Giáo trình giải tích 1 part 10 pot

n →∞
2sin
x
2
(sin x + ···+sinnx)=cos
x
2
− cos
2n +1
2
x
f [a, b]
lim
n→∞
b − a
n
n

k=1
f(a +
k(b −a)
n
)=

b
a
f(x)dx
lim
n→∞
1
n

((
1
n
)
2
+(
2
n
)
2
+ ···+(
n
n
)
2
) lim
n→∞
1
n
(e
3
n
+ e
3.2
n
+ ···+ e
3n
n
)
lim

n→∞
(
1
n +1
+
1
n +2
+ ···+
1
2n
) lim
n→∞
1
p
+2
p
+ ···+ n
p
n
p+1
S
n
=
1
n

(1 +
1
n
)sin

2π
n
+(1+
2
n
)sin
4π
n
+ ···+(1+
n
n
)sin
2nπ
n

lim
n→+∞
S
n
lim
n→+∞
S
n
f [0, 1]

1
0
f −
1
n

n

k=1
f(
k
n
)=O(
1
n
)

b
a
f(x)dx =

b
a
f(t)dt =

b
a
f(u)du

2
0
(3x
2
− 5)dx =3

2

0
x
2
dx − 5

2
0
dx =3(
2
3
3
− 0) − 5(2 − 0).
|f| f
[0, 1]
{0,
1
10
,
2
10
,
3
10
, ··· , 1} f(x)=sin
1
x
,f(0) = 7
f(x)=
1
n

x =
1
n
,n∈ N f(x)=0
D(x)=0 x D(x)=1 x
f [a, b] f(x)=g(x)
g
f [a, b] f(x)=g(x)
g
f [a, b] F (x)=

x
a
f
[a, b] |F (x) −F(y)|≤max
t∈[a,b]
|f(t)||x −y|
0 ≤ x ≤ 1
x
2
√
2
≤
x
2
√
1+x
≤ x
2
1

3
√
2
≤

1
0
x
2
√
1+x
dx ≤
2
3
2π
2
9
≤

π
2
π
6
2x
sin x
dx ≤
4
9
π
2

f [a, b] f ≥ 0
c f(c) > 0

b
a
f>0

b
a
f =0 f ≡ 0
f,g [a, b]


b
a
f(x)g(x)dx

2
≤

b
a
f
2
(x)dx

b
a
g
2

(x)dx
n =1, 2, 3, ···

n
1
ln xdx < ln n! <

n+1
1
ln xdx
e

n
e

n
<n! <e

n +1
e

n+1
n!=

n
e

n
O(n)
f :[1, +∞) → R

S
n
=
n

k=1
f(k) I
n
=

n
1
f(x)dx
f(k) <

k
k−1
f(x)dx < f(k − 1) (k =2, 3, ···)
(S
n
− I
n
)
n∈N
[0,f(1)]
1+
1
2
+ ···+
1

n
− ln n
f(x)=x +
n

k=1
(a
k
cos kx + b
k
sin kx) (−π, π)
f [a, b]
d
dx


x
a
f(t)dt

= f(x)
f(x)=

x
0

t + t
6
dt
df

dx
df
dt
f F (x)=

x
2
0
f F

(x)
ϕ [a, b] f ϕ([a, b])
d
dx
(

ϕ(x)
ϕ(a)
f(t)dt)=f(ϕ(x))ϕ

(x).
◦

a
0
x
2

a
2

− x
2
dx

a
1
√
a
2
− x
2
x
dx
◦

1
0
xe
x
dx

π/2
0
x cos xdx

π/2
0
e
x
cos xdx

◦

1
0
dx
x
2
− 5x +6

1
0
xdx
(1 + x)
2

1
0
x
5
dx
1+x
2

1
0
dx
x
4
+4x
2

+3
◦

√
2
√
2/3
dx
x
√
x
2
− 1

7
2
dx
√
2+x +1
◦

π
0
sin xdx
cos
2
x − 3

π
0

sin
4
xdx

π/4
0
tan
6
xdx

π/4
0
dx
cos
4
x
f [−a, a]
f

a
−a
f =2

a
0
f f

a
−a
f =0

f T [0,T] a ∈ R

a+T
a
f =

T
0
f
f [0, 1]

π
0
f(sin x)dx =2

π/2
0
f(sin x)dx

π
0
xf(sin x)dx =
π
2

π
0
f(sin x)dx

π

0
x sin x
1+cos
2
x
dx

π
0
x
3
sin x
1+cos
2
x
dx
f(x)=
1
(1 + x)
2
f(x)=
d
dx
(
−1
1+x
)

+∞
−∞

f(x)dx = lim
a→+∞

a
−a
f(x)dx = lim
a→+∞
(
−1
1+a
−
1
1+a
)=0.

+∞
1
dx
x
2/3

+∞
1
dx
x
4/3

1
0
dx

x
2/3

1
0
dx
x
4/3

+∞
0
x
2
+1
x
4
+1
dx

+∞
0
x cos xdx

+∞
0
x ln xdx

+∞
1
xdx

√
x − 1

+∞
0
x
2
dx
x
4
+ x
2
+1

+∞
1
x
n
dx
1+x
m
n ≥ 0

+∞
1
ln(1 + x)dx
x
n

+∞

0
cos xdx
2+x
n

1
0
dx
√
1 − x
2

1
0
√
xdx
√
1 − x
4

1
0
√
xdx
e
sin x
− 1

1
0

ln xdx
1 − x
2

+∞
0
sin x
x
3/2
dx
p, q, p
1
, ··· ,p
n

b
a
dx
(x − a)
p

+∞
0
x
p
1+x
dx

+∞
1

dx
x
p
ln
q
x
Γ(p)=

+∞
0
e
−x
x
p−1
dx
B(p, q)=

1
0
x
p
(1 − x)
q
dx

π/2
0
dx
cos
p

x sin
q
x

+∞
−∞
dx
|x − a
1
|
p
1
|x − a
2
|
p
2
···|x −a
n
|
p
n

+∞
0
cos x
x
dx
f [0, 1]


1
0
f(x)
√
1 − x
2
dx

1
0
f(x)
√
1 − x
2
dx =

π/2
0
f(sin u)du
F (x)=

x
0
sin t
t
3/2
dt (0 <x<∞) max x = π

1
0

ln xdx lim
n→∞
n
√
n!
n
R
2
y = x
2
+4,y= x +4
x
2
a
2
+
y
2
b
2
=1,y= x
2
y =ln(
k
x
),y=0,x=1,x= e (k>0) k ∈ N <e− 2
x = a(t − sin t),y= a(1 −cos t) y =0
r
2
= a

2
cos 2ϕ
r = a(1 + cos ϕ)
I(t)=

1
0
|e
x
− t|dx
R
3
y =
b
a

a
2
− x
2
, −a ≤ x ≤ a
y
2
=4− x, 0 ≤ x ≤ 4
R
2
y
2
=
1

9
x
3
, 0 ≤ x ≤ 1,y >0 y
2
=2px, a ≤ x ≤ b
x = a cos
3
t, y = a sin
3
t
r =sin
3
ϕ
3
, 0 ≤ ϕ ≤ π/2
0, 61111 ··· 1, 33333 ··· −2, 343434 ··· e π ln 2
S =1+2+4+8+16+··· 2S =2+4+8+···= S − 1
S = −1
a
1
+ a
2
+ a
3
+ ··· S a
2
+ a
3
+ ···

S − a
1
1
1.2
+
1
2.3
+
1
3.4
+
1
4.5
+
1
5.6
+ ···
1
1.4
+
1
4.7
+
1
7.10
+
1
10.13
+ ···
1

1.3
+
1
4.6
+
1
7.9
+
1
10.12
+
1
13.15
+ ···
1
2
−
1
4
+
1
8
−
1
16
+
1
32
+ ···
∞


k=0
2
k
+3
k
6
k
∞

k=0
(
1 − x
1+x
)
k
∞

k=0
(
√
k +2− 2
√
k +1+
√
k)
∞

k=1
1

k(k +1)(k +2)
∞

k=1
1
k(k + m)
(m ∈ N)
∞

k=0
k
4
k!
∞

k=0
1+k
1+k
2
∞

k=0
3
4+2
k
∞

k=0
k ln k
k

2
+2k +3
∞

k=1
k!3
k
k
k
∞

k=0
(k!)
2
(2k)!
∞

k=2
1
(ln k)
k
∞

k=0
(1 +
1
k
)
2k
e

k
∞

k=1
1
k
p
∞

k=2
1
k ln k
∞

k=2
1
k
p
ln
q
k
∞

k=0
sin kx
a
k
=
1
√

k
+
(−1)
k
k
∞

k=1
(−1)
k
a
k
a
k
> 0 a
k
→ 0
(
1
2
)
0
+(
1
4
)
1
+(
1
2

)
2
+(
1
4
)
3
+(
1
2
)
4
+ ···
S =
∞

k=1
1
k
p
n S
n
p>0
1
(k +1)
p
<

k+1
k

1
x
p
dx <
1
k
p
(k =1, 2, ···)
p>1
p>1 S
n−1
+

+∞
n
1
x
p
dx < S < S
n
+

+∞
n
1
x
p
dx
1
(p − 1)(n +1)

p−1
<S− S
n
<
1
(p − 1)n
p−1
a
k
,b
k
> 0
∞

k=0
a
k
∞

k=0
b
k
∞

k=0
a
k
b
k
,

∞

k=0
a
2
k
,
∞

k=0
(a
k
+ b
k
)
2
,
∞

k=0
√
a
k
k
1 −
1
2
+
1
3

−
1
4
+
1
5
−
1
6
+ ··· =1+(
1
2
− 1) +
1
3
+(
1
4
−
1
2
)+
1
5
+(
1
6
−
1
3

)+···
=(1+
1
2
+
1
3
+
1
4
+ ···) − 1 −
1
2
−
1
3
−
1
4
−···
=(1+
1
2
+
1
3
+
1
4
+ ···) − (1 +

1
2
+
1
3
+
1
4
+ ···)
=0.
1+x
2
+ x + x
4
+ x
6
+ x
3
+ x
8
+ x
10
+ x
5
+ ···=
1
1 − x
|x| < 1