\"f&f3#*)G<$o!#>$\"3It
C!fNUbo%!#?7$$\"3%zshZRJnN\"!#=$\"3[ewJkV#)p=F-7$$\"3*HW$>L2aF>F4$\"3f
0k<hv)y$QF-7$$\"3e[)QA#GlxCF4$\"3G`dC#4c#)['F-7$$\"3N];'p'[<3OF4$\"3U)
*eoFsr$[\"F47$$\"3Bv;DAJ:VXF4$\"3,&4Nb\\*oJEF47$$\"3v0(QF'>A6_F4$\"3H%
HWmze&QSF47$$\"3W<yGmi5^bF4$\"3iu\"4A!\\0LbF47$$\"39JIRn%GAe&F4$\"3-%H
JXqB.A(F47$$\"35?:!R5L5C&F4$\"3M=94+tq^!*F47$$\"3'Q/()=vjK]%F4$\"3u)y;
^&eT*3\"!#<7$$\"3iR=K>(G@K$F4$\"3E/HfnR&GF\"F^o7$$\"3\"Q;;5`Y9$>F4$\"3
_#Q7/1CGU\"F^o7$$!3_+$Qd$)**GQ*!#@$\"3q4!yO:$Rr:F^o7$$!3-%po*)H(REBF4$
\"3u=#4$[qa\"p\"F^o7$$!3e,+4g>Gl[F4$\"37]^JqS1t<F^o7$$!3Y9wAQpc'Q(F4$
\"31fjFez)H\"=F^o7$$!3wv4b&GF&e5F^o$\"3)z;8.;zG\"=F^o7$$!3MRReg$4.M\"F
^o$\"3gr?\">d(yo<F^o7$$!33ExXVF@s;F^o$\"39jv^bFfj;F^o7$$!3#o%R+sP0i>F^
o$\"3WE)**3:k0_\"F^o7$$!3'[!3Frr$oE#F^o$\"3@'oc=lJ$48F^o7$$!3q0DT%3&fN
DF^o$\"3Q\\PJ0(oj0\"F^o7$$!3[d0*=#)eXy#F^o$\"3.MDB\\*H2T(F47$$!3!)>?jB
$)=xHF^o$\"3E-8_\"*)\\#)3%F47$$!3\"*>?&fCw%QJF^o$\"34#p`l!GVL(*F07$$!3
KC+S%Q9qC$F^o$!30=S1<$3WU%F47$$!3K*GQLtooG$F^o$!3y3#[DT1Pg)F47$$!37;x>
n\"QzE$F^o$!3t\\nAE#>!H8F^o7$$!3[*[K6TKo<$F^o$!3W`ga$yKB#=F^o7$$!3\\Yz
KR9w9IF^o$!3??9u)*yW.BF^o7$$!3i&3K;mH$)y#F^o$!3%e')z9:lsv#F^o7$$!3E/s#
)y+UdCF^o$!3::A@W;MNKF^o7$$!3Wf!QI6J54#F^o$!3Zb([JHz2j$F^o7$$!3\\_?on#
H>j\"F^o$!3E*pneawe+%F^o7$$!3k'yg9'oXg6F^o$!3_?DU>/W%H%F^o7$$!3s$>wYE$
3>fF4$!3>\"GtOG&[WXF^o7$$!3XzmLPK#)R9F-$!3wL@o\"e4$4ZF^o7$$\"3YqIMIx`M
iF4$!3P6v>x-@,[F^o7$$\"3=#zURC\\+F\"F^o$!3]A!*G.&od![F^o7$$\"3'fZ%*G5
\"ee>F^o$!31]EKeW[:ZF^o7$$\"3+AQAhF\"*>EF^o$!3W,Z')G-$Q`%F^o7$$\"3a%>(
R*=&\\!G$F^o$!3a.qk&oK5D%F^o7$$\"3B!Q5$eW)e!RF^o$!3_p@ujivuQF^o7$$\"3%
RZg$*yP%zTF^o$!3afl#y,m1n$F^o7$$\"3xHF5/(\\CW%F^o$!3sbJmL(4vW$F^o7$$\"
3$3$[\"))zK#HZF^o$!3wr;19o**oJF^o7$$\"3yl(*HU!4&)*\\F^o$!3)e'Hn0F5nGF^
o7$$\"3EIK7em)HA&F^o$!3OX.w/x<yDF^o7$$\"3H+x@eJmIaF^o$!33R)p8zBBF#F^o7
$$\"3.KQ(>#\\=KcF^o$!3[w+:\\GoG>F^o7$$\"3Dqb#\\#=w6eF^o$!3eA)[XHB#o:F^
o7$$\"3=n`*ogR='fF^o$!3N9W[N9x37F^o7$$\"3Idki$et%*3'F^o$!3)31[w'*4aO)F
47$$\"3S:*)eS;X+iF^o$!3,ZWvco-RUF47$$\"3?+++3`=$G'F^o$\"3qF=n%R6[:&!#E
-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q!6\"F\\^l-%%VIEWG6$%(
DEFAULTGFa^l" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 
"Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf(approxle
n(f1,g1,0,2*Pi,50));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+<D$Q7#!\")
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "evalf(approxlen(f1,g1,0
,2*Pi,100));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+!3!=D@!\")" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "evalf(approxlen(f1,g1,0,2*Pi
,500));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+R9hD@!\")" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 58 "The length of the curve appears to be app
roximately 21.25." }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "With the power of " }{TEXT 259 5 "
Maple" }{TEXT -1 26 ", the numerical procedure " }{TEXT 260 9 "approxl
en" }{TEXT -1 425 " is all we need to find arclengths as accurately as
 we want.  Very often a numerical procedure of this type is the only p
ractical way of finding arclength, but it is possible to write down an
 integral which gives the exact length of the curve.  (In practice, th
e integral will have to be evaluated numerically for most curves of in
terest.)  As explained in class, an argument with the Mean Value Theor
em leads to the integral" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "I
nt(sqrt(diff(f,t)^2+diff(g,t)^2),t = a .. b);" "6#-%$IntG6$-%%sqrtG6#,
&*$-%%diffG6$%\"fG%\"tG\"\"#\"\"\"*$-F,6$%\"gGF/F0F1/F/;%\"aG%\"bG" }
{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "The following \+
procedure returns this integral.  Since many arclength integrals canno
t be evaluated synbolically, " }{TEXT 261 9 "arclength" }{TEXT -1 39 "
 returns the integral unevaluated; use " }{TEXT 262 5 "evalf" }{TEXT 
-1 45 " to get a decimal approximation to its value." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 117 "arclength := proc(f,g,a,b) \n           \+
    Int(sqrt(diff(f(t),t)^2 + diff(g(t),t)^2), t=a..b);\n             \+
end proc:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 153 "For the example we \+
have been using, we can confirm that the integral formula agrees with \+
the arclength we found from the piecewise-linear approximations:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f := t-> t^2 ; g := t-> t^3 \+
- 3*t ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"tG6\"6$%)operat
orG%&arrowGF(*$)9$\"\"#\"\"\"F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"gGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"$\"\"\"F1*&F0F1F/F
1!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(arcle
ngth(f,g,-2,2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+j5#4_\"!\")" }
}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 10 "Question 3" }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 131 "Confirm that the integral formula agrees with the
 results you obtained with piecewise-linear approximations in Question
s 1 and 2.\n\n" }{TEXT 266 9 "Solution." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "evalf(arclength(f,g,-sqrt(3),sqrt(3)));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+og]t5!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "evalf(arclength(f1,g1,0,2*Pi));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+:%Hc7#!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 224 
"These two values agree with the approximations obtained in Questions \+
1 and 2, at least in the first few decimal places.  If we were to use \+
more subintervals in Questions 1 and 2, we would of course get better \+
approximations." }}}}}{MARK "8 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 
1 }{PAGENUMBERS 0 1 2 33 1 1 }