{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 297 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 298 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "C\341lculo II" }}{PARA 257 "" 0 "" {TEXT -1 31 "Li\347\343o 13a: Integrais Impr\363prias" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "At\351 ent\343o, em nosso estudo d e integra\347\343o, temos considerado " }{XPPEDIT 18 0 "Int(f(x),x = a .. b);" "6#-%$IntG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 9 " , send o " }{TEXT 268 1 "f" }{TEXT -1 36 " \351 uma fun\347\343o no intervalo limitado " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 57 ". \+ Deseja-se agora analisar o que acontece quando ambos, " }{TEXT 269 1 "f" }{TEXT -1 15 " ou o intervalo" }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG %\"bG" }{TEXT -1 56 " , tornam-se ilimitados. Em ambos os casos, tem- se uma " }{TEXT 270 18 "integral impr\363pria" }{TEXT -1 49 " (as inte grais estudadas at\351 agora s\343o integrais " }{TEXT 271 8 "pr\363pr ias" }{TEXT -1 157 "). Como veremos mais adiante, uma integral impr \363pria n\343o \351 definida diretamente em termos de parti\347\365es e somas, mas sim como o limite de integrais pr\363prias." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 28 "Tipo 1: I ntervalos Infinitos" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Iniciaremos com o caso onde o intervalo de integra\347\343o, [" }{XPPEDIT 18 0 "a ,infinity;" "6$%\"aG%)infinityG" }{TEXT -1 16 "), \351 ilimitado \340 " }{TEXT 272 1 " " }{TEXT -1 32 "esquerda. Neste caso, definimos" }}} {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "Int(f(x),x = a \+ .. infinity) := limit(Int(f(x),x = a .. b),b = infinity);" "6#>-%$IntG 6$-%\"fG6#%\"xG/F*;%\"aG%)infinityG-%&limitG6$-F%6$-F(6#F*/F*;F-%\"bG/ F8F." }{TEXT -1 2 " ," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "contanto que o limite exista. Se " }{TEXT 273 1 "f" }{TEXT -1 99 " \351 uma f un\347\343o positiva, este limite pode ser interpretado como sendo a \+ \341rea total sob o gr\341fico de " }{TEXT 274 1 "f" }{TEXT -1 15 ", \+ \340 direita de " }{TEXT 275 1 "a" }{TEXT -1 72 "; em geral, tem-se as mesmas interpreta\347\365es que para a integral pr\363pria." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "O " }{TEXT 276 5 "Maple" }{TEXT -1 67 " \351 capaz de resolver muitas integrais impr \363prias como, por exemplo," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Int(1/(2*x - 5)^3, x=4..infi nity) = int(1/(2*x - 5)^3, x=4..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$),&*&\"\"#F(%\"xGF(F(\"\"&!\"\"\" \"$F(F0/F.;\"\"%%)infinityG#F(\"#O" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := x->1/(2*x - 5)^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&\"\"\"F-*$),&*&\"\"#F -9$F-F-\"\"&!\"\"\"\"$F-F5F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "p1 := plot(f(x), x=3.5..8,thickness=2):\np2 := plott ools[line]([4,0],[4,f(4)],color=blue,thickness=2):\nplots[display](p1, p2);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVE SG6%7Y7$$\"3++++++++N!#<$\"3+++++++]7!#=7$$\"38+vV)z@X_$F*$\"35M\"3?Qv B;\"F-7$$\"3E+](ofV!\\NF*$\"3%onGO,`F3\"F-7$$\"3S+DJ&RlNd$F*$\"3r$f<0' )R-,\"F-7$$\"33++v$>(3)f$F*$\"32=g#*4')fS%*!#>7$$\"3+]i:l(f2k$F*$\"3I' fb!RtH?%)FB7$$\"3!**\\ilLKMo$F*$\"3j%QWGs\")=a(FB7$$\"3.]P41@UJPF*$\"3 ))3GDbM0%p'FB7$$\"39+]iv=TzPF*$\"3#Qi!)f.)oofFB7$$\"3=++]_(>x#QF*$\"3# Hl0Qw31M&FB7$$\"3y**\\PHw-wQF*$\"31mX_r)\\wz%FB7$$\"3'**\\7`=%=sRF*$\" 3w**RWr8jFB7$$\"3*)***\\Pn_@W%F*$\"3El\"4Hw(Fbr7$$\"3=+Dc12N*=&F*$\"3.^$Q93*REkFbr7$$ \"3U****\\A-\"yF&F*$\"3m&3W(\\czJeFbr7$$\"3k+DcJV'[P&F*$\"3%z&z)3fu3E& Fbr7$$\"3o+vo%z#GnaF*$\"3[36krHY%y%Fbr7$$\"3O+]il&>Fbr7$$\"3I++vo^$zf'F*$\"3S#fo]n8k\"=Fbr7$$ \"3y*\\iST\")fo'F*$\"3zw'QWW!>/$[npF*$\"3 4YD`UN\">S\"Fbr7$$\"3/++vVK/gqF*$\"3(>voxBm#=8Fbr7$$\"3M*\\i!R]%p:(F*$ \"3eE(yC;vwB\"Fbr7$$\"3]+++&)HF]sF*$\"3=a!H2r\\h;\"Fbr7$$\"3g**\\P*G9d M(F*$\"3EM=]%>,IM )F_[l7$$\"3))*\\P%eWA-zF*$\"3Gm@#pA?&GzF_[l7$$\"\")\"\"!$\"3-yd,4![J^( F_[l-%'COLOURG6&%$RGBG$\"#5!\"\"$Fg\\lFg\\lFa]l-%*THICKNESSG6#\"\"#-F$ 6%7$7$$\"\"%Fg\\lFa]l7$Fj]l$\"+/Pq.P!#6-F[]l6&F]]lFa]lFa]l$\"*++++\"! \")Fb]l-%+AXESLABELSG6$Q\"x6\"Q!Fi^l-%%VIEWG6$;$\"#NF`]lFe\\l%(DEFAULT G" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" " Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 197 "Contudo, o uso deste s comando nos oferecem pouca clareza. Desta forma, resolveremos a mes ma integral passo a passo. Inicialmente, \351 preciso resolver a inte gral de 4 a um limite superior vari\341vel " }{TEXT 277 1 "b" }{TEXT -1 117 ". Manualmente, \351 preciso encontrar uma anti-derivada para \+ a fun\347\343o e aplicar ent\343o o Teorema Fundamental do C\341lculo: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "ad := int(f(x), x);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%#adG,$*&\"\"\"F'*&\"\"%F'),&*&\"\"#F'%\"xGF'F'\"\"&!\"\"F-F'F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "inttob := subs(x=b, ad) - subs(x=4, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,&*&\"\" \"F'*&\"\"%F'),&*&\"\"#F'%\"bGF'F'\"\"&!\"\"F-F'F0F0#F'\"#OF'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Pode-se obter o mesmo resuldado di retamente com o " }{TEXT 278 5 "Maple" }{TEXT -1 34 ", no c\341lculo d a integral de 4 at\351 " }{TEXT 279 1 "b" }{TEXT -1 78 ". (A resposta aparece de forma diferente. Por\351m, verifica-se que \351 a mesma.) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "int(f(x), x=4..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #,$*(\"\"*!\"\",(\"\"%\"\"\"*$)%\"bG\"\"#F)F)*&\"\"&F)F,F)F&F),&*&F-F) F,F)F)F/F&!\"#F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplif y(inttob - int(f(x), x=4..b));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\" !" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Tendo encontrado a integral \+ at\351 " }{TEXT 280 1 "b" }{TEXT -1 66 ", de uma forma ou de outra, po de-se ent\343o tormar o limite quando " }{XPPEDIT 18 0 "proc (b) opti ons operator, arrow; infinity end;" "6#f*6#%\"bG7\"6$%)operatorG%&arro wG6\"%)infinityGF*F*F*" }{TEXT -1 2 " :" }{TEXT 281 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "Int(1/(2*x - 5)^3, x=4..infinity) = limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$Int G6$*&\"\"\"F(*$),&*&\"\"#F(%\"xGF(F(\"\"&!\"\"\"\"$F(F0/F.;\"\"%%)infi nityG#F(\"#O" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Esta resposta enc ontra-se em acordo com a resposta original do " }{TEXT 282 5 "Maple" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "Se o intervalo for ilimitado \340 esquerda, procedemos de forma similar, definindo" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = -infinity .. b) := limit (Int(f(x),x = a .. b),a = -infinity);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;, $%)infinityG!