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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 21 "Numerical Integration" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 525 "Even if it is not possible to find an an
tiderivative of a given function in terms of elementary functions, it \+
usually is possible to approximate values of definite integrals by the
 fact that definite integrals correspond to area of certain regions.  \+
These area can be approximated by covering them with a large number of
 small rectangles, for instance.  This is, in fact, what is being done
 in numerical integration.  Provided that the function to be integrate
d is not too bad, this simple observation can give good results. " }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "Left approximations" }}{PARA 0 "
" 0 "" {TEXT -1 89 "We need rountines that are containedi n the studen
t -package.  So we start by loading it." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "with(student);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Consider the integral \+
" }{XPPEDIT 18 0 "Int(sin(sin(x))/x,x = 0 .. 1);" "6#-%$IntG6$*&-%$sin
G6#-F(6#%\"xG\"\"\"F,!\"\"/F,;\"\"!F-" }{TEXT -1 98 ".  This integral \+
cannot be expressed in terms of elementary functions.   Let us try to \+
compute it:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "f:=unapply(s
in(sin(x))/x,x):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Int(f(x
),x=1..2): % = value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Thew graph of the function in
 question is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(f(x),x
=1..2,y=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "and the value of the integral is t
he area below the graph and above the interval [0,1]." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 440 "Following the idea gi
ven above, we divide the interval [0,1] into 4 subintervals [0,1/4], [
1/4,2/4], [2/4,3/4], [3/4,1], and we approximate the area of the domai
n in question by the sum of the areas of the rectangles having the int
evals [j/4,(j+1)/4], j= 0 .. 4, as their base, and having the height c
orresponding to the value of the function in the left end point of eac
h interval.  This approximation of the above domain looks as follows:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "leftbox(f(x),x=1..2,4);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 280 "The areas of the rectangles in the picture are easy
 to compute, andt his gives already a rather good estimate of the area
 of the domain bounded by the graph of the original function.  This ca
n be improved by taking more division points.  For 20 points we get th
e following picture" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "left
box(f(x),x=1..2,20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "This way of approximating the are
 corresponding to a given definite integral is called " }{TEXT 256 18 
"left approximation" }{TEXT -1 235 ".  Clearly, when increasing the nu
mber of division points, the approximation gets better.  The sum of th
e values of the areas of the corresponding rectangles is an approximat
ion of the integral.  It can readily be computed by Maple:   " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "leftsum(f(x),x=1..2,4);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 102 "That is the exact formula.  To get a floating point appr
oximation, you must evaluate the above by the " }{MPLTEXT 1 0 6 "evalf
 " }{TEXT -1 10 "command.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
30 "evalf(leftsum(f(x),x=1..2,4));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 131 "To get an idea of the accuracy of our computations, we may com
pue leftsums for several values of the number of the division points:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "for j from 4 to 20 do e
valf(leftsum(f(x),x=1..2,j)); od;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "You observe \+
that, as the number of the division points increase, the approximation
s get smaller.  Also the approximation is not really very good.  " }
{TEXT 257 148 "What is the reason for this phenomena, i.e., why do the
 values of the approximation always get smaller as the number of divis
ion points increases?  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 8 "Problems" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 177 "P
lot left approximations for the following integrals.  Discuss the accu
racy of the approximation.  What happens to the approximations as the \+
number of division points increases?" }}{PARA 4 "" 0 "" {TEXT -1 0 "" 
}}{SECT 1 {PARA 257 "" 0 "" {XPPEDIT 18 0 "Int(exp(x^2),x=0..1)" "6#-%
$IntG6$-%$expG6#*$%\"xG\"\"#/F*;\"\"!\"\"\"" }{TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{SECT 1 {PARA 258 "" 0 "" {XPPEDIT 18 0 "Int(exp(-x^2),x=0..1)" "6#-%$
IntG6$-%$expG6#,$*$%\"xG\"\"#!\"\"/F+;\"\"!\"\"\"" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 
259 "" 0 "" {XPPEDIT 18 0 "Int(sin(x^3),x=0..2);" "6#-%$IntG6$-%$sinG6
#*$%\"xG\"\"$/F*;\"\"!\"\"#" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 91 "Give \+
an example of a function  f  for which the left-point approximations o
f the integral  " }{XPPEDIT 18 0 "Int(f(x),x=0..1)" "6#-%$IntG6$-%\"fG
6#%\"xG/F);\"\"!\"\"\"" }{TEXT -1 6 " give " }}{SECT 1 {PARA 265 "" 0 
"" {TEXT -1 22 "always the exact value" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 
{PARA 266 "" 0 "" {TEXT -1 68 "always a value that is smaller than the
 actual value of the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 267 "" 0 "" {TEXT -1 67 "
always a value that is larger than the actual value of the integral" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 20 "Right a
pproximations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 232 "We can replace the left-end point by the right end-point
 in the aboe left approximations to get right approximations of the do
main and the value of the inregral.  Look at our original example:  ap
proximate the value of the integral " }{XPPEDIT 18 0 "int(sin(sin(x))/
x,x=1..2)" "6#-%$intG6$*&-%$sinG6#-F(6#%\"xG\"\"\"F,!\"\"/F,;F-\"\"#" 
}{TEXT -1 46 ".     The right approximaitons look as follows" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "rightbox(f(x),x=1..2,4);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 105 "Here you can increase the number of division points by c
hanging the last argument of the above command.  " }{TEXT 258 100 "Wha
t happens to the value of the right approximation as the number of div
ision points increases?    " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 69 "The values of the approximations can again be c
omputed by the command" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "r
ightsum(sin(sin(x))/x,x=1..2,10);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Evaluation gi
ves" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "evalf(rightsum(sin(s
in(x))/x,x=1..2,10));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Inreasing the number of division
 points we get" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "for j fro
m 10 to 20 do evalf(rightsum(sin(sin(x))/x,x=1..2,j));od;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Problems" }}{SECT 1 {PARA 4 "" 
0 "" {TEXT -1 54 "Give examples and plot pictures of functions for whi
ch" }}{SECT 1 {PARA 260 "" 0 "" {TEXT -1 106 "the right sum gives alwa
ys an approximation of the integral that is smaller than the actual va
lue integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{SECT 1 {PARA 261 "" 0 "" {TEXT -1 112 "the right sum gives
 always an approximation of the integral that is larger than the actua
l value of the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 66 "Give examples of functions for which the left approxima
tion always" }}{SECT 1 {PARA 262 "" 0 "" {TEXT -1 38 "is larger than t
he right approximation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 263 "" 0 
"" {TEXT -1 39 "is smaller than the right approximation" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 264 
"" 0 "" {TEXT -1 35 "is equal to the right approximation" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 106 "How many division poin
ts one must take so that the error of the left point approximation of \+
the integral  " }{XPPEDIT 18 0 "int(exp(-x^2),x=0..1)" "6#-%$intG6$-%$
expG6#,$*$%\"xG\"\"#!\"\"/F+;\"\"!\"\"\"" }{TEXT -1 92 " is at most 0.
01?\nExplain you reasoning and illustrate the situation with appropria
te plots." }}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 4 "" 0 "" 
{TEXT -1 2 "  " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 22 "Middle  approximations" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 223 "Instead of considering the end
points of the subintervals in question, one can take the value of the \+
function at the midpoints as the height of the approximating rectangle
.  One usually gets better approximations that way.  " }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Consider the integral \+
" }{XPPEDIT 18 0 "int(exp(-x^2),x=0..1)" "6#-%$intG6$-%$expG6#,$*$%\"x
G\"\"#!\"\"/F+;\"\"!\"\"\"" }{TEXT -1 159 ".   In the problem above on
e finds out that to ensure  0.01 accuracy for the approximation (using
 left or right sums) one has to take over 60 dividion points. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The mid p
oint approximation looks as follows" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 
"with(student):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "g:=unapp
ly(exp(-x^2),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "middleb
ox(g(x),x=0..1,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "Just by looking at the correspond
ing pictures of the left and right approximations, one sees immediatel
y that the mid approximation gives clearly better results in this case
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "We \+
may verify this  by numerical computations.  The following line comput
es floating point values for the left- , middle- and the rightsum resp
ectively.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 9 "Problem. " }{TEXT -1 203 " By c
hanging the number of division points in the left- and rightsums, esti
mate how many points you must choose to get the same accuracy as what \+
you get with the middle sum using just 10 division points." }}{PARA 0 
"" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "L
eftsum=evalf(leftsum(g(x),x=0..1,10)); Middlesum=evalf(middlesum(g(x),
x=0..1,10));Rightsum=evalf(rightsum(g(x),x=0..1,10));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Problems" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 116 "Give an exa
mple of a function for which the left sum gives a better approximation
 of the integral as the middle sum." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 
117 "Give an example of a function for which the right sum gives a bet
ter approximation of the integral as the middle sum." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Trapezo
id approximations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 258 "In the above we have considered three different ways to \+
approximate a definite integral.  They all are based on the various wa
ys to approximate a domain by rectangles.  The three different ways we
re the  following.   Here we apply the method to the function " }
{XPPEDIT 18 0 "f(x) = exp(x^2);" "6#/-%\"fG6#%\"xG-%$expG6#*$F'\"\"#" 
}{TEXT -1 313 " in the interval  [1,2], and cover that interval with o
ne box only.    The purpose of this example is to illustrate the error
s involved in the numerical approximation.  In the numerical approxima
tion, we approximate the area under the red curve by the area of the r
ectangles indicated in the respective pictures. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{GLPLOT2D 177 154 154 
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}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 191 "We conclude that the left approximation is pro
bably worst of all the three approximations.  But in this case, all th
e approximations give rather bad results.  That can be checked  as fol
lows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 
"with(student):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "Next set n to
 be the number of intervals into which the original interval of integr
ation is to be divided." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "n
:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "D
efine the function to be integrated numerically:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 23 "f:=unapply(exp(x^2),x):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "Next
 let Maple compute a numerical approximation of the integral Int(exp(x
^2),x) and compare that to the results of the left, right and  middle \+
approximations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 176 "Int(f(x),x=1..2): % = evalf
(value(%)); `Left sum` = evalf(leftsum(f(x),x=1..2,n));`Right sum` = e
valf(rightsum(f(x),x=1..2,n)); `Middle sum` = evalf(middlesum(f(x),x=1
..2,n)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$expG6#*$)%\"
xG\"\"#\"\"\"/F,;F.F-$\"+,w**)\\\"!\")" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/%)Left~sumG$\"+G=G=F!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%*
Right~sumG$\"+.]\")fa!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%+Middle
~sumG$\"+Oet([*!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 253 "Clearly the errors are rather lar
ge as was to be expected.  That can be improved if we use, instead of \+
a rectangular approximation, a \"trapezoidal\" approximation.  In this
 method we approximate the domain in question by the polygon of the pi
cture below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }{GLPLOT2D 155 142 142 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$\"\"\"
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-$\"1hprXqtrLFes7$$\"1+++&Qk\\*=F-$\"1r;$\\$RjEOFes7$$\"1LL3dg6<>F-$\"
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!*eCYFes7$$\"1+](=5s#y>F-$\"1&Qn4V$p2]Fes7$$\"\"#F*$\"1CWJ.]\")faFes-%
'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%)POLYGONSG6#7&7$F(F*7$F($\"+G=G=F!\"*7
$Fez$\"+.]\")fa!\")7$FezF*-%+AXESLABELSG6$Q\"x6\"%!G-%&TITLEG6#Q8Trape
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4 2 1.000000 45.000000 46.000000 0 0 "" "" }}}{PARA 0 "" 0 "" {TEXT 
-1 170 "It is clear that, in this case, the area of the polygon is lar
ger than the area under the red curve, but the approxiation is better \+
than the right or left approximations." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 21 "It is immediate that " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 268 "" 0 "" {TEXT 260 69 "Trapzoid approxim
ation = (left approximation + right approximation)/2" }{TEXT -1 1 "." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 40 "Let us compare the above approximations:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "n:=10;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"nG\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 223 "Int(f(x),x=1..2): % = evalf(value(%)); `Left sum` = evalf(lefts
um(f(x),x=1..2,n));`Right sum` = evalf(rightsum(f(x),x=1..2,n)); `Midd
le sum` = evalf(middlesum(f(x),x=1..2,n));  Trapezoidal = evalf(trapez
oid(f(x),x=1..2,n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$e
xpG6#*$)%\"xG\"\"#\"\"\"/F,;F.F-$\"+,w**)\\\"!\")" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%)Left~sumG$\"+u!zsD\"!\")" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%*Right~sumG$\"+cx2w<!\")" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%+Middle~sumG$\"+D<=!\\\"!\")" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%,TrapezoidalG$\"+:%ym^\"!\")" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "Left an
d right sums gave miserable approximations while the middle sum and th
e trapezoidal approximations were much better.  All the above methods \+
yield better results, when one increases the value of  " }{XPPEDIT 18 
0 "n" "6#%\"nG" }{TEXT -1 150 " but in order to get, by the left and t
he right sums,  results comparable to the middle sum or to the trapezo
idal approximation, one has to increase  " }{XPPEDIT 18 0 "n" "6#%\"nG
" }{TEXT -1 6 "  to  " }{XPPEDIT 18 0 "10*n" "6#*&\"#5\"\"\"%\"nGF%" }
{TEXT -1 62 " as one can check by performing the computations in the a
bove." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 89 "The trapezoid approximation picture in the abov
e was produced by the following procedure:" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "trapezoidgraph :=
 proc(f::symbol, a::numeric, b::numeric, n::posint)" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 30 "local Polygon,PLOT1, PLOT2, i;" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "Polygon := [[a,0],[a,f(a)]];" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "for i to n do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "P
olygon := [op(Polygon),[a+i*(b-a)/n,f(a+i*(b-a)/n)]]:" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Polygon
 := [op(Polygon),[b,0]]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(pl
ots):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "PLOT1 := polygonplot(Polyg
on):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "PLOT2 := plot(f(x),x=a..b):
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "display(\{PLOT1,PLOT2\},title=
\"Trapezoid approximation\");" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "trapezoidgraph(f,1,2,1
);" }}{PARA 13 "" 1 "" {GLPLOT2D 187 155 155 {PLOTDATA 2 "6'-%'CURVESG
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.A@W$*))F-7$$\"1nmT5k]*\\\"F-$\"1l2M)R*pt%*F-7$$\"1n;zRQb@:F-$\"1i\"GC
'*>E,\"!#97$$\"1+](=>Y2a\"F-$\"1UJ[)4=R2\"Fes7$$\"1nm\"zXu9c\"F-$\"1uk
yFOCX6Fes7$$\"1+++&y))Ge\"F-$\"1TlrOM,D7Fes7$$\"1++DE&QQg\"F-$\"1U7/Z%
)e48Fes7$$\"1+]7y%3Ti\"F-$\"1&zGDmS\")R\"Fes7$$\"1++v.[hY;F-$\"1\\Ex$H
V\\]\"Fes7$$\"1LLLQx$om\"F-$\"1M;x/:C4;Fes7$$\"1++]P+V)o\"F-$\"1lonC[:
I<Fes7$$\"1n;zpe*zq\"F-$\"1gqH&pO!\\=Fes7$$\"1++]#\\'QH<F-$\"1=;nT_6!*
>Fes7$$\"1L$e9S8&\\<F-$\"1**G!H]cW8#Fes7$$\"1+]i?=bq<F-$\"1D;)))RF&)H#
Fes7$$\"1LL$3s?6z\"F-$\"1Imb(HQKZ#Fes7$$\"1+]7`Wl7=F-$\"1i:SC>\"Gn#Fes
7$$\"1nmm'*RRL=F-$\"1pM(G#3w#)GFes7$$\"1nmTvJga=F-$\"1K8))y=I<JFes7$$
\"1L$e9tOc(=F-$\"1hprXqtrLFes7$$\"1+++&Qk\\*=F-$\"1r;$\\$RjEOFes7$$\"1
LL3dg6<>F-$\"1*H,vR$=YRFes7$$\"1nmmw(Gp$>F-$\"19r@#[8$fUFes7$$\"1+]7oK
0e>F-$\"1n?AE!*eCYFes7$$\"1+](=5s#y>F-$\"1&Qn4V$p2]Fes7$$\"\"#F*$\"1CW
J.]\")faFes-%'COLOURG6&%$RGBG$\"#5!\"\"F*F*-%)POLYGONSG6#7&7$F(F*7$F($
\"+G=G=F!\"*7$Fez$\"+.]\")fa!\")7$FezF*-%+AXESLABELSG6$Q\"x6\"%!G-%&TI
TLEG6#Q8Trapezoid~approximationFb\\l-%%VIEWG6$;F(Fez%(DEFAULTG" 1 2 0 
1 10 0 2 9 1 4 2 1.000000 45.000000 46.000000 0 0 "" "" }}}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Problems" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "C
onsider  " }{XPPEDIT 18 0 "Int(exp(x^2),x=1..2)" "6#-%$IntG6$-%$expG6#
*$%\"xG\"\"#/F*;\"\"\"F+" }{TEXT -1 210 ".  Explain why the trapezoid \+
approximation for this integral always gives upper estimates for the v
alue of the integral while the middle approximation always gives lower
 estimates (lower than the actual value)." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 
-1 9 "Estimate " }{XPPEDIT 18 0 "Int(1/(1+x^2),x=0..1)" "6#-%$IntG6$*&
\"\"\"F',&F'F'*$%\"xG\"\"#F'!\"\"/F*;\"\"!F'" }{TEXT -1 179 " using th
e left, right, middle and trapezoid approximations with varying number
 of intervals into which the interval [0,1] is to be divided.  The exa
ct value of this integral is  " }{XPPEDIT 18 0 "Pi/4" "6#*&%#PiG\"\"\"
\"\"%!\"\"" }{TEXT -1 93 ".  Hence by the above approximations one get
s an approximation for the value of the constant " }{XPPEDIT 18 0 "Pi
" "6#%#PiG" }{TEXT -1 34 ".  Is the value thus obtained for " }
{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 79 " by the left approximation \+
too large or too small?  Explain your reasoning.    " }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 144 "Give examples of func
tions  f  such that when approximating the integral  Int(f(x),x=0..1) \+
by dividing the interval [0,1] into two subintervals " }}{SECT 1 
{PARA 5 "" 0 "" {TEXT 261 66 "the left approximation is better than th
e trapezoid approximation " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 5 "" 0 "" 
{TEXT 262 66 "the right approximation is better than the trapezoid app
roximation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 5 "" 0 "
" {TEXT 263 68 "the trapezoid approximation is better than the middle \+
approximation " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 269 "
" 0 "" {TEXT -1 67 "the middle approximation is better than the trapez
oid approximation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 14 "Simpson's rule" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 15 "Simpson's rule:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 264 0 "Simpso
n*approximation = (2*`Middle approximation`+Trapezoid*approximation)/3
;" "6#/*&%(SimpsonG\"\"\"%.approximationGF&*&,&*&\"\"#F&%5Middle~appro
ximationGF&F&*&%*TrapezoidGF&F'F&F&F&\"\"$!\"\"" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 71 "usually gives better approximations than the above described me
thods.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
128 "Let us experiment with these approximations, and compare errors. \+
 We perform the same computations as in the text (on page 353)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student
):" }}{PARA 7 "" 1 "" {TEXT -1 29 "Warning, new definition for D" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=unapply(1/x,x);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&\"\"\"F&f*6#%\"xG6\"6$%)operatorG%&a
rrowGF*9$F*F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "n:=
2;Digits:=10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"#" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"#5" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "`Maple's value`:=evalf(int(1/x,x=1..2)):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 100 "for j from 1 to 3 do `Trapezoid error`||j:=`Maple's \+
value` - evalf(trapezoid(f(x),x=1..2,j*50)): od:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 105 "for j from 1 to 2 do `Ratio of trapezoid err
ors`||j := `Trapezoid error`||(j+1)/`Trapezoid error`||j; od;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%;Ratio~of~trapezoid~errors1G$\"+/+3+
D!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%;Ratio~of~trapezoid~errors2G
$\"+.6\"RW%!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "`Maple's \+
value`;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"_qvo=k*pKjeg:Zpp>i$R$\\4
+o?TDb-OM,]v!ol<e97KsT4`%*f0=ZJp!$+\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0 0" 7 }
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