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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT 257 27 "TP7 : Int\351grales et \+
s\351ries." }{TEXT 348 9 "(corrig\351)" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 68 "Premi\350re pa
rtie : Int\351gration par parties et changement de variable." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 16 "Dans le package " }{TEXT 258 7 "student" 
}{TEXT -1 37 " on trouve deux fonctions de Maple ( " }{TEXT 259 8 "int
parts" }{TEXT -1 6 "  et  " }{TEXT 260 9 "changevar" }{TEXT -1 206 " )
 qui permettent d'int\351grer par parties et de changer de variable da
ns une int\351grale. Nous allons nous int\351resser dans cette premi
\350re partie aux diff\351rentes applications de ces deux m\351thodes \+
d'int\351gration." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
258 "" 1 "" {TEXT 261 33 "1. Int\351gration par parties(IPP): " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "On
 charge  le package " }{TEXT 262 7 "student" }{TEXT -1 1 " " }
{MPLTEXT 0 21 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restar
t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%\"DG%%DiffG%*DoubleintG%$IntG%&Limi
tG%(LineintG%(ProductG%$SumG%*TripleintG%*changevarG%/completesquareG%
)distanceG%'equateG%*integrandG%*interceptG%)intpartsG%(leftboxG%(left
sumG%)makeprocG%*middleboxG%*middlesumG%)midpointG%(powsubsG%)rightbox
G%)rightsumG%,showtangentG%(simpsonG%&slopeG%(summandG%*trapezoidG" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 10 "
Exemple 1." }{TEXT -1 30 " - Calculer une primitive de  " }{TEXT 266 
3 "y =" }{TEXT -1 1 " " }{TEXT 264 9 "2x ln(x) " }{TEXT -1 10 " avec I
PP:" }}}{EXCHG {PARA 262 "" 0 "" {TEXT 273 46 " Dans la formule de l'i
nt\351gration par parties:" }}{PARA 261 "" 0 "" {TEXT 272 27 "Int(uv')
 = [uv] - Int(u'v)." }}{PARA 260 "" 0 "" {TEXT 274 17 "il faut pr\351c
iser " }{TEXT 276 1 "u" }{TEXT 277 27 " - l'expression \340 d\351river
. " }{TEXT -1 8 "[Ex.:   " }{TEXT 275 1 "u" }{TEXT -1 7 "=ln(x)]" }
{TEXT 278 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "J:=Int(2*
x*ln(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"JG-%$IntG6$,$*&%\"x
G\"\"\"-%#lnG6#F*F+\"\"#F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "intparts(J,ln(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&-%#lnG6#
%\"xG\"\"\")F(\"\"#F)F)-%$IntG6$F(F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 41 "On peut donc dire que la primitive J de  " }{TEXT 270 3 "
y =" }{TEXT -1 1 " " }{TEXT 269 8 "2x ln(x)" }{TEXT -1 15 " est \351ga
le  \340 :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "J=ln(x)*x^2-i
nt(x,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$,$*&%\"xG\"\"\"-
%#lnG6#F)F*\"\"#F),&*&F+F*)F)F.F*F**&#F*F.F**$F1F*F*!\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 34 "
Exercice 1.1 - It\351ration de l'IPP." }}{PARA 0 "" 0 "" {TEXT -1 28 "
1. -  Calculer la primitive " }{XPPEDIT 18 0 "J[0]:=Int(exp(x),x)" "6#
>&%\"JG6#\"\"!-%$IntG6$-%$expG6#%\"xGF." }{TEXT -1 53 " . Appliquer la
 m\351thode d'int\351gration par parties \340  " }{XPPEDIT 18 0 "J[1]:
=Int(x*exp(x),x)" "6#>&%\"JG6#\"\"\"-%$IntG6$*&%\"xGF'-%$expG6#F,F'F,
" }}{PARA 0 "" 0 "" {TEXT -1 18 "- Une conclusion ?" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 34 "J[0]:=Int(exp(x),x)=int(exp(x),x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"JG6#\"\"!/-%$IntG6$-%$expG6#%\"xG
F/F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "J[1]:=Int(x*exp(x),
x);intparts(J[1],x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"JG6#\"\"
\"-%$IntG6$*&%\"xGF'-%$expG6#F,F'F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#,&*&%\"xG\"\"\"-%$expG6#F%F&F&-%$IntG6$F'F%!\"\"" }}}{EXCHG {PARA 
259 "" 1 "" {TEXT -1 5 "2. - " }{TEXT 268 1 " " }{TEXT -1 45 "Trouver,
 avec la m\352me m\351thode, la primitive  " }{TEXT 267 1 " " }
{XPPEDIT 18 0 "J[2]:=Int(x^2*exp(x),x)" "6#>&%\"JG6#\"\"#-%$IntG6$*&%
\"xGF'-%$expG6#F,\"\"\"F," }{TEXT 271 1 " " }{TEXT -1 34 "   et \351ta
blir une relation entre  " }{XPPEDIT 18 0 "J[1]" "6#&%\"JG6#\"\"\"" }
{TEXT -1 4 " et " }{XPPEDIT 18 0 "J[2]" "6#&%\"JG6#\"\"#" }{TEXT -1 2 
" ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "J[2]:=Int(x^2*exp(x)
,x);intparts(J[2],x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"JG6#\"
\"#-%$IntG6$*&)%\"xGF'\"\"\"-%$expG6#F-F.F-" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*&)%\"xG\"\"#\"\"\"-%$expG6#F&F(F(-%$IntG6$,$*&F&F(F)
F(F'F&!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "3. - Soit la suite
 de primitives  " }{XPPEDIT 18 0 "J:=n->Int(x^n*exp(x),x)" "6#>%\"JGR6
#%\"nG7\"6$%)operatorG%&arrowG6\"-%$IntG6$*&)%\"xGF'\"\"\"-%$expG6#F2F
3F2F,F,F," }{TEXT -1 39 ". Calculer   J(n)  avec    n = 0,1,2,.." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "J:=n->Int(x^n*exp(x),x);J(n)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"JGR6#%\"nG6\"6$%)operatorG%&a
rrowGF(-%$IntG6$*&)%\"xG9$\"\"\"-%$expG6#F1F3F1F(F(F(" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG%\"nG\"\"\"-%$expG6#F(F*F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "4. - " }{TEXT 280 83 "Nous allons m
aintenant chercher une relation de r\351currence entre  J(n+1)  et  J(
n)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "  \+
    a)  Calculer J(n) et J(n+1).  Int\351grer  J(n+1)  par parties  ( \+
penser \340 " }{TEXT 279 3 " u " }{TEXT -1 2 "et" }{TEXT 281 3 " v'" }
{TEXT -1 6 "... )." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "J(n);
J(n+1);intparts(%,x^(n+1));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG%\"nG\"\"\"-%$expG6#F
(F*F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG,&%\"nG\"\"
\"F+F+F+-%$expG6#F(F+F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&)%\"xG,
&%\"nG\"\"\"F)F)F)-%$expG6#F&F)F)-%$IntG6$*&*(F%F)F'F)F*F)F)F&!\"\"F&F
2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "      b)  Une " }{TEXT 282 
14 "simplification" }{TEXT -1 24 " s'impose. Appellons    " }{XPPEDIT 
18 0 "J[n+1]" "6#&%\"JG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 16 "   son r\351
sultat:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "J[n+1]=simplify(
%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/&%\"JG6#,&%\"nG\"\"\"F)F),(*&)
%\"xGF'F)-%$expG6#F-F)F)*&-%$IntG6$*&)F-F(F)F.F)F-F)F(F)!\"\"F2F7" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "     c)  Voil\340 la r\351currence
. Il n'y a plus qu'\340 remplacer (" }{TEXT 283 4 "subs" }{TEXT -1 45 
"tituer) le J(n) par J[n] dans l'\351quation du  " }{XPPEDIT 18 0 "J[n
+1]" "6#&%\"JG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 3 " : " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 18 "subs(J(n)=J[n],%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/&%\"JG6#,&%\"nG\"\"\"F)F),(*&)%\"xGF'F)-%$expG6#F-F)F)
*&&F%6#F(F)F(F)!\"\"F2F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "     \+
d)  En for\347ant  Maple \340 r\351soudre  (" }{MPLTEXT 0 21 5 "solve
" }{TEXT -1 33 ")  l'\351quation ci-dessus (avec le " }{XPPEDIT 18 0 "
J[n]" "6#&%\"JG6#%\"nG" }{TEXT -1 31 " pour l'inconnue ) la  relation
" }}{PARA 0 "" 0 "" {TEXT -1 16 "        entre   " }{XPPEDIT 18 0 "J[n
]" "6#&%\"JG6#%\"nG" }{TEXT -1 6 " et   " }{XPPEDIT 18 0 "J[n+1]" "6#&
%\"JG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 34 "  appara\356t encore plus clai
rement:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(%,\{J[n]\}
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/&%\"JG6#%\"nG,$*&,&&F&6#,&F(
\"\"\"F/F/F/*&)%\"xGF.F/-%$expG6#F2F/!\"\"F/F.F6F6" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 18 "Comme on conna\356t  " }{XPPEDIT 18 0 "J[0]:=Int(
exp(x),x)=exp(x)" "6#>&%\"JG6#\"\"!/-%$IntG6$-%$expG6#%\"xGF/-F-6#F/" 
}{TEXT -1 38 " , la  relation de r\351currence entre   " }{XPPEDIT 18 
0 "J[n]" "6#&%\"JG6#%\"nG" }{TEXT -1 6 " et   " }{XPPEDIT 18 0 "J[n+1]
" "6#&%\"JG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 61 "  donne ais\351ment  les
 primitives successives des fonctions   " }{XPPEDIT 18 0 "x^n*exp(x)" 
"6#*&)%\"xG%\"nG\"\"\"-%$expG6#F%F'" }{TEXT -1 40 "   pour  n = 0, 1, \+
2, ... . La commande " }{TEXT 286 6 "rsolve" }{TEXT -1 159 " du Maple \+
permet souvent de r\351soudre les \351quations r\351currentes et d'y t
rouver une formule explicite du terme g\351n\351ral de la suite. Ce se
ra la question suivante." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 285 41 "Exercice 1.1*- Solution de la r\351curre
nce." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "
Nous allons chercher une formule explicite du terme g\351n\351ral de l
a suite \{ J(n) \} \351tudi\351e dans l'exercice pr\351c\350dent:" }}
{PARA 0 "" 0 "" {TEXT -1 22 "Appliquer la commande " }{TEXT 288 6 "rso
lve" }{TEXT -1 37 " \340 une des \351quations r\351currentes de  " }
{XPPEDIT 18 0 "J(n+1)" "6#-%\"JG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 29 "  a
vec la condition initiale " }{XPPEDIT 18 0 "J[0]:=exp(x)" "6#>&%\"JG6#
\"\"!-%$expG6#%\"xG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 8 "- F
aire " }{TEXT 289 14 " copier-coller" }{TEXT -1 57 "  avec une des \+
\351quations de  la fin de l'Exercice 1.1. ( " }{TEXT 290 45 "en metta
nt tout d'abord  les compteurs \340 z\351ro" }{TEXT -1 6 " ! ). " }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;
" }}}{EXCHG {PARA 12 "" 1 "" {TEXT -1 5 "     " }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 67 "rsolve(\{J(n) = (-J(n+1)+x^(n+1)*exp(x))/(n+1)
,J(0)=exp(x)\},\{J(n)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/-%\"JG
6#%\"nG,(*()%\"xGF(\"\"\"),$F,!\"\",$F(F0F--%&GAMMAG6$,&F(F-F-F-F/F-F-
*&)F0F(F--F36#F5F-F-*(F8F-F+F-F.F-F0" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 62 "rsolve(\{J(n+1)=x^(n+1)*exp(x)-J(n)*(n+1),J(0)=exp(x)
\},\{J(n)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/-%\"JG6#%\"nG,(*()
%\"xGF(\"\"\"),$F,!\"\",$F(F0F--%&GAMMAG6$,&F(F-F-F-F/F-F-*&)F0F(F--F3
6#F5F-F-*(F8F-F+F-F.F-F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "La so
lution de l'Exercice 1.1*  contient  la fonction   " }{XPPEDIT 18 0 "G
amma(a)" "6#-%&GammaG6#%\"aG" }{TEXT -1 33 "   et  sa version  g\351n
\351ralis\351e   " }{XPPEDIT 18 0 "Gamma(a,z) = Int( exp(-t)*t^(a-1), \+
t=z..infinity )" "6#/-%&GammaG6$%\"aG%\"zG-%$IntG6$*&-%$expG6#,$%\"tG!
\"\"\"\"\")F1,&F'F3F3F2F3/F1;F(%)infinityG" }{TEXT -1 2 ", " }}{PARA 
0 "" 0 "" {TEXT -1 78 "- objets math\351matiques reconnus par Maple. P
enchons-nous un instant sur ces  \"" }{TEXT 287 25 "factorielles g\351
n\351ralis\351es" }{TEXT -1 97 "\" tr\350s utilis\351es  en physique, \+
en \351lectronique et en particulier dans le tra\356tement des signaux
.(" }{TEXT 293 59 "Renseignez-vous sur la fonction  GAMMA avec l'aide \+
du Maple" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "#?GAMMA" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 291 26 "Ex
ercice 1.2 -  Fonction  " }{XPPEDIT 18 0 "GAMMA" "6#%&GAMMAG" }{TEXT 
292 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 32 "On va s'occuper de la foncti
on  " }{XPPEDIT 18 0 "Gamma(a)" "6#-%&GammaG6#%\"aG" }{TEXT -1 71 "  d
ans sa version de base, d\351finie pour tout r\351el  a>0  par l'int
\351grale" }}{PARA 0 "" 0 "" {TEXT -1 20 "impropre suivante:  " }
{XPPEDIT 18 0 "Gamma:=a->Int(x^(a-1)*exp(-x),x=0..infinity)" "6#>%&Gam
maGR6#%\"aG7\"6$%)operatorG%&arrowG6\"-%$IntG6$*&)%\"xG,&F'\"\"\"F4!\"
\"F4-%$expG6#,$F2F5F4/F2;\"\"!%)infinityGF,F,F," }{TEXT -1 4 "  . " }}
{PARA 0 "" 0 "" {TEXT -1 59 "   a)   Calculer quelques valeurs de   GA
MMA(a)   ( ex.:   " }{TEXT 294 38 "a = 1, 2, 3, ...,5, ...1/2,... etc.
 ) " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 39 "Utiliser ces r
\351sultats pour comparer   " }{XPPEDIT 18 0 "Gamma(n+1)" "6#-%&GammaG
6#,&%\"nG\"\"\"F(F(" }{TEXT -1 13 "   et   n!  (" }{TEXT 296 16 "facto
rielle de n" }{TEXT -1 43 ", d\351finie pour tout entier  n  non-n\351
gatif)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "GAMMA(1/2);GAMMA(1);1!;GAMMA(3);2!;GAMMA(5);4!;GAMMA(
10);9!;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*$-%%sqrtG6#%#PiG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#C" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#\"#C" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"'!)GO" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"'!)GO" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "G
AMMA(n+1)-n!;simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%&GAMM
AG6#,&%\"nG\"\"\"F)F)F)-%*factorialG6#F(!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "   b)  Re
taper la formule de Gamma ,  en introduisant la  " }{TEXT 295 12 "form
e inerte" }{TEXT -1 1 " " }{TEXT 284 3 "Int" }{TEXT -1 19 "  de l'int
\351grale . " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Gamma:=a->I
nt(t^(a-1)*exp(-t),t=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%&GammaGR6#%\"aG6\"6$%)operatorG%&arrowGF(-%$IntG6$*&)%\"tG,&9$\"\"\"
F4!\"\"F4-%$expG6#,$F1F5F4/F1;\"\"!%)infinityGF(F(F(" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 18 "   c)  Comparer   " }{XPPEDIT 18 0 "Gamma(a)" "
6#-%&GammaG6#%\"aG" }{TEXT -1 5 " et  " }{XPPEDIT 18 0 "Gamma(a+1)" "6
#-%&GammaG6#,&%\"aG\"\"\"F(F(" }{TEXT -1 32 " en testant quelques vale
urs de " }{TEXT 314 1 "a" }{TEXT -1 75 ">0 .  C'est encore l'IPP appli
qu\351e cette fois \340 l'int\351grale impropre de    " }{XPPEDIT 18 
0 "Gamma(a+1)" "6#-%&GammaG6#,&%\"aG\"\"\"F(F(" }{TEXT -1 52 "   qui n
ous menera vers une solution ( - le package " }{TEXT 297 7 "student" }
{TEXT -1 24 " doit \352tre  recharg\351 ) :" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 14 "with(student);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#
7@%\"DG%%DiffG%*DoubleintG%$IntG%&LimitG%(LineintG%(ProductG%$SumG%*Tr
ipleintG%*changevarG%/completesquareG%)distanceG%'equateG%*integrandG%
*interceptG%)intpartsG%(leftboxG%(leftsumG%)makeprocG%*middleboxG%*mid
dlesumG%)midpointG%(powsubsG%)rightboxG%)rightsumG%,showtangentG%(simp
sonG%&slopeG%(summandG%*trapezoidG" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "a:=6;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG\"\"'" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Gamma(a);value(%);Gamma(a
+1);intparts(Gamma(a+1),t^a);value(%);a:='a';" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$IntG6$*&)%\"tG\"\"&\"\"\"-%$expG6#,$F(!\"\"F*/F(;\"
\"!%)infinityG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$?\"" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"tG\"\"'\"\"\"-%$expG6#,$F(!\"\"F*
/F(;\"\"!%)infinityG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%$IntG6$,$*
&)%\"tG\"\"&\"\"\"-%$expG6#,$F*!\"\"F,!\"'/F*;\"\"!%)infinityGF1" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$?(" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"aGF$" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
258 "" 1 "" {TEXT 298 48 "2. Int\351gration par changement de variable
(ChV): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 39 "On recommence et on charge  le package " }{TEXT 299 7 "st
udent" }{TEXT -1 1 " " }{MPLTEXT 0 21 1 "." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "with(student);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%\"DG%%DiffG%*
DoubleintG%$IntG%&LimitG%(LineintG%(ProductG%$SumG%*TripleintG%*change
varG%/completesquareG%)distanceG%'equateG%*integrandG%*interceptG%)int
partsG%(leftboxG%(leftsumG%)makeprocG%*middleboxG%*middlesumG%)midpoin
tG%(powsubsG%)rightboxG%)rightsumG%,showtangentG%(simpsonG%&slopeG%(su
mmandG%*trapezoidG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 300 10 "Exemple 2." }{TEXT -1 31 " - Calculer une  primi
tive de  " }{TEXT 302 3 "y =" }{TEXT -1 1 " " }{XPPEDIT 18 0 "2*x*exp(
x^2)" "6#*(\"\"#\"\"\"%\"xGF%-%$expG6#*$F&F$F%" }{TEXT 301 1 " " }
{TEXT -1 10 " avec ChV:" }}}{EXCHG {PARA 265 "" 0 "" {TEXT 304 1 " " }
}{PARA 265 "" 0 "" {TEXT 315 57 "On effectue un changement de variable
 dans une int\351grale:" }}{PARA 264 "" 0 "" {TEXT 303 36 "Int(f(u(x))
u'(x) dx) = Int(f(u)du) ." }}{PARA 263 "" 0 "" {TEXT 308 19 "avec la f
onction   " }{TEXT 309 20 "changevar ( u=u(x) ," }{TEXT 310 20 " Int(f
(u(x))u'(x),x)" }{TEXT 311 5 " , u)" }{TEXT 306 2 ". " }}{PARA 263 "" 
0 "" {TEXT 312 3 "   " }{TEXT -1 9 "[ Ex.:   " }{TEXT 305 1 "u" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "x^2" "6#*$%\"xG\"\"#" }{TEXT -1 13 " \+
,  u' = 2x ]" }{TEXT 307 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "K:=Int(2*x*exp(x^2),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"K
G-%$IntG6$,$*&%\"xG\"\"\"-%$expG6#*$)F*\"\"#F+F+F1F*" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 25 "K1:=changevar(u=x^2,K,u);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#K1G-%$IntG6$-%$expG6#%\"uGF+" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 47 "On demande alors la valeur de cette primitive :
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "value(K1);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#-%$expG6#%\"uG" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 67 "Puis, on revient \340 nos variables d'origine \340 l'aide
 de la fonction " }{TEXT 313 4 "subs" }{TEXT -1 2 " :" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "K=subs(u=x^2,%);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/-%$IntG6$,$*&%\"xG\"\"\"-%$expG6#*$)F)\"\"#F*F*F0F)F
+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 316 54 "Exercice 2.1 -  Calcul des \+
primitives avec ChV et IPP." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 97 "Appliquer l'une des deux m\351thodes d'int\351g
ration ChV et/ou IPP au calcul des primitives suivantes:" }}{PARA 0 "
" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "A:=Int(x/(1+x^2),x)" "6#>%\"AG-%
$IntG6$*&%\"xG\"\"\",&F*F**$F)\"\"#F*!\"\"F)" }{TEXT -1 12 " ,        \+
  " }{XPPEDIT 18 0 "B:=Int(2*x/(1+x^2)^n,x)" "6#>%\"BG-%$IntG6$*(\"\"#
\"\"\"%\"xGF*),&F*F**$F+F)F*%\"nG!\"\"F+" }{TEXT -1 13 " ,           \+
" }{XPPEDIT 18 0 "C:=Int(tan(x),x)" "6#>%\"CG-%$IntG6$-%$tanG6#%\"xGF+
" }{TEXT -1 16 " ,              " }{XPPEDIT 18 0 "De:=Int(sqrt(1-x^2),
x)" "6#>%#DeG-%$IntG6$-%%sqrtG6#,&\"\"\"F,*$%\"xG\"\"#!\"\"F." }{TEXT 
-1 15 "    (x=sin(u) ," }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "E:=Int(arctan(x),x);" "6#>%\"EG-%$IntG6$-%'arctanG6#%\"xGF+" }
{TEXT -1 32 "  (d'abord IPP ensuite ChV),   ," }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "A:=Int(x/(1
+x^2),x);changevar(u=1+x^2,A,u);value(%);A=subs(u=1+x^2,%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%$IntG6$*&%\"xG\"\"\",&F*F**$)F)\"\"#
F*F*!\"\"F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&\"\"\"F(%
\"uG!\"\"#F(\"\"#F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#%\"uG
#\"\"\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&%\"xG\"\"
\",&F)F)*$)F(\"\"#F)F)!\"\"F(,$-%#lnG6#F*#F)F-" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 83 "B:=Int(2*x/(1+x^2)^n,x);value(%);changevar(u=1
+x^2,B,u);value(%);B=subs(u=1+x^2,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"BG-%$IntG6$,$*&%\"xG\"\"\"),&F+F+*$)F*\"\"#F+F+%\"nG!\"\"F0F*" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&*&*$)%\"xG\"\"#\"\"\"F*,&!\"\"F*
%\"nGF*F,F,*&F*F*F+F,F,F*-%$expG6#*&F-F*-%#lnG6#,&F*F**$F'F*F*F*F," }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F')%\"uG%\"nG!\"\"F)
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"uG\"\"\"*&,&!\"\"F&%\"nGF&F
&-%$expG6#*&F*F&-%#lnG6#F%F&F&F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
/-%$IntG6$,$*&%\"xG\"\"\"),&F*F**$)F)\"\"#F*F*%\"nG!\"\"F/F),$*&F,F**&
,&F1F*F0F*F*-%$expG6#*&F0F*-%#lnG6#F,F*F*F1F1" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 69 "C:=Int(tan(x),x);changevar(u=cos(x),C,u);value(%
);C=subs(u=cos(x),%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG-%$IntG
6$-%$tanG6#%\"xGF+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&\"
\"\"F(%\"uG!\"\"F*F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#%\"u
G!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$tanG6#%\"xGF*,$
-%#lnG6#-%$cosGF)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "
De:=Int(sqrt(1-x^2),x);changevar(x=sin(u),De,u);value(%);simplify(%);D
e=subs(u=arcsin(x),%);simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%#DeG-%$IntG6$*$-%%sqrtG6#,&\"\"\"F-*$)%\"xG\"\"#F-!\"\"F-F0" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&-%%sqrtG6#,&\"\"\"F+*$)-%$s
inG6#%\"uG\"\"#F+!\"\"F+-%$cosGF0F+F1" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#,&*&-%$sinG6#%\"uG\"\"\"-%%sqrtG6#,&F)F)*$)F%\"\"#F)!\"\"F)#F)F0*&F
2F)-%'arcsinG6#F%F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(-%$sinG6#
%\"uG\"\"\"-%%csgnG6#-%$cosGF'F)F-F)#F)\"\"#*&F/F)-%'arcsinG6#F%F)F)" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*$-%%sqrtG6#,&\"\"\"F,*$)%
\"xG\"\"#F,!\"\"F,F/,&*(-%$sinG6#-%'arcsinG6#F/F,-%%csgnG6#-%$cosGF6F,
F=F,#F,F0*&F?F,-F86#F4F,F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG
6$*$-%%sqrtG6#,&\"\"\"F,*$)%\"xG\"\"#F,!\"\"F,F/,&*&F/F,F(F,#F,F0*&F4F
,-%'arcsinG6#F/F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "E:=I
nt(arctan(x),x);intparts(E,arctan(x));changevar(u=1+x^2,%,u); value(%)
;E=subs(u=1+x^2,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG-%$IntG6$
-%'arctanG6#%\"xGF+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&-%'arctanG6
#%\"xG\"\"\"F(F)F)-%$IntG6$*&F(F),&F)F)*$)F(\"\"#F)F)!\"\"F(F2" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&-%'arctanG6#%\"xG\"\"\"F(F)F)-%$In
tG6$,$*&F)F)%\"uG!\"\"#F)\"\"#F/F0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
,&*&-%'arctanG6#%\"xG\"\"\"F(F)F)*&#F)\"\"#F)-%#lnG6#%\"uGF)!\"\"" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%'arctanG6#%\"xGF*,&*&F'\"
\"\"F*F-F-*&#F-\"\"#F--%#lnG6#,&F-F-*$)F*F0F-F-F-!\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 245 "Les deux m\351thodes : int\351gration pa
r parties IPP et changement de variable ChV sont \351galement  applica
bles aux int\351grales d\351finies ainsi qu'aux int\351grales impropre
s. Attention tout de m\352me aux changements de bornes de l'int\351gra
le  pour le ChV !!!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "resta
rt;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%\"DG%%DiffG%*DoubleintG%$IntG%&Limi
tG%(LineintG%(ProductG%$SumG%*TripleintG%*changevarG%/completesquareG%
)distanceG%'equateG%*integrandG%*interceptG%)intpartsG%(leftboxG%(left
sumG%)makeprocG%*middleboxG%*middlesumG%)midpointG%(powsubsG%)rightbox
G%)rightsumG%,showtangentG%(simpsonG%&slopeG%(summandG%*trapezoidG" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 317 26 "
Exercice 2.2 -  Fonction  " }{XPPEDIT 18 0 "beta(m,n)" "6#-%%betaG6$%
\"mG%\"nG" }{TEXT 318 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 164 "La fonction Beta(x,y) est une fonction de Mapl
e d\351finissable avec la fonction GAMMA (voir aide  Maple). On  donne
 ici une d\351finition \351quivalente de cette fonction  " }{XPPEDIT 
18 0 "beta(m,n)" "6#-%%betaG6$%\"mG%\"nG" }{TEXT -1 35 " , valable pou
r tout r\351el  m, n >0." }}{PARA 266 "" 0 "" {XPPEDIT 18 0 "beta:= (m
,n)->int(x^(m-1)*(1-x)^(n-1),x=0..1)" "6#>%%betaGR6$%\"mG%\"nG7\"6$%)o
peratorG%&arrowG6\"-%$intG6$*&)%\"xG,&F'\"\"\"F5!\"\"F5),&F5F5F3F6,&F(
F5F5F6F5/F3;\"\"!F5F-F-F-" }}{PARA 0 "" 0 "" {TEXT -1 31 "   a) Utilis
er la forme inerte " }{TEXT 326 3 "Int" }{TEXT -1 14 " pour d\351finir
 " }{XPPEDIT 18 0 "beta" "6#%%betaG" }{TEXT -1 82 " . Tester la nature
 de cette int\351grale impropre avec quelques valeurs r\351elles de  \+
" }{TEXT 319 1 "m" }{TEXT -1 6 "  et  " }{TEXT 320 1 "n" }{TEXT -1 52 
" .Que remarquez-vous en permutant les deux valeurs  " }{TEXT 321 5 "m
 , n" }{TEXT -1 2 " ?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "beta:= (m,n)->Int(x^(m-1)*(1-x)^(n
-1),x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%betaGR6$%\"mG%\"nG6
\"6$%)operatorG%&arrowGF)-%$IntG6$*&)%\"xG,&9$\"\"\"F5!\"\"F5),&F5F5F2
F6,&9%F5F5F6F5/F2;\"\"!F5F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 38 "beta(1,2);value(%);beta(2,1);value(%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$IntG6$,&\"\"\"F'%\"xG!\"\"/F(;\"\"!F'" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
-%$IntG6$%\"xG/F&;\"\"!\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"
\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "   b)  En admettant qu
e  " }{TEXT 322 1 "m" }{TEXT -1 6 "  et  " }{TEXT 323 1 "n" }{TEXT -1 
40 "  sont positives, comparer les valeurs (" }{TEXT 327 5 "value" }
{TEXT -1 22 ") des deux fonctions  " }{TEXT 324 9 "Beta(m,n)" }{TEXT 
-1 6 "  et  " }{TEXT 325 9 "beta(m,n)" }{TEXT -1 2 " ." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "assume(m,positive) ;additionally(n,
positive); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Beta(m,n)-be
ta(m,n);value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%%BetaG6$%#m|i
rG%#n|irG\"\"\"-%$IntG6$*&)%\"xG,&F'F)F)!\"\"F)),&F)F)F/F1,&F(F)F)F1F)
/F/;\"\"!F)F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "   c)  Ca
lculer " }{XPPEDIT 18 0 "beta(n,m)" "6#-%%betaG6$%\"nG%\"mG" }{TEXT 
-1 74 ". Effectuer ensuite le changement de variable  X=1-x   dans la \+
formule de " }{XPPEDIT 18 0 "beta(m,n)" "6#-%%betaG6$%\"mG%\"nG" }
{TEXT -1 46 " pour en d\351duire une relation \351vidente entre  " }
{XPPEDIT 18 0 "beta(m,n)" "6#-%%betaG6$%\"mG%\"nG" }{TEXT -1 7 "  et  \+
 " }{XPPEDIT 18 0 "beta(n,m)" "6#-%%betaG6$%\"nG%\"mG" }{TEXT -1 3 "  \+
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "beta(n,m);changevar(X=
1-x,beta(m,n),X);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG
,&%#n|irG\"\"\"F+!\"\"F+),&F+F+F(F,,&%#m|irGF+F+F,F+/F(;\"\"!F+" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&),&\"\"\"F)%\"XG!\"\",&%#m|
irGF)F)F+F))F*,&%#n|irGF)F)F+F)/F*;\"\"!F)" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 44 "   d)  Donner une forme trigonom\351trique de  " }
{XPPEDIT 18 0 "beta(m,n)" "6#-%%betaG6$%\"mG%\"nG" }{TEXT -1 33 "  en \+
passant par  le ChV :   x = " }{XPPEDIT 18 0 "(sin(t))^2." "6#)-%$sinG
6#%\"tG$\"\"#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "chang
evar(x=sin(t)^2,beta(m,n),t);subs(1-sin(t)^2=cos(t)^2,%);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#-%$IntG6$,$**)*$)-%$sinG6#%\"tG\"\"#\"\"\",&%#m
|irGF0F0!\"\"F0),&F0F0F)F3,&%#n|irGF0F0F3F0F+F0-%$cosGF-F0F//F.;\"\"!,
$%#PiG#F0F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$**)*$)-%$sin
G6#%\"tG\"\"#\"\"\",&%#m|irGF0F0!\"\"F0)*$)-%$cosGF-F/F0,&%#n|irGF0F0F
3F0F+F0F7F0F//F.;\"\"!,$%#PiG#F0F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 173 "   e)   Avec l'aide du Maple on peut avoir l'expression de  Be
ta(x,y)  en fonction de GAMMA. Modifier cette expression afin d'\351ta
blir une relation entre les coefficients :  " }{TEXT 328 14 "binomial(
n,k) " }{TEXT -1 19 " et les valeurs de " }{TEXT 329 9 "Beta(n,k)" }
{TEXT -1 27 " . Rapellons qu'on a d\351j\340  " }{XPPEDIT 18 0 "Gamma(
n+1) = n!" "6#/-%&GammaG6#,&%\"nG\"\"\"F)F)-%*factorialG6#F(" }{TEXT 
-1 39 "  avec les r\351sultats de l'Exercice 1.2." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 14 "n:='n';m:='m';" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"nGF$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mGF$" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 74 "Avec copier-coller (et des modifications \+
de noms de variables) on obtient:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "Beta(x,y) :=(GAMMA(x) * GAMMA(y))/GAMMA(x+y);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%%BetaG6$%\"xG%\"yG*&*&-%&GAMMAG6#F'
\"\"\"-F,6#F(F.F.-F,6#,&F'F.F(F.!\"\"" }}}{EXCHG {PARA 267 "" 1 "" 
{TEXT -1 54 "Soit une version symbolique de la fonction Beta(x,y) ." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "BETA := (x,y)-> (GAMMA(x) \+
* GAMMA(y))/GAMMA(x+y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%BETAGR6$
%\"xG%\"yG6\"6$%)operatorG%&arrowGF)*&*&-%&GAMMAG6#9$\"\"\"-F06#9%F3F3
-F06#,&F2F3F6F3!\"\"F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "1/BETA(n-k,k+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%&GAMMAG6#,
&%\"nG\"\"\"F)F)F)*&-F%6#,&F(F)%\"kG!\"\"F)-F%6#,&F.F)F)F)F)F/" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "binomial(n,k):=n!/k!/(n-k)!=
GAMMA(n+1)/GAMMA(k+1)/GAMMA(n-k+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>-%)binomialG6$%\"nG%\"kG/*&-%*factorialG6#F'\"\"\"*&-F,6#F(F.-F,6#,&
F'F.F(!\"\"F.F5*&-%&GAMMAG6#,&F'F.F.F.F.*&-F86#,&F(F.F.F.F.-F86#,(F'F.
F(F5F.F.F.F5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 141 "Visiblement le p
roduit   binomial(n,k)* (n-k+1)* Beta(n-k, k+1)  est \351gal \340  1  \+
  ( m\352me si la simplification avec Maple me r\351siste un peut)." }
}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 330 41 "Deux
i\350me partie :  Int\351grales et s\351ries ." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 125 "Reprenons  quelques m\351thodes \+
 de  TP5 et TP6 afin de trouver des liens entre les sommes infinies et
 les int\351grales impropres." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 4 "" 0 "" {TEXT -1 36 "1. S\351ries et  int\351grales de Riemann
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 187 "On \+
consid\351re les s\351ries num\351riques de la forme :  Sum(1/(n^a),n \+
= 1 .. infinity) pour les comparer graphiquement avec les int\351grale
s impropres de la forme Int(1/(x^a),x = 1 .. infinity) :" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "serie[a]:= Sum(1/n^a,n=1..infinity)
; integral[a]:=Int(1/x^a,x=1..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>&%&serieG6#%\"aG-%$SumG6$*&\"\"\"F,)%\"nGF'!\"\"/F.;F,%)infinit
yG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%)integralG6#%\"aG-%$IntG6$*&
\"\"\"F,)%\"xGF'!\"\"/F.;F,%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 333 11 "Exemple 3. " }{TEXT -1 112 "-Pour d\351terminer la natur
e d'une s\351rie ou d'une int\351grale impropre on fait appel \340 des
 limites correspondantes. " }}{PARA 0 "" 0 "" {TEXT -1 20 "La somme  S
  de la  " }{TEXT 331 9 "serie[a] " }{TEXT -1 117 "ci-dessus prendra d
onc la valeur (finie ou infinie) correspondante \340 la limite des som
mes partielles s[n]  suivantes:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 75 "s[n]=Sum(1/i^a,i=1..n);S:=Sum(1/i^a,i=1..infinity)= Limit(s[n]
,n=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/&%\"sG6#%\"nG-%$SumG
6$*&\"\"\"F,)%\"iG%\"aG!\"\"/F.;F,F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"SG/-%$SumG6$*&\"\"\"F*)%\"iG%\"aG!\"\"/F,;F*%)infinityG-%&LimitG
6$&%\"sG6#%\"nG/F8F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "De la m
\352me fa\347on on aura pour l'int\351grale " }{TEXT 332 13 " integral
[a] " }{TEXT -1 44 " la valeur J obtenue par le calcul suivant :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "j[n]=Int(1/x^a,x=1..n);J:=In
t(1/x^a,x=1..infinity)= Limit(j[n],n=infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/&%\"jG6#%\"nG-%$IntG6$*&\"\"\"F,)%\"xG%\"aG!\"\"/F.;F,
F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"JG/-%$IntG6$*&\"\"\"F*)%\"xG
%\"aG!\"\"/F,;F*%)infinityG-%&LimitG6$&%\"jG6#%\"nG/F8F1" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 68 "Pour comparerer (graphiquement et alg\350
briquement)  les deux suites \{" }{XPPEDIT 18 0 "s[n]" "6#&%\"sG6#%\"n
G" }{TEXT -1 7 "\} et  \{" }{XPPEDIT 18 0 "j[n]" "6#&%\"jG6#%\"nG" }
{TEXT -1 75 "\}on fixe le param\350tre a (a=1, a=2, a=1/2, ...) et on \+
utilise des fonctions " }{TEXT 334 16 "rightbox/leftbox" }{TEXT -1 3 "
 , " }{TEXT 335 17 "rightsum/leftsum " }{TEXT -1 12 " du package " }
{TEXT 336 7 "student" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "with(student);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7@%
\"DG%%DiffG%*DoubleintG%$IntG%&LimitG%(LineintG%(ProductG%$SumG%*Tripl
eintG%*changevarG%/completesquareG%)distanceG%'equateG%*integrandG%*in
terceptG%)intpartsG%(leftboxG%(leftsumG%)makeprocG%*middleboxG%*middle
sumG%)midpointG%(powsubsG%)rightboxG%)rightsumG%,showtangentG%(simpson
G%&slopeG%(summandG%*trapezoidG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
34 "Prenons les valeurs concretes de  " }{TEXT 337 1 "a" }{TEXT -1 21 
"   et   d'un entier  " }{TEXT 338 2 " k" }{TEXT -1 73 " - pour l'inte
rvale d'int\351gration [1,k+1] ainsi que pour la la taille (n=" }
{TEXT 339 1 "k" }{TEXT -1 40 ") de sa subdivision puis pour l'indice  \+
" }{TEXT 340 1 "n" }{TEXT -1 5 "  de " }{XPPEDIT 18 0 "s[n]" "6#&%\"sG
6#%\"nG" }{TEXT -1 8 "  et de " }{XPPEDIT 18 0 "j[n]" "6#&%\"jG6#%\"nG
" }{TEXT -1 15 "  \340 comparer.  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "a:=1.1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "k:=10;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"#6!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"kG\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
44 "rightbox(1/x^a, x=1..k+1, k, shading=green);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 
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r$\"36:k`BP'H8\"F[s7$$\"3+++]P![hY(Fhr$\"3CB=1')HX&4\"F[s7$$\"3iKLL$Qx
$owFhr$\"3q\"*Qd#=<P1\"F[s7$$\"3Y+++v.I%)yFhr$\"35Z`(p%orJ5F[s7$$\"3?m
m\"zpe*z!)Fhr$\"3ZS1ET(oU+\"F[s7$$\"3;,++D\\'QH)Fhr$\"3'QLNIxY\"e(*!#>
7$$\"31KLe9S8&\\)Fhr$\"37pJL&=QT]*Fd_l7$$\"3h,+D1#=bq)Fhr$\"3'4&)=\"QI
z^#*Fd_l7$$\"3!QLL$3s?6*)Fhr$\"3#*)Q&)\\dgr,*Fd_l7$$\"3a***\\7`Wl7*Fhr
$\"3AA[)4b1My)Fd_l7$$\"3enmmm*RRL*Fhr$\"3%f:&39v'*o&)Fd_l7$$\"3%zmmTvJ
ga*Fhr$\"3-LVF&zz(f$)Fd_l7$$\"3]MLe9tOc(*Fhr$\"3'4Xy1SX<;)Fd_l7$$\"31,
++]Qk\\**Fhr$\"37)H'Q[a^()zFd_l7$$\"3[LL3dg6<5!#;$\"3eAFtT*pjz(Fd_l7$$
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7F8F+F+-%*THICKNESSG6#F2-%&STYLEG6#%%LINEG-%+AXESLABELSG6$Q\"x6\"Q!6\"
-%%VIEWG6$;F(F]r%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve
 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" }}}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sr:=rightsum(1/x^a, x=1..k+1);evalf
(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#srG,$-%$SumG6$*&\"\"\"F**$)
,&F*F**&#\"\"&\"\"#F*%\"iGF*F*$\"#6!\"\"F*F5/F2;F*\"\"%F/" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"+%[nZR\"!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "j[k+1]:=Int(1/x^a,x=1..k+1)=evalf(%);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>&%\"jG6#\"#6/-%$IntG6$*&\"\"\"F-*$)%\"xG$F'!\"\"F
-F2/F0;F-F'/$\"+yb1K@!\"*$\"+%[nZR\"F8" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 341 11 "Question 1." }{TEXT -1 98 "   D\351duire de ces op\351ra
tions  une majoration des   s[n]  avec  j[n]  ( et puis avec  J,  quan
d  J " }{TEXT 343 8 "converge" }{TEXT -1 2 ")." }}}{EXCHG {PARA 11 "" 
1 "" {TEXT 342 11 "Reponse:   " }{TEXT -1 56 " s[n]  < 1 +  j[n]      \+
(et   s[n]  < 1 + J ,  quand  J-" }{TEXT 344 3 "cv " }{TEXT -1 1 ")" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "leftbox(1/x^a, x=1..k+1,k,
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$F]`lF^]lF_^l-Fh]l6$7&F_`l7$F]`l$\"+X)zEq\"Fg^l7$$\"\"'F*Fd`l7$Fg`lF^]
lF_^l-Fh]l6$7&Fi`l7$Fg`l$\"+qYE$R\"Fg^l7$$\"\"(F*F^al7$FaalF^]lF_^l-Fh
]l6$7&Fcal7$Faal$\"+M*ef<\"Fg^l7$$\"\")F*Fhal7$F[blF^]lF_^l-Fh]l6$7&F]
bl7$F[bl$\"+&\\:`,\"Fg^l7$$\"\"*F*Fbbl7$FeblF^]lF_^l-Fh]l6$7&Fgbl7$Feb
l$\"+'o]$>*)!#67$$\"#5F*F\\cl7$F`clF^]lF_^l-Fh]l6$7&Fbcl7$F`cl$\"+ZBGV
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F(Fc\\l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve
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" 1 "" {XPPMATH 20 "6#>%#slG,$-%$SumG6$*&\"\"\"F**$),&F*F**&#\"\"*\"\"
%F*%\"iGF*F*$\"#6!\"\"F*F5/F2;\"\"!\"\"$F/" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+E2(oW$!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 37 "j[k+1]:=Int(1/x^a,x=1..k+1);evalf(%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"jG6#\"#6-%$IntG6$*&\"\"\"F,*$)%\"xG$F'!\"\"F,F1/F/
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{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 345 10 "
Question2." }{TEXT -1 77 "   Retrouver  une minoration des   s[n]  par
 les int\351grales corr\351spondantes. " }}}{EXCHG {PARA 11 "" 1 "" 
{TEXT 346 11 "Reponse:   " }{TEXT -1 21 "  j[n+1] < s[n]      " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 347 10 "Question3." }{TEXT -1 81 "   En d
\351duire  des relations minoration/majorations entre les limites de s
uites \{" }{XPPEDIT 18 0 "s[n]" "6#&%\"sG6#%\"nG" }{TEXT -1 7 "\} et  \+
\{" }{XPPEDIT 18 0 "j[n]" "6#&%\"jG6#%\"nG" }{TEXT -1 92 "\}. Conclure
 par une comparaison \351vidente de la nature de sommes et d'int\351gr
ales de Riemann. " }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{MARK "109 0 0
" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }