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{SECT 0 {SECT 0 {PARA 3 "" 0 "" {TEXT -1 26 "Chapter Improper Integral
s" }}{EXCHG {PARA 18 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
190 "Up to now all of the functions that have been studied with regard
 to integration have been bounded and defined on intervals of finite l
ength.  In this section you will learn how to deal with" }}{PARA 0 "" 
0 "" {TEXT -1 29 "integrals like the following:" }}{PARA 0 "" 0 "" 
{TEXT -1 17 "1.  The integral " }{XPPEDIT 18 0 "int(exp(-x^2),x=0..inf
inity) " "6#-%$intG6$-%$expG6#,$*$%\"xG\"\"#!\"\"/F+;\"\"!%)infinityG
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 17 "2.  The integral " }
{XPPEDIT 18 0 "int(1/(sqrt(x*(1-x))),x=0..1)" "6#-%$intG6$*&\"\"\"\"\"
\"-%%sqrtG6#*&%\"xGF(,&\"\"\"F(F-!\"\"F(F0/F-;\"\"!\"\"\"" }}{PARA 0 "
" 0 "" {TEXT -1 39 "Integrals like the last two are called " }{TEXT 
256 18 "improper integrals" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 62 "For example, consider the problem of integrating the function \+
" }{XPPEDIT 18 0 "exp(-x^2)" "6#-%$expG6#,$*$%\"xG\"\"#!\"\"" }{TEXT 
-1 42 " over variousfinite intervals of the form " }{TEXT 258 5 "[0,T]
" }{TEXT -1 11 ",    where " }{TEXT 257 5 "T > 0" }{TEXT -1 1 "." }}}}
{SECT 0 {PARA 3 "" 0 "" {TEXT -1 15 "Maple Try these" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(int(exp(-x^2),x=0..2));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(int(exp(-x^2),x=0..4));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(int(exp(-x^2),x=0..5))
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(int(exp(-x^2),x=
0..6));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(int(exp(-x
^2),x=0..7));" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 12 "Introduction" 
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Observe,  that the values of the
 integral for  " }{TEXT 259 5 "T = 5" }{TEXT -1 9 " through " }{TEXT 
260 5 "T = 7" }{TEXT -1 18 " are unchanged to " }{TEXT 261 2 "10" }
{TEXT -1 36 " digits of accuracy.  We know that  " }{XPPEDIT 18 0 "0<e
xp(-x^2) " "6#2\"\"!-%$expG6#,$*$%\"xG\"\"#!\"\"" }{TEXT -1 56 "for al
l  x.  Thus we know that the area under the curve " }{XPPEDIT 18 0 "y \+
= exp(-x^2)" "6#/%\"yG-%$expG6#,$*$%\"xG\"\"#!\"\"" }{TEXT -1 10 " bet
ween  " }{TEXT 262 3 "x=5" }{TEXT -1 7 "  and  " }{TEXT 263 3 "x=7" }
{TEXT -1 38 " must be positive.  We also know that " }}{PARA 258 "" 0 
"" {TEXT -1 3 "   " }{XPPEDIT 18 0 "int( exp(-x^2),x=0..7) = int(exp(-
x^2),x=0..5) + int(e^(-x^2),x=5..7)" "6#/-%$intG6$-%$expG6#,$*$%\"xG\"
\"#!\"\"/F,;\"\"!\"\"(,&-F%6$-F(6#,$*$F,\"\"#F./F,;F1\"\"&\"\"\"-F%6$)
%\"eG,$*$F,\"\"#F./F,;\"\"&\"\"(F>" }}{PARA 0 "" 0 "" {TEXT -1 7 "and \+
so " }}{PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 
0 "int( exp(-x^2),x=0..5)  <  int( exp(-x^2),x=0..7)" "6#2-%$intG6$-%$
expG6#,$*$%\"xG\"\"#!\"\"/F,;\"\"!\"\"&-F%6$-F(6#,$*$F,\"\"#F./F,;F1\"
\"(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 67 "Looking at the si
tuation from a geometric point of view we now plot" }}{PARA 0 "" 0 "" 
{TEXT -1 62 "the indefinite integral of  exp(-x^2) over the interval [
