calc 7.8(10)
DESCRIPTION
mmTRANSCRIPT
Improper Integrals
Objective: Evaluate integrals that become infinite within the interval
of integration
Improper Integrals
• Our main objective in this section is to extend the concept of a definite integral to allow for infinite intervals of integration and integrands with vertical asymptotes within the interval of integration or one of the bounds.
• We will call the vertical asymptotes infinite discontinuities and we will call integrals with infinite intervals of integration or infinite discontinuities within the interval of integration improper integrals.
Improper Integrals
• Examples of such integrals are:• Infinite intervals of integration.
0
dxex
12x
dx
21 x
dx
Improper Integrals
• Examples of such integrals are:• Infinite discontinuities in the interval of integration.
2
1 1x
dx
3
32x
dx
0
tan xdx
Improper Integrals
• Examples of such integrals are:• Both Infinite discontinuities in the interval of
integration and infinite intervals of integration.
92x
dx
0 x
dx
1
sec xdx
Definition
• The improper integral of f over the interval is defined to be
• In the case where the limit exists, the improper integral is said to converge, and the limit is defined to be the value of the integral. In the case where the limit does not exist, the improper integral is said to diverge, and is not assigned a value.
b
ab
a
dxxfdxxf )(lim)(
),[ a
Example 1a
• Evaluate
12x
dx
Example 1a
• Evaluate
12x
dx
b
b x
dx
x
dx
12
12
lim
Example 1a
• Evaluate
12x
dx
b
b x
dx
x
dx
12
12
lim
b
b
b
b xx
dx
112
1limlim
11
11
b
Example 1b
• Evaluate
1 x
dx
Example 1b
• Evaluate
1 x
dx
b
b x
dx
x
dx
11
lim
Example 1b
• Evaluate
Diverges
1 x
dx
b
b x
dx
x
dx
11
lim
bb
b
bx
x
dx1
1
lnlimlim
1lnlnb
Example 1c
• Evaluate
13x
dx
Example 1c
• Evaluate
13x
dx
b
b x
dx
x
dx
13
13
lim
Example 1c
• Evaluate
13x
dx
b
b x
dx
x
dx
13
13
lim
b
b
b
b xx
dx
12
13 2
1limlim
2
1
2
1
2
12
b
Improper Integrals
• These examples lead us to this theorem.
if p > 1 if p < 1
divergesp
x
dxp
11
1
Example 1
• On the surface, the graphs of the last three examples seem very much alike and there is nothing to suggest why one of the areas should be infinite and the other two finite. One explanation is that 1/x3 and 1/x2 approach zero more rapidly than 1/x as x approaches infinity so that the area over the interval [1, b] accumulates less rapidly under the curves y = 1/x3 and y = 1/x2 than under y = 1/x. This slight difference is just enough that two areas are finite and one infinite.
Example 3
• Evaluate
0
)1( dxex x
Example 3
• Evaluate
• We need to use integration by parts.
xu 1
0
)1( dxex x
dxdu dxedv xxev
Example 3
• Evaluate
• We need to use integration by parts.
xu 1
0
)1( dxex x
dxdu dxedv xxev
dxexedxex xxx
)1()1(0
Example 3
• Evaluate
• We need to use integration by parts.
xu 1
0
)1( dxex x
dxdu dxedv xxev
dxexedxex xxx
)1()1(0
xxxxx xeexeedxex
0
)1(
Example 3
• Evaluate
0
)1( dxex x
bxb
x xedxex 0
0
lim)1(
Example 3
• Evaluate
0
)1( dxex x
bxb
x xedxex 0
0
lim)1(
bb e
b
lim
Example 3
• Evaluate
• This is of the form so we will use L’Hopital’s Rule
0
)1( dxex x
bb e
b
lim
Example 3
• Evaluate
• We can interpret this to mean that the net signed area between the graph of and the interval is 0.
0
)1( dxex x
01
lim bb e
xexy )1(),0[
bb e
b
lim
Definition 7.8.3
• The improper integral of f over the interval is defined to be
• The integral is said to converge if the limit exists and diverge if it does not.
b
aa
b
dxxfdxxf )(lim)(
],( b
Definition 7.8.3
• The improper integral of f over the interval is defined to be
where c is any real number (we will usually choose 0 to make it easier). The improper integral is said to converge if both terms converge and diverge if either term diverges.
c
c
dxxfdxxfdxxf )()()(
),(
Example 4
• Evaluate
21 x
dx
Example 4
• Evaluate
21 x
dx
2
)(tanlimtanlim1
lim1
10
1
02
02
bxx
dx
x
dxb
b
b
b
b
Example 4
• Evaluate
21 x
dx
2
)(tanlimtanlim1
lim1
10
1
02
02
bxx
dx
x
dxb
b
b
b
b
2
)tan(limtanlim1
lim1
1010
2
0
2
axx
dx
x
dxa
aa
aa
Example 4
• Evaluate
21 x
dx
2
)(tanlimtanlim1
lim1
10
1
02
02
bxx
dx
x
dxb
b
b
b
b
2
)tan(limtanlim1
lim1
1010
2
0
2
axx
dx
x
dxa
aa
aa
22
Definition 7.8.4
• If f is continuous on the interval [a, b], except for an infinite discontinuity at b, then the improper integral of f over the interval [a, b] is defined as
• In the case where the limit exists, the improper integral is said to converge, and the limit is defined to be the value of the integral. If the limit does not exist, the integral is said to diverge.
k
abk
b
a
dxxfdxxf )(lim)(
Definition 7.8.5
• If f is continuous on the interval [a, b], except for an infinite discontinuity at a, then the improper integral of f over the interval [a, b] is defined as
• In the case where the limit exists, the improper integral is said to converge, and the limit is defined to be the value of the integral. If the limit does not exist, the integral is said to diverge.
b
kak
b
a
dxxfdxxf )(lim)(
Definition 7.8.5
• If f is continuous on the interval [a, b], except for an infinite discontinuity at a point c in (a, b), then the improper integral of f over the interval [a, b] is defined as
• The improper integral is said to converge if both terms converge and diverge if either term diverges.
b
c
c
a
b
a
dxxfdxxfdxxf )()()(
Example 6a
• Evaluate
2
1 1 x
dx
Example 6a
• Evaluate
2
1 1 x
dx
2
1
2
1 1lim
1 kk x
dx
x
dx
Example 6a
• Evaluate
2
1 1 x
dx
21
2
1
2
1
|1|lnlim1
lim1 k
kk
kx
x
dx
x
dx
Example 6a
• Evaluate
2
1 1 x
dx
21
2
1
2
1
|1|lnlim1
lim1 k
kk
kx
x
dx
x
dx
|1|ln|1|ln|1|lnlim 2
1kx k
k
Example 6b
• Evaluate
4
13/2)2(x
dx
Example 6b
• Evaluate
4
13/2)2(x
dx
4
23/2
2
13/2
4
13/2 )2()2()2( x
dx
x
dx
x
dx
Example 6b
• Evaluate
4
13/2)2(x
dx
4
23/2
2
13/2
4
13/2 )2()2()2( x
dx
x
dx
x
dx
3/12
13/22
2
13/2
)2(3lim)2(
lim)2(
x
x
dx
x
dxk
k
k
Example 6b
• Evaluate
4
13/2)2(x
dx
4
23/2
2
13/2
4
13/2 )2()2()2( x
dx
x
dx
x
dx
3/12
13/22
2
13/2
)2(3lim)2(
lim)2(
x
x
dx
x
dxk
k
k
3)21(3)2(3lim 3/13/1
2
k
k
Example 6b
• Evaluate
4
13/2)2(x
dx
4
23/2
2
13/2
4
13/2 )2()2()2( x
dx
x
dx
x
dx
3/12
4
3/22
4
23/2
)2(3lim)2(
lim)2(
x
x
dx
x
dxk
kk
Example 6b
• Evaluate
4
13/2)2(x
dx
4
23/2
2
13/2
4
13/2 )2()2()2( x
dx
x
dx
x
dx
3/12
4
3/22
4
23/2
)2(3lim)2(
lim)2(
x
x
dx
x
dxk
kk
33/13/1
223)2(3)24(3lim
k
k
Example 6b
• Evaluate
4
13/2)2(x
dx 3 233
Homework
• Section 7.8• Page 554• 1-21 odd• Skip 5