Journal of Inequalities in Pure and
Applied Mathematics
SOME INTEGRAL INEQUALITIES INVOLVING TAYLOR’S
REMAINDER
volume 3, issue 2, article 26,
2002.
HILLEL GAUCHMAN
Department of Mathematics,
Eastern Illinois University,
Charleston, IL 61920, USA
EMail: cfhvg@ux1.cts.eiu.edu
Received 17 September, 2001;;
accepted 23 January, 2001.
Communicated by: G. Anastassiou
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2000
Victoria University
ISSN (electronic): 1443-5756
068-01
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Abstract
In this paper, using Steffensen’s inequality we prove several inequalities involving Taylor’s remainder. Among the simplest particular cases we obtain Iyengar’s
inequality and one of Hermite-Hadamard’s inequalities for convex functions.
2000 Mathematics Subject Classification: 26D15.
Key words: Taylor’s remainder, Steffensen’s inequality, Iyengar’s inequality, HermiteHadamard’s inequality.
Contents
1
Introduction and Statement of Main Results . . . . . . . . . . . . . . 3
2
Proofs of Theorems 1.1 and 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Applications of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4
Applications of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
References
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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1.
Introduction and Statement of Main Results
In this paper, using Steffensen’s inequality we prove several inequalities (Theorems 1.1 and 1.2) involving Taylor’s remainder. In Sections 3 and 4 we give
several applications of Theorems 1.1 and 1.2. Among the simplest particular
cases we obtain Iyengar’s inequality and one of Hermite-Hadamard’s inequalities for convex functions. We prove Theorems 1.1 and 1.2 in Section 2.
In what follows n denotes a non-negative integer, I ⊆ R is a generic interval, and I ◦ is the interior of I. We will denote by Rn,f (c, x) the nth Taylor’s
remainder of function f (x) with center c, i.e.
Rn,f (c, x) = f (x) −
n
X
f (k) (c)
k=0
k!
(x − c)k .
The following two theorems are the main results of the present paper.
Theorem 1.1. Let f : I → R and g : I → R be two mappings, a, b ∈ I ◦ with
a < b, and let f ∈ C n+1 ([a, b]), g ∈ C ([a, b]). Assume that m ≤ f (n+1) (x) ≤
M , m 6= M , and g(x) ≥ 0 for all x ∈ [a, b]. Set
(n)
1
λ=
f (b) − f (n) (a) − m(b − a) .
M −m
Then
Z b
1
(x − b + λ)n+1 g(x)dx
(i)
(n + 1)! b−λ
Z b
1
(x − a)n+1
≤
Rn,f (a, x) − m
g(x)dx
M −m a
(n + 1)!
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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1
≤
(n + 1)!
b
Z
a
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
Z
(−1)n+1 a+λ
(a + λ − x)n+1 g(x)dx;
+
(n + 1)! a
and
1
(n + 1)!
(ii)
Z
a
a+λ
(a + λ − x)n+1 g(x)dx
Z (−1)n+1 b
(x − b)n+1
≤
Rn,f (b, x) − m
g(x)dx
M −m a
(n + 1)!
Z b
1
≤
(b − x)n+1 − (b − λ − x)n+1 g(x)dx
(n + 1)! a
Z
(−1)n+1 b
+
(x − b + λ)n+1 g(x)dx.
(n + 1)! b−λ
Theorem 1.2. Let f : I → R and g : I → R be two mappings, a, b ∈ I ◦
with a < b, and let f ∈ C n+1 ([a, b]), g ∈ C ([a, b]). Assume that f n+1 (x) is
increasing on [a, b] and m ≤ g(x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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λ1
λ2
Z b
1
m
b−a
=
(x − a)n+1 g(x)dx −
·
,
n+1
(M − m)(b − a)
M −m n+2
a
Z b
1
m
b−a
=
(b − x)n+1 g(x)dx −
·
.
n+1
(M − m)(b − a)
M −m n+2
a
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Then
(i)
f (n) (a − λ1 ) − f (n) (a)
Z b
(n + 1)!
≤
Rn,f (a, x)(g(x) − m)dx
(M − m)(b − a)n+1 a
≤ f (n) (b) − f (n) (b − λ1 );
and
Some Integral Inequalities
Involving Taylor’s Remainder. I
(ii)
Hillel Gauchman
f (n) (a + λ2 ) − f (n) (a)
(n + 1)!
≤ (−1)
(M − m)(b − a)n+1
≤ f (n) (b) − f (n) (b − λ2 ).
n+1
Z
b
Rn,f (b, x) (g(x) − m) dx
a
Remark 1.1. It is easy to verify that the inequalities in Theorems 1.1 and 1.2
become equalities if f (x) is a polynomial of degree ≤ n + 1.
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2.
Proofs of Theorems 1.1 and 1.2
The following is well-known Steffensen’s inequality:
Theorem 2.1. [4]. Suppose the f and g are integrable functions defined on
(a, b), f is decreasing and for each x ∈ (a, b), 0 ≤ g(x) ≤ 1. Set λ =
Rb
g(x)dx. Then
a
Z
b
Z
f (x)dx ≤
b−λ
b
Z
f (x)g(x)dx ≤
a
a+λ
f (x)dx.
a
Proposition 2.2. Let f : I → R and g : I → R be two maps, a, b ∈ I ◦ with
a < b and let f ∈ C n+1 ([a, b]), g ∈ C[a, b]. Assume that 0 ≤ f (n+1) (x) ≤ 1 for
Rb
all x ∈ [a, b] and x (t − x)n g(t)dt is a decreasing function of x on [a, b]. Set
λ = f (n) (b) − f (n) (a). Then
(2.1)
1
(n + 1)!
Z
b
(x − b + λ)n+1 g(x)dx
b−λ
Z
≤
b
Rn,f (a, x)g(x)dx
Z b
1
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
≤
(n + 1)! a
Z
(−1)n+1 a+λ
+
(a + λ − x)n+1 g(x)dx.
(n + 1)! a
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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Proof. Set
Z
1 b
Fn (x) =
(t − x)n g(t)dt,
n! x
Gn (x) = f n+1 (x),
Z b
λ =
Gn (x)dx = f (n) (b) − f (n) (a).
a
Then Fn (x), Gn (x), and λ satisfy the conditions of Theorem 2.1. Therefore
Z
b
Z
Fn (x)dx ≤
(2.2)
b−λ
b
Z
a+λ
Fn (x)Gn (x)dx ≤
a
Fn (x)dx.
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
a
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It is easy to see that Fn0 (x) = −Fn−1 (x). Hence
Contents
Z
b
Z
b
Fn (x)df (n) (x)
a
b Z b
(n)
= f (x)Fn (x) +
f (n) (x)Fn−1 (x)dx
Fn (x)Gn (x)dx =
a
a
Z
b
a
Z
b
(a)
(x − a)n g(x)dx +
Fn−1 (x)Gn−1 (x)dx
n!
a
a
Z
f (n) (a) b
=−
(x − a)n g(x)dx
n!
a
Z
Z b
(n−1)
f
(a) b
n−1
−
(x − a) g(x)dx +
Fn−2 (x)Gn−2 (x)dx
(n − 1)! a
a
=−
f
(n)
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= ...
Z
f (n−1) (a) b
(x − a) g(x)dx −
(x − a)n−1 g(x)dx
(n
−
1)!
a
a
Z b
Z b
− · · · − f (a)
g(x)dx +
f (x)g(x)dx.
f (n) (a)
=−
n!
Z
b
n
a
a
Thus
b
Z
Z
Fn (x)Gn (x)dx =
(2.3)
a
Some Integral Inequalities
Involving Taylor’s Remainder. I
b
Rn,f (a, x)g(x)dx.
Hillel Gauchman
a
In addition
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Z
a+λ
Fn (x)dx =
a
1
n!
Z
a
a+λ
Z
b
n
(t − x) g(t)dt dx.
x
Changing the order of integration, we obtain
Z a+λ
Fn (x)dx
a
Z t
Z a+λ
Z
Z
1 a+λ
1 b
n
n
=
(t − x) g(t)dx dt +
(t − x) g(t)dx dt
n! a
n! a+λ a
a
x=t
x=a+λ
Z a+λ
Z
1
(t − x)n+1 1 b
(t − x)n+1 =−
g(t)
dt −
g(t)
dt
n! a
n + 1 x=a
n! a+λ
n + 1 x=a
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1
=
(n + 1)!
Z
a+λ
(t − a)n+1 g(t)dt
a
Z b
1
−
(t − a − λ)n+1 − (t − a)n+1 g(t)dt
(n + 1)! a+λ
Z b
Z b
1
1
n+1
=
(t − a) g(t)dt −
(t − a − λ)n+1 g(t)dt
(n + 1)! a
(n + 1)! a
Z a+λ
1
+
(t − a − λ)n+1 g(t)dt.
(n + 1)! a
Thus,
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
Z
(2.4)
a
a+λ
1
Fn (x)dx =
(n + 1)!
Similarly we obtain
Z b
(2.5)
Fn (x)dx =
b−λ
Z
b
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
a
Z
(−1)n+1 a+λ
(a + λ − x)n+1 g(x)dx.
+
(n + 1)! a
1
(n + 1)!
Z
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b
n+1
(x − b + λ)
g(x)dx
b−λ
Substituting (2.3), (2.4), and (2.5) into (2.2), we obtain (2.1).
Proposition 2.3. Let f : I → R and g : I → R be two maps, a, b ∈ I ◦ with
a < b and let f ∈ C n+1 ([a, b]), g ∈ C ([a, b]). Assume that m ≤ f (n+1) (x) ≤
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Rb
M for all x ∈ [a, b] and x (t − x)n g(t)dt is adecreasing function of x on [a, b].
1
Set λ = M −m
f (n) (b) − f (n) (a) − m(b − a) . Then
(2.6)
1
(n + 1)!
Z
b
(x − b + λ)n+1 g(x)dx
b−λ
Z b
1
(x − a)n+1
≤
Rn,f (a, x) − m
g(x)dx
M −m a
(n + 1)!
Z b
1
(x − a)n+1 − (x − a − λn+1 ) g(x)dx
≤
(n + 1)! a
Z
(−1)n+1 a+λ
+
(a + λ − x)n+1 g(x)dx.
(n + 1)! a
Hillel Gauchman
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Proof. Set
n+1
1
(x
−
a)
f˜(x) =
f (x) − m
.
M −m
(n + 1)!
Then 0 ≤ f˜(n+1) (x) ≤ 1 and
λ=
Some Integral Inequalities
Involving Taylor’s Remainder. I
(n)
1
f (b) − f (n) (a) − m(b − a) = f˜(n) (b) − f˜(n) (a).
M −m
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Hence f˜(x), g(x), and λ satisfy the conditions of Proposition 2.2. Substituting
f˜(x) instead of f (x) into (2.1), we obtain (2.6).
Rb
Proof of Theorem 1.1(i). If g(x) ≥ 0 for all x ∈ [a, b], then x (t − x)n g(t)dt
is a decreasing function of x on [a, b]. Hence Proposition 2.3 implies Theorem
1.1(i).
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Proof of Theorems 1.1(ii), 1.2(i), and 1.2(ii). Proofs of Theorems 1.1(ii), 1.2(i),
and 1.2(ii) are similar to the above proof of Theorem 1.1(i). For the proof of
Theorem 1.1(ii) we take
Z
1 x
Fn (x) = −
(x − t)n g(t)dt, Gn (x) = f n+1 (x).
n! a
For the proof of Theorem 1.2(i) we take
Fn (x) = −f
(n+1)
1
(x), Gn (x) =
n!
b
Z
(t − x)n g(t)dt.