0,7]." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(int(exp(-t^2),t=0..x)
,x=0..7);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 9 "Exmaple 1" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "According to  the graph of the ant
iderivative,  " }{TEXT 265 4 "F(x)" }{TEXT -1 6 ",  of " }{XPPEDIT 18 
0 "exp(-x^2)" "6#-%$expG6#,$*$%\"xG\"\"#!\"\"" }{TEXT -1 5 "  it " }}
{PARA 0 "" 0 "" {TEXT -1 13 "appears that " }{TEXT 266 4 "F(x)" }
{TEXT -1 26 " has horizontal asymptote." }}{PARA 0 "" 0 "" {TEXT 264 
9 "Example 1" }{TEXT -1 32 " Determine that the integral of " }
{XPPEDIT 18 0 "1/(t^2* ln(t))" "6#*&\"\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#F'
F%!\"\"" }{TEXT -1 20 " over the half-line " }{XPPEDIT 18 0 "[2,infini
ty)" "6#7$\"\"#%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 30 "converges. \+
 Find the limit to " }{TEXT 267 2 "10" }{TEXT -1 20 " digits of accura
cy." }}{PARA 0 "" 0 "" {TEXT 268 9 "Solution:" }{TEXT -1 79 "  To prov
e the improper integral converges, we have the following two methods: \+
" }}{PARA 0 "" 0 "" {TEXT 273 8 "Method 1" }{TEXT -1 27 ". (Comparison
 Test) Since  " }{TEXT 272 2 "ln" }{TEXT -1 26 "   is increasing we ha
ve  " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "ln(t)>=ln(2) " "6#1-%#lnG6#\"\"
#-F%6#%\"tG" }{TEXT -1 6 " for  " }{XPPEDIT 18 0 "t >=  2" "6#1\"\"#%
\"tG" }{TEXT -1 18 ". This means that " }{XPPEDIT 18 0 "1/(t^2*ln(t)) \+
<=1/(ln(2)*t^2)" "6#1*&\"\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#F(F&!\"\"*&\"\"
\"F&*&-F+6#\"\"#F&*$F(\"\"#F&F-" }}{PARA 0 "" 0 "" {TEXT -1 8 "for all
 " }{XPPEDIT 18 0 "t >= 2" "6#1\"\"#%\"tG" }{TEXT -1 5 ". So " }
{XPPEDIT 18 0 "int(1/(t^2*ln(t)),t = 2 .. infinity);" "6#-%$intG6$*&\"
\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#F*F(!\"\"/F*;\"\"#%)infinityG" }{TEXT 
-1 23 " is convergent because " }{XPPEDIT 18 0 "int(1/(t^2*ln(2)),t = \+
2 .. infinity);" "6#-%$intG6$*&\"\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#\"\"#F(
!\"\"/F*;\"\"#%)infinityG" }{TEXT -1 4 " is." }}{PARA 0 "" 0 "" {TEXT 
274 9 "Method 2." }{TEXT -1 105 " (If the antiderivative function is i
ncreasing and bounded above, then the antiderivative is convergent.)" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "Let   F
(x) be " }{XPPEDIT 18 0 "int(1/(t^2*ln(t)),t = 2 .. x);" "6#-%$intG6$*
&\"\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#F*F(!\"\"/F*;\"\"#%\"xG" }{TEXT -1 
10 ".  Since  " }{TEXT 269 2 "ln" }{TEXT -1 26 "   is increasing we ha
ve  " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "ln(t)>=ln(2) " "6#1-%#lnG6#\"\"
#-F%6#%\"tG" }{TEXT -1 6 " for  " }{XPPEDIT 18 0 "t >=  2" "6#1\"\"#%
\"tG" }{TEXT -1 18 ". This means that " }{XPPEDIT 18 0 "1/(t^2*ln(t)) \+
<=1/(ln(2)*t^2)" "6#1*&\"\"\"\"\"\"*&%\"tG\"\"#-%#lnG6#F(F&!\"\"*&\"\"
\"F&*&-F+6#\"\"#F&*$F(\"\"#F&F-" }}{PARA 0 "" 0 "" {TEXT -1 8 "for all
 " }{XPPEDIT 18 0 "t >= 2" "6#1\"\"#%\"tG" }{TEXT -1 2 ". " }}{PARA 0 
"" 0 "" {TEXT -1 16 "It follows that " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 6 "F(x) =" }{XPPEDIT 18 0 " int(1/(t^2*ln
