Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page43of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page44of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
anda≤x1 ≤b,a ≤α1 ≤x1 ≤α2 ≤b. Then we have
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
−(a−α1)jf(j−1)(a) +n
(x1−α1)j (6.2)
−(x1−α2)jo
f(j−1)(x1) + (b−α2)jf(j−1)(b)
≤
f(n) ∞
(n+ 1)! (α1 −a)n+1+ (x1−α1)n+1 + (α2−x1)n+1+ (b−α2)n+1
≤
f(n) ∞
(n+ 1)! (x1−a)n+1+ (b−x1)n+1
≤
f(n) ∞
(n+ 1)! (b−a)n+1, f(n) ∈L∞[a, b].
Proof. Consider the divisiona = x0 ≤ x1 ≤ x2 =b and the numbersα0 = a, α1 ∈[a, x1),α2 ∈(x1, b]andα3 =b.
From the left hand side of (3.1) we obtain
n
X
j=1
(−1)j j!
2
X
i=0
n
(xi−αi)j −(xi−αi+1)jo
f(j−1)(xi)
=
n
X
j=1
(−1)j j!
−(a−α1)jf(j−1)(a) +
n
(x1−α1)j −(x1−α2)j o
f(j−1)(x1) + (b−α2)jf(j−1)(b)
.
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page45of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
From the right hand side of (3.1) we obtain f(n)
∞ (n+ 1)!
1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1
=
f(n) ∞
(n+ 1)! (α1−a)n+1+ (x1−α1)n+1+ (α2−x1)n+1+ (b−α2)n+1 and hence the first line of the inequality (6.2) follows.
Notice that if we chooseα1 = aandα2 = b in Theorem6.1we obtain the inequality (1.2).
The following proposition embodies a number of results, including the Os-trowski inequality, the midpoint and Simpson’s inequalities and the three-eighths Newton-Cotes inequality including its generalisation.
Proposition 6.2. Let f be defined as in Theorem6.1and leta ≤ x1 ≤ b, and a ≤ (m−1)a+bm ≤ x1 ≤ a+(m−1)bm ≤ bform a natural number,m ≥ 2, then we have the inequality
|Pm,n|:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
"
b−a m
j
f(j−1)(b) (6.3)
− (−1)jf(j−1)(a)o +
(
(x1−a)− b−a m
j
−
b−a
m −(b−x1) j)
f(j−1)(x1)
#
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page46of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤
f(n) ∞ (n+ 1)! 2
b−a m
n+1
+
x1−a−
b−a m
n+1
+
b−x1−
b−a m
n+1!
if f(n)∈L∞[a, b]. Proof. From Theorem6.1we note that
α1 = (m−1)a+b
m and α2 = a+ (m−1)b m so that
a−α1 = −
b−a m
, b−α2 = b−a m , x1−α1 = x1−a−
b−a m
and x1−α2 =
b−a m
−(b−x1). From the left hand side of (6.2) we have
−(a−α1)jf(j−1)(a) + n
(x1−α1)j
− (x1−α2)jo
f(j−1)(x1) + (b−α2)jf(j−1)(b)
=
b−a m
j
f(j−1)(b)−f(j−1)(a) +
(
x1−a−
b−a m
j
−
b−a m
−(b−x1) j)
f(j−1)(x1).
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page47of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
From the right hand side of (6.2),
(α1−a)n+1+ (x1−α1)n+1+ (α2−x1)n+1+ (b−α2)n+1
= 2
b−a m
n+1
+
x1−a−
b−a m
n+1
+
b−x1−
b−a m
n+1
and the inequality (6.3) follows, hence the proof is complete.
The following corollary points out that the optimum of Proposition 6.2 oc-curs atx1 = α1+α2 2 = a+b2 in which case we have:
Corollary 6.3. Letfbe defined as in Proposition6.2and letx1 = a+b2 in which case we have the inequality
Pm,n
a+b 2
(6.4)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
"
b−a m
j
×n
f(j−1)(b)−(−1)jf(j−1)(a)o +
(m−2) (b−a) 2m
j
1−(−1)j f(j−1)
a+b 2
#
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page48of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤ 2 f(n)
∞ (n+ 1)!
b−a m
n+1
1 +
m−2 2
n+1!
if f(n) ∈L∞[a, b].
The proof follows directly from (6.3) upon substitutingx1 = a+b2 .
A number of other corollaries follow naturally from Proposition 6.2 and Corollary6.3and will now be investigated.
The following two corollaries generalise the Simpson inequality and follow directly from (6.3) and (6.4) form = 6.
Corollary 6.4. Let the conditions of Corollary6.3 hold and putm = 6. Then we have the inequality
|P6,n| (6.5)
:=
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
"
b−a 6
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o
+ (
x1− 5a+b 6
j
−
x1− a+ 5b 6
j)
f(j−1)(x1)
#
≤
f(n) ∞ (n+ 1)! 2
b−a 6
n+1
+
x1 −5a+b 6
n+1
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page49of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
+
−x1+ a+ 5b 6
n+1!
, f(n)∈L∞[a, b], which is the generalised Simpson inequality.
Corollary 6.5. Let the conditions of Corollary6.3 hold and putm = 6,. Then at the midpointx1 = a+b2 we have the inequality
P6,n
a+b 2
(6.6)
:=
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
"
b−a 6
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o
+
b−a 3
j
1−(−1)j f(j−1)
a+b 2
#
≤ 2 f(n)
∞ (n+ 1)!
b−a 6
n+1
1 + 2n+1
, f(n) ∈L∞[a, b]. Remark 6.1. Choosingn = 1in (6.6) we have
(6.7)
P6,1
a+b 2
:=
Z b a
f(t)dt−
b−a
6 f(a) + 4f
a+b 2
+f(b)
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page50of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤ 5
36kf0k∞(b−a)2, f0 ∈L∞[a, b]. Remark 6.2. Choosing n = 2 in (6.6) we have a perturbed Simpson type in-equality,
P6,2
a+b 2
(6.8)
:=
Z b a
f(t)dt−
b−a
6 f(a) + 4f
a+b 2
+f(b)
+
b−a 6
2
f0(b)−f0(a) 2
≤ (b−a)3
72 kf00k∞, f00∈L∞[a, b].
Corollary 6.6. Let f be defined as in Corollary 6.3 and let m = 4, then we have the inequality
P4,n
a+b 2
(6.9)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
b−a 4
j
×h
f(j−1)(b)−(−1)jf(j−1)(a) +
1−(−1)j f(j−1)
a+b 2
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page51of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤
f(n) ∞
(n+ 1)!4n(b−a)n+1, f(n) ∈L∞[a, b]. Remark 6.3. From (6.9) we choosen= 2and we have the inequality
P4,2
a+b 2
(6.10)
:=
Z b a
f(t)dt−
b−a
4 f(b) +f(a) + 2f
a+b 2
+
b−a 4
2
f0(b)−f0(a) 2
≤ kf00k∞
96 (b−a)3, f00 ∈L∞[a, b].
Theorem 6.7. Letf : [a, b]→Rbe an absolutely continuous mapping on[a, b]
and leta < x1 ≤x2 ≤bandα1 ∈[a, x1),α2 ∈[x1, x2)andα3 ∈[x2, b]. Then we have the inequality
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
h−(a−α1)jf(j−1)(a) (6.11)
+
(x1−α1)j −(x1 −α2)j
f(j−1)(x1) +
(x2−α2)j −(x2 −α3)j
f(j−1)(x2) + (b−α3)jf(j−1)(b)i
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page52of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤
f(n) ∞
(n+ 1)! (α1−a)n+1 + (x1−α1)n+1+ (α2 −x1)n+1 + (x2−α2)n+1+ (α3 −x2)n+1+ (b−α3)n+1 , f(n) ∈L∞[a, b].
Proof. Consider the division a = x0 < x1 < x2 = b, α1 ∈ [a, x1), α2 ∈ [x1, x2), α3 ∈ [x2, b], α0 = a, x0 = a, x3 = b and put α4 = b. From the left hand side of (3.1) we obtain
n
X
j=1
(−1)j j!
3
X
i=0
n
(xi−αi)j −(xi−αi+1)jo
f(j−1)(xi)
=
n
X
j=1
(−1)j j!
h−(a−α1)jf(j−1)(a)
+
(x1−α1)j −(x1−α2)j
f(j−1)(x1) +
(x2−α2)j−(x2−α3)j
f(j−1)(x2) + (b−α3)jf(j−1)(b)i . From the right hand side of (3.1) we obtain
f(n) ∞ (n+ 1)!
2
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1
=
f(n) ∞
(n+ 1)! (α1−a)n+1 + (x1−α1)n+1+ (α2−x1)n+1
+ (x2−α2)n+1+ (α3−x2)n+1+ (b−α3)n+1
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page53of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
and hence the first line of the inequality (6.11) follows and Theorem 6.7 is proved.
Corollary 6.8. Letf be defined as in Theorem6.7and consider the division a≤α1 ≤ (m−1)a+b
m ≤α2 ≤ a+ (m−1)b
m ≤α3 ≤b forma natural number,m≥2. Then we have the inequality
|Qm,n| (6.12)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
h−(a−α1)jf(j−1)(a)
+ (
(m−1)a+b
m −α1
j
−
(m−1)a+b
m −α2
j)
f(j−1)(x1)
+ (
a+ (m−1)b
m −α2
j
−
a+ (m−1)b
m −α3
j)
f(j−1)(x2) + (b−α3)jf(j−1)(b)i
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page54of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤
f(n) ∞ (n+ 1)!Rm,n :=
f(n) ∞
(n+ 1)! (α1−a)n+1+
(m−1)a+b
m −α1
n+1
+
α2− (m−1)a+b m
n+1
+
a+ (m−1)b
m −α2
n+1
+
α3− a+ (m−1)b m
n+1
+ (b−α3)n+1
! ,
f(n) ∈L∞[a, b].
Proof. Choose in Theorem 6.7, x1 = (m−1)a+bm , andx2 = a+(m−1)bm , hence the theorem is proved.
Remark 6.4. For particular choices of the parametersmandn, Corollary6.8 contains a generalisation of the three-eighths rule of Newton and Cotes.
The following corollary is a consequence of Corollary6.8.
Corollary 6.9. Let f be defined as in Theorem 6.7 and choose α2 = a+b2 =
x1+x2
2 , then we have the inequality
Q¯m,n
(6.13)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
h−(a−α1)jf(j−1)(a)
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page55of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
+ (
(x1−α1)j −(−1)j
(m−2) (b−a) 2m
j)
f(j−1)(x1)
+ (
(m−2) (b−a) 2m
j
−(x2−α3)j )
f(j−1)(x2) + (b−α3)jf(j−1)(b)i
≤
f(n) ∞
(n+ 1)!
R¯m,n :=
f(n) ∞
(n+ 1)! (α1−a)n+1+
(m−1)a+b
m −α1
n+1
+ 2
(b−a)(m−2) 2m
n+1
+
α3 −a+ (m−1)b m
n+1
+ (b−α3)n+1
, f(n) ∈L∞[a, b].
Proof. If we putα2 = a+b2 = x1+x2 2 into (6.12), we obtain the inequality (6.13) and the corollary is proved.
The following corollary contains an optimum estimate for the inequality (6.13).
Corollary 6.10. Letf be defined as in Theorem6.7and make the choices α1 =
3m−4 2m
a+
4−m 2m
b and
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page56of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
α3 =
4−m 2m
a+
3m−4 2m
b then we have the best estimate
Qˆm,n (6.14)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
×
"
(b−a) (4−m) 2m
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o
+
(b−a) (m−2) 2m
j
n
1−(−1)jo
f(j−1)(x1) +f(j−1)(x2)
#
≤ 2 f(n)
∞ (n+ 1)!
b−a 2m
n+1
(4−m)n+1+ 2 (m−2)n+1 , f(n)∈L∞[a, b].
Proof. Using the choiceα2 = a+b2 = x1+x2 2, x1 = (m−1)a+bm andx2 = a+(m−1)bm we may calculate
(α1−a) = (4−m) (b−a)
2m = (b−α3)
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page57of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
and
(x1−α1) = (α2−x1) = (x2−α2) = (α3−x2)
= (m−2) (b−a)
2m .
Substituting in the inequality (6.13) we obtain the proof of (6.14).
Remark 6.5. Form = 3, we have the best estimation of (6.14) such that
Qˆ3,n (6.15)
:=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
×
"
b−a 6
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o +
b−a 6
j
× n
1−(−1)jo f(j−1)
2a+b 2
+f(j−1)
a+ 2b 2
≤
f(n) ∞
6n(n+ 1)!(b−a)n+1, f(n) ∈L∞[a, b]. Proof. From the right hand side of (6.14), consider the mapping
Mm,n :=
4−m 2m
n+1
+ 2
m−2 2m
n+1
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page58of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
then
Mm,n0 = 2 (n+ 1) m2
− 2
m − 1 2
n
+ 1
2− 1 m
n
andMm,nattains its optimum when 2 m − 1
2 = 1 2 − 1
m,
in which casem = 3. Substitutingm= 3into (6.14), we obtain (6.15) and the corollary is proved.
Whenn = 2, then from (6.15) we have
Qˆ3,2 :=
Z b a
f(t)dt−
b−a 6
(f(b) +f(a)) +1
2
b−a 6
2
(f0(b)−f0(a)) +
b−a
3 f
2a+b 2
+f
a+ 2b 2
≤ kf00k∞
216 (b−a)3, f00∈L∞[a, b].
The next corollary encapsulates the generalised Newton-Cotes inequality.
Corollary 6.11. Letf be defined as in Theorem6.7and choose x1 = 2a+b
3 , x2 = a+ 2b
3 , α2 = a+b 2 , α1 = (r−1)a+b
r andα3 = a+ (r−1)b
r .
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page59of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
Then for ra natural number,r≥3, we have the inequality
|Tr,n| (6.16)
:=
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
"
b−a r
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o
+ (
(b−a) (r−3) 3r
j
−(−1)j
b−a 6
j)
f(j−1)(x1)
+ (
b−a 6
j
−(−1)j
(b−a) (r−3) 3r
j)
f(j−1)(x2)
#
≤ 2 f(n)
∞
(n+ 1)! (b−a)n+1 1 rn+1 +
r−3 3r
n+1
+ 1 6n+1
! , f(n) ∈L∞[a, b].
Proof. From Theorem 6.7, we put x1 = 2a+b3 , x2 = a+2b3 , α2 = a+b2 , α1 =
(r−1)a+b
r andα3 = a+(r−1)br . Then (6.16) follows.
Remark 6.6. The optimum estimate of the inequality (6.16) occurs whenr= 6.
from (6.16) consider the mapping
Mr,n:= 1 rn+1 +
r−3 3r
n+1
+ 1
6n+1
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page60of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
the Mr,n0 = −(n+ 1)r−n−2 + n+1r2
1
3 −1rn
and Mr,n attains its optimum when 1r = 13 − 1r, in which caser = 6. In this case, we obtain the inequality (6.15) and specifically forn = 1, we obtain a Simpson type inequality
Qˆ3,1 :=
Z b a
f(t)dt−
b−a 6
(f(a) +f(b)) (6.17)
−
b−a
3 f
2a+b 3
+f
a+ 2b 3
≤ kf0k∞
12 (b−a)2, f0 ∈L∞[a, b], which is better than that given by (6.7).
Corollary 6.12. Let f be defined as in Theorem6.7 and choose m = 8 such thatα1 = 7a+b8 ,α2 = a+b8 andα3 = a+7b8 withx1 = 2a+b3 andx2 = a+2b3 . Then we have the inequality
|T8,n| (6.18)
:=
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
"
b−a 8
j
f(j−1)(b)−(−1)jf(j−1)(a)
+
b−a 6
j 5 4
j
−(−1)j
!
×
f(j−1)(x1)−(−1)jf(j−1)(x2)i
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page61of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
≤ 2 f(n)
∞ (n+ 1)!
b−a 24
n+1
3n+1+ 4n+1+ 5n+1 , f(n) ∈L∞[a, b]. Proof. From Theorem6.7we put
x1 = 2a+b
3 , x2 = a+ 2b
3 , α1 = 7a+b
8 , α3 = a+ 7b 8 andα2 = a+b2 and the inequality (6.18) is obtained.
Whenn = 1we obtain from (6.18) the ‘three-eighths rule’ of Newton-Cotes.
Remark 6.7. From (6.16) withr= 3we have
|T3,n| :=
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
"
b−a 3
j
f(j−1)(b)−(−1)jf(j−1)(a)
+
b−a 6
j
f(j−1)(x2)−f(j−1)(x1)
#
≤ 2 f(n)
∞
(n+ 1)! (b−a)n+1 1
3n+1 + 1 6n+1
, f(n) ∈L∞[a, b]. In particular, forn= 2, we have the inequality
|T3,2|:=
Z b a
f(t)dt−
b−a 3
(f(b) +f(a)) +
b−a 3
2
f0(b)−f0(a) 2
−
b−a
6 f
a+ 2b 3
+f
2a+b 3
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page62of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
+
b−a 6
2
f0 a+2b3
−f0 2a+b3 2
!
≤ kf00k∞
72 (b−a)3, f00 ∈L∞[a, b].
The following theorem encapsulates Boole’s rule.
Theorem 6.13. Let f : [a, b] → R be an absolutely continuous mapping on [a, b] and leta < x1 < x2 < x3 < b andα1 ∈ [a, x1), α2 ∈ [x1, x2), α3 ∈ [x2, x3)andα4 ∈[x3, b]. Then we have the inequality
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
h−(a−α1)jf(j−1)(a) (6.19)
+
(x1−α1)j −(x1 −α2)j
f(j−1)(x1) +
(x2−α2)j −(x2 −α3)j
f(j−1)(x2) +
(x3−α3)j −(x3 −α4)j
f(j−1)(x3) + (b−α4)jf(j−1)(b)i
≤
f(n) ∞
(n+ 1)! (α1−a)n+1 + (x1−α1)n+1+ (α2−x1)n+1 + (x2−α2)n+1+ (α3−x2)n+1+ (x3−α3)n+1 + (α4−x3)n+1+ (b−α4)n+1
if f(n)∈L∞[a, b].
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page63of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au
Proof. Follows directly from (3.1) with the pointsα0 = x0 = a, x4 = α5 = b and the division a = x0 < x1 < x2 < x3 = b, α1 ∈ [a, x1), α2 ∈ [x1, x2), α3 ∈[x2, x3)andα4 ∈[x3, b].
The following inequality arises from Theorem6.13.
Corollary 6.14. Let f be defined as in Theorem 6.7 and choose α1 = 11a+b12 , α2 = 11a+7b18 , α3 = 7a+11b18 , α4 = a+11b12 , x1 = 7a+2b9 , x3 = 2a+7b9 and x2 =
x1+x3
2 = a+b2 , then we can state:
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
"
b−a 12
j
n
f(j−1)(b)−(−1)jf(j−1)(a)o
+
b−a 6
j( 5
6 j
−(−1)j )
×
f(j−1)
7a+ 2b 9
−(−1)jf(j−1)
2a+ 7b 9
+
b−a 9
j
n
1−(−1)jo f(j−1)
a+b 2
#
≤ 2 f(n)
∞
(n+ 1)!
b−a 36
n+1
3n+1+ 4n+1+ 5n+1+ 6n+1 , iff(n)∈L∞[a, b].
Integral Inequalities of the Ostrowski Type
A. Sofo
Title Page Contents
JJ II
J I
Go Back Close
Quit Page64of83
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
http://jipam.vu.edu.au