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IMPROPER INTEGRALS ELECTRONIC VERSION OF LECTURE
HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics
HCMC — 2019. (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
1 / 33
OUTLINE
1
T YPE 1: INFINITE INTERVALS
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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OUTLINE
1
T YPE 1: INFINITE INTERVALS
2
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
2 / 33
OUTLINE
1
T YPE 1: INFINITE INTERVALS
2
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
3
MATL AB
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.1 Let f (x) be defined for every number x Ê a and be integrable on every interval [a, b]. Then Φ(b) =
b
Z a
f (x)d x is defined on the interval [a, +∞).
The limit I = lim Φ(b) = lim b→+∞
Z
b→+∞ a
b
(1)
f (x)d x
is called an improper integral of type 1 of function f (x) on the interval [a, +∞) and denoted by +∞
Z
f (x)d x. a (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.2 1
If the limit I = lim
Z
b
f (x)d x exists (as a finite Z +∞ number) then the improper integrals f (x)d x b→+∞ a
are called convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
a
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.2 1
If the limit I = lim
Z
b
f (x)d x exists (as a finite Z +∞ number) then the improper integrals f (x)d x b→+∞ a
a
are called convergent. 2
If the limit I = lim
Z
b→+∞ a
b
f (x)d x does not exist or is
equal to ∞ then the improper integrals Z +∞
f (x)d x are called divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
4 / 33
Type 1: Infinite intervals
Geometric meaning
GEOMETRIC MEANING If f (x) Ê 0, ∀x ∈ [a, +∞) and the integral
Z
convergent then the improper integrals
+∞
f (x)d x is aZ
+∞
f (x)d x a
can be interpreted as an area of the region S = {(x, y)|x Ê a, 0 É y É f (x)}.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Geometric meaning
According to geometric meaning of an improper Z +∞
integral of type 1:
f (x)d x , if
a
lim f (x) = A 6= 0
x→+∞
and f (x) is integrable on every interval Z[a, b] ⊂ [a, +∞), then the improper integrals +∞
f (x)d x are divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Geometric meaning
According to geometric meaning of an improper Z +∞
integral of type 1:
f (x)d x , if
a
lim f (x) = A 6= 0
x→+∞
and f (x) is integrable on every interval Z[a, b] ⊂ [a, +∞), then the improper integrals +∞
f (x)d x are divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
6 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
THEOREM 1.1 (NEWTON-LEIBNIZ’S FORMULA) Suppose that f (x) has antiderivative F (x) on the interval [a, +∞) and is intergrable on Zevery interval +∞
[a, b]. The improper integral of type 1
f (x)d x is
a
convergent if and only if lim F (b) = F (+∞) exists as a b→+∞
finite number. Then +∞
Z a
(HCMUT-OISP)
¯+∞ ¯ f (x)d x = F (+∞) − F (a) = F (x)¯ a
IMPROPER INTEGRALS
HCMC — 2019.
(2)
7 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
SOLUTION ¯+∞ ¯ I = sin x ¯ = lim sin b − sin 0 = lim sin b. 0
(HCMUT-OISP)
b→+∞
IMPROPER INTEGRALS
b→+∞
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
SOLUTION ¯+∞ ¯ I = sin x ¯ = lim sin b − sin 0 = lim sin b. 0
b→+∞
b→+∞
The limit lim sin b does not exist. Therefore the b→+∞
improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.2 Evaluate I =
Z
+∞
2
xe −x d x
0
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.2 Evaluate I =
Z
+∞
2
xe −x d x
0
SOLUTION ¯ 1 −x 2 ¯¯+∞ e d (−x ) = − e ¯ 2 0 0 1 −b 2 1 1 = lim − e + = b→+∞ 2 2 2
1 I =− 2
Z
+∞
−x 2
2
So the given integral I is convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent?
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent? If α 6= 1 then 1 I =− α−1
(HCMUT-OISP)
µ lim
1
b→+∞ b α−1
IMPROPER INTEGRALS
−
1
¶
1α−1
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent? If α 6= 1 then 1 I =− α−1
If α > 1, then lim
1
b→+∞ b α−1
µ lim
1
b→+∞ b α−1
¶
1α−1
= 0. Therefore I =
the integral I converges. (HCMUT-OISP)
−
1
IMPROPER INTEGRALS
1 and so α−1 HCMC — 2019.
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Type 1: Infinite intervals
If α < 1, then lim diverges.
(HCMUT-OISP)
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges. SUMMARY 1
If α > 1 then I =
Z 1
(HCMUT-OISP)
+∞
dx converges. xα
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges. SUMMARY 1
2
+∞
dx converges. xα 1 Z +∞ dx If α É 1 then I = diverges. xα 1
If α > 1 then I =
(HCMUT-OISP)
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
A COMPARISON TEST FOR IMPROPER INTEGRALS OF TYPE 1 THEOREM 1.2 Suppose that f and g are continuous functions on every interval [a, b] ⊂ [a, +∞) with 0 É g (x) É f (x), ∀x Ê a. Z 1
If
+∞
f (x)d x is convergent, then
a
g (x)d x is a
convergent.
(HCMUT-OISP)
+∞
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
A COMPARISON TEST FOR IMPROPER INTEGRALS OF TYPE 1 THEOREM 1.2 Suppose that f and g are continuous functions on every interval [a, b] ⊂ [a, +∞) with 0 É g (x) É f (x), ∀x Ê a. Z 1
+∞
If
f (x)d x is convergent, then
a
If
+∞
Z
g (x)d x is divergent then a
+∞
Z
f (x)d x is a
divergent. (HCMUT-OISP)
g (x)d x is a
convergent. 2
+∞
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
(HCMUT-OISP)
A comparison test for improper integrals of type 1
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
1
A comparison test for improper integrals of type 1
If the area under the top curve y = f (x) is finite, then so is the area under the bottom curve y = g (x).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
1
A comparison test for improper integrals of type 1
If the area under the top curve y = f (x) is finite, then so is the area under the bottom curve y = g (x).
2
If the area under y = g (x) is infinite, then so is the area under y = f (x). (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.4 DetermineZ whether the integral is convergent or +∞
divergent
1
(HCMUT-OISP)
1 + e −x dx x
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.4 DetermineZ whether the integral is convergent or +∞
divergent
1
SOLUTION
Since +∞
Z 1
Z 1
+∞
1 + e −x dx x
1 + e −x 1 > x x
1 d x is divergent, so the integral x
1 + e −x d x is divergent. x (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
If λ = 0 and converges.
(HCMUT-OISP)
∞
Z
g (x)d x converges then a
∞
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
2
If λ = 0 and converges.
∞
Z
If λ > 0 then
g (x)d x converges then a
g (x)d x and a
converges or diverges.
(HCMUT-OISP)
f (x)d x a
∞
Z
∞
Z
∞
Z
f (x)d x either a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
2
If λ = 0 and converges.
∞
Z
If λ > 0 then
g (x)d x converges then a
g (x)d x and a
converges or diverges. 3
If λ = +∞ and diverges. (HCMUT-OISP)
f (x)d x a
∞
Z
∞
Z
∞
Z
f (x)d x either a
∞
Z
g (x)d x diverges then a
∞
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
INDETERMINATE FORM
1
2
A comparison test for improper integrals of type 1
∞ ∞
xα lim = 0 (a > 1) x→+∞ a x lnα x lim = 0 (∀α > 0, β > 0) x→+∞ x β
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
INDETERMINATE FORM
1
2
A comparison test for improper integrals of type 1
∞ ∞
xα lim = 0 (a > 1) x→+∞ a x lnα x lim = 0 (∀α > 0, β > 0) x→+∞ x β lnα x << x β << a x , (α, β > 0, a > 1)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
DEFINITION 1.3 Functions f (x) and g (x) are called equivalent as x → a, if f (x) =1 x→a g (x)
(3)
lim
x→a
Denote by f (x) ∼ g (x).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.5 Determine whether the integral is convergent or Z +∞
divergent I =
1
(HCMUT-OISP)
p
dx
x 2 − 2x + 3
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.5 Determine whether the integral is convergent or Z +∞
divergent I =
1
We have p
p
where
x 2 − 2x + 3
1 x 2 − 2x + 3 p
Z
dx
+∞
1
(HCMUT-OISP)
> 0, ∀x Ê 1 and 1
x 2 − 2x + 3
x→+∞
∼
1 , x
dx diverges, therefore I diverges. x IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.6 Determine whether the integral is convergent or Z +∞
divergent I =
2
(HCMUT-OISP)
(x + 1)d x p 3 x 7 − 3x − 2
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.6 Determine whether the integral is convergent or Z +∞
divergent I =
2
We have p 3
x +1 x 7 − 3x − 2 p 3
where
Z
(x + 1)d x p 3 x 7 − 3x − 2
+∞
2
(HCMUT-OISP)
> 0, ∀x Ê 2 and
x +1
x→+∞
x 7 − 3x − 2
∼
x x 7/3
=
1 x 4/3
,
dx converges, therefore I converges. x 4/3 IMPROPER INTEGRALS
HCMC — 2019.
19 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
Suppose that f is defined on a finite interval [a, b) but has a vertical asymptote as x → b − and f is integrable on every interval [a, η] ⊂ [a, b). Then Z Φ(η) =
η
f (x)d x is defined on the interval [a, b).
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
Suppose that f is defined on a finite interval [a, b) but has a vertical asymptote as x → b − and f is integrable on every interval [a, η] ⊂ [a, b). Then Z Φ(η) =
η
f (x)d x is defined on the interval [a, b).
a
DEFINITION 2.1 The limit of function Φ(η) as η → b − is called an improper integral of type 2 on the interval [a, b) b
Z a
(HCMUT-OISP)
f (x)d x = lim− Φ(η) = lim− η→b
η→b
IMPROPER INTEGRALS
η
Z
f (x)d x
(4)
a
HCMC — 2019.
20 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
DEFINITION 2.2 1
If lim− Φ(η) = lim− η→b
η→b
η
Z
f (x)d x exists (as a finite a
number) then the improper integral of type 2 b
Z
f (x)d x converges. a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
DEFINITION 2.2 1
If lim− Φ(η) = lim− η→b
η→b
η
Z
f (x)d x exists (as a finite a
number) then the improper integral of type 2 b
Z
f (x)d x converges. Z η If lim− Φ(η) = lim− f (x)d x does not exist or is a
2
η→b
η→b
a
equal to ∞ then the improper integral of type 2 b
Z
f (x)d x diverges. a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Geometric meaning
GEOMETRIC MEANING If f (x) Ê 0, ∀x ∈ [a, b) and the integral
b
Z
f (x)d x is Z b convergent then the improper integrals f (x)d x a
a
can be interpreted as an area of the region S = {(x, y)|a É x < b, 0 É y É f (x)}, where x = b is the vertical asymptote of the graph of function f (x)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
NEWTON-LEIBNIZ’S FORMULA Suppose that f (x) has antiderivative F (x) on every intervals [a, η] ⊂ [a, b) and lim− f (x) = ∞. The x→b
improper integral of type 2
b
Z
f (x)d x is convergent if a
and only if lim− F (η) = F (b − 0) exists as a finite η→b
number. Then b
Z a
(HCMUT-OISP)
¯b − ¯ f (x)d x = F (b − 0) − F (a) = F (x)¯ . a
IMPROPER INTEGRALS
HCMC — 2019.
(5)
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z 0
(HCMUT-OISP)
1
dx x
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z
1
0
dx x
SOLUTION 1 We have lim = +∞. Therefore x = 0 is vertical x→0+
asymptote.
(HCMUT-OISP)
x
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z
1
0
dx x
SOLUTION 1 We have lim = +∞. Therefore x = 0 is vertical x→0+
x
asymptote. Since ¯1 ¯ I = ln |x|¯ = ln 1 − lim ln |a| = +∞. 0
a→0+
so improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
(HCMUT-OISP)
1
arccos x dx p 2 −1 1−x
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.In other hand, x = 1 is arccos x not vertical asymptote because lim p = 1. x→1− 1 − x2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.In other hand, x = 1 is arccos x not vertical asymptote because lim p = 1. x→1− 1 − x2
Since
1
¯1 1 ¯ 2 I =− arccos xd (arccos x) = − · (arccos x)¯ −1 2 −1 2 1 π = − (arccos2 1 − lim arccos2 a) = a→−1+ 2 2 so improper integral I is convergent. Z
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
(HCMUT-OISP)
b
Z a
dx , (a < b) (b − x)α
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
If α < 1 then lim− (b − η)−α+1 = 0. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
If α < 1 then lim− (b − η)−α+1 = 0. η→b
If α > 1 then lim− (b − η)−α+1 = ∞. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
SUMMARY 1
If α < 1, then improper integral converges.
(HCMUT-OISP)
IMPROPER INTEGRALS
b
Z a
dx (b − x)α
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
SUMMARY 1
2
If α < 1, then improper integral converges.
Z
If α Ê 1, then improper integral
Z
diverges. (HCMUT-OISP)
IMPROPER INTEGRALS
b
dx (b − x)α
b
dx (b − x)α
a
a
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges.
(HCMUT-OISP)
a
a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges. 2
If λ > 0 then
a
a
b
Z
g (x)d x and a
converges or diverges.
(HCMUT-OISP)
b
Z
f (x)d x either a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges. 2
If λ > 0 then
a
a
b
Z
g (x)d x and a
converges or diverges. 3
If λ = +∞ and diverges. (HCMUT-OISP)
b
Z
f (x)d x either a
b
Z
g (x)d x diverges then a
b
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
As x → 0 the following functions are equivalent 1
2
3
4
5
6
1 sin x ∼ x, tan x ∼ x, 1 − cos x ∼ x 2 2 arctan x ∼ x, arcsin x ∼ x a x − 1 ∼ x. ln a, (a > 0, a 6= 1), e x − 1 ∼ x x loga (1 + x) ∼ x loga e = , (a > 0, a 6= 1), ln(1 + x) ∼ x ln a p x (1 + x)µ − 1 ∼ µ.x, (µ ∈ R), 1 + x − 1 ∼ , 2 p x n 1 + x − 1 ∼ , (n ∈ N) n x2 sinh x ∼ x, cosh x − 1 ∼ 2 (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
(HCMUT-OISP)
cos2 x dx p 3 1 − x2
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
cos2 x dx p 3 1 − x2
cos2 x cos2 x 1 =∞ lim− p = lim . p x→1 3 1 − x 2 x→1− 3 1 + x (1 − x)1/3
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
cos2 x dx p 3 1 − x2
cos2 x cos2 x 1 =∞ lim− p = lim . p x→1 3 1 − x 2 x→1− 3 1 + x (1 − x)1/3 2 cos2 x 1 cos2 x 1 x→1− cos 1 =p ∼ . . . p p 3 3 3 1 + x (1 − x)1/3 2 (1 − x)1/3 1 − x2 1 We have α = < 1, therefore I converges. 3 (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
(HCMUT-OISP)
1
ln(1 + 3 x) dx e sin x − 1
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
1
ln(1 + 3 x) dx e sin x − 1
p ln(1 + 3 x) x 1/3 1 lim+ sin x = lim+ = lim+ 2/3 = ∞ x→0 x→0 x→0 x e −1 x
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
1
ln(1 + 3 x) dx e sin x − 1
p ln(1 + 3 x) x 1/3 1 lim+ sin x = lim+ = lim+ 2/3 = ∞ x→0 x→0 x→0 x e −1 x p ln(1 + 3 x) x→0+ x 1/3 1 ∼ = e sin x − 1 x x 2/3 Due to α = 32 < 1 so I converges. (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
3
>> s yms x; >> i nt (1/sqr t (x), 0, 1) ⇒ Ans = 2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
3
>> s yms x; >> i nt (1/sqr t (x), 0, 1) ⇒ Ans = 2
4
>> s yms x; >> i nt (1/x, 0, 1) ⇒ Ans = I n f
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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MatLab
THANK YOU FOR YOUR ATTENTION
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
33 / 33
HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics
HCMC — 2019. (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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OUTLINE
1
T YPE 1: INFINITE INTERVALS
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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OUTLINE
1
T YPE 1: INFINITE INTERVALS
2
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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OUTLINE
1
T YPE 1: INFINITE INTERVALS
2
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
3
MATL AB
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.1 Let f (x) be defined for every number x Ê a and be integrable on every interval [a, b]. Then Φ(b) =
b
Z a
f (x)d x is defined on the interval [a, +∞).
The limit I = lim Φ(b) = lim b→+∞
Z
b→+∞ a
b
(1)
f (x)d x
is called an improper integral of type 1 of function f (x) on the interval [a, +∞) and denoted by +∞
Z
f (x)d x. a (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.2 1
If the limit I = lim
Z
b
f (x)d x exists (as a finite Z +∞ number) then the improper integrals f (x)d x b→+∞ a
are called convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
a
HCMC — 2019.
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Type 1: Infinite intervals
R Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.2 1
If the limit I = lim
Z
b
f (x)d x exists (as a finite Z +∞ number) then the improper integrals f (x)d x b→+∞ a
a
are called convergent. 2
If the limit I = lim
Z
b→+∞ a
b
f (x)d x does not exist or is
equal to ∞ then the improper integrals Z +∞
f (x)d x are called divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Geometric meaning
GEOMETRIC MEANING If f (x) Ê 0, ∀x ∈ [a, +∞) and the integral
Z
convergent then the improper integrals
+∞
f (x)d x is aZ
+∞
f (x)d x a
can be interpreted as an area of the region S = {(x, y)|x Ê a, 0 É y É f (x)}.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Geometric meaning
According to geometric meaning of an improper Z +∞
integral of type 1:
f (x)d x , if
a
lim f (x) = A 6= 0
x→+∞
and f (x) is integrable on every interval Z[a, b] ⊂ [a, +∞), then the improper integrals +∞
f (x)d x are divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
6 / 33
Type 1: Infinite intervals
Geometric meaning
According to geometric meaning of an improper Z +∞
integral of type 1:
f (x)d x , if
a
lim f (x) = A 6= 0
x→+∞
and f (x) is integrable on every interval Z[a, b] ⊂ [a, +∞), then the improper integrals +∞
f (x)d x are divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
6 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
THEOREM 1.1 (NEWTON-LEIBNIZ’S FORMULA) Suppose that f (x) has antiderivative F (x) on the interval [a, +∞) and is intergrable on Zevery interval +∞
[a, b]. The improper integral of type 1
f (x)d x is
a
convergent if and only if lim F (b) = F (+∞) exists as a b→+∞
finite number. Then +∞
Z a
(HCMUT-OISP)
¯+∞ ¯ f (x)d x = F (+∞) − F (a) = F (x)¯ a
IMPROPER INTEGRALS
HCMC — 2019.
(2)
7 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
8 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
SOLUTION ¯+∞ ¯ I = sin x ¯ = lim sin b − sin 0 = lim sin b. 0
(HCMUT-OISP)
b→+∞
IMPROPER INTEGRALS
b→+∞
HCMC — 2019.
8 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1 Evaluate I =
Z
+∞
cos xd x. 0
SOLUTION ¯+∞ ¯ I = sin x ¯ = lim sin b − sin 0 = lim sin b. 0
b→+∞
b→+∞
The limit lim sin b does not exist. Therefore the b→+∞
improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.2 Evaluate I =
Z
+∞
2
xe −x d x
0
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.2 Evaluate I =
Z
+∞
2
xe −x d x
0
SOLUTION ¯ 1 −x 2 ¯¯+∞ e d (−x ) = − e ¯ 2 0 0 1 −b 2 1 1 = lim − e + = b→+∞ 2 2 2
1 I =− 2
Z
+∞
−x 2
2
So the given integral I is convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent?
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent? If α 6= 1 then 1 I =− α−1
(HCMUT-OISP)
µ lim
1
b→+∞ b α−1
IMPROPER INTEGRALS
−
1
¶
1α−1
HCMC — 2019.
10 / 33
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3 For what values of α is the integral +∞
Z I= 1
dx xα
convergent? If α 6= 1 then 1 I =− α−1
If α > 1, then lim
1
b→+∞ b α−1
µ lim
1
b→+∞ b α−1
¶
1α−1
= 0. Therefore I =
the integral I converges. (HCMUT-OISP)
−
1
IMPROPER INTEGRALS
1 and so α−1 HCMC — 2019.
10 / 33
Type 1: Infinite intervals
If α < 1, then lim diverges.
(HCMUT-OISP)
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
IMPROPER INTEGRALS
HCMC — 2019.
11 / 33
Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
11 / 33
Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges. SUMMARY 1
If α > 1 then I =
Z 1
(HCMUT-OISP)
+∞
dx converges. xα
IMPROPER INTEGRALS
HCMC — 2019.
11 / 33
Type 1: Infinite intervals
If α < 1, then lim
1
b→+∞ b α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges. If α = 1, then I = lim ln |b| − ln 1 = +∞ and so the b→+∞
integral I diverges. SUMMARY 1
2
+∞
dx converges. xα 1 Z +∞ dx If α É 1 then I = diverges. xα 1
If α > 1 then I =
(HCMUT-OISP)
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
A COMPARISON TEST FOR IMPROPER INTEGRALS OF TYPE 1 THEOREM 1.2 Suppose that f and g are continuous functions on every interval [a, b] ⊂ [a, +∞) with 0 É g (x) É f (x), ∀x Ê a. Z 1
If
+∞
f (x)d x is convergent, then
a
g (x)d x is a
convergent.
(HCMUT-OISP)
+∞
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
A COMPARISON TEST FOR IMPROPER INTEGRALS OF TYPE 1 THEOREM 1.2 Suppose that f and g are continuous functions on every interval [a, b] ⊂ [a, +∞) with 0 É g (x) É f (x), ∀x Ê a. Z 1
+∞
If
f (x)d x is convergent, then
a
If
+∞
Z
g (x)d x is divergent then a
+∞
Z
f (x)d x is a
divergent. (HCMUT-OISP)
g (x)d x is a
convergent. 2
+∞
Z
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
(HCMUT-OISP)
A comparison test for improper integrals of type 1
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
1
A comparison test for improper integrals of type 1
If the area under the top curve y = f (x) is finite, then so is the area under the bottom curve y = g (x).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
1
A comparison test for improper integrals of type 1
If the area under the top curve y = f (x) is finite, then so is the area under the bottom curve y = g (x).
2
If the area under y = g (x) is infinite, then so is the area under y = f (x). (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.4 DetermineZ whether the integral is convergent or +∞
divergent
1
(HCMUT-OISP)
1 + e −x dx x
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.4 DetermineZ whether the integral is convergent or +∞
divergent
1
SOLUTION
Since +∞
Z 1
Z 1
+∞
1 + e −x dx x
1 + e −x 1 > x x
1 d x is divergent, so the integral x
1 + e −x d x is divergent. x (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
If λ = 0 and converges.
(HCMUT-OISP)
∞
Z
g (x)d x converges then a
∞
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
2
If λ = 0 and converges.
∞
Z
If λ > 0 then
g (x)d x converges then a
g (x)d x and a
converges or diverges.
(HCMUT-OISP)
f (x)d x a
∞
Z
∞
Z
∞
Z
f (x)d x either a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
THEOREM 1.3 Suppose f (x), g (x) be integrable on every [a, b] ⊂ [a, +∞) and f (x), g (x) Ê 0, ∀x Ê a. Evaluate f (x) =λ x→∞ g (x) lim 1
2
If λ = 0 and converges.
∞
Z
If λ > 0 then
g (x)d x converges then a
g (x)d x and a
converges or diverges. 3
If λ = +∞ and diverges. (HCMUT-OISP)
f (x)d x a
∞
Z
∞
Z
∞
Z
f (x)d x either a
∞
Z
g (x)d x diverges then a
∞
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
INDETERMINATE FORM
1
2
A comparison test for improper integrals of type 1
∞ ∞
xα lim = 0 (a > 1) x→+∞ a x lnα x lim = 0 (∀α > 0, β > 0) x→+∞ x β
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
INDETERMINATE FORM
1
2
A comparison test for improper integrals of type 1
∞ ∞
xα lim = 0 (a > 1) x→+∞ a x lnα x lim = 0 (∀α > 0, β > 0) x→+∞ x β lnα x << x β << a x , (α, β > 0, a > 1)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
DEFINITION 1.3 Functions f (x) and g (x) are called equivalent as x → a, if f (x) =1 x→a g (x)
(3)
lim
x→a
Denote by f (x) ∼ g (x).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.5 Determine whether the integral is convergent or Z +∞
divergent I =
1
(HCMUT-OISP)
p
dx
x 2 − 2x + 3
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.5 Determine whether the integral is convergent or Z +∞
divergent I =
1
We have p
p
where
x 2 − 2x + 3
1 x 2 − 2x + 3 p
Z
dx
+∞
1
(HCMUT-OISP)
> 0, ∀x Ê 1 and 1
x 2 − 2x + 3
x→+∞
∼
1 , x
dx diverges, therefore I diverges. x IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.6 Determine whether the integral is convergent or Z +∞
divergent I =
2
(HCMUT-OISP)
(x + 1)d x p 3 x 7 − 3x − 2
IMPROPER INTEGRALS
HCMC — 2019.
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Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.6 Determine whether the integral is convergent or Z +∞
divergent I =
2
We have p 3
x +1 x 7 − 3x − 2 p 3
where
Z
(x + 1)d x p 3 x 7 − 3x − 2
+∞
2
(HCMUT-OISP)
> 0, ∀x Ê 2 and
x +1
x→+∞
x 7 − 3x − 2
∼
x x 7/3
=
1 x 4/3
,
dx converges, therefore I converges. x 4/3 IMPROPER INTEGRALS
HCMC — 2019.
19 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
Suppose that f is defined on a finite interval [a, b) but has a vertical asymptote as x → b − and f is integrable on every interval [a, η] ⊂ [a, b). Then Z Φ(η) =
η
f (x)d x is defined on the interval [a, b).
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
20 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
Suppose that f is defined on a finite interval [a, b) but has a vertical asymptote as x → b − and f is integrable on every interval [a, η] ⊂ [a, b). Then Z Φ(η) =
η
f (x)d x is defined on the interval [a, b).
a
DEFINITION 2.1 The limit of function Φ(η) as η → b − is called an improper integral of type 2 on the interval [a, b) b
Z a
(HCMUT-OISP)
f (x)d x = lim− Φ(η) = lim− η→b
η→b
IMPROPER INTEGRALS
η
Z
f (x)d x
(4)
a
HCMC — 2019.
20 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
DEFINITION 2.2 1
If lim− Φ(η) = lim− η→b
η→b
η
Z
f (x)d x exists (as a finite a
number) then the improper integral of type 2 b
Z
f (x)d x converges. a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
21 / 33
Type 2: Infinity discontinuous integrands
R Definition of an improper integral of type 2 ab f (x)d x on [a, b)
DEFINITION 2.2 1
If lim− Φ(η) = lim− η→b
η→b
η
Z
f (x)d x exists (as a finite a
number) then the improper integral of type 2 b
Z
f (x)d x converges. Z η If lim− Φ(η) = lim− f (x)d x does not exist or is a
2
η→b
η→b
a
equal to ∞ then the improper integral of type 2 b
Z
f (x)d x diverges. a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
21 / 33
Type 2: Infinity discontinuous integrands
Geometric meaning
GEOMETRIC MEANING If f (x) Ê 0, ∀x ∈ [a, b) and the integral
b
Z
f (x)d x is Z b convergent then the improper integrals f (x)d x a
a
can be interpreted as an area of the region S = {(x, y)|a É x < b, 0 É y É f (x)}, where x = b is the vertical asymptote of the graph of function f (x)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
22 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
NEWTON-LEIBNIZ’S FORMULA Suppose that f (x) has antiderivative F (x) on every intervals [a, η] ⊂ [a, b) and lim− f (x) = ∞. The x→b
improper integral of type 2
b
Z
f (x)d x is convergent if a
and only if lim− F (η) = F (b − 0) exists as a finite η→b
number. Then b
Z a
(HCMUT-OISP)
¯b − ¯ f (x)d x = F (b − 0) − F (a) = F (x)¯ . a
IMPROPER INTEGRALS
HCMC — 2019.
(5)
23 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z 0
(HCMUT-OISP)
1
dx x
IMPROPER INTEGRALS
HCMC — 2019.
24 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z
1
0
dx x
SOLUTION 1 We have lim = +∞. Therefore x = 0 is vertical x→0+
asymptote.
(HCMUT-OISP)
x
IMPROPER INTEGRALS
HCMC — 2019.
24 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1 Evaluate I =
Z
1
0
dx x
SOLUTION 1 We have lim = +∞. Therefore x = 0 is vertical x→0+
x
asymptote. Since ¯1 ¯ I = ln |x|¯ = ln 1 − lim ln |a| = +∞. 0
a→0+
so improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
24 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
(HCMUT-OISP)
1
arccos x dx p 2 −1 1−x
Z
IMPROPER INTEGRALS
HCMC — 2019.
25 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
25 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.In other hand, x = 1 is arccos x not vertical asymptote because lim p = 1. x→1− 1 − x2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
25 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2 Evaluate I =
1
arccos x dx p 2 −1 1−x
Z
arccos x
SOLUTION We have lim p x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.In other hand, x = 1 is arccos x not vertical asymptote because lim p = 1. x→1− 1 − x2
Since
1
¯1 1 ¯ 2 I =− arccos xd (arccos x) = − · (arccos x)¯ −1 2 −1 2 1 π = − (arccos2 1 − lim arccos2 a) = a→−1+ 2 2 so improper integral I is convergent. Z
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
25 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
(HCMUT-OISP)
b
Z a
dx , (a < b) (b − x)α
IMPROPER INTEGRALS
HCMC — 2019.
26 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
26 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
If α < 1 then lim− (b − η)−α+1 = 0. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
26 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3 Evaluate I =
b
Z a
dx , (a < b) (b − x)α
SOLUTION η
¯η dx 1 −α+1 ¯ I = lim− =− lim (b − x) ¯ α a η→b −α + 1 η→b − a (b − x) 1 1 lim− (b − η)−α+1 + (b − a)−α+1 = α − 1 η→b −α + 1 Z
If α < 1 then lim− (b − η)−α+1 = 0. η→b
If α > 1 then lim− (b − η)−α+1 = ∞. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
26 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
27 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
SUMMARY 1
If α < 1, then improper integral converges.
(HCMUT-OISP)
IMPROPER INTEGRALS
b
Z a
dx (b − x)α
HCMC — 2019.
27 / 33
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then Zη I = lim− η→b
a
¯η dx ¯ = − lim− ln |b − x|¯ a η→b b−x
= − lim− ln |b − η| + ln(b − a) = ∞. η→b
SUMMARY 1
2
If α < 1, then improper integral converges.
Z
If α Ê 1, then improper integral
Z
diverges. (HCMUT-OISP)
IMPROPER INTEGRALS
b
dx (b − x)α
b
dx (b − x)α
a
a
HCMC — 2019.
27 / 33
Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges.
(HCMUT-OISP)
a
a
IMPROPER INTEGRALS
HCMC — 2019.
28 / 33
Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges. 2
If λ > 0 then
a
a
b
Z
g (x)d x and a
converges or diverges.
(HCMUT-OISP)
b
Z
f (x)d x either a
IMPROPER INTEGRALS
HCMC — 2019.
28 / 33
Type 2: Infinity discontinuous integrands
A comparison test for improper integrals of type 2
THEOREM 2.1 Suppose f (x), g (x) be integrable on every [a, η] ⊂ [a, b) and not be bounded as x → b − . Moreover, for every f (x) =λ x→b g (x) Z b Z b f (x)d x g (x)d x converges then If λ = 0 and
x ∈ [a, b), we have 0 É f (x), 0 É g (x), lim− 1
converges. 2
If λ > 0 then
a
a
b
Z
g (x)d x and a
converges or diverges. 3
If λ = +∞ and diverges. (HCMUT-OISP)
b
Z
f (x)d x either a
b
Z
g (x)d x diverges then a
b
Z
f (x)d x a
IMPROPER INTEGRALS
HCMC — 2019.
28 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
As x → 0 the following functions are equivalent 1
2
3
4
5
6
1 sin x ∼ x, tan x ∼ x, 1 − cos x ∼ x 2 2 arctan x ∼ x, arcsin x ∼ x a x − 1 ∼ x. ln a, (a > 0, a 6= 1), e x − 1 ∼ x x loga (1 + x) ∼ x loga e = , (a > 0, a 6= 1), ln(1 + x) ∼ x ln a p x (1 + x)µ − 1 ∼ µ.x, (µ ∈ R), 1 + x − 1 ∼ , 2 p x n 1 + x − 1 ∼ , (n ∈ N) n x2 sinh x ∼ x, cosh x − 1 ∼ 2 (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
29 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
(HCMUT-OISP)
cos2 x dx p 3 1 − x2
IMPROPER INTEGRALS
HCMC — 2019.
30 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
cos2 x dx p 3 1 − x2
cos2 x cos2 x 1 =∞ lim− p = lim . p x→1 3 1 − x 2 x→1− 3 1 + x (1 − x)1/3
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
30 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.4 Determine whether the integral is convergent or Z 1
divergent I =
0
cos2 x dx p 3 1 − x2
cos2 x cos2 x 1 =∞ lim− p = lim . p x→1 3 1 − x 2 x→1− 3 1 + x (1 − x)1/3 2 cos2 x 1 cos2 x 1 x→1− cos 1 =p ∼ . . . p p 3 3 3 1 + x (1 − x)1/3 2 (1 − x)1/3 1 − x2 1 We have α = < 1, therefore I converges. 3 (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
30 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
(HCMUT-OISP)
1
ln(1 + 3 x) dx e sin x − 1
IMPROPER INTEGRALS
HCMC — 2019.
31 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
1
ln(1 + 3 x) dx e sin x − 1
p ln(1 + 3 x) x 1/3 1 lim+ sin x = lim+ = lim+ 2/3 = ∞ x→0 x→0 x→0 x e −1 x
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
31 / 33
Type 2: Infinity discontinuous integrands
Some basic equivalent functions
EXAMPLE 2.5 Determine whether thepintegral is convergent or divergent I =
Z
0
1
ln(1 + 3 x) dx e sin x − 1
p ln(1 + 3 x) x 1/3 1 lim+ sin x = lim+ = lim+ 2/3 = ∞ x→0 x→0 x→0 x e −1 x p ln(1 + 3 x) x→0+ x 1/3 1 ∼ = e sin x − 1 x x 2/3 Due to α = 32 < 1 so I converges. (HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
31 / 33
MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
32 / 33
MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
32 / 33
MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
3
>> s yms x; >> i nt (1/sqr t (x), 0, 1) ⇒ Ans = 2
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
32 / 33
MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
3
>> s yms x; >> i nt (1/sqr t (x), 0, 1) ⇒ Ans = 2
4
>> s yms x; >> i nt (1/x, 0, 1) ⇒ Ans = I n f
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
32 / 33
MatLab
THANK YOU FOR YOUR ATTENTION
(HCMUT-OISP)
IMPROPER INTEGRALS
HCMC — 2019.
33 / 33
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