(t)),t = 2 .. x) <=  int(1/(t^2*ln(2)),t=2..x)" "6#1-%$intG6$*&\"\"\"
\"\"\"*&%\"tG\"\"#-%#lnG6#F+F)!\"\"/F+;\"\"#%\"xG-F%6$*&\"\"\"F)*&F+\"
\"#-F.6#\"\"#F)F0/F+;\"\"#F4" }{TEXT -1 4 "  = " }{XPPEDIT 18 0 "(1/2 \+
- 1/x)/ln(2)<=1/(2*ln(2))" "6#1*&,&*&\"\"\"\"\"\"\"\"#!\"\"F(*&\"\"\"F
(%\"xGF*F*F(-%#lnG6#\"\"#F**&\"\"\"F(*&\"\"#F(-F/6#\"\"#F(F*" }{TEXT 
-1 11 ",  for all " }{XPPEDIT 18 0 "x >=  2" "6#1\"\"#%\"xG" }{TEXT 
-1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
15 "Since  F'(x) = " }{XPPEDIT 18 0 "1/(x^2*ln x)" "6#*&\"\"\"\"\"\"*(
%\"xG\"\"#%#lnGF%F'F%!\"\"" }{TEXT -1 183 " > 0   for  x >2,  F(x)  is
 an increasing function which is bounded above by  1/(2*ln(2)).  Thus \+
the integral converges.  Its value to 10 digits of accuracy is given b
y the following." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "I1 := Int(1/(x^
2*ln(x)),x=2..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 
"I1 := evalf(I1);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 9 "Example 2" 
}}{EXCHG {PARA 0 "" 0 "" {TEXT 270 11 "Example 2. " }{TEXT -1 31 "  De
termine that the integral  " }{XPPEDIT 18 0 "int(1/(t^2+sqrt(t)),t = 0
 .. 1);" "6#-%$intG6$*&\"\"\"\"\"\",&*$%\"tG\"\"#F(-%%sqrtG6#F+F(!\"\"
/F+;\"\"!\"\"\"" }{TEXT -1 21 " converges.  Find the" }}{PARA 0 "" 0 "
" {TEXT -1 31 "limit to 10 digits of accuracy." }}{PARA 0 "" 0 "" 
{TEXT 271 8 "Solution" }{TEXT -1 7 ":  Let " }}{PARA 0 "" 0 "" {TEXT 
-1 18 "                  " }{XPPEDIT 18 0 "F(x) = int(1/(t^2+sqrt(t)),
t=x..1)" "6#/-%\"FG6#%\"xG-%$intG6$*&\"\"\"\"\"\",&*$%\"tG\"\"#F--%%sq
rtG6#F0F-!\"\"/F0;F'\"\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 
-1 53 "Since  t^2 > 0  for all t  not equal to  0,  we have " }}{PARA 
0 "" 0 "" {TEXT -1 17 "                 " }{XPPEDIT 18 0 "1/(t^2+ sqrt
(t)) < 1/sqrt(t))" "6#2*&\"\"\"\"\"\",&*$%\"tG\"\"#F&-%%sqrtG6#F)F&!\"
\"*&\"\"\"F&-F,6#F)F." }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 11 "
This means " }{XPPEDIT 18 0 "int(1/(t^2+ sqrt(t)),t=x..1) <=  int(1/sq
rt(t),t=x..1) " "6#1-%$intG6$*&\"\"\"\"\"\",&*$%\"tG\"\"#F)-%%sqrtG6#F
,F)!\"\"/F,;%\"xG\"\"\"-F%6$*&\"\"\"F)-F/6#F,F1/F,;F4\"\"\"" }{TEXT 
-1 3 " = " }{XPPEDIT 18 0 "2 - 2 sqrt(x) <= 2" "6#1,&\"\"#\"\"\"*&\"\"
#F&-%%sqrtG6#%\"xGF&!\"\"\"\"#" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" 
{TEXT -1 2 "  " }}{PARA 0 "" 0 "" {TEXT -1 161 "for all 0 < x <1.  Not
ice that F'(x) >0 for all  0 <x <1. Thus, F(x) is increasing and bound
ed above for all 0<x<1. Thus, we conclude that the improper integral \+
" }{XPPEDIT 18 0 "int(1/(t^2+sqrt(t)),t = 0 .. 1);" "6#-%$intG6$*&\"\"
\"\"\"\",&*$%\"tG\"\"#F(-%%sqrtG6#F+F(!\"\"/F+;\"\"!\"\"\"" }{TEXT -1 
12 "  converges." }}{PARA 0 "" 0 "" {TEXT -1 98 "The following Maple V
 segment obtains the value of the improper integral to 10 digits of ac
curacy." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "I2 := Int(1/(x^2+sqrt(x)),x=0..1);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 16 "I2 := evalf(I2);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "plot(1/(x^2+sqrt(x)),x=0..1);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0" 0 }{VIEWOPTS 1 1 0 1 1 
1803 }