IRf:I→R
a, b a < b I = [a, b]
σ[a, b]
(x0, x1, . . . , xn)
a=x0< x1<··· < xn=b.
σ h(σ)
h(σ) = max
0≤i≤n−1xi+1 −xi.
[a, b]E
σ= (x0, x1, . . . , xn)E
S(σ) =
n−1
X
i=0
(xi+1 −xi)f(xi).
n≥1h=b−a
nxi=a+ih i = 0, . . . , n
σc
n:= (x0, x1, . . . , xn)∈E.
f[a, b]
J∈Rσ∈E
S(σ)→J h(σ)→0
ε > 0hε>0σ∈E
h(σ)≤hε=⇒ |S(σ)−J| ≤ ε.
J f [a, b]
Rb
af(x) dx
f[a, b]J∈R
σ∈E
S(σ)→J h(σ)→0.
f[a, b]
(x0,··· , xn) [a, b]i= 0,··· , n −1f
]xi, xi+1[ [xi, xi+1]
f[a, b]
i= 0,··· , n −1gi: [xi, xi+1]→R[xi, xi+1]
f]xi, xi+1[f[a, b]
Zb
a
f(x) dx=
n−1
X
i=0 Zxi+1
xi
gi(x) dx.
E
I:= [a, b]f:I→E
I
(e1, . . . , eN) (e0
1, . . . , e0
N)E
x∈I(f1(x), . . . , fN(x)) (g1(x), . . . , gN(x))
f(x) (e1, . . . , eN) (e0
1, . . . , e0
N)
i= 1, . . . , N figiI
N
X
i=1 ZI
fi(x)x ei=
N
X
i=1 ZI
gi(x)x e0
i.
N
X
i=1 ZI
fi(x)x eif I Zb
a
f(x) dx
E=C
Zb
a
f(x) dx=Zb
a
Re f (x) dx+iZb
a
Im f(x) dx.
f
I[c, d]I f [c, d]
[c, d]
f: [0,1] →R, x 7→ (1
xx6= 0
1x= 0 ,
[0,1] ]0,1[
]0,1[
a∈Rb∈]a, ∞]I= [a, b[
f
[a, b[f[a, b[
I→R, x 7→ Zx
a
f(t)dt
x→bRb
af(t)dt
f[a, b[
Zb
a
f(t)dt = lim
x→bZx
a
f(t)dt.
[0,∞[f: [0,∞[→Rt7→ e−t
Z∞
0
e−tdt = 1.
f
IRb
af(t)dt
ε > 0xε∈I
x, y ∈I, xε≤x < y < b =⇒Zy
x
f(t)dt≤ε.
f I
Rb
af(t)dt Rb
a|f(t)|dt
f I Rb
af(t)dt
Rb
af(t)dt
I= [a, b[f, g :I→[0,∞[
I f g b
Zb
a
f(t)dt Zb
a
g(t)dt
Z∞
1
√1 + t
t2dt 0≤√1+t
t2∼1
t3/2
t→ ∞
a, b a < b I = [a, b]
n∈Nfn: [a, b]→R
[a, b] (fn) [a, b]
[a, b]→R, x 7→ lim
n→∞ fn(x)
[a, b]
lim
n→∞ Zb
a
fn(x)dx =Zb
a
lim
n→∞ fn(x)dx.
U
R
f: [a, b]×U−→ R
(x, t)7−→ f(x, t),
[a, b]×U
U−→ R
t7−→ Zb
a
f(x, t)dx
U
J
Rf: [a, b]×J→R
t∈J[a, b]→Rx7→ f(x, t)
[a, b]
x∈[a, b]J→Rt7→ f(x, t)J
[a, b]×J→R,(x, t)7→ ∂
∂tf(x, t)
[a, b]×J
J→Rt7→ Rb
af(x, t)dx J
t∈J
d
dt Zb
a
f(x, t)dx =Zb
a
∂
∂tf(x, t)dx.
a, b a < b I = [a, b]f:I→R
n∈Na≤x0< x1<··· < xn−1< xn≤b
Pnn
Pn(xi) = f(xi)∀i= 0,··· , n.
Pn(x) =
n
X
i=0
f(xi)`i(x),
`i
`i(x) = Y
j=0,···,n
j6=i
x−xj
xi−xj∀x∈R.
Pnn`eme
f x0,··· , xn
n= 1
P1(x) = f(x0) + f(x1)−f(x0)
x1−x0
(x−x0).
I(f) = Rb
af(x)dx In(f) = Rb
aPn(x)dx
Zb
a
Pn(x)dx =
n
X
i=0
f(xi)Zb
a
`i(x)dx,
ai=Rb
a`i(x)dx i = 0,··· , n
n= 0, x0=a+b
2, `0(x) = 1, a0=b−a,
I0(f) = (b−a)fa+b
2.
n= 1, x0=a, x1=b,
`0(x) = b−x
b−a, `1(x) = a−x
a−b, a0=a1=b−a
2,
I1(f) = f(a) + f(b)
2(b−a).
n= 2, x0=a, x1=a+b
2, x2=b,
a0=b−a
6, a1=2
3(b−a), a2=b−a
6,
I2(f) = b−a
6f(a)+4fa+b
2+f(b).
1 / 6 100%
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans l'interface ou les textes ? Ou savez-vous comment améliorer l'interface utilisateur de StudyLib ? N'hésitez pas à envoyer vos suggestions. C'est très important pour nous!