674 0
CHAPTER8
TECHNIQUES
OFINTEGRATION
8.8 Improper Integrals
1. (a) Since Jl(X)x4e-Z4 dx has an infinite interval of integration, it is an improper integral of Type I.
(b) Since y
= sec x has an infinite
(c) Since y
=
(x
integral.
- 2)(;
discontinuity at x
= ~, Jo'/f/2 sec x dx is a Type II improper integral.
- 3) lias an intinite discontinuityat x
= 2,L2
+
(d) Since fO
5 dx hasan infinite intervalof integration,it is an improper integral of1:ypeI.
J-<x>x +
2. (a) Since y
(b) Sincey
f
(c) Since
= 1/(2x =
IX)
-IX)
~
~
1) is
and continuous on [1,2], the integral is proper.
hasaninfinitediscontinuity
at x = ~, 11
dx has an infinite
interval of integration,
it is an improper
integral of Type I.
+x
(c) The total areaunderthe graphof f is lim F(t) = lim 10(1 - t-O.l) = 10.
t-.cxo
t-.cxo
The total areaunderthe graphof 9 doesnot exist, since lim G(t) = lim 10(tO,l-
t-+cxo
t-+cxo
1) = 00.
SECTION
8.8 IMPROPER
INTEGRAlS 0
I
1
=it
rOO
{t
(3:1:+ 1)2dz ==t~
1
it
f (".3;+"1)2
1
I f 1
dz ==3" ~ du
1
==-",3; +c
(3:1:+ 1)2 dz. Now
[u = 3:1:+ 1, du = 3d:z:)
1
==-3(3:1: + 1) +c,
SOI=t~[-~):==t~~[-~+A)
10
1
lim
-dx=
-002x - 5
675
1 0 -2
t-c+-OO t
==o+A==A.
1 5 dx =
lim
X -
[i lnl2x
- 51]~ =
lim
t-c+-oo
[i In5 -
i lnl2t
- 51] =
-00.
t-c+oo
Divergent
1 -1
1
1t -1-~=:=dw
V2-
lim
dw=
t--oo
-oo~
= t--~
lim
1
[-2-./"3+ 2 V2=1] =00.
i
Xdx=Um
00
0 (x2+ 2)2
t-~
Xdx=lim-
t
= t-+~
lim
4
it e-1//2 dy
4
Z
10
xdx
-00 1""'+-;2 =
1 0 xdx
- 001""'+-;2=
1 =J:::'(X)(2
Jim
t--oo
xdx
t
lim
(X)
1
=
] t
-2e-1//2
4
1lD1
. [--e1
z
(
)
-1
1
1
= '2t~~
t2+2
+ '2
(X)
-2t
]
2
Jim (-2e-t/2
+ 2e~2)
= 0 + 2e-2 = 2e-2.
t-+~
[-2e
1 2 + 1 -2Z
2e ] = 00.
lim
= Z-"-(X)
-1
Z
Divergent
xdx
(X)
-00 i""+-;2 +
t-oo 2 x2 + 2 0
[
= t-+~
lim
Ii m J,
r-l e-2t dt =
100
-dw)
Convergent
Convergent
z-"-(X)
[u=2-w,du=
t
Divergent
1 [ -=- 1 ] t
0 (x2 + 2)2
1) _1
--1 2"(0+ 2"
- 4'
L~ e~1//2 dy
= t-+-oo
Um [-2~]-1
W
-1
0
+x 2and
[!In(1+x2)]~=
t
lim
(X)
[O-!ln(1+t2)]
=-00.
Divergent
- V4) dv = It + 12= J~(X)(2- v4) dv + Jo(X)(2- v4) dv, but
[211 -
k1l5]~
= -00.
lim (-2t + itS)
=
t-+-oo
Since
II is divergent,I is divergent,andthereis no need
toevaluate12. Divergent
r
oo
-00
xe-z2
-'-
U;I;
f-o xe-z2-,-
U;I;
00
-- f -~
o xe -z2 U;I;
-'- + r~
-'Jo xe -z2 U;I;.
=
11m
. ( --1 )[e-z2
t--~
dx = t-~
lim
]
=
0
2
t
(--1 )(1 - e-t2 ) = --1 . 1 = --, 1 and
lim
t---~
2
2
(_1 ) [e-z2 ]t = lim (_1 ) (e-t2
2
0
t--~
-1
) =_1 . (-1)
2
2
2
= 1.
2
Convergent
J~
00
x2e-Z3
dx = t-+-OO
lim
[
-!e-Z3
]
O
= -!
t
lim J; 1r sin fJd() = t-+CXJ
lim
t--+CXJ
Divergent
+!
(
lim e-t3
t-+-OO
[- cosfJ] ~ =
W
)=
00.
Divergent
lim (-COBt + 1). This limit doesnot exist, so the integral
t-+CXJ
&78 0
TECHNIQUES
OFINTEGRATION
Jo1(x-l)-1/5dx=
.
-
f33(x
1
1im j;t(X-l)-1/5dx=
t-l0
=
1)~1/5 dx
t
= 1/(4y -
-
.§.(x
4
1)4/5
[i(t - 1)4/5 :.- ~4] = -~4
lim
t-1-
0
[
t--1+
= -~ + 20 =~.
]t
4
lim 133(x- 1)-1/5dx = lim
t-+1+
Thus, J:3(X -1)-1/5&
34. f(y)
[~(X-l)4/5
1im
t-l-
] 33
t
Convergent
1) has an infinite discontinuity at y = ~.
1
.
1
1
lim
t--+(1/4)+
t
1:4~dY=
lim [14 ln3 = t-o(1/4)+
[i1 In 14y-
lim
t-(1/4}+
-d- 1 y41/-
11J:
i1 In(4t - 1)] = 00
1
4Y-=l dy diverges. Divergent
35.
Ii
t
(,,/2 secxdx = t--+.%sec x dx = fowl:}secx dx + f:/2 secx dx. Jo
fo sec x dx
J~n-
[In Isecx + tan xl ]
t
= t--/2lim
36.
14
dx
14
0 x2+x-6-
12
dx
-
(x + 3)(x
dx
(x - 2)(x + 3) --
1im
12
.!.
[
so
ez
1_
ez -dx-
1
"
eX
so -lex-
1
1
-
eX
eX
-
j
lim
t-+O-
t + oS/
t
3
]=
2
Divergent
dx
4
+
+ 3)
(x -
2)(x
1im
'-+2-"5
1
1
J o e%
-dx=
1
-dx+
-1 e%- 1
1
0
t~-
[
[lnle~ -11]~1
r~._l
} 2x-3ax-2
x-3
-dx=
13/2
2x-3
1
dx.
lim [In let
dx also divergessince
[[
lim
lim InleX-I!
lnle"'-ll
t-+O+
t-+O+
X- 3
2x - 3
j ~ax=2
2X-6.
lim
e'"
e-;-=l
]]t =- t-+O+
lim
In Ie -II-Inlet
lim [Inle1
1
t
t-+o+
!a 2
X - 3
3/2 -dxand
2x - 3
I
j
r
3]
[l-~JaX=~x-tinl'2x-SI'+C,mJ
i [ 2X - 3ln 12x-
31
]
t
0
=00.
Divergent
-11
]=
I]
'
0
X
-!!-
11
eX
-dx+
t-3/2-
~
t-+O-
Oivp.rup.nt
0
2:-2
0117 eX-]
1
!o 3/2
I
Divergent
-00.
e%
-1 e%- 1
lln
[
0:
~dx=
-1 eO:- 1
e'"
lim
Ii
-dx=
0 ex t-+O+
t -dx=
ex - 1
;;;-=-]:dx=o= t-+~+ t e'" - 1
!o 2 X - 3
38. 0 -dx=
2x - 3
and
+ 3)'
[partial fractions]
J1
1 dx is divergent.The integral
I dx=
2)(x
ln l ~l-ln~
37. Thereis an infinite discontinuityat x = o.
-1
(x -
= 00.
1
dx
- 2) --
1/5 - m
1/5 ] dx
Ii
ft [ ;--=2
t~10
t-+2- (j
10
In lsect + tantl
lim
t-+wj2-
0
00.
IMPROPER
INTEGRALS 0
[§.(3ln2-1)_lt3(3lnt-l)
9
9
:; t-+O+
Urn
NowL =
lim
[t3(3lnt
t-+O+
1= §3 ln2-§.
-1
=§.ln2-§._11im
3
9
9
]
)] = lim 3lnt -1
t-+O+
dx -
1
1im {I ~
- t-+O+it
0 -IX
H t-+O+
O+
-3/t4
lim~
[t3(3lnt-l)
]
9
= O.Thus,L = 0 and
lim (-t3)
=
=§.ln2-§'_lL.
3
9
t-+O+
9'
by partswith u = lnx, dv = dx/.JX
Integrate
1~
-
t-3
t
"*
du
(
[2VXlnX]~
JX dx= lim
= dx/:1:.v = 2-.IX.
1
-2
1
t-+O+
t
=t-+O+
lim (
~) =
yX
) =-4
-2Vtlnt-4+4Vt
lim -t-3/3
lit 12 =
t-+O+
lim (-2 Vi) = O.
t-+O+
42.
Area = f:e-Z/2dx
Area= f~tX>ex dx =
lim
=e- t--CX)
et
lim
[eZ ] l
t-+-~
t
=
-2
lim
t-a>
=e
e-t/3
[ e-Z/2 ] t
= -2limt-+~
-2
+ 2e = 2e
44.
IS
7.
Area
=1
~
2
X2
1
-dx
0
X2
~
-dx=2.2
-(X)
+9
1
Area =
+9
r !:tan-l.=
=4lim
.t~J~
= ~ lim
rt~dx=4lim
.3' roY'
l 3
~
- 0] = ~ . ~ = ~
[ tan-l!
3 t-+(X)
J~
3
3 2
]t
3
r
10
x
(X)
-dx
r2x
t
X2+ 9
t-+(X) 10 ;2+9dx
X2 i
= lim [lln(x2+9 )] t
t-+(X) 2
JO
= lim
0
= ~t~ [In(r + 9) -ln9] = 00
Inti .
mlearea
679