\"\"%\"bG-%&limitG6$-F%6$-F(6#F*/F*;%\"aGF0/F:,$F.F/" } {TEXT -1 2 " ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "Int(exp(3 *x), x=-infinity..2) = int(exp(3*x), x=-infinity..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$expG6#,$*&\"\"$\"\"\"%\"xGF-F-/F.;,$% )infinityG!\"\"\"\"#,$*&#F-F,F--F(6#\"\"'F-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "inttoa := int(exp(3*x), x=a..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoaG,&*&#\"\"\"\"\"$F(-%$expG6#\"\"'F(F(*&#F (F)F(-F+6#,$*&F)F(%\"aGF(F(F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "limit(inttoa, a=-infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"\"\"\"\"$F&-%$expG6#\"\"'F&F&" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Finalmente, se a integral \351 ilimitada de " } {TEXT 283 5 "ambos" }{TEXT -1 44 " os lados, escolhe-se um ponto conve niente, " }{TEXT 284 2 "c " }{TEXT -1 11 "(em geral, " }{TEXT 285 1 "c " }{TEXT -1 17 " = 0) e define-se" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = -infini ty .. infinity) := Int(f(x),x = -infinity .. c)+Int(f(x),x = c .. infi nity);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;,$%)infinityG!\"\"F.,&-F%6$-F(6# F*/F*;,$F.F/%\"cG\"\"\"-F%6$-F(6#F*/F*;F8F.F9" }{TEXT -1 2 " ," }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 286 55 "contanto que ambas as integrais do lado direito existam" }{TEXT -1 66 ". Esta condi\347 \343o \351 crucial. 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Deve-se ter um m\355nimo de cuidado, para que n\343o s e calcule esta integral como " }{XPPEDIT 18 0 "Limit(Int((1+(1+x^2)*si n(x))/(1+x^2),x = -a .. a),a = infinity);" "6#-%&LimitG6$-%$IntG6$*&,& \"\"\"F+*&,&F+F+*$%\"xG\"\"#F+F+-%$sinG6#F/F+F+F+,&F+F+*$F/F0F+!\"\"/F /;,$%\"aGF6F:/F:%)infinityG" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "finiteint := int((1 + (1+x^2)*sin(x))/(1+x^2),x = \+ -a .. a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*finiteintG,$*&\"\"#\" \"\"-%'arctanG6#%\"aGF(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Int(f(x) ,x = -infinity..infinity) = limit(finiteint, a=infinity); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&,&\"\"\"F)*&,&*$)%\"xG \"\"#F)F)F)F)F)-%$sinG6#F.F)F)F)F+!\"\"/F.;,$%)infinityGF3F7%#PiG" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 287 26 "ESTA RESPOSTA EST \301 ERRADA!" }{TEXT -1 96 " Para calcular corretamente a integral, d eve-se fazer com que as duas extremidades tendam a +/-" }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 32 " separadamente. Ao escolh ermos " }{XPPEDIT 18 0 "c = 0;" "6#/%\"cG\"\"!" }{TEXT -1 27 ", a inte gral impr\363pria at\351 " }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 17 " \351 calculada por " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "inttob := int(f(x), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,(-%'arctanG6#%\"bG\"\"\"-%$cosGF(!\"\"F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Int(f(x), x=0..infinity) \+ = limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$In tG6$*&,&\"\"\"F)*&,&*$)%\"xG\"\"#F)F)F)F)F)-%$sinG6#F.F)F)F)F+!\"\"/F. ;\"\"!%)infinityG;,$*&F/F3%#PiGF)F),&F/F)*&F/F3F;F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "A apar\352ncia um tanto estranha da resp osta mostra que algo de errado aconteceu. De fato, tem-se que a expres s\343o " }{TEXT 288 6 "inttob" }{TEXT -1 26 " n\343o possui limite qua ndo " }{XPPEDIT 18 0 "proc (b) options operator, arrow; infinity end; " "6#f*6#%\"bG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 105 " . N\343o h\341 necessidade de continuar: quando uma das integra is impr\363pria semi-infinitas n\343o existe, ent\343o " }{XPPEDIT 18 0 "Int((1+(1+x^2)*sin(x))/(1+x^2),x = -infinity .. infinity);" "6#-%$I ntG6$*&,&\"\"\"F(*&,&F(F(*$%\"xG\"\"#F(F(-%$sinG6#F,F(F(F(,&F(F(*$F,F- F(!\"\"/F,;,$%)infinityGF3F7" }{TEXT -1 19 " tamb\351m n\343o existe. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 9 "Quest\343o 1" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "Encontrar as i ntegrais impr\363pria a seguir. Caso n\343o seja poss\355vel, mostre \+ que n\343o existem. (N\343o \351 aconselh\341vel calcular com um \372 nico passo do " }{TEXT 289 5 "Maple" }{TEXT -1 49 ": calcule-as como l imites de integrais pr\363prias.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Int((2*x + 1)/(x^4 + 1), x=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&,&*&\"\"#\"\"\"%\"xGF*F*F*F*F*,&*$)F +\"\"%F*F*F*F*!\"\"/F+;\"\"!%)infinityG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttob := int((2*x + 1)/(x^4 + 1), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,**&#\"\"\"\"\")F(*&\"\"##F(F+-%#l nG6#*&,(*$)%\"bGF+F(F(*&F4F(F+F,F(F(F(F(,(F2F(F5!\"\"F(F(F7F(F(F(*&#F( \"\"%F(*&F+F,-%'arctanG6#,&F5F(F(F(F(F(F(*&F9F(*&F+F,-F=6#,&F5F(F(F7F( F(F(-F=6#F2F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "limit(intt ob, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&#\"\"\"\"\"%F& *&\"\"##F&F)%#PiGF&F&F&*&F*F&F+F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Int(1/(x*ln(x)), x=2..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F'*&%\"xGF'-%#lnG6#F)F'!\"\"/F);\"\"#% )infinityG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "inttob := int (1/(x*ln(x)), x=2..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,& -%#lnG6#-F'6#%\"bG\"\"\"-F'6#-F'6#\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "(Ve rifica-se que esta integral impr\363pria n\343o existe.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int((2*x + 1)/(x^4 + 1), x=-infinit y..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&,&*&\"\"# \"\"\"%\"xGF*F*F*F*F*,&*$)F+\"\"%F*F*F*F*!\"\"/F+;,$%)infinityGF0F4" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "part1 := int((2*x + 1)/(x^ 4 + 1), x=-a..0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&part1G,**&#\" \"\"\"\")F(*&\"\"##F(F+-%#lnG6#*&,(*$)%\"aGF+F(F(*&F4F(F+F,!\"\"F(F(F( ,(F2F(F5F(F(F(F6F(F(F6*&#F(\"\"%F(*&F+F,-%'arctanG6#,&F5F(F(F6F(F(F(*& F9F(*&F+F,-F=6#,&F5F(F(F(F(F(F(-F=6#F2F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 := limit(part1, a=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G,&*&#\"\"\"\"\"%F(*&\"\"##F(F+%#PiGF(F(F(* &#F(F+F(F-F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "part2 \+ := int((2*x + 1)/(x^4 + 1), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%&part2G,**&#\"\"\"\"\")F(*&\"\"##F(F+-%#lnG6#*&,(*$)%\"bGF+F(F(*&F 4F(F+F,F(F(F(F(,(F2F(F5!\"\"F(F(F7F(F(F(*&#F(\"\"%F(*&F+F,-%'arctanG6# ,&F5F(F(F(F(F(F(*&F9F(*&F+F,-F=6#,&F5F(F(F7F(F(F(-F=6#F2F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit2 := limit(part2, b=infinity); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit2G,&*&#\"\"\"\"\"%F(*&\"\" ##F(F+%#PiGF(F(F(*&F,F(F-F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "Como ambas as partes das integral impr\363pria possuem limites, separ adamente, deve-se adicion\341-las para que se obtenha a resposta final ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "Int((2*x + 1)/(x^4 + 1 ), x=-infinity..infinity) = limit1 + limit2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&,&*&\"\"#\"\"\"%\"xGF+F+F+F+F+,&*$)F,\"\"%F +F+F+F+!