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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 78 "Lab # 7 Definite Integral
s as area, Riemann Integrals and Antidifferentiation " }}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 29 "1. Definite Integrals as Area" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 442 "In this section you will learn how to us
e Maple to visualize the approximate area under a positive function us
ing boxes (rectangles really).   There is another command to find the \+
summed area of these boxes (again rectangles).  This command will be u
sed with the limit command to find the area under the curve.  The mapl
e commands that draw boxes and compute the areas are part of the \"stu
dent\" package which you must load in order to access." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "restart:    # this clears all varia
bles" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "with(student);  # t
his loads the student library" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%\"
DG%%DiffG%*DoubleintG%$IntG%&LimitG%(LineintG%(ProductG%$SumG%*Triplei
ntG%*changevarG%/completesquareG%)distanceG%'equateG%*integrandG%*inte
rceptG%)intpartsG%(leftboxG%(leftsumG%)makeprocG%*middleboxG%*middlesu
mG%)midpointG%(powsubsG%)rightboxG%)rightsumG%,showtangentG%(simpsonG%
&slopeG%(summandG%*trapezoidG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 
"Note, the student library has many commands.  Some of which are redef
ined from the original.  Make sure to \"restart\" after you are done u
sing the student packages.  " }}{PARA 0 "" 0 "" {TEXT -1 39 "To approx
imate the area under the cuve " }{XPPEDIT 18 0 "f(x) = x^2;" "6#/-%\"f
G6#%\"xG*$F'\"\"#" }{TEXT -1 102 " over the interval from 1 to 3, seve
ral boxes can be used.   The following command draws the graph of " }
{XPPEDIT 18 0 "f(x) = x^2;" "6#/-%\"fG6#%\"xG*$F'\"\"#" }{TEXT -1 41 "
 and six boxes using the left end points." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "leftbox(x^2, x = 1..3,6);" }}{PARA 13 "" 1 "" 
{GLPLOT2D 400 300 300 {PLOTDATA 2 "6+-%'CURVESG6&7S7$$\"\"\"\"\"!F(7$$
\"3ALLL3VfV5!#<$\"3EGdQ!3*3*3\"F.7$$\"3smm\"H[D:3\"F.$\"3CH\")>qtpp6F.
7$$\"3XLL$e0$=C6F.$\"3krF-Vvyj7F.7$$\"3QLL$3RBr;\"F.$\"3%yHI%4q<i8F.7$
$\"3imm\"zjf)47F.$\"3iZ)fMMgPY\"F.7$$\"3WLLe4;[\\7F.$\"3gdpo#H/7c\"F.7
$$\"3-++Dmy]!H\"F.$\"3+GJ&Gb5am\"F.7$$\"3>LLezs$HL\"F.$\"3)o,.8z@nx\"F
.7$$\"31++D@1Bv8F.$\"3?hOih#f7*=F.7$$\"3pmmm@Xt=9F.$\"3PX(oHk2G,#F.7$$
\"3MLL$3y_qX\"F.$\"3r\\T81G+B@F.7$$\"3'******\\1!>+:F.$\"3+/Z7J-d]AF.7
$$\"3*******\\Z/Na\"F.$\"3yD]MkgS#Q#F.7$$\"35+++NfC&e\"F.$\"3kCSVuY+8D
F.7$$\"3LLLez6:B;F.$\"3`mFy^(>Yj#F.7$$\"3_mmm\"=C#o;F.$\"3->uH?>(Hy#F.
7$$\"3gmmmEpS1<F.$\"3PyfP*fC=\"HF.7$$\"3%)***\\i`A3v\"F.$\"33#)3W`&z`1
$F.7$$\"3Ymmmwy8!z\"F.$\"3&4kwuh$f/KF.7$$\"3/++DOIFL=F.$\"3!RHTa-!*3O$
F.7$$\"3!****\\(3zMu=F.$\"3y(\\NI3!=8NF.7$$\"3emm;H_?<>F.$\"3e5Su!*env
OF.7$$\"3mmm\"zihl&>F.$\"3%p\"f$QSL\"GQF.7$$\"3KLL$3#G,**>F.$\"3CSg&yD
_g*RF.7$$\"3<LLezw5V?F.$\"3'=Rs.**)GuTF.7$$\"3o***\\PQ#\\\"3#F.$\"37c#
4Oa5EL%F.7$$\"3BLL$e\"*[H7#F.$\"3CDzB*47p]%F.7$$\"3#*******pvxl@F.$\"3
'[5:F[#f!p%F.7$$\"3z****\\_qn2AF.$\"3u'3N\"oz$Q([F.7$$\"3%)***\\i&p@[A
F.$\"3)Q,qB[zW0&F.7$$\"3#)****\\2'HKH#F.$\"3W.YrK?!*e_F.7$$\"3_mmmwanL
BF.$\"3c$R&RI7/YaF.7$$\"3u*****\\2goP#F.$\"3G/!Hh\"QY\\cF.7$$\"3CLLeR<
*fT#F.$\"3F+\\t&3;q$eF.7$$\"3'******\\)HxeCF.$\"3I5ew\"fkb/'F.7$$\"3Cm
m\"H!o-*\\#F.$\"3y$*eph\\8XiF.7$$\"3))***\\7k.6a#F.$\"3U3Sd:x?dkF.7$$
\"3emmmT9C#e#F.$\"3/X21j3(zm'F.7$$\"3\"****\\i!*3`i#F.$\"3=rbB`oC#*oF.
7$$\"3;LLL$*zymEF.$\"3YDoQ,#e<6(F.7$$\"30LL$3N1#4FF.$\"3zkcR^!*zRtF.7$
$\"3kmm\"HYt7v#F.$\"3wm%\\xm0&pvF.7$$\"3%*******p(G**y#F.$\"3!Hrt;a-Py
(F.7$$\"3Umm;9@BMGF.$\"3M_O(pnrG.)F.7$$\"3/LLL`v&Q(GF.$\"35^5&oBd!f#)F
.7$$\"30++DOl5;HF.$\"3,s*f2LxO])F.7$$\"3/++v.UacHF.$\"3<@xsGO:T()F.7$$
\"\"$F*$\"\"*F*-%'COLOURG6&%$RGBG$\"*++++\"!\")$F*F*F][l-%*THICKNESSG6
#\"\"#-%&STYLEG6#%%LINEG-%)POLYGONSG6$7&7$F(F][lF'7$$\"+LLLL8!\"*F(7$F
\\\\lF][l-%&COLORG6&Fiz$\"\"(!\"\"$FezFe\\lFc\\l-Fg[l6$7&F_\\l7$F\\\\l
$\"+yxxx<F^\\l7$$\"+nmmm;F^\\lF[]l7$F^]lF][lF`\\l-Fg[l6$7&F`]l7$F^]l$
\"+yxxxFF^\\l7$$Fa[lF*Fe]l7$Fh]lF][lF`\\l-Fg[l6$7&Fi]l7$Fh]l$\"\"%F*7$
$\"+LLLLBF^\\lF^^l7$Fa^lF][lF`\\l-Fg[l6$7&Fc^l7$Fa^l$\"+WWWWaF^\\l7$$
\"+nmmmEF^\\lFh^l7$F[_lF][lF`\\l-Fg[l6$7&F]_l7$F[_l$\"+6666rF^\\l7$Fbz
Fb_l7$FbzF][lF`\\l-%+AXESLABELSG6$Q\"x6\"Q!Fj_l-%%VIEWG6$;F(Fbz%(DEFAU
LTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1
" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "To find the combined area of these
 boxes we use the \"leftsum\" command" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "leftsum(x^2,x=1..3,6);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$-%$SumG6$*$),&\"\"\"F**&#F*\"\"$F*%\"iGF*F*\"\"#F*/F.;\"\"!\"
\"&F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+p.Pqt!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 95 "You can assume that this value smaller than the actual ar
ea under the curve.  Using rightboxes:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "rightbox(x^2,x=1..3,6);" }}{PARA 13 "" 1 "" 
{GLPLOT2D 400 300 300 {PLOTDATA 2 "6+-%)POLYGONSG6$7&7$$\"\"\"\"\"!$F*
F*7$F($\"+yxxx<!\"*7$$\"+LLLL8F/F-7$F1F+-%&COLORG6&%$RGBG$\"\"(!\"\"$
\"\"*F:F8-F$6$7&F37$F1$\"+yxxxFF/7$$\"+nmmm;F/FA7$FDF+F4-F$6$7&FF7$FD$
\"\"%F*7$$\"\"#F*FK7$FNF+F4-F$6$7&FP7$FN$\"+WWWWaF/7$$\"+LLLLBF/FU7$FX
F+F4-F$6$7&FZ7$FX$\"+6666rF/7$$\"+nmmmEF/Fin7$F\\oF+F4-F$6$7&F^o7$F\\o
$F<F*7$$\"\"$F*Fco7$FeoF+F4-%'CURVESG6&7S7$F(F(7$$\"3ALLL3VfV5!#<$\"3E
GdQ!3*3*3\"F`p7$$\"3smm\"H[D:3\"F`p$\"3CH\")>qtpp6F`p7$$\"3XLL$e0$=C6F
`p$\"3krF-Vvyj7F`p7$$\"3QLL$3RBr;\"F`p$\"3%yHI%4q<i8F`p7$$\"3imm\"zjf)
47F`p$\"3iZ)fMMgPY\"F`p7$$\"3WLLe4;[\\7F`p$\"3gdpo#H/7c\"F`p7$$\"3-++D
my]!H\"F`p$\"3+GJ&Gb5am\"F`p7$$\"3>LLezs$HL\"F`p$\"3)o,.8z@nx\"F`p7$$
\"31++D@1Bv8F`p$\"3?hOih#f7*=F`p7$$\"3pmmm@Xt=9F`p$\"3PX(oHk2G,#F`p7$$
\"3MLL$3y_qX\"F`p$\"3r\\T81G+B@F`p7$$\"3'******\\1!>+:F`p$\"3+/Z7J-d]A
F`p7$$\"3*******\\Z/Na\"F`p$\"3yD]MkgS#Q#F`p7$$\"35+++NfC&e\"F`p$\"3kC
SVuY+8DF`p7$$\"3LLLez6:B;F`p$\"3`mFy^(>Yj#F`p7$$\"3_mmm\"=C#o;F`p$\"3-
>uH?>(Hy#F`p7$$\"3gmmmEpS1<F`p$\"3PyfP*fC=\"HF`p7$$\"3%)***\\i`A3v\"F`
p$\"33#)3W`&z`1$F`p7$$\"3Ymmmwy8!z\"F`p$\"3&4kwuh$f/KF`p7$$\"3/++DOIFL
=F`p$\"3!RHTa-!*3O$F`p7$$\"3!****\\(3zMu=F`p$\"3y(\\NI3!=8NF`p7$$\"3em
m;H_?<>F`p$\"3e5Su!*envOF`p7$$\"3mmm\"zihl&>F`p$\"3%p\"f$QSL\"GQF`p7$$
\"3KLL$3#G,**>F`p$\"3CSg&yD_g*RF`p7$$\"3<LLezw5V?F`p$\"3'=Rs.**)GuTF`p
7$$\"3o***\\PQ#\\\"3#F`p$\"37c#4Oa5EL%F`p7$$\"3BLL$e\"*[H7#F`p$\"3CDzB
*47p]%F`p7$$\"3#*******pvxl@F`p$\"3'[5:F[#f!p%F`p7$$\"3z****\\_qn2AF`p
$\"3u'3N\"oz$Q([F`p7$$\"3%)***\\i&p@[AF`p$\"3)Q,qB[zW0&F`p7$$\"3#)****
\\2'HKH#F`p$\"3W.YrK?!*e_F`p7$$\"3_mmmwanLBF`p$\"3c$R&RI7/YaF`p7$$\"3u
*****\\2goP#F`p$\"3G/!Hh\"QY\\cF`p7$$\"3CLLeR<*fT#F`p$\"3F+\\t&3;q$eF`
p7$$\"3'******\\)HxeCF`p$\"3I5ew\"fkb/'F`p7$$\"3Cmm\"H!o-*\\#F`p$\"3y$
*eph\\8XiF`p7$$\"3))***\\7k.6a#F`p$\"3U3Sd:x?dkF`p7$$\"3emmmT9C#e#F`p$
\"3/X21j3(zm'F`p7$$\"3\"****\\i!*3`i#F`p$\"3=rbB`oC#*oF`p7$$\"3;LLL$*z
ymEF`p$\"3YDoQ,#e<6(F`p7$$\"30LL$3N1#4FF`p$\"3zkcR^!*zRtF`p7$$\"3kmm\"
HYt7v#F`p$\"3wm%\\xm0&pvF`p7$$\"3%*******p(G**y#F`p$\"3!Hrt;a-Py(F`p7$
$\"3Umm;9@BMGF`p$\"3M_O(pnrG.)F`p7$$\"3/LLL`v&Q(GF`p$\"35^5&oBd!f#)F`p
7$$\"30++DOl5;HF`p$\"3,s*f2LxO])F`p7$$\"3/++v.UacHF`p$\"3<@xsGO:T()F`p
Fdo-%'COLOURG6&F7$\"*++++\"!\")F+F+-%*THICKNESSG6#FO-%&STYLEG6#%%LINEG
-%+AXESLABELSG6$Q\"x6\"Q!Fj_l-%%VIEWG6$;F(Feo%(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" }}}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 30 "evalf(rightsum(x^2,x=1..3,6));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+/Pq.5!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "a
nd we may safely assume this value is larger than the actual area unde
r the curve.  " }}{PARA 0 "" 0 "" {TEXT -1 92 "Question: How do we fin
d the exact area (without using the fundemental theorem of calculus)?
" }}{PARA 0 "" 0 "" {TEXT -1 94 "Answer: By taking the limit of either
 of the above as the number of boxes goes to infinity.   " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 23 "First lets try 50 boxes" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 25 "rightbox(x^2, x=1..3,50);" }}{PARA 13 "" 
1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6W-%)POLYGONSG6$7&7$$\"\"\"\"
\"!$F*F*7$F($\"+++g\"3\"!\"*7$$\"++++S5F/F-7$F1F+-%&COLORG6&%$RGBG$\"
\"(!\"\"$\"\"*F:F8-F$6$7&F37$F1$\"+++Sm6F/7$$\"++++!3\"F/FA7$FDF+F4-F$
6$7&FF7$FD$\"+++Sa7F/7$$\"++++?6F/FK7$FNF+F4-F$6$7&FP7$FN$\"+++gX8F/7$
$\"++++g6F/FU7$FXF+F4-F$6$7&FZ7$FX$\"++++S9F/7$$\"+++++7F/Fin7$F\\oF+F
4-F$6$7&F^o7$F\\o$\"+++gP:F/7$$\"++++S7F/Fco7$FfoF+F4-F$6$7&Fho7$Ffo$
\"+++SQ;F/7$$\"++++!G\"F/F]p7$F`pF+F4-F$6$7&Fbp7$F`p$\"+++SU<F/7$$\"++
++?8F/Fgp7$FjpF+F4-F$6$7&F\\q7$Fjp$\"+++g\\=F/7$$\"++++g8F/Faq7$FdqF+F
4-F$6$7&Ffq7$Fdq$\"++++g>F/7$$\"+++++9F/F[r7$F^rF+F4-F$6$7&F`r7$F^r$\"
+++gt?F/7$FinFer7$FinF+F4-F$6$7&Fhr7$Fin$\"+++S!>#F/7$$\"++++![\"F/F]s
7$F`sF+F4-F$6$7&Fbs7$F`s$\"+++S5BF/7$$\"++++?:F/Fgs7$FjsF+F4-F$6$7&F\\
t7$Fjs$\"+++gLCF/7$$\"++++g:F/Fat7$FdtF+F4-F$6$7&Fft7$Fdt$\"++++gDF/7$
$\"+++++;F/F[u7$F^uF+F4-F$6$7&F`u7$F^u$\"+++g*o#F/7$$\"++++S;F/Feu7$Fh
uF+F4-F$6$7&Fju7$Fhu$\"+++SAGF/7$$\"++++!o\"F/F_v7$FbvF+F4-F$6$7&Fdv7$
Fbv$\"+++SeHF/7$$\"++++?<F/Fiv7$F\\wF+F4-F$6$7&F^w7$F\\w$\"+++g(4$F/7$
$\"++++g<F/Fcw7$FfwF+F4-F$6$7&Fhw7$Ffw$\"++++SKF/7$$\"+++++=F/F]x7$F`x
F+F4-F$6$7&Fbx7$F`x$\"+++g&Q$F/7$$\"++++S=F/Fgx7$FjxF+F4-F$6$7&F\\y7$F
jx$\"+++SMNF/7$$\"++++!)=F/Fay7$FdyF+F4-F$6$7&Ffy7$Fdy$\"+++S'o$F/7$$
\"++++?>F/F[z7$F^zF+F4-F$6$7&F`z7$F^z$\"+++gTQF/7$F[rFez7$F[rF+F4-F$6$
7&Fhz7$F[r$\"\"%F*7$$\"\"#F*F][l7$F`[lF+F4-F$6$7&Fb[l7$F`[l$\"+++ghTF/
7$$\"++++S?F/Fg[l7$Fj[lF+F4-F$6$7&F\\\\l7$Fj[l$\"+++SEVF/7$$\"++++!3#F
/Fa\\l7$Fd\\lF+F4-F$6$7&Ff\\l7$Fd\\l$\"+++S%\\%F/7$$\"++++?@F/F[]l7$F^
]lF+F4-F$6$7&F`]l7$F^]l$\"+++glYF/7$$\"++++g@F/Fe]l7$Fh]lF+F4-F$6$7&Fj
]l7$Fh]l$\"++++S[F/7$$\"+++++AF/F_^l7$Fb^lF+F4-F$6$7&Fd^l7$Fb^l$\"+++g
<]F/7$$\"++++SAF/Fi^l7$F\\_lF+F4-F$6$7&F^_l7$F\\_l$\"+++S)>&F/7$$\"+++
+!G#F/Fc_l7$Ff_lF+F4-F$6$7&Fh_l7$Ff_l$\"+++S#Q&F/7$$\"++++?BF/F]`l7$F`
`lF+F4-F$6$7&Fb`l7$F``l$\"+++gpbF/7$$\"++++gBF/Fg`l7$Fj`lF+F4-F$6$7&F
\\al7$Fj`l$\"++++gdF/7$$\"+++++CF/Faal7$FdalF+F4-F$6$7&Ffal7$Fdal$\"++
+g`fF/7$$\"++++SCF/F[bl7$F^blF+F4-F$6$7&F`bl7$F^bl$\"+++S]hF/7$$\"++++
![#F/Febl7$FhblF+F4-F$6$7&Fjbl7$Fhbl$\"+++S]jF/7$$\"++++?DF/F_cl7$Fbcl
F+F4-F$6$7&Fdcl7$Fbcl$\"+++g`lF/7$F[uFicl7$F[uF+F4-F$6$7&F\\dl7$F[u$\"
++++gnF/7$$\"+++++EF/Fadl7$FddlF+F4-F$6$7&Ffdl7$Fddl$\"+++gppF/7$$\"++
++SEF/F[el7$F^elF+F4-F$6$7&F`el7$F^el$\"+++S#=(F/7$$\"++++!o#F/Feel7$F
helF+F4-F$6$7&Fjel7$Fhel$\"+++S)R(F/7$$\"++++?FF/F_fl7$FbflF+F4-F$6$7&
Fdfl7$Fbfl$\"+++g<wF/7$$\"++++gFF/Fifl7$F\\glF+F4-F$6$7&F^gl7$F\\gl$\"
++++SyF/7$$\"+++++GF/Fcgl7$FfglF+F4-F$6$7&Fhgl7$Ffgl$\"+++gl!)F/7$$\"+
+++SGF/F]hl7$F`hlF+F4-F$6$7&Fbhl7$F`hl$\"+++S%H)F/7$$\"++++!)GF/Fghl7$
FjhlF+F4-F$6$7&F\\il7$Fjhl$\"+++SE&)F/7$$\"++++?HF/Fail7$FdilF+F4-F$6$
7&Ffil7$Fdil$\"+++gh()F/7$$\"++++gHF/F[jl7$F^jlF+F4-F$6$7&F`jl7$F^jl$F
<F*7$$\"\"$F*Fejl7$FgjlF+F4-%'CURVESG6&7S7$F(F(7$$\"3ALLL3VfV5!#<$\"3E
GdQ!3*3*3\"Fb[m7$$\"3smm\"H[D:3\"Fb[m$\"3CH\")>qtpp6Fb[m7$$\"3XLL$e0$=
C6Fb[m$\"3krF-Vvyj7Fb[m7$$\"3QLL$3RBr;\"Fb[m$\"3%yHI%4q<i8Fb[m7$$\"3im
m\"zjf)47Fb[m$\"3iZ)fMMgPY\"Fb[m7$$\"3WLLe4;[\\7Fb[m$\"3gdpo#H/7c\"Fb[
m7$$\"3-++Dmy]!H\"Fb[m$\"3+GJ&Gb5am\"Fb[m7$$\"3>LLezs$HL\"Fb[m$\"3)o,.
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "evalf(rightsum(x^2,x=1..3,50
));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++?F))!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 157 "It seems as though this is getting close
r to the exact area by inspection of the graph, but is still too big. \+
 We create a function of the number of boxes by" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "boxarea := n -> rightsum(x^2, x=1..3,n);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(boxareaGf*6#%\"nG6\"6$%)operatorG%&
arrowGF(-%)rightsumG6%*$)%\"xG\"\"#\"\"\"/F1;F3\"\"$9$F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(boxarea(100));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++!ou)!\"*" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 98 "This should be closer still.  To get the limit as \"
n\" goes to infinity we use the \"limit\" command:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 29 "limit(boxarea(n),n=infinity);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6##\"#E\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"#E\"\"
$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "This is the exact area whic
h we can check by the fundemental theorem of calculus.  The decimal ap
proximation is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+nmmm')!\"*" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 47 "We can see that our error in using 100 boxes is" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "evalf(boxarea(100) - Limi
t(boxarea(n),n=infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\")LL8!
)!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "Not bad. Notice that us
ing right boxes to approximate the area under an increasing function w
ill always be greater than the exact area. " }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 21 "# end of this section" }}}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 20 "2. Riemann Integrals" }}{PARA 0 "" 0 "" {TEXT -1 603 "I
n the last section we saw how the rightsums or leftsums can converge t
o the area under a curve as the number of rectangles gets large. This \+
is a type of Riemann sum, specifically the function was positive so th
e definite integral was the area below the curve and above the x-axis \+
over a certain interval.  A Riemann integral is a more general version
 of this where the resulting sum defines the net difference = (area ab
ove the x-axis, below the curve) - (area below the x-axis, above the c
urve)  over an interval of x values.  Maple has two commands (Int, int
) that will calculate this integral.      " }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 40 "restart;   # erase \"student\" definitions" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Similar to the \"Diff\" (big D), \+
\"Int\" (big I) gives an expression for the integral but does not eval
uate it. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Int(x^2,x = 1 \+
.. 3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*$)%\"xG\"\"#\"\"\"
/F(;F*\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6##\"#E\"\"$" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 152 "Notice, this is the same answer we obtained by the limit
ing process in the previous section.  You may replace the above two st
eps with the \"int\" command" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "int(x^2,x = 1 .. 3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"#E\"
\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Try the \"int\" command o
n an odd function over a symmetric interval about x=0. (odd implies f(
-x) = -f(x))" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "int(3*x^3 +
 x, x = -2 .. 2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 66 "The reason this value is zero can be seen
 by plotting the function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
28 "plot(3*x^3 + x,x = -2 .. 2);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 
300 300 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$!\"#\"\"!$!#EF*7$$!3MLLL$Q6G\"
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3;LLL$)G[k6F0$\"3K$Q(po/n,fF07$$\"3#)****\\7yh]7F0$\"3a)*o_&4&o=rF07$$
\"3xmmm')fdL8F0$\"3[d;sd-d[%)F07$$\"3bmmm,FT=9F0$\"3#=Y\\QB.&z**F07$$
\"3FLL$e#pa-:F0$\"3A(Ru&z4#z;\"F37$$\"3!*******Rv&)z:F0$\"3E6)G%4$f4M
\"F37$$\"3ILLLGUYo;F0$\"3]1LZkSBg:F37$$\"3_mmm1^rZ<F0$\"3q]'*\\#\\%Hw<
F37$$\"34++]sI@K=F0$\"3QdEk*pX%G?F37$$\"34++]2%)38>F0$\"39\\&=LoE=H#F3
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Q\"x6\"Q!Fh[l-%%VIEWG6$;F(Fhz%(DEFAULTG" 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 "" 
{TEXT -1 217 "The area above the curve below the x-axis is cancelled o
ut the area above the x-axis and below the curve. Hence the signed are
a is zero.  This is a property of all odd functions integrated over a \+
symmetric interval.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "# \+
end of this section" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 22 "3. Antid
ifferentiation" }}{PARA 0 "" 0 "" {TEXT -1 169 "The commands for anitd
ifferentiation are the same as those for integration: \"int\" and \"In
t\" only the limits of integration are replaced by the variable of int
egration.  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "restart;  #Th
is clears all previous variables" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "Int(x^2,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$Int
G6$*$)%\"xG\"\"#\"\"\"F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "The \+
\"Int\" (big I) command produces an expression for the antiderivative.
 You can evaluate this indefinite integral by" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 9 "value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$
)%\"xG\"\"$\"\"\"#F(F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "Notice
 this gives only one antiderivative, where the constant of integration
 = zero.  The above two commands can be replace by the single \"int\" \+
(small i) command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "int(x
^2,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%\"xG\"\"$\"\"\"#F(F'" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "This is a fairly easy example \+
and hopefully one that you could do in your head.  Let's try more diff
icult example" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "h := x -> \+
exp(x)/(exp(x) + 1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hGf*6#%\"x
G6\"6$%)operatorG%&arrowGF(*&-%$expG6#9$\"\"\",&F-F1F1F1!\"\"F(F(F(" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "recall that with the substituti
on u = 1 + e^x,  the integrand has the form du/u and we found the foll
owing antiderivative" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "int
(h(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%#lnG6#,&-%$expG6#%\"xG
\"\"\"F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "However, Mikki sug
gested we first multiply the numerator and denominator of \"h\" by exp
(-x) to get the equivalent function" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "g := x -> 1/(1 + exp(-x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"gGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&\"\"\"F-,&F-
F--%$expG6#,$9$!\"\"F-F4F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 
"Here, no u-substitution is obvious. So we let Maple do it" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "int(g(x),x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&-%#lnG6#,&\"\"\"F(-%$expG6#,$%\"xG!\"\"F(F(-F%6#F)F.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "Is this the same answer?  It
 should be. Careful use of the properties of logarithms will result in
 the same answer. You may wish to avoid such trivial tasks by requesti
ng Maple to simplify the expression by" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%#
lnG6#,&\"\"\"F(-%$expG6#,$%\"xG!\"\"F(F(-F%6#F)F." }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 96 "This didn't help at all.  The \"simplify\" command
 can only do limited simpifications for example:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 34 "simplify((cos(x))^2 + (sin(x))^2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
260 "The simplify command is useful for some trig expressions, some ex
ponential and logarithmic expressions, and some algebraic expressions,
 however it by no means defines the simplest from of all expressions. \+
Lesson:  Maple will not solve all of your problems.    " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "# end of this section" }}}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 10 "Assignment" }}{EXCHG {PARA 256 "" 0 "" 
{TEXT -1 0 "" }{TEXT 256 13 "Problem # 1: " }}{PARA 0 "" 0 "" {TEXT 
-1 164 "This problem involves using commands in the \"student\" librar
y so make sure to type \"with(student)\". Recall that the solution to \+
the initial value problem:  y'(x) = " }{XPPEDIT 18 0 "e^(-x^2);" "6#)%
\"eG,$*$%\"xG\"\"#!\"\"" }{TEXT -1 31 "  ;  y(0) = 2 can be written as
" }}{PARA 0 "" 0 "" {TEXT -1 7 "y(x) = " }{XPPEDIT 18 0 "int(e^(-t^2),
t = 0 .. x);" "6#-%$intG6$)%\"eG,$*$%\"tG\"\"#!\"\"/F*;\"\"!%\"xG" }
{TEXT -1 50 "  + 2.    As a check, notice y(0) = 2 and y'(x) = " }
{XPPEDIT 18 0 "e^(-x^2);" "6#)%\"eG,$*$%\"xG\"\"#!\"\"" }{TEXT -1 11 "
  by FTC.  " }}{PARA 0 "" 0 "" {TEXT -1 812 "While this counts as a so
lution, we cannot integrate this explicitly. However, we can approxima
te values for y(x) using the boxes technique from section 4 of this la
b.  Your job is to estimate y(2), using 20 boxes for the integral part
 of y(x).  Display the graph of the integral with the 20 boxes (with t
he \"leftbox\" or \"rightbox\" command).  Evaluate the sum of the area
s of these boxes (with the \"leftsum\" or \"rightsum\" command) and hy
pothesize whether this approximation for the integral results in an es
timation for y(2) that is larger or smaller than the exact value of y(
2).  Using the \"limit\" command, find the exact value of y(2) by lett
ing the number of boxes go to infinity. The resulting answer will be g
iven in terms of the \"erf\" function, use \"evalf\" on this answer to
 get a decimal approximation.  " }}{PARA 0 "" 0 "" {TEXT -1 28 "Was yo
ur hypothesis correct?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 257 0 "" }{TEXT 258 0 "" }{TEXT 259 0 "" }{TEXT 260 11 "Problem \+
#2:" }}{PARA 0 "" 0 "" {TEXT -1 195 " The general notion is that cardi
ac output averages around 5000 ml/min.  However what does that say abo
ut cardiac output on a smaller time scale (seconds)?  A \"first order
\"  approximation of the " }{TEXT 261 4 "rate" }{TEXT -1 34 " of cardi
ac output is 80(1 + sin(2" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 
14 "t))  ml/sec.  " }{TEXT -1 162 "Call this function \"f\" and plot i
t over one second.  Use the int command to define a function (F) of to
tal cardiac output, in ml., by  (don't use the # symbol):  " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "# F := x -> int(f(t), t=0..x
); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "Plot F over one second an
d again over one minute.  What is the total cardiac output, in ml., af
ter one minute? Does this agree with the general notion of cardiac out
put in ml/min?    " }}}}}{MARK "0 0 0" 7 }{VIEWOPTS 1 1 0 1 1 1803 1 
1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }