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{SECT 0 {PARA 0 "" 0 "" {TEXT 260 1 "(" }{TEXT 262 1 "1" }{TEXT 261 3 
")  " }{TEXT 258 34 "Si calcolino tutte le primitive di" }{TEXT 256 1 
" " }{TEXT -1 2 "  " }{XPPEDIT 18 0 "f := x->1/(x^2-x)" ">%\"fG:6#%\"x
G7\"6$%)operatorG%&arrowG6\"*&\"\"\"\"\"\",&*$F&\"\"#F.F&!\"\"F2F+F+" 
}{TEXT -1 2 "  " }{TEXT 257 65 "per x in (0, 1) e si determini la prim
itiva F tale che F(1/2) = 0" }{TEXT -1 47 ".\nPotremmo usare direttame
nte Maple e ottenere;" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f:=
x->1/(x^2-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Int(f(x),x
) = int(f(x),x);" }}}{PARA 0 "" 0 "" {TEXT 259 2 "ma" }{TEXT -1 246 " \+
in casi come questo, in cui f(x) \350 definito in un'uniione di interv
alli, occorre stare attenti: potrebbe produrre primitive buone su uno \+
e non su tutti.\nControlliamo a mano (e, comunque, in un compito dobbi
amo giustificare l'intergale ottenuto):" }}{PARA 0 "" 0 "" {TEXT -1 
121 "idea: cerco di espriemere l'integranda come somma di frazioni ave
nti per denominatore i fattori del denominatore attuale:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "a/x+b/(x-1)=1/(x^2-x); simplify(a/x
+b/(x-1)=1/(x^2-x));" }}}{PARA 0 "" 0 "" {TEXT -1 43 "Posso prendere  \+
a = -1  e  b = 1. Verifica:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "simplify(-1/x+1/(x-1));" }}}{PARA 0 "" 0 "" {TEXT -1 6 " OK   " }
{XPPEDIT 18 0 "int(f(x),x) = int(-1/x+1/(x-1),x)" "/-%$intG6$-%\"fG6#%
\"xGF)-F$6$,&*&\"\"\"\"\"\"F)!\"\"F0*&\"\"\"F/,&F)F/\"\"\"F0F0F/F)" }
{TEXT -1 169 " \n\nSo che le primitive di 1/t hanno la forma log(t)+c1
 per t > 0  e  log(-t)+c2  per t < 0.\nPer x in (0,1)   x > 0  e  x-1 \+
< 0, per cui, a meno di costanti additive,\n   " }{XPPEDIT 18 0 "int(1
/x,x) = log(x)" "/-%$intG6$*&\"\"\"\"\"\"%\"xG!\"\"F)-%$logG6#F)" }
{TEXT -1 7 "   e   " }{XPPEDIT 18 0 "int(1/(x-1),x) = log(1-x)" "/-%$i
ntG6$*&\"\"\"\"\"\",&%\"xGF(\"\"\"!\"\"F,F*-%$logG6#,&\"\"\"F(F*F," }
{TEXT -1 10 " \nQuindi  " }{XPPEDIT 18 0 "int(f(x),x) = -ln(x)+ln(1-x)
" "/-%$intG6$-%\"fG6#%\"xGF),&-%#lnG6#F)!\"\"-F,6#,&\"\"\"\"\"\"F)F.F3
" }{TEXT -1 69 "   ( +c)\nOra cerco F(x) = -log(x) + log(1-x) + c  tal
e che F(1/2) = 0" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "solve(-l
og(1/2)+log(1-1/2)+c = 0,c);" }}}{PARA 0 "" 0 "" {TEXT -1 40 " F(x) = \+
 -log(x) - log(1-x).   Verifica:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 59 "F := x -> -log(x)+log(1-x): F(1/2); simplify(diff(F(x),x));" }
}}{PARA 0 "" 0 "" {TEXT -1 2 "OK" }}{PARA 0 "" 0 "" {TEXT -1 25 "-----
-------------------\n" }{TEXT 263 4 "(2) " }{TEXT 267 1 " " }{TEXT 
268 34 "Si calcolino al variare di x in R:" }{TEXT -1 6 "\n(a)  " }
{XPPEDIT 18 0 "int(abs(t^2-1),t=2..x)" "-%$intG6$-%$absG6#,&*$%\"tG\"
\"#\"\"\"\"\"\"!\"\"/F*;\"\"#%\"xG" }{TEXT -1 13 "    e  (b)   " }
{XPPEDIT 18 0 "int(1/(t^2+9),t=0..x)" "-%$intG6$*&\"\"\"\"\"\",&*$%\"t
G\"\"#F'\"\"*F'!\"\"/F*;\"\"!%\"xG" }{TEXT -1 2 "  " }}{PARA 0 "" 0 "
" {TEXT -1 1 "(" }{TEXT 266 1 "a" }{TEXT -1 40 ")  Pensiamo l'andament
o dell'integranda:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f:= t \+
-> abs(t^2-1): plot (f,-3..3);" }}}{PARA 0 "" 0 "" {TEXT -1 12 "Per x \+
>= 1  " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "int(abs(t^2-1),t=2..x) = int(
t^2-1,t=2..x) " "/-%$intG6$-%$absG6#,&*$%\"tG\"\"#\"\"\"\"\"\"!\"\"/F+
;\"\"#%\"xG-F$6$,&*$F+\"\"#F-\"\"\"F//F+;\"\"#F3" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "x^3/3-x-2/3" ",(*&%\"xG\"\"$\"\"$!\"\"\"\"\"F$F'*&\"\"#
F(\"\"$F'F'" }}{PARA 0 "" 0 "" {TEXT -1 44 "in quanto t varia in un in
tervallo in cui   " }{XPPEDIT 18 0 "abs(t^2-1)= t^2-1" "/-%$absG6#,&*$
%\"tG\"\"#\"\"\"\"\"\"!\"\",&*$F(\"\"#F*\"\"\"F," }{TEXT -1 20 "  \nPe
r x  in [-1,1]\n" }{XPPEDIT 18 0 "int(abs(t^2-1),t=2..x) = int(abs(t^2
-1),t=2..1)+int(abs(t^2-1),t=1..x) " "/-%$intG6$-%$absG6#,&*$%\"tG\"\"
#\"\"\"\"\"\"!\"\"/F+;\"\"#%\"xG,&-F$6$-F'6#,&*$F+\"\"#F-\"\"\"F//F+;
\"\"#\"\"\"F--F$6$-F'6#,&*$F+\"\"#F-\"\"\"F//F+;\"\"\"F3F-" }{TEXT -1 
4 "  = " }{XPPEDIT 18 0 "1^3/3-1-2/3 + int(-t^2+1,t=1..x)" ",**&\"\"\"
\"\"$\"\"$!\"\"\"\"\"\"\"\"F'*&\"\"#F(\"\"$F'F'-%$intG6$,&*$%\"tG\"\"#
F'\"\"\"F(/F2;\"\"\"%\"xGF(" }{TEXT -1 3 "\n= " }{XPPEDIT 18 0 "1/3-1-
2/3-x^3/3+x-2/3" ",.*&\"\"\"\"\"\"\"\"$!\"\"F%\"\"\"F'*&\"\"#F%\"\"$F'
F'*&%\"xG\"\"$\"\"$F'F'F-F%*&\"\"#F%\"\"$F'F'" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "x-x^3/3-2" ",(%\"xG\"\"\"*&F#\"\"$\"\"$!\"\"F(\"\"#F(" 
}{TEXT -1 86 "\nAbbiamo separato quanto accade per t tra 2 e 1 e per t
 che varia in [-1,1], in cui   " }{XPPEDIT 18 0 "abs(t^2-1)=-(t^2-1)" 
"/-%$absG6#,&*$%\"tG\"\"#\"\"\"\"\"\"!\"\",$,&*$F(\"\"#F*\"\"\"F,F," }
{TEXT -1 13 "\nPer x <= -1\n" }{XPPEDIT 18 0 "int(abs(t^2-1),t=2..x) =
 int(abs(t^2-1),t=2..1)+int(abs(t^2-1),t=1..-1) +int(abs(t^2-1),t=-1..
x) " "/-%$intG6$-%$absG6#,&*$%\"tG\"\"#\"\"\"\"\"\"!\"\"/F+;\"\"#%\"xG
,(-F$6$-F'6#,&*$F+\"\"#F-\"\"\"F//F+;\"\"#\"\"\"F--F$6$-F'6#,&*$F+\"\"
#F-\"\"\"F//F+;\"\"\",$\"\"\"F/F--F$6$-F'6#,&*$F+\"\"#F-\"\"\"F//F+;,$
\"\"\"F/F3F-" }{TEXT -1 6 "  \n=  " }{XPPEDIT 18 0 "1^3/3-1-2/3 + (-1)
 - (-1)^3/3-2 +  int(t^2-1,t=-1..x)" ",0*&\"\"\"\"\"$\"\"$!\"\"\"\"\"
\"\"\"F'*&\"\"#F(\"\"$F'F',$\"\"\"F'F(*&,$\"\"\"F'\"\"$\"\"$F'F'\"\"#F
'-%$intG6$,&*$%\"tG\"\"#F(\"\"\"F'/F:;,$\"\"\"F'%\"xGF(" }{TEXT -1 4 "
\n=  " }{XPPEDIT 18 0 "-4+x^3/3-x-2/3" ",*\"\"%!\"\"*&%\"xG\"\"$\"\"$F
$\"\"\"F&F$*&\"\"#F)\"\"$F$F$" }{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "x^3
/3-x-10/3" ",(*&%\"xG\"\"$\"\"$!\"\"\"\"\"F$F'*&\"#5F(\"\"$F'F'" }
{TEXT -1 25 "\n  I calcoli con Maple:  " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 144 "int(t^2-1,t = 2 .. x);\nint(t^2-1,t = 2 .. 1)+int(-(
t^2-1),t = 1 .. x);\nint(t^2-1,t = 2 .. 1)+int(-(t^2-1),t = 1 .. -1)+i
nt((t^2-1),t = -1 .. x);" }}}{PARA 0 "" 0 "" {TEXT -1 143 "Facciamo un
 controllo grafico; dovremmo ottenere una funzione continua e un grafi
co con una simmetria centrale rispetto al pun to di ascissa 0:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "F := x -> piecewise(x>=1,1/3
*x^3-x-2/3,x<1 and x>=-1,-2-1/3*x^3+x,x<-1,-10/3+1/3*x^3-x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "plot (F,-3..3);" }}}{PARA 0 
"" 0 "" {TEXT -1 5 "OK\n\n(" }{TEXT 265 1 "b" }{TEXT -1 78 ") Facilmen
te fattibile con la sostituzione  t = 3u, u=t/3, t^2 = 9u, dt = 3 du" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "Int(1/(t^2+9),t=0..x) = 1
/3*Int(1/(u^2+1),u=0..x/3);\n1/3*Int(1/(u^2+1),u=0..x/3) = 1/3*int(1/(
u^2+1),u=0..x/3);" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Controllo con Maple
:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Int(1/(t^2+9),t=0..x) =
 int(1/(t^2+9),t=0..x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT 264 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "30 0" 0 }{VIEWOPTS 1 1 0 1 
1 1803 }