\"\"/F,;,$%)infinityGF1F5,$*(F*F1F*#F+F*%#PiGF+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Int(sin(x)/(x^2 + 1), x=-infinity..infinity);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&-%$sinG6#%\"xG\"\"\",&*$)F* \"\"#F+F+F+F+!\"\"/F*;,$%)infinityGF0F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "part1 := int(sin(x)/(x^2 + 1), x=-a..0);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&part1G,0*(-%#SiG6#^#\"\"\"F+-%%coshG6#F+F+F* F+F+*&-%#CiGF)F+-%%sinhGF.F+F+*(^##!\"\"\"\"#F+F2F+%#PiGF+F+*(F5F+-F(6 #,&%\"aGF+F*F+F+F,F+F+*&#F+F8F+*&-F16#,&F>F7^#F7F+F+F2F+F+F7*(^##F+F8F +-F(6#,&F>F+FEF+F+F,F+F+*&#F+F8F+*&-F16#,&F>F7F*F+F+F2F+F+F7" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 := limit(part1, a=inf inity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G,(*(-%#SiG6#^#\" \"\"F+-%%coshG6#F+F+F*F+F+*&-%#CiGF)F+-%%sinhGF.F+F+*(^##!\"\"\"\"#F+F 2F+%#PiGF+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "part2 := in t(sin(x)/(x^2 + 1), x=0..b);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&par t2G,0*(^##!\"\"\"\"#\"\"\"-%#SiG6#,&%\"bGF+^#F)F+F+-%%coshG6#F+F+F+*&# F+F*F+*&-%#CiGF.F+-%%sinhGF4F+F+F+*(^#F6F+-F-6#,&F0F+^#F+F+F+F2F+F+*&F 6F+*&-F9F?F+F:F+F+F+*(F1F+-F-6#FAF+F2F+F+*&-F9FGF+F:F+F)*(F=F+F:F+%#Pi GF+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit2 := limit(pa rt2, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit2G,(*(^#! \"\"\"\"\"-%#SiG6#^#F)F)-%%coshG6#F)F)F)*&-%#CiGF,F)-%%sinhGF0F)F(*(^# #F)\"\"#F)F4F)%#PiGF)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Estes limites envolvem algumas fun\347\365es n\343o familiares. Por\351m, ( ambas) existem, e podem ser adicionadas a fim de obter-se a resposta. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "Int(sin(x)/(x^2 + 1), x =-infinity..infinity) = limit1 + limit2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&-%$sinG6#%\"xG\"\"\",&*$)F+\"\"#F,F,F,F,!\"\"/F+;,$% )infinityGF1F5\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "A respost a final \351 menos surpreendente quando compreendemos que a fun\347 \343o integrada \351 \355mpar, e os limites da integral s\343o sim\351 tricos em torno de 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "In t(x*sin(x)/(x^2 + 1), x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*(%\"xG\"\"\"-%$sinG6#F'F(,&*$)F'\"\"#F(F(F(F( !\"\"/F';,$%)infinityGF0F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "part1 := int(x*sin(x)/(x^2 + 1), x=-a..0);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&part1G,,*&#\"\"\"\"\"#F(*&-%%sinhG6#F(F(%#PiGF(F(F(* &F'F(*&-%#SiG6#,&%\"aGF(^#F(F(F(-%%coshGF-F(F(F(*(^##!\"\"F)F(-%#CiG6# ,&F5F<^#F6#,&F5 F " 0 "" {MPLTEXT 1 0 35 "limit1 := limit(part1, a=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit 1G,&*&#\"\"\"\"\"#F(*&-%%sinhG6#F(F(%#PiGF(F(!\"\"*&#F(F)F(*&F.F(-%%co shGF-F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "part2 := int (x*sin(x)/(x^2 + 1), x=0..b);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&pa rt2G,,*&#\"\"\"\"\"#F(*&-%#SiG6#,&%\"bGF(^#!\"\"F(F(-%%coshG6#F(F(F(F( *(^#F'F(-%#CiGF-F(-%%sinhGF4F(F(*&F'F(*&-F,6#,&F/F(^#F(F(F(F2F(F(F(*(^ ##F1F)F(-F8F>F(F9F(F(*&#F(F)F(*&F9F(%#PiGF(F(F1" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 35 "limit2 := limit(part2, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit2G,&*&#\"\"\"\"\"#F(*&-%%sinhG6#F(F( %#PiGF(F(!\"\"*&#F(F)F(*&F.F(-%%coshGF-F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Int(x*sin(x)/(x^2 + 1), x=-infinity..infinity) = limit1 + limit2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*(% \"xG\"\"\"-%$sinG6#F(F),&*$)F(\"\"#F)F)F)F)!\"\"/F(;,$%)infinityGF1F5, &*&-%%sinhG6#F)F)%#PiGF)F1*&F;F)-%%coshGF:F)F)" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 42 "Int(2*x/(x^2 + 5), x=-infinity..infinity);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*(\"\"#\"\"\"%\"xGF),&*$)F* F(F)F)\"\"&F)!\"\"F)/F*;,$%)infinityGF/F3" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 37 "part1 := int(2*x/(x^2 + 5), x=-a..0);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%&part1G,&-%#lnG6#\"\"&\"\"\"-F'6#,&*$)%\"aG\" \"#F*F*F)F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 \+ := limit(part1, a=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lim it1G,$%)infinityG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Como um a das duas partes da integral impr\363pria n\343o existe, deve-se ence rrar os c\341lculos: " }{XPPEDIT 18 0 "Int(2*x/(x^2+5),x = -infinity \+ .. infinity);" "6#-%$IntG6$*(\"\"#\"\"\"%\"xGF(,&*$F)F'F(\"\"&F(!\"\"/ F);,$%)infinityGF-F1" }{TEXT -1 13 " n\343o existe." }}}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 26 "Tipo 2: Fun\347\365es Ilimitadas" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "O segundo tipo de integral impr \363pria \351 aquele em que a fun\347\343o \351 ilimitada no intervalo " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 237 " . Um exe mplo comum ocorre quando a fun\347\343o torna-se ilimitada em uma das \+ extremidades do intervalo. Examinaremos este caso. (Como explicado ant eriormente, outros casos podem ser reduzidos a este.) Suponhamos inic ialmente que, para todo " }{XPPEDIT 18 0 "0 < epsilon;" "6#2\"\"!%(eps ilonG" }{TEXT -1 10 " pequeno, " }{TEXT 290 1 "f" }{TEXT -1 35 " \351 \+ integr\341vel em todo o intervalo (" }{XPPEDIT 18 0 "a+epsilon,b;" "6$ ,&%\"aG\"\"\"%(epsilonGF%%\"bG" }{TEXT -1 50 "] . Portanto, a integral impr\363pria \351 definida como" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = a .. b) := limit(Int(f(x),x = a+epsilon .. b),epsilon = 0,right);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG-%&li mitG6%-F%6$-F(6#F*/F*;,&F-\"\"\"%(epsilonGF9F./F:\"\"!%&rightG" } {TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Apresentamos um exemplo a seguir. Verifica-se que o " }{TEXT 291 5 "Maple" }{TEXT -1 54 " tamb\351m pode resolver este tipo de integral impr\363pria." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int(1/sqrt(x), x=0..2) = i nt(1/sqrt(x), x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*& \"\"\"F(*$%\"xG#F(\"\"#!\"\"/F*;\"\"!F,,$*&F,F(F,#F(F,F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Para verificar a resposta, resolvemos a i ntegral de " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 17 " a t\351 2 e fazemos " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 59 " aproximar-se de 0 pela direita. (Note o uso do argumento " } {TEXT 292 5 "right" }{TEXT -1 19 " no comando limit.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ad := int(1/sqrt(x), x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#adG,$*&\"\"#\"\"\"%\"xG#F(F'F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "properint := subs(x=2, ad) - subs(x=epsilon, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*properintG ,&*&\"\"#\"\"\"F'#F(F'F(*&F'F(%(epsilonGF)!\"\"" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 59 "Int(1/sqrt(x), x=0..2) = limit(properint, epsi lon=0,right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(* $%\"xG#F(\"\"#!\"\"/F*;\"\"!F,,$*&F,F(F,#F(F,F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "A prop\363sito, note que o limite de dois lados, " } {XPPEDIT 18 0 "limit(sqrt(epsilon),epsilon = 0);" "6#-%&limitG6$-%%sqr tG6#%(epsilonG/F)\"\"!" }{TEXT -1 45 ", n\343o existe. Por\351m, ao m enos neste caso, o " }{TEXT 293 6 "Maple " }{TEXT -1 16 "n\343o nos in forma:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "limit(properint, \+ epsilon=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"#\"\"\"F%#F&F%F &" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "\n\nNo caso em que " }{TEXT 294 1 "f" }{TEXT -1 53 " torna-se ilimitado na extremidade direita, de fine-se" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = a .. \+ b) := limit(Int(f(x),x = a .. b-epsilon),epsilon = 0,right);" "6#>-%$I ntG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG-%&limitG6%-F%6$-F(6#F*/F*;F-,&F.\"\" \"%(epsilonG!\"\"/F:\"\"!%&rightG" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int(1/(x-2)^2, x=0..2) = int(1/(x-2)^2, x=0 ..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$),&%\"x GF(\"\"#!\"\"F-F(F./F,;\"\"!F-%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "A resposta do " }{TEXT 295 5 "Maple" }{TEXT -1 58 " suger e que a integral impr\363pria n\343o exite. Verifiquemos:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ad := int(1/(x-2)^2, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#adG,$*&\"\"\"F',&%\"xGF'\"\"#!\"\"F+F+" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "properint := subs(x=2-epsi lon, ad) - subs(x=0, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*proper intG,&*&\"\"\"F'%(epsilonG!\"\"F'#F'\"\"#F)" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 34 "Pode-se notar que o limite quando " }{XPPEDIT 18 0 "pro c (epsilon) options operator, arrow; 0 end;" "6#f*6#%(epsilonG7\"6$%)o peratorG%&arrowG6\"\"\"!F*F*F*" }{TEXT -1 19 " n\343o existe. Pelo " }{TEXT 296 5 "Maple" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit(properint, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 " \n\nFinalmente, se " }{TEXT 297 1 "f" }{TEXT -1 95 " \351 ilimitada em ambas as extremidades, deve-se dividir o intervalo em algum ponto con veniente, " }{TEXT 298 1 "c" }{TEXT -1 13 ", e define-se" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = a .. b) := Int(f(x),x = \+ a .. c)+Int(f(x),x = c .. b);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;%\"aG%\"b G,&-F%6$-F(6#F*/F*;F-%\"cG\"\"\"-F%6$-F(6#F*/F*;F6F.F7" }{TEXT -1 2 " \+ ," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "contanto que ambas as integr ais impr\363prias do lado direito existam." }}}{SECT 0 {PARA 4 "" 0 " " {TEXT -1 9 "Quest\343o 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "Reso lver as integrais impr\363prias a seguir. Caso n\343o seja poss\355ve l, mostrar que n\343o existem." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Int(1/sqrt(1-x), x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#- %$IntG6$*&\"\"\"F'*$,&F'F'%\"xG!\"\"#F'\"\"#F+/F*;\"\"!F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttoe := int(1/sqrt(1-x), x=0..1-e psilon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,&*&\"\"#\"\"\"% (epsilonG#F(F'!\"\"F'F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 " Int(1/sqrt(1-x), x=0..1) = limit(inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$,&F(F(%\"xG!\"\"#F(\" \"#F,/F+;\"\"!F(F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Int(1 /sqrt(4-x^2), x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*& \"\"\"F'*$,&\"\"%F'*$)%\"xG\"\"#F'!\"\"#F'F.F//F-;\"\"!F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "inttoe := int(1/sqrt(4-x^2), x=0..2 -epsilon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,$-%'arcsinG6# ,&\"\"\"!\"\"*&\"\"#F+%(epsilonGF*F*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Int(1/sqrt(4-x^2), x=0..2) = limit(inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$,&\" \"%F(*$)%\"xG\"\"#F(!\"\"#F(F/F0/F.;\"\"!F/,$*&F/F0%#PiGF(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 25 "Int(1/(x^2 - 1), x=1..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F',&*$)%\"xG\"\"#F'F'F'!\"\"F-/F+;F'F, " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttoe := int(1/(x^2 - \+ 1), x=1+epsilon..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,(*& ^##\"\"\"\"\"#F)%#PiGF)F)*&#F)F*F)-%#lnG6#\"\"$F)!\"\"-%(arctanhG6#,&F )F)%(epsilonGF)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "limit( inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infini tyG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Esta integral impr\363pria n\343o existe." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Int(exp( sqrt(x))/sqrt(x), x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6 $*&-%$expG6#*$%\"xG#\"\"\"\"\"#F-F+#!\"\"F./F+;\"\"!F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "inttoe := int(exp(sqrt(x))/sqrt(x), x=epsilon..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,&*&\"\"# \"\"\"-%$expG6#F(F(F(*&F'F(-F*6#*$%(epsilonG#F(F'F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Int(exp(sqrt(x))/sqrt(x), x=0..1) = limit(inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%$IntG6$*&-%$expG6#*$%\"xG#\"\"\"\"\"#F.F,#!\"\"F//F,;\"\"!F.,&*&F/ F.-F)6#F.F.F.F/F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Int(1/ (x^2 - 5*x + 6), x=1..4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$ *&\"\"\"F',(*$)%\"xG\"\"#F'F'*&\"\"&F'F+F'!\"\"\"\"'F'F//F+;F'\"\"%" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Note que o denominador desta fun \347\343o \351 igual a 0 em dois lugares: " }{TEXT 299 1 "x" }{TEXT -1 7 " = 2 e " }{TEXT 300 1 "x" }{TEXT -1 196 " = 3. Portanto, deve-s e dividir a integral em (pelo menos) 3 partes, resolv\352-las separada mente e ent\343o adicion\341-las. Primeiramente, deve-se tratar a int egral de 1 a 2, sendo esta impr\363pria em 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "inttoe := int(1/(x^2 - 5*x + 6), x=1..2-epsilon) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,(-%#lnG6#,&\"\"\"!\"\" %(epsilonGF+F*-F'6#,$F,F+F+-F'6#\"\"#F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "part1 := limit(inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&part1G%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Deve-se parar neste momento: a primeira parte da in tegral n\343o existe e, portanto, a integral completa tamb\351m n\343o existe." }}}}}}{MARK "5 22 20 0 0" 120 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }