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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 24 "MAPLE Worksheet Number 8" }}
{PARA 256 "" 0 "" {TEXT -1 16 "Integral as Area" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "We recall that       " }
{XPPEDIT 18 0 " int(f,x=a..b)" "6#-%$intG6$%\"fG/%\"xG;%\"aG%\"bG" }
{TEXT -1 382 "   is called the \"definite integral of f from a to b\" \+
and is the \"net area\" between the graph of f and the x-axis between \+
a and b, in other words the area above the x-axis minus the area below
 the x-axis.  If the graph of f is made up of straight lines only, we \+
can compute this definite integral using the formulae for the areas of
 triangles and rectangles.  For example compute   " }{XPPEDIT 18 0 "in
t(f,x=-4..2)" "6#-%$intG6$%\"fG/%\"xG;,$\"\"%!\"\"\"\"#" }{TEXT -1 22 
"    for the function  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "f
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-%*PIECEWISEG6$7$,&%\"xG\"\"\"F+F+1F*!\"\"7$,&F*F-\"\"%F+%*otherwiseG
" }}}{PARA 0 "" 0 "" {TEXT -1 64 "First plot the graph and then use th
e appropriate area formulae." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "plot(f,x=-5..5,discont=true);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 
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2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 156 "Use the formula for the area of a triang
le, one half the base time the height for the left triangle.  Subtract
 the area of triangles for the right-hand part" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "
(1/2)*(-1-(-4))*subs(x=-4,x+1)+ (1/2)*(4-(-1))*subs(x=-1, -x+4)-(1/2)*
(4-2)*subs(x=2, -x+4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 15 "The area equals" }{TEXT 
257 23 "                       " }{TEXT 258 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 147 "The MAPLE command syntax for computing the definite inte
gral of a function g with respect to the variable var between the valu
es var=a and var=b is" }}{PARA 256 "" 0 "" {TEXT -1 16 "int(g,var=a..b
);" }}{PARA 0 "" 0 "" {TEXT -1 18 "Try it to compute " }{XPPEDIT 18 0 
" int(f,x=-4..2)" "6#-%$intG6$%\"fG/%\"xG;,$\"\"%!\"\"\"\"#" }{TEXT 
-1 123 "  for our f defined above. Notice that x is the integration va
riable. (Convert command to math symbolism on your worksheet." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "int(f,x=-4..2);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#\"\"'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 126 "MAPLE has a very nice package designed t
o help students of Calculus visualize such area problems.  It is loade
d by the command" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(stu
dent);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%\"DG%%DiffG%*DoubleintG%$
IntG%&LimitG%(LineintG%(ProductG%$SumG%*TripleintG%*changevarG%/comple
tesquareG%)distanceG%'equateG%*integrandG%*interceptG%)intpartsG%(left
boxG%(leftsumG%)makeprocG%*middleboxG%*middlesumG%)midpointG%(powsubsG
%)rightboxG%)rightsumG%,showtangentG%(simpsonG%&slopeG%(summandG%*trap
ezoidG" }}}{PARA 0 "" 0 "" {TEXT -1 163 "Many of these commands should
 have a familiar ring to them.   Let's try the *box and *sum commands.
  After loading the student package, try the following commands." }}
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++BFg_l7$$\"++++g<Fg_lF^jo7$FajoFg^lFi_l-F`_l6$7&Fcjo7$Fajo$\"++++SAFg
_l7$$\"++++?=Fg_lFhjo7$F[[pFg^lFi_l-F`_l6$7&F][p7$F[[p$\"++++!=#Fg_l7$
$\"++++!)=Fg_lFb[p7$Fe[pFg^lFi_l-F`_l6$7&Fg[p7$Fe[p$\"++++?@Fg_l7$$\"+
+++S>Fg_lF\\\\p7$F_\\pFg^lFi_l-F`_l6$7&Fa\\p7$F_\\p$\"++++g?Fg_l7$F^^l
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8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" "C
urve 15" "Curve 16" "Curve 17" "Curve 18" "Curve 19" "Curve 20" "Curve
 21" "Curve 22" "Curve 23" "Curve 24" "Curve 25" "Curve 26" "Curve 27
" "Curve 28" "Curve 29" "Curve 30" "Curve 31" "Curve 32" "Curve 33" "C
urve 34" "Curve 35" "Curve 36" "Curve 37" "Curve 38" "Curve 39" "Curve
 40" "Curve 41" "Curve 42" "Curve 43" "Curve 44" "Curve 45" "Curve 46
" "Curve 47" "Curve 48" "Curve 49" "Curve 50" "Curve 51" "Curve 52" "C
urve 53" "Curve 54" "Curve 55" "Curve 56" "Curve 57" "Curve 58" "Curve
 59" "Curve 60" "Curve 61" "Curve 62" "Curve 63" "Curve 64" "Curve 65
" "Curve 66" "Curve 67" "Curve 68" "Curve 69" "Curve 70" "Curve 71" "C
urve 72" "Curve 73" "Curve 74" "Curve 75" "Curve 76" "Curve 77" "Curve
 78" "Curve 79" "Curve 80" "Curve 81" "Curve 82" "Curve 83" "Curve 84
" "Curve 85" "Curve 86" "Curve 87" "Curve 88" "Curve 89" "Curve 90" "C
urve 91" "Curve 92" "Curve 93" "Curve 94" "Curve 95" "Curve 96" "Curve
 97" "Curve 98" "Curve 99" "Curve 100" "Curve 101" }}}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 53 "leftsum(f,x=-4..2,100)=value(leftsum(f,x=-4
..2,100));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-%$SumG6$-%*PIECEWISE
G6$7$,&!\"$\"\"\"*&#\"\"$\"#]F.%\"jGF.F.1,&!\"%F.*&F0F.F3F.F.!\"\"7$,&
\"\")F.*&#F1F2F.F3F.F8%*otherwiseG/F3;\"\"!\"#**F0#\"#d\"#5" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "Notice how the
 more boxes we use the closer the answer agrees with the actual answer
 above.  Continue with the following command sequences using rightbox.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "rightbox(f,x=-4..4,7);" 
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$\"3C)*****\\=BQ<Fdo7$$\"\"%F*$F*F*-%'COLOURG6&%$RGBG$\"*++++\"!\")F`^
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($!+dG9d=!\"*7$$!+dG9dGFh_lFf_l7$Fj_lF`^l-%&COLORG6&Fd^l$\"\"(!\"\"$\"
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6$7&Fcbl7$Fabl$\"+'G9dG#Fh_l7$$\"+9dG9<Fh_lFhbl7$F[clF`^lF]`l-Fa_l6$7&
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LTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1
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" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rightsum(f,x=-4..2,7)=
value(rightsum(f,x=-4..2,7));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-%
$SumG6$-%*PIECEWISEG6$7$,&!\"$\"\"\"*&#\"\"'\"\"(F.%\"jGF.F.1,&!\"%F.*
&F0F.F3F.F.!\"\"7$,&\"\")F.*&#F1F2F.F3F.F8%*otherwiseG/F3;F.F2F0#\"$!R
\"#\\" }}}{PARA 0 "" 0 "" {TEXT -1 151 "Notice this is a lot closer th
an was leftsum using only 7 boxes.  Let's use the average of the leftb
ox and rightbox idea.  In other words we will use  " }{XPPEDIT 18 0 "(
area of the leftbox + the area of the right box)/2" "6#*&,&**%%areaG\"
\"\"%#ofGF'%$theGF'%(leftboxGF'F'*.F)F'F&F'F(F'F)F'%&rightGF'%$boxGF'F
'F'\"\"#!\"\"" }{TEXT -1 3 ".  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 101 "(leftsum(f,x=-4..2,7)+rightsum(f,x=-4..2,7))/2=value((leftsum
(f,x=-4..2,7)+rightsum(f,x=-4..2,7))/2);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/,&-%$SumG6$-%*PIECEWISEG6$7$,&!\"$\"\"\"*&#\"\"'\"\"(F.%\"jGF.F
.1,&!\"%F.*&F0F.F3F.F.!\"\"7$,&\"\")F.*&#F1F2F.F3F.F8%*otherwiseG/F3;
\"\"!F1#\"\"$F2*&FBF.-F&6$F(/F3;F.F2F.F.#\"$&G\"#\\" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "This is pretty close. \+
 I seem to recall that the average of the leftbox area and the rightbo
x area is the area of the obvious trapezoid. Try the trapezoid command
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "trapezoid(f,x=-4..2,7)=
value(trapezoid(f,x=-4..2,7));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&#
!\"$\"\"(\"\"\"*&#\"\"'F'F(-%$SumG6$-%*PIECEWISEG6$7$,&F&F(*&F*F(%\"jG
F(F(1,&!\"%F(*&F*F(F5F(F(!\"\"7$,&\"\")F(*&#F+F'F(F5F(F:%*otherwiseG/F
5;F(F+F(F(#\"$&G\"#\\" }}}{PARA 0 "" 0 "" {TEXT -1 66 "I guess I remem
bered correctly. Now we try the middlebox approach." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 21 "middlebox(f,x=-4..2);" }}{PARA 13 "" 1 "" 
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KCnuFhp$\"3D+++vcF`KF07$$\"3A(***\\(=n#f()Fhp$\"3G++D\"GtS7$F07$$\"3P+
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\"3N++v)Q?QD\"F0$\"3l***\\7hzhu#F07$$\"3G+++5jyp8F0$\"3s*******o8-j#F0
7$$\"3<++]Ujp-:F0$\"3#)****\\dOI(\\#F07$$\"3++++gEd@;F0$\"3********RtU
yBF07$$\"39++v3'>$[<F0$\"3')***\\7R!o^AF07$$\"37++D6Ejp=F0$\"3()***\\(
)Qn.8#F07$$\"\"#F*F^^l-%'COLOURG6&%$RGBG$\"*++++\"!\")$F*F*Fg^l-%&STYL
EG6#%%LINEG-%*THICKNESSG6#F_^l-%)POLYGONSG6$7&7$F(Fg^l7$F($!++++]A!\"*
7$$!+++++DFg_lFe_l7$Fi_lFg^l-%&COLORG6&Fc^l$\"\"(!\"\"$\"\"*Fa`lF_`l-F
`_l6$7&F[`l7$Fi_l$!+++++v!#57$$Fa`lF*Fh`l7$F\\alFg^lF\\`l-F`_l6$7&F]al
7$F\\al$\"++++]UFg_l7$$\"+++++]Fj`lFbal7$FealFg^lF\\`l-F`_l6$7&Fgal7$F
eal$\"++++]FFg_l7$F^^lF\\bl7$F^^lFg^lF\\`l-%+AXESLABELSG6$Q\"x6\"Q!Fdb
l-%%VIEWG6$;F(F^^l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve
 5" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "middlesum(f,x=-4..2)
=value(middlesum(f,x=-4..2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-%
$SumG6$-%*PIECEWISEG6$7$,&#!\"*\"\"%\"\"\"*&#\"\"$\"\"#F0%\"jGF0F01,&#
!#8F/F0*&F2F0F5F0F0!\"\"7$,&#\"#HF/F0*&#F3F4F0F5F0F;%*otherwiseG/F5;\"
\"!F3F2\"\"'" }}}{PARA 0 "" 0 "" {TEXT -1 208 "Look at the correct pic
ture and say why this \"approximation\" to the definite integral actua
lly gives the exact answer. Apparently middlebox drew the picture inco
rrectly but middlesum computed the correct sum." }}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 121 "If you cut off the triangles that stick up above or
 down below the graph,  you can put them into the holes in the figure.
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 471 "Of \+
course the more interesting area problems occur when the graphs are no
t straight line segments, but are curves. For each of the following fu
nctions find the area of the region in the first quadrant, to the left
 of the line x=10, above the x-axis, and below the graph.  Then use on
e of the \"box\" commands to plot the graph with the indicated area ap
proximated by boxes.  Use the corresponding \"sum\" and evalf commands
 to approximate the area to within 2 decimal places." }}{PARA 0 "" 0 "
" {TEXT -1 7 "a.     " }{XPPEDIT 18 0 "3*sin(2*x)" "6#*&\"\"$\"\"\"-%$
sinG6#*&\"\"#F%%\"xGF%F%" }{TEXT -1 3 "   " }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 16 "f := 3*sin(2*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"fG,$-%$sinG6#,$%\"xG\"\"#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "plot(f, x=0..10);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 
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6$Q\"x6\"Q!Fahq-%%VIEWG6$;F(F_go%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+
4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve
 11" "Curve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17
" "Curve 18" "Curve 19" "Curve 20" "Curve 21" "Curve 22" "Curve 23" "C
urve 24" "Curve 25" "Curve 26" "Curve 27" "Curve 28" "Curve 29" "Curve
 30" "Curve 31" "Curve 32" "Curve 33" "Curve 34" "Curve 35" "Curve 36
" "Curve 37" "Curve 38" "Curve 39" "Curve 40" "Curve 41" "Curve 42" "C
urve 43" "Curve 44" "Curve 45" "Curve 46" "Curve 47" "Curve 48" "Curve
 49" "Curve 50" "Curve 51" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 255 "We can use this to estimate the total possible
 error in using the Taylor polynomial to approximate the values of a f
unction near a point.  For example to compute the error from using the
 3rd degree Taylor polynomial to approximate sin(x) in the interval  \+
" }{XPPEDIT 18 0 "-1/10" "6#,$*&\"\"\"F%\"#5!\"\"F'" }{TEXT -1 5 " to \+
 " }{XPPEDIT 18 0 "1/10" "6#*&\"\"\"F$\"#5!\"\"" }{TEXT -1 26 " we com
pute the integral. " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 " int(abs(sin(x
)-taylor(sin(x),x=0,4)),x=-0.1..0.1)" "6#-%$intG6$-%$absG6#,&-%$sinG6#
%\"xG\"\"\"-%'taylorG6%-F+6#F-/F-\"\"!\"\"%!\"\"/F-;,$-%&FloatG6$F.F7F
7-F<6$F.F7" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 83 "Compute  this error, remember to convert the taylor expression \+
to a polynomial.    " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "merr
or:=sin(x)-convert(taylor(sin(x),x=0,4),polynom);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%'merrorG,(-%$sinG6#%\"xG\"\"\"F)!\"\"*&#F*\"\"'F*)F)
\"\"$F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 58 "Int(abs(merror),x=-0.1..0.1)=int(abs(merror),x=-0.1
..0.1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$absG6#,(-%$sin
G6#%\"xG\"\"\"F.!\"\"*&#F/\"\"'F/)F.\"\"$F/F//F.;$F0F0$F/F0$\"+,=GxF!#
=" }}}{PARA 0 "" 0 "" {TEXT -1 40 "Next try computing the exact error \+
from " }{XPPEDIT 18 0 "-1/10" "6#,$*&\"\"\"F%\"#5!\"\"F'" }{TEXT -1 4 
" to " }{XPPEDIT 18 0 "1/10" "6#*&\"\"\"F$\"#5!\"\"" }{TEXT -1 1 "." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Int(abs(merror),x = -1/10 .
. 1/10)=int(abs(merror),x=-1/10..1/10);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$IntG6$-%$absG6#,(-%$sinG6#%\"xG\"\"\"F.!\"\"*&#F/\"\"'F/)F.
\"\"$F/F//F.;#F0\"#5#F/F9-%$intGF&" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We
ll Duh!  Try evaluating the answer." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "Int(abs(merror),x = -1/10 .. 1/10)=evalf(int(abs(merr
or),x=-1/10..1/10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$a
bsG6#,(-%$sinG6#%\"xG\"\"\"F.!\"\"*&#F/\"\"'F/)F.\"\"$F/F//F.;#F0\"#5#
F/F9$\"+,=GxF!#=" }}}{PARA 0 "" 0 "" {TEXT 259 65 "Explain why this sh
ould be the same as the following calculations" }{XPPEDIT 18 0 "2*int(
error,x=0..h)" "6#%#%?G" }{TEXT -1 94 "   The function inside the abso
lute value signs is odd.  Taking the absolute value produces an" }}
{PARA 0 "" 0 "" {TEXT -1 14 "even function." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "2*Int(merror,x=0..
1/10)=2*int(merror,x=0..1/10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$-
%$IntG6$,(-%$sinG6#%\"xG\"\"\"F,!\"\"*&#F-\"\"'F-)F,\"\"$F-F-/F,;\"\"!
#F-\"#5\"\"#,&-%$cosG6#F7!\"##\"',)Q#\"'++7F-" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 31 "evalf(2*int(merror,x=0..1/10));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"\"#!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 291 "Now we consider some interesting example
s in which MAPLE  shows us some of its limitations.  In each  case plo
t the function and compute the definite integral for x=0 to x=10 using
 the int(f,x=a..b) command.  For which of these does MAPLE give the co
rrect answer and for which does it not.  " }{TEXT 262 23 "Explain your
 reasoning." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {XPPEDIT 19 1 "f[1]:=floor(sin(x));" "6#>&%\"fG6#\"\"\"-%&floorG6#-%
$sinG6#%\"xG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"fG6#\"\"\"-%&floo
rG6#-%$sinG6#%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(
f[1], x=0..10);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 
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54 "Really the same as the last function on this interval." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(f[4], x=0..10);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#-%$intG6$-%$absG6#-%&floorG6#%\"xG/F,;\"\"!\"#5" }
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{PARA 0 "" 0 "" {TEXT -1 286 "The moral of this story is \"When workin
g with piecewise defined functions, be very careful to check the valid
ity of the MAPLE results.\"  We saw other examples in exercise number \+
5 where MAPLE had trouble with piecewise defined functions and floor i
s defined with infinitely many pieces." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 80 "Suppose we want to find the area of t
he finite regions bounded by f and g below." }}{EXCHG {PARA 0 "> " 0 "
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{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 70 "Ploting \+
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"" {TEXT -1 45 "We need to solve for the intersection points." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "fsolve(f-g);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"\"!F$" }}}{PARA 0 "" 0 "" {TEXT -1 246 "This yi
eld only one of the three points, what about the other two.  Looking a
t the graphs we see there should be intersection between x=-1 and x=1,
  between x=1 and x=2, and between x=2 and x=4. We can include bounds \+
on x in the fsolve command.  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 59 "fsolve(f-g,x=-1..1); fsolve(f-g,x=1..2);fsolve(f-g,x=2..4);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"!F$" }}{PARA 11 "" 1 "" {XPPMATH 
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" 1 "" {XPPMATH 20 "6#>%%AreaG#\"#7\"\"&" }}}{PARA 0 "" 0 "" {TEXT -1 
113 "Let's try some \"definite\" integrals with \"indefinite\" limits \+
of integration.  For example compute using MAPLE    " }{XPPEDIT 18 0 "
int(x^3,x=1..a)" "6#-%$intG6$*$%\"xG\"\"$/F';\"\"\"%\"aG" }{TEXT -1 5 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 15 "In fact try    " }
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s result should be suggestive.  We'll return to it at the end of the e
xercise.  But first let's consider some \"improper integrals.\"   In p
articular we'll look at some integrals with one, or both, limits of in
tegration being  " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" }{TEXT 
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ormal distribution\" and its graph is the usual \"bell shaped\" curve \+
in statistics.  The total area under its graph and above the x-axis is
 1,  the area under its graph and between x=-1 and x=1 is the portion \+
of the total population that is 1 \"standard deviation\" form the mean
.  Plot the graph with suitable range on x to generate a nice bell sha
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o_+#Fap7$$\"3qKLLo,\"QD\"F1$\"3m<_l%H*z<=Fap7$$\"35mmm1>qM8F1$\"3U(R**
\\94rj\"Fap7$$\"3%)*******HSu]\"F1$\"3&>W*z?hw!G\"Fap7$$\"3'HLL$ep'Rm
\"F1$\"3;wN3\">QD***FX7$$\"3')******R>4N=F1$\"3SH<6IZ=2uFX7$$\"3#emm;@
2h*>F1$\"3N-!*QjXDTaFX7$$\"3]*****\\c9W;#F1$\"3A#yB)**[\"R$QFX7$$\"3Km
mmmd'*GBF1$\"3mG:[qH-\\EFX7$$\"3j*****\\iN7]#F1$\"3:SLCoCUZ<FX7$$\"3`L
LLt>:nEF1$\"3S,!f%4Y7Q6FX7$$\"35LLL.a#o$GF1$\"3)Q]$=S#4^8(FC7$$\"3ammm
^Q40IF1$\"3z(=^:5\"ekVFC7$$\"3y******z]rfJF1$\"3D5hw7W\"*4FFC7$$\"3gmm
mc%GpL$F1$\"3la,=tr*Q_\"FC7$$\"3/LLL8-V&\\$F1$\"3<#zn%[cVn))F-7$$\"3=+
++XhUkOF1$\"3![))p3H7B%[F-7$$\"3=+++:o<EQF1$\"3Q17-q]8UEF-7$$\"\"%F*F+
-%'COLOURG6&%$RGBG$\"#5!\"\"$F*F*Fa`l-%+AXESLABELSG6$Q\"x6\"Q!Ff`l-%%V
IEWG6$;F(Fh_l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 222 "Then compute
 the probability of being within 1 standard deviation from the mean. (
You will have to use the evalf command to get an answer that makes sen
se. More about this in the next exercise on indefinite integration.)  \+
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "int(p, x=-1..1); evalf(%
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$erfG6#,$*$-%%sqrtG6#\"\"#\"\"
\"#F,F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+?\\*o#o!#5" }}}{PARA 0 
"" 0 "" {TEXT -1 78 "What is the probability of being within 2 standar
d deviations from the mean?  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "int(p, x=-2..2); evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%
$erfG6#*$-%%sqrtG6#\"\"#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+
gt*\\a*!#5" }}}{PARA 0 "" 0 "" {TEXT -1 95 "Finally confirm that the p
robability of being in the population is 1 (ie the total area from   \+
" }{XPPEDIT 18 0 "-infinity" "6#,$%)infinityG!\"\"" }{TEXT -1 10 "    \+
to    " }{XPPEDIT 18 0 "+infinity" "6#%)infinityG" }{TEXT -1 9 "   is \+
1)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "int(p, x=-infinity..i
nfinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "Execute the following M
APLE command sequences and see what you can conjecture." }}{EXCHG 
{PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(int(x^3,x=2..t),t);" "6#-%%diffG
6$-%$intG6$*$%\"xG\"\"$/F*;\"\"#%\"tGF/" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#*$)%\"tG\"\"$\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "
diff(int(x^3,x=-6..t),t);" "6#-%%diffG6$-%$intG6$*$%\"xG\"\"$/F*;,$\"
\"'!\"\"%\"tGF1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*$)%\"tG\"\"$\"\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "diff(int(sin(t),t=a..x),x
);" "6#-%%diffG6$-%$intG6$-%$sinG6#%\"tG/F,;%\"aG%\"xGF0" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#-%$sinG6#%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "diff(int(sin(s^2),s=q..u),u);" "6#-%%diffG6$-%$intG6$-%
$sinG6#*$%\"sG\"\"#/F-;%\"qG%\"uGF2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#-%$sinG6#*$)%\"uG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 
19 1 "diff(int(exp(x^2),x=10..t),t);" "6#-%%diffG6$-%$intG6$-%$expG6#*
$%\"xG\"\"#/F-;\"#5%\"tGF2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$expG6
#*$)%\"tG\"\"#\"\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 64 "The derivative with respect to the upper limit \+
is the integrand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 70 "One version of the FUNDAMENTAL THEOREM OF CALCULUS is the
 following:  " }}{PARA 256 "" 0 "" {TEXT -1 40 "If f is a continuous f
unction then      " }{XPPEDIT 18 0 "diff(int(f(t),t=a..x),x)   =   f(x
)" "6#/-%%diffG6$-%$intG6$-%\"fG6#%\"tG/F-;%\"aG%\"xGF1-F+6#F1" }
{TEXT -1 4 " .  " }}{PARA 0 "" 0 "" {TEXT -1 36 "Notice the result is \+
independent of " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 69 ".  This ca
n be shown to be equivalent to the following other version:" }}{PARA 
256 "" 0 "" {TEXT -1 5 "    ." }}{PARA 256 "" 0 "" {TEXT -1 4 "If  " }
{XPPEDIT 18 0 "diff(F(x),x)=f(x)" "6#/-%%diffG6$-%\"FG6#%\"xGF*-%\"fG6
#F*" }{TEXT -1 16 "     then       " }{XPPEDIT 18 0 "int(f(x),x=a..b)=
(F(b)-F(a)" "6#/-%$intG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG,&-%\"FG6#F.\"\"\"
-F16#F-!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 100 "In this statement  the function  F(x)   is cal
led the (indefinite) integral of    f(x) and denoted  " }{XPPEDIT 18 
0 "int(f(x),x)" "6#-%$intG6$-%\"fG6#%\"xGF)" }{TEXT -1 414 ".         \+
 We will study techniques for computing indefinite integrals, and see \+
how they are used to solve simple initial valued differential equation
s in exercise number 8. In the mean time see if MAPLE knows the genera
l statements of the Fundamental Theorem.  Perform the following MAPLE \+
commands, note that F(x) and a are undefined symbols. In each case con
vert the MAPLE input to the corresponding math symbol." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Diff(int(F(x),x=a..t),t)=diff(int(F
(x),x=a..t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%DiffG6$-%$intG6
$-%\"FG6#%\"xG/F-;%\"aG%\"tGF1-F+6#F1" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 50 "Int(diff(F(x),x),x=a..b)=int(diff(F(x),x),x=a..b);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%%diffG6$-%\"FG6#%\"xGF-/F
-;%\"aG%\"bG,&-F+6#F1\"\"\"-F+6#F0!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 14 "Yup, it knows." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "213 \+
0 0" 14 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }