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Section 8.8 – Improper Integrals. The Fundamental Theorem of Calculus. If f is continuous on the interval [ a , b ] and F is any function that satisfies F '( x ) = f ( x ) throughout this interval then. REMEMBER: [ a,b ] is a closed interval. Improper Integral. - PowerPoint PPT Presentation
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Section 8.8 – Improper Integrals
The Fundamental Theorem of Calculus
If f is continuous on the interval [a,b] and F is any function that satisfies F '(x) = f(x) throughout this interval then
b
af x dx F b F a
REMEMBER: [a,b] is a closed interval.
-1 1 2 3 4 5 6 7 8 9 10-1
1
2
3
4
5
6
7
8
9
10
Improper IntegralAreas of unbounded regions also arise in applications and are represented by improper integrals. Consider the following integral:
0x dx
Intuitively it appears any unbounded
region should have infinite
area.
Numerical InvestigationComplete the table for with the table below.
b1
2
4
6
8
10
20
40
50
100
Numerically, it appears
0
∞
𝑥 𝑑𝑥=¿
1/2 ∙12− 0=1 /21/2 ∙ 22− 0=21/2 ∙ 42− 0=81/2 ∙62 −0=181/2 ∙ 82 −0=321/2 ∙102− 0=501/2 ∙202− 0=2001/2 ∙402− 0=800
1250
000
∞
Definition: The Integral Diverges
If the limit fails to exist, the improper integral diverges.
For instance:
Improper IntegralAreas of unbounded regions also arise in applications and are represented by improper integrals. Consider the following integral:
0
xe dx
Can the unbounded region have finite area?
Numerical InvestigationComplete the table for with the table below.
b1
2
3
5
8
9
10
20
50
100
Numerically, it appears
0
∞
𝑒−𝑥𝑑𝑥=¿
0.6320.8650.950
0.993
1.000
1.000
1.000
1.0001.0001.000
1
Definition: The Integral Converges
If the limit is finite, the improper integral converges and the limit is the value of the improper integral.
For instance:
“Horizontal” Improper IntegralsIn a horizontal improper integral, the left limit of integration vanishes into or the right limit vanishes into , or both limits vanish in respective directions and we integrate over the whole x-axis.
If is continuous on the entire interval, then
limb b
aaf x dx f x dx
limb
a abf x dx f x dx
c
cf x dx f x dx f x dx
Note: It can be shown the value of c above is unimportant. You can
evaluate the integral with any choice.
Both integrals must converge for
the sum to converge.
Example 1Analytically evaluate .
0 0lim
bx x
be dx e dx
0
limbx
be
0lim b
be e
lim 1 b
be
1
Improper Integral TechniqueThe technique for evaluating an improper integral “properly” is to evaluate the integral on a bounded closed interval where the function is continuous and the Fundamental Theorem of Calculus applies, then take the offending end of the interval to the limit.
On any free-response question, always use the limit notation to evaluate improper integrals. While the statement below may involve less writing, it is mathematically incorrect and will lose you points:
00
x xe dx e DO NOT
WRITE THIS!
Example 2Analytically evaluate .
1 1lim
bx x
bxe dx xe dx
1lim
bx x
bxe e
1 1lim 1b b
bbe e e e
1lim 1 2b
be b e
2e
u dv x xe dx
du v 1 dx xe
1lim
bx x
bxe e dx
1 2lim bb
be e
1 2lim bb e e
Use L'Hôpital's Rule
Example 2Analytically evaluate .
2 2 2
01 1 1
1 1 10x x xdx dx dx
lim arctan 0 arctan lim arctan arctan 0a b
a b
0 02 2
0
0lim arctan lim arctan b
aa bx x
2 2
01 1
1 10lim lim
b
x xaa bdx dx
White Board Challenge
Evaluate:
20
24 3dx
x x
ln 3
The Fundamental Theorem of Calculus
If f is continuous on the interval [a,b] and F is any function that satisfies F '(x) = f(x) throughout this interval then
b
af x dx F b F a
REMEMBER: f must be a continuous function.
1
1
2
3
4
Improper IntegralAreas of unbounded regions also arise in applications and are represented by improper integrals. Consider the following integral:
1
0
dxx
Can the unbounded region have finite area?
Numerical InvestigationComplete the table for with the table below.
b0.5
0.4
0.3
0.2
0.1
0.05
0.01
0.001
0.0001
0.00001
Numerically, it appears
0
1 1√𝑥
𝑑𝑥=¿
0.5860.7350.905
2 −2√0. 2 ≈ 1.1061.368
2 −2√0. 0 5 ≈ 1.5532 −2√0. 01=1.8
1.9371.981.994
2
-1 1
1
2
3
4
5
1 2
1
2
3
4
5
6
7
8
9
10
1 2
1
2
3
4
5
6
7
8
9
10
“Vertical” Improper IntegralsIn a vertical improper integral, we integrate over a closed interval but the function has a vertical asymptote at one or both ends of the interval.
If is continuous on the entire interval except one or both endpoints, then
limb c
a ac bf x dx f x dx
limb b
a cc af x dx f x dx
b c b
a a cf x dx f x dx f x dx
Note: It can be shown the value of c above is unimportant. You can
evaluate the integral with any choice.
Both integrals must converge for
the sum to converge.
Example 1Analytically evaluate .
1 1
0 0
1 1limcc
dx dxx x
1
0lim 2
ccx
0
lim 2 1 2c
c
2
Example 2Analytically evaluate .
0.5 0.5
0 0
1 1limcc
dx dxx x
0.5
0lim ln
ccx
0
lim ln 0.5 lnc
c
Since the integral is infinite, it diverges (does not exist).
Example 3Analytically evaluate .
2 3 2 3 2 3
3 1 31 1 11 1 10 0 1x x x
dx dx dx
1 3 1 3 1 3 1 3
1 1lim 3 1 3 0 1 lim 3 3 1 3 1c c
c c
30 3 3 2 0 33 3 2
31 3 1 3
1 10lim 3 1 lim 3 1
c
c c cx x
2 3 2 3
31 11 101 1
lim limc
x xcc cdx dx
White Board Challenge
Evaluate:1
21 1
dx
x
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