Culegere de probleme de Analiz˘a numeric˘a

9.2. Formule de integrare numerică de tip Newton-Cotes 129
Problema 9.2.4 Deducet¸i formula lui Newton s¸i restul ei
b
f(x)dx = b−a
2a+b a+2b
f(a)+3f +3f
8 3 3
a
R3(f) = − (b−a)5
f
648
(4) (ξ)
Solut¸ie. Este o formulă Newton-Cotes închisă pentru m = 3.
Ak = (−1) m−k
m
h t
k!(m−k)! 0
[m+1]
t−k dt
A0 = A3 = (−1) 3b−a
3
1!
(t−1)(t−2)(t−3)dt =
3 0!3! 0
b−a
8
A1 = A2 = (−1) 2b−a
3
1!
t(t−2)(t−3)dt =
3 1!2!
3(b−a)
8
⎪⎩
b
0
+f(b) +R3(f)
R3(f) = K3(t)f
a
(4) (t)dt
K3(t) = 1
(b−t)
3!
4
−
4
b−a
(a−t)
8
3
+ 2a+b
+3
0 3 −t
3 +
+
a+2b
+3
3 −t
3 +(b−t)
+
3
+ =
= 1
⎧
(b−t)
⎪⎨
3!
4
b−a − 4 8 (b−t)3 t ∈ a, 2a+b
3
(b−t) 4
b−a
− (b−t) 4 8
3 +3 2a+b
3 −t 3
t ∈ 2a+b
, 3 a+2b
3
(b−t) 4
b−a
− (b−t) 4 8
3 +3 2a+b
3 −t 3
a+2b
+ +3 3 −t
3
+3 a+2b
3 −t 3
t ∈ a+b
3 ,b
K3(t) ≤ 0
R3(f) = 1
4! f(4) (ξ)R(e4) = 1
24 f(4) (ξ)R(e4)
b
R(e4) = x
a
4 dx− b−a
a
8
4 4
2a+b a+2b
+3 +3
3 3
= b5 −a5
b−a
− a
5 8
4 + (2a+b)4
+
27
(a+2b)4
+b
27
4
=
b
= (b−a)
4 +ab3 +a2b2 +ab3 +a4 −
5
1
8 a4 − 1
8 b4− − (2a+b)4
8·27
− (a+2b)4
8·27
= b−a
8·27·5 ·40(b−a)4
4
+b 4
=