Giáo trình giải tích 2 part 7 doc

u, v x, y x =1,y = −1,u=1,v = −1
x =0,y =1,u=0,v=0
∂u
∂x
x =1,y = −1 x =0,y =1
f(x, y)=x
2
+ y
2
− 1=0 y = g(x)
g
F (x, y)=x
3
− y
3
=0
y = x
∂F
∂y
(0, 0) = 0
n u =(u
0
, ··· ,u
n−1
)
P
u
(x)=x
n
+ u
n−1

x
n−1
+ ···+ u
1
x + u
0
u = a x
0
P
a
P
a
(x
0
)=0,P

a
(x
0
) =0
U a V x
0
u ∈ U x(u) ∈ V P
u
(x)=0
C
∞
3 x
3
+ px + q =0 p, q

∆=4p
3
+27q
2
∆ > 0 x
∗
(p, q)
∆ < 0 x
−
(p, q) <x
0
(p, q) <x
+
(p, q)
∆=0,q >0 x
−
(p, q) < 0 x
0+
(p, q)
∆=0,q <0 x
0−
(p, q) < 0 x
+
> 0
(p, q)=(0, 0) x =0
x
∗
C
∞
x

−
,x
0
,x
+
C
∞
✲
x
✻
y
abS
inf f
sup f
R
n
A =[a
1
,b
1
] ×···[a
n
,b
n
].
A v(A)=(b
1
− a
1
) ···(b

n
− a
n
)
P A [a
i
,b
i
],i=1, ··· ,n
a
i
= c
i0
<c
i1
< ···<c
im
i
= b
i
m
1
m
2
···m
n
A
S =[c
1i
1

,c
1i
1+1
] ×···×[c
ni
n
,c
ni
n+1
].
S ∈ P
f : A → R P A
L(f,P)=

S∈P
inf
x∈S
f(x) v(S)
U(f,P)=

S∈P
sup
x∈S
f(x) v(S)
L(f,P) ≤ U(f,P) P

P
P

P P


P
L(f,P) ≤ L(f,P

) U(f, P

) ≤ U(f,P).
I
(f)=sup
P
L(f,P) ≤ inf
P
U(f,P)=I(f)
f
A I(f)=I(f)
f A

A
f

A
f(x)dx

b
1
a
1
···

b

n
a
n
f(x
1
, ··· ,x
n
)dx
1
···dx
n
f A
>0 P A U(f,P) −L(f,P) <.
f ≡ c U(f,P)=L(f,P)=cv(A) P
f A

A
f = cv(A)
D(x)=

0 x
1 x
[0, 1] P
L(D,P)=0,U(D,P)=1.
P A ξ
P
=(ξ
S
,S ∈ P ) ξ
S

∈ S
S(f, P,ξ
P
)=

S∈P
f(ξ
S
) v(S)
|P | S ∈ P
f : A → R A ⊂ R
n
f A

A
f = I
lim
|P |→0
S(f, P,ξ
P
)=I, ∀ξ
P
>0 δ>0
P A |P | <δ
|S(f, P,ξ
P
) −I| < ∀ ξ
P
P
0

A >0 δ>0
P A |P | <δ P
P <
n
n =1 A =[a, b] P
0
N δ = /N P
|P | <δ P P
0
≤ × ≤ N × δ = 
n>1 P
0
V
1
, ··· ,V
k
T
δ = /T P A |P | <δ
S ∈ P S ⊂ V
i
,i=1, ··· ,k S P
0
v(S) ≤ δD D V
1
, ··· ,V
k
S

S∈P,S⊂V
i

,∀i
v(S) <δT = 
⇒ |f(x)| <M,∀x ∈ A
P
0
U(f,P
0
) −I</2 ,I− L(f,P
0
) </2
 := /2M δ>0 P
|P | <δ P
1
P P
0
P
2
P P
0
ξ
P

S∈P
f(ξ
S
) v(S) ≤

S∈P
1
f(ξ

S
) v(S)+

S∈P
2
f(ξ
S
) v(S)
≤ U(f,P
0
)+M/M < I + 

S∈P
f(ξ
S
) v(S) ≥ L(f,P
0
) −/2 >I−
|S(f, P,ξ
P
) −I| <
⇒ >0 δ>0 P N P
S ∈ P ξ
S
∈ S |f(ξ
S
) −sup
S
f| </v(S)N
|U(f,P) −I|≤|U(f,P) −


S∈P
f(ξ
S
)v(S)| + |

S∈P
f(ξ
S
)v(S)
I
|
<

S∈P
v(S)/v(S)N =  U(f,P) −I| < 2
|L(f,P) − I| < 2 |U(f,P) − L(f,P)| < 4
f A 
f : A =[a
1
,b
1
] ×[a
n
,b
n
] → R
P
N
[a

i
,b
i
],i =1, ··· ,n N +1
c
ik
= a
i
+
k(b
i
− a
i
)
N
,k=0, ··· ,N N →∞
(b
1
− a
1
) ···(b
n
− a
n
)
N
n
N

k

1
,···,k
n
=1
f(c
1k
1
, ··· ,c
nk
n
) −→

A
f.

1
0
xdx = lim
N→∞
1
N
N

k=1
k
N
= lim
N→∞
1
N

2
N(N +1)
2
=
1
2
.
C ⊂ R
n
C
χ
C
(x)=

1 x ∈ C
0 x ∈ C
A C C
χ
C
A
C
v(C)=

A
χ
C
.
A C
n =1,n=2
U(χ

C
,P) P C
L(χ
C
,P) P C
C A
A
C C
C ⊂ R
n
f : A → R
A C f
C fχ
C
A f C

C
f =

A
fχ
C
.
R(C) C
B ⊂ R
n
µ(B)=0
µ
n
(B)=0 >0 S

1
,S
2
, ···
B B ⊂∪
i
S
i

i
v(S
i
) <.
N ⊂ R R
R
2
µ(B)=0 f : B → R
m
µ(f(B)) = 0
µ(B
i
)=0,i=1, 2, ··· µ(∪
i
B
i
)=0
>0 i S
i1
,S
i2

, ··· B
i

j
v(S
ij
) </2
i
{S
ij
} ∪
i
B
i

ij
v(S
ij
) <

i
1
2
i
< 
R
n
f : A → R A ⊂ R
n
f

f
f S o(f, S)=sup
x∈S
f(x) − inf
x∈S
f(x)
f a
o(f,a) = lim
r→0
+
o(f,B(a, r)).
r
o(f,a)=0 f a
B = {x : o(f,x) > 0} f
(⇒) µ(B)=0 >0 B

= {x : o(f,x) ≥ } B

⊂ B
µ(B

)=0 B

S
1
, ··· ,S
N
B

N


i=1
v(S
i
) <
P A S ∈ P S ∩ B

= ∅ S ⊂ S
i
i ∈{1, ··· ,N}
P
1
= {S ∈ P : S ∩B

= ∅} P
2
= {S ∈ P : ∃iS⊂ S
i
}
S ∈ P
1
o(f,x) <,x∈ S S P
S

∈ P
1
sup
S

f − inf

S

f ≤ 2
U(f,P) −L(f,P)=(

S∈P
1
+

S∈P
2
)

(sup
S
f − inf
S
f)v(S)

≤

S∈P
1
2v(S)+

S∈P
2
Mv(S), (M =sup
A
f − inf

A
f)
≤ 2v(A)+M
N

i=1
v(S
i
) < (2v(A)+M).
f A
(⇐) f A B =

k∈N
B
1
k
µ(B
1
k
)=0, ∀k ∈ N k
>0 P U(f,P) −L(f,P)=

S∈P
(sup
S
f − inf
S
f)v(S) <

k

.

S∩B
1
k
=∅
1
k
v(S) ≤

S∩B
1
k
=∅
(sup
S
f − inf
S
f)v(S) <

k
{S ∈ P : S ∩B
1
k
= ∅} B
1
k

S∩B
1

k
=∅
v(S) < µ(B
1
k
)=0 
C ⊂ R
n
µ(∂C)=0.
C ⊂ R
n
f : C −→ R
f C
f :[a, b] → R f
C χ
C
χ
C
∂C
k ∈ N
D
k
= {x ∈ [a, b]:o(f, x) ≥|f(a) − f(b)|/k} k
f B =

k
D
k
f 
f(x)=sin

1
x
x =0 f(0) = 0 [−1, 1]
f(x, y)=x
2
+sin
1
y
y =0 f(x, 0) = 0 A = {x
2
+y
2
≤ 1}
A R
n
f,g
A
α, β ∈ R αf + βg A

A
(αf + βg)=α

A
f + β

A
g
A
1
,A

2
⊂ A f A
1
,A
2

A
1
∪A
2
f =

A
1
f +

A
2
f −

A
1
∩A
2
f.
f ≤ g A

A
f ≤


A
g.
|f| A |

A
f|≤

A
|f|.
f A c ∈ A

A
f = f(c)v(A).
inf
A
fv(A) ≤

A
f ≤ sup
A
fv(A)

A, B A ∪B v(A ∪B)=v(A)+v(B) −v(A ∩B).
f [a, b] c ∈ [a, b] f(c)=
1
b −a

b
a
f(x)dx

f :[a, b] −→ R
d
dx

x
a
f = f, x ∈ [a, b].
F f [a, b]
F

(x)=f(x),x∈ [a, b]

b
a
f = F (b) −F(a).
g :[a, b] −→ R f g([a, b])

g(b)
g(a)
f =

b
a
f ◦ gg

.
u, v [a, b]

b
a

uv

= u(b)v(b) −u(a)v(a) −

b
a
u

v.
•
•
✲
x
✻
y
g
2
g
1
x
C
a b
C g
1
,g
2
[a, b] g
1
≤ g
2

C x ∈ [a, b] d(x) C
x
=
x ×R ∩ C = x ×[g
1
(x),g
2
(x)]
(C)=

C
dxdy =

b
a
d(x)dx =

b
a
(

g
2
(x)
g
1
(x)
dy)dx
f [a, b] ×[c, d]
f [a, b] ×[c, d] V = {a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z ≤

f(x, y)} x ∈ [a, b]
S(x)= {(y, z):c ≤ y ≤ d, 0 ≤ z ≤ f(x, y)} =

d
c
f(x, y)dy
(V )=

V
dxdydz =

b
a
S(x)dx =

b
a
(

d
c
f(x, y)dy)dx
C ⊂ R
n
× R
m
f : C → R
Ω={x ∈ R
n
: ∃y ∈ R

m
, (x, y) ∈ C} C R
n
C
x
= {y ∈ R
m
:(x, y) ∈ C} C x

C
x
f(x, y)dy x ∈ Ω

C
f(x, y)dxdy =

Ω
(

C
x
f(x, y)dy)dx.
C = A×B A, B R
n
, R
m
P, P

A, B P ×P


A ×B
S × S

,S ∈ P, S

∈ P

L(f,P × P

)=

S×S

∈P ×P

inf{f(x, y); x ∈ S, y ∈ S

}v(S × S

)
=

S∈P
(

S

∈P

inf{f(x, y); x ∈ S, y ∈ S


}v(S

))v(S)
≤

S∈P
inf{

S

∈P

inf{f(x, y); y ∈ S

}v(S); x ∈ S}v(S)
≤

S∈P
inf
x∈S
(

B
f(x, y)dy)v(S)
≤ L(

B
f(x, y)dy, P ).
L(f,P × P


) ≤ L(

B
f(x, y)dy, P ) ≤ U(

B
f(x, y)dy, P ) ≤ U(f,P × P

).

A×B
f =

A
(

B
f(x, y)dy)dx.
C A, B C ⊂ A × B f
A ×B 0 C 
g
1
,g
2
:[a, b] → R g
1
≤ g
2
C = {(x, y):a ≤ x ≤ b, g

1
(x) ≤ y ≤ g
2
(x)} f C
C

C
f(x, y)dxdy =

b
a
(

g
2
(x)
g
1
(x)
f(x, y)dy)dx
h
1
,h
2
:Ω→ R Ω ⊂ R
2
h
1
≤ h
2

C = {(x, y, z):(x, y) ∈ Ω,h
1
(x, y) ≤ z ≤ h
2
(x, y)} f
C C

C
f(x, y, z)dxdydz =

Ω
(

h
2
(x,y)
h
1
(x,y)
f(x, y, z)dz)dxdy
E = {
x
2
a
2
+
y
2
b
2

≤ 1}
E Ox [−a, a] C
x
= {y : −
b
a

a
2
− x
2
≤ y ≤
b
a

a
2
− x
2
}
v(E)=

a
−a
(

b
a
√
a

2
−x
2
−
b
a
√
a
2
−x
2
dy)dx =

a
−a
2
b
a

a
2
− x
2
dx
=2ab(arcsin
x
a
+
x
2

a

a
2
− x
2
)|
a
−a
= πab.