21 improper integrals
TRANSCRIPT
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Improper Integrals
![Page 2: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/2.jpg)
Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx 0
∞
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, 0
1
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x).
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define ∫a
∞
f(x) dx as lim f(x) dx
∫a
u
u ∞
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define ∫a
∞
f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)
∫a
u
u ∞ x ∞a
∞
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define
Example: Find
∫0
∞
e-x dx
∫a
∞
f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)
∫a
u
u ∞ x ∞a
∞
![Page 9: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/9.jpg)
Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define
Example: Find
∫0
∞
e-x dx
∫0
∞
e-x dx = -e-x|
0
∞
∫a
∞
f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)
∫a
u
u ∞ x ∞a
∞
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define
Example: Find
∫0
∞
e-x dx
∫0
∞
e-x dx = -e-x| = lim (-e-x) – (-e0)
x ∞0
∞
∫a
∞
f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)
∫a
u
u ∞ x ∞a
∞
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Improper Integrals
Integrals over an infinite interval such as ∫ e-x dx or
0
∞
integrals of f(x) that goes to ∞ within the interval of
integration such as ∫ Ln(x) dx, are called
improper integrals. 0
1
Let F(x) be the antiderivative of f(x). We define
Example: Find
∫0
∞
e-x dx
∫0
∞
e-x dx = -e-x| = lim (-e-x) – (-e0) = 0 – (-1) = 1
x ∞0
∞
∫a
∞
f(x) dx as lim f(x) dx = F(x)| = lim F(x) – F(a)
∫a
u
u ∞ x ∞a
∞
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
x a
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x aa
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 |
1
0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2)
1
0 x 0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
∫0
1
x-2 dx = –x–1 | 1
0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
∫0
1
x-2 dx = –x–1 | = -1 – lim(-x-1)
1
0 x 0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
∫0
1
x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞
1
0 x 0
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
∫0
1
x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞
1
0 x 0
If the improper integral exists, we say it converges.
a
b
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Improper Integrals
Let F(x) be the antiderivative of f(x) and lim f(x) ∞.
We define x a
∫a
b
f(x) dx = F(x)| = F(b) – limF(x) x a
Example: Find
∫0
1
x-1/2 dx
∫0
1
x-1/2 dx = 2x1/2 | = 2 – lim(2x1/2) = 2
1
0 x 0
Example: Find
∫0
1
x-2 dx
∫0
1
x-2 dx = –x–1 | = -1 – lim(-x-1) = -1 – (-∞) = ∞
1
0 x 0
If the improper integral exists, we say it converges. If the improper integral fails to exist or its infinite, we say it diverges.
a
b
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Improper Integrals(The Floor theorem)
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Improper Integrals(The Floor theorem)
y = f(x)
y = g(x)∞
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, ∫
a
b
y = f(x)
y = g(x)∞
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫
a
b
∫a
b
y = f(x)
y = g(x)∞
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫
a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫
a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
y = f(x)
y = g(x)
N
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫
a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
If f(x) > g(x) > 0 and f(x) dx = N converges ∫a
b
y = f(x)
y = g(x)
N
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Improper Integrals(The Floor theorem) If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫
a
b
∫a
b
y = f(x)
y = g(x)∞
(The Ceiling theorem)
If f(x) > g(x) > 0 and f(x) dx = N converges then
g(x) dx converges also.∫a
b
∫a
b
y = f(x)
y = g(x)
N
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Improper Integrals
Note that no conclusion may be drawn if
f(x) > g(x) > 0 and g(x) dx converges, ∫a
b
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Improper Integrals
Note that no conclusion may be drawn if
f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a
b
∫a
b
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Improper Integrals
The function y = serves as an important
"boundary" for divergence and convergence.
x1
Note that no conclusion may be drawn if
f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a
b
∫a
b
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Improper Integrals
The function y = serves as an important
"boundary" for divergence and convergence. The following are the graphs of y = and
x ,1
x1
x2 ,1
x3 1
for x > 1.
Note that no conclusion may be drawn if
f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a
b
∫a
b
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Improper Integrals
The function y = serves as an important
"boundary" for divergence and convergence. The following are the graphs of y = and
x ,1
x1
x2 ,1
x3 1
for x > 1.
(1, 1)
y = 1/x
y = 1/x2
y = 1/x3
Note that no conclusion may be drawn if
f(x) > g(x) > 0 and g(x) dx converges, (or f(x) dx = ∞).∫a
b
∫a
b
1 ∞
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Improper Integrals
∫1
∞
One checks that
(1, 1)
y = 1/x
y = 1/x2
y = 1/x3
x1 dx = Ln(x)| = ∞,
1
∞
1 ∞
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Improper Integrals
∫1
∞
One checks that
(1, 1)
y = 1/x
y = 1/x2
y = 1/x3
x1 dx = Ln(x)| = ∞, but ∫1
∞
x21 dx and ∫1
∞
x31 dx are finite.
1
∞
1 ∞
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Improper Integrals
∫1
∞
One checks that
(1, 1)
y = 1/x
y = 1/x2
y = 1/x3
x1 dx = Ln(x)| = ∞, but ∫1
∞
x21 dx and ∫1
∞
x31 dx are finite.
The p-Theorem:
∫1 xp1
∞
dx converges for p > 1, A.
1
∞
1 ∞
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Improper Integrals
∫1
∞
One checks that
(1, 1)
y = 1/x
y = 1/x2
y = 1/x3
x1 dx = Ln(x)| = ∞, but ∫1
∞
x21 dx and ∫1
∞
x31 dx are finite.
The p-Theorem:
∫1 xp1
∞
dx converges for p > 1, diverges for p < 1.A.
1
∞
1 ∞
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Improper Integrals
y = 1/x
y = 1/x1/2
y = 1/x1/3
0 1
![Page 42: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/42.jpg)
Improper Integrals
∫0
1
x1
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
y = 1/x
y = 1/x1/2
y = 1/x1/3
0 1
![Page 43: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/43.jpg)
Improper Integrals
∫0
1
x1
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
and that x2
1 > x1 over (0, 1]
y = 1/x
y = 1/x1/2
y = 1/x1/3
0 1
![Page 44: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/44.jpg)
Improper Integrals
∫0
1
x1
So ∫0
1
x2
1 dx = ∞
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
and that x2
1 > x1 over (0, 1]
y = 1/x
y = 1/x1/2
y = 1/x1/3
0 1
![Page 45: 21 improper integrals](https://reader036.vdocuments.mx/reader036/viewer/2022081504/558e93ea1a28ab46108b4655/html5/thumbnails/45.jpg)
Improper Integrals
∫0
1
x1
So ∫0
1
x2
1 dx = ∞
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
and that x2
1 > x1 over (0, 1]
y = 1/x
y = 1/x1/2
y = 1/x1/3
∫0
1
x1/2
1 dx and ∫ x1/3
1 dx are finite.
One checks easily that
0
1
0 1
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Improper Integrals
∫0
1
x1
So ∫0
1
x2
1 dx = ∞
The p-Theorem:
∫ xp1 dx converges for p < 1, B.
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
and that x2
1 > x1 over (0, 1]
y = 1/x
y = 1/x1/2
y = 1/x1/3
∫0
1
x1/2
1 dx and ∫ x1/3
1 dx are finite.
One checks easily that
0
1
0
1
0 1
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Improper Integrals
∫0
1
x1
So ∫0
1
x2
1 dx = ∞
The p-Theorem:
∫ xp1 dx converges for p < 1, diverges for p > 1.B.
On the other hand, over (0, 1]
dx = Ln(x)| = ∞0
1
and that x2
1 > x1 over (0, 1]
y = 1/x
y = 1/x1/2
y = 1/x1/3
∫0
1
x1/2
1 dx and ∫ x1/3
1 dx are finite.
One checks easily that
0
1
0
1
0 1