Some Integral Inequalities
Involving Taylor’s Remainder. I
x
Hillel Gauchman
For the proof of Theorem 1.2(ii) we take
Fn (x) = −f
(n+1)
1
(x), Gn (x) =
n!
Z
a
x
(x − t)n g(t)dt.
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3.
Applications of Theorem 1.1
Theorem 3.1. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
f ∈ C n+1 ([a, b]). Assume that m ≤ f (n+1) (x) ≤ M , m 6= M , for all x ∈ [a, b].
Set
(n)
1
λ=
f (b) − f (n) (a) − m(b − a) .
M −m
Then
(i)
1
m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
≤
Rn,f (a, x)dx
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
a
≤
1
M (b − a)n+2 − (M − m)(b − a − λ)n+2 ;
(n + 2)!
and
(ii)
1
m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
n+1
≤ (−1)
Rn,f (b, x)dx
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a
≤
1
M (b − a)n+2 − (M − m)(b − a − λ)n+2 .
(n + 2)!
Proof. Take g(x) ≡ 1 on [a, b] in Theorem 1.1.
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Two inequalities of the form A ≤ X ≤ B and A ≤ Y ≤ B imply two
new inequalities A ≤ 21 (X + Y ) ≤ B and |X − Y | ≤ B − A. Applying this
construction to inequalities (i) and (ii) of Theorem 3.1, we obtain the following
two more symmetric with respect to a and b inequalities:
Theorem 3.2. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
f ∈ C n+1 ([a, b]). Assume that m ≤ f n+1 (x) ≤ M , m 6= M , for all x ∈ [a, b].
Set
(n)
1
λ=
f (b) − f (n) (a) − m(b − a) .
M −m
Then
(i)
1
[m(b − a)n+2 + (M − m)λn+2 ]
(n + 2)!
Z b
1
≤
Rn,f (a, x) + (−1)n+1 Rn,f (b, x) dx
a 2
1
≤
M (b − a)n+2 − (M − m)(b − a − λ)n+2 ;
(n + 2)!
and
Z b
n
(ii) [Rn,f (a, x) + (−1) Rn,f (b, x)] dx
a
M −m ≤
(b − a)n+2 − λn+2 − (b − a − λ)n+2 .
(n + 2)!
We now consider the simplest cases of inequalities (i) and (ii) of Theorem
3.2, namely the cases when n = 0 or 1.
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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Case 3.1. n = 0
Inequality (i) of Theorem 3.2 for n = 0 gives us the following result.
Theorem 3.3. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b and let
f ∈ C 1 ([a, b]). Assume that m ≤ f 0 (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
λ=
1
[f (b) − f (a) − m(b − a)] .
M −m
Then
2
m+
2
(M − m)λ
f (b) − f (a)
(M − m)(b − a − λ)
≤
≤M−
.
2
(b − a)
b−a
(b − a)2
Remark 3.1. Theorem 3.3 is an improvement of a trivial inequality m ≤
f (b)−f (a)
≤ M.
b−a
For n = 0, inequality (ii) of Theorem 3.2 gives the following result:
Theorem 3.4. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
f ∈ C 1 ([a, b]). Assume that m ≤ f 0 (x) ≤ M , m 6= M for all x ∈ [a, b]. Then
Some Integral Inequalities
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Z b
f
(a)
+
f
(b)
(b − a)
f (x)dx −
2
a
≤
[f (b) − f (a) − m(b − a)] [M (b − a) − f (b) + f (a)]
.
2(M − m)
Theorem 3.4 is a modification of Iyengar’s inequality due to Agarwal and
Dragomir [1]. If |f 0 (x)| ≤ M , then taking m = −M in Theorem 3.4, we
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obtain
Z b
M (b − a)2
f (a) + f (b)
1
≤
f
(x)dx
−
(b
−
a)
−
[f (b) − f (a)]2 .
2
4
4M
a
This is the original Iyengar’s inequality [2]. Thus, inequality (ii) of Theorem
3.2 can be considered as a generalization of Iyengar’s inequality.
Case 3.2. n = 1
In the case n = 1, inequality (i) of Theorem 3.2 gives us the following result:
Theorem 3.5. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b and let
f ∈ C 2 ([a, b]). Assume that m ≤ f 00 (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
λ=
1
[f 0 (b) − f 0 (a) − m(b − a)] .
M −m
Then
1
m(b − a)3 + (M − m)λ3
6
Z b
f 0 (b) − f 0 (a)
f (a) + f (b)
(b − a) +
(b − a)2
≤
f (x)dx −
2
4
a
1
M (b − a)3 − (M − m)(b − a − λ)3 .
≤
6
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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In the case n = 1, inequality (ii) of Theorem 3.2 implies that if f ∈ C 2 ([a, b])
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and m ≤ f 00 (x) ≤ M , then
f (b) − f (a) f 0 (a) + f 0 (b) −
b−a
2
[f 0 (b) − f 0 (a) − m(b − a)] [M (b − a) − f 0 (b) + f 0 (a)]
≤
.
2(b − a)(M − m)
This result follows readily from Iyengar’s inequality if we take f 0 (x) instead of
f (x) in Theorem 3.4.
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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4.
Applications of Theorem 1.2
Take g(x) = M on [a, b] in Theorem 1.2. Then λ1 = λ2 =
1.2 implies
b−a
n+2
and Theorem
Theorem 4.1. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
f ∈ C n+1 ([a, b]). Assume that f n+1 (x) is increasing on [a, b]. Then
b−a
(b − a)n+1 (n)
(n)
f
a+
− f (a)
(i)
(n + 1)!
n+2
Z b
(b − a)n+1 (n)
b−a
(n)
≤
Rn,f (a, x)dx ≤
f (b) − f
b−
;
(n + 1)!
n+2
a
and
(ii)
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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(b − a)n+1 (n)
b−a
(n)
f
a+
− f (a)
(n + 1)!
n+2
Z b
n+1
≤ (−1)
Rn,f (b, x)dx
a
(b − a)n+1 (n)
b−a
(n)
≤
f (b) − f
b−
.
(n + 1)!
n+2
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The next theorem follows from Theorem 4.1 in exactly the same way as
Theorem 3.2 follows from Theorem 3.1.
Page 17 of 20
Theorem 4.2. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
J. Ineq. Pure and Appl. Math. 3(2) Art. 26, 2002
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f ∈ C n+1 ([a, b]). Assume that f n+1 (x) is increasing on [a, b]. Then
(b − a)n+1 (n)
b−a
(n)
(i)
f
a+
− f (a)
(n + 1)!
n+2
Z
1 b
≤
Rn,f (a, x) + (−1)n+1 Rn,f (b, x) dx
2 a
(b − a)n+1 (n)
b−a
(n)
≤
f (b) − f
b−
;
(n + 1)!
n+2
Some Integral Inequalities
Involving Taylor’s Remainder. I
Z b
n
(ii) [Rn,f (a, x) + (−1) Rn,f (b, x)] dx
a
b−a
b−a
(b − a)n+1 (n)
(n)
(n)
(n)
≤
f (b) − f
b−
−f
a+
+ f (a) .
(n + 1)!
n+2
n+2
We now consider inequalities (i) and (ii) of Theorem 4.2 in the simplest cases
when n = 0 or 1.
Case 4.1. n = 0.
Inequality (i) of Theorem 4.2 gives a trivial fact: If f 0 (x) increases then f a+b
≤
2
f (a)+f (b)
0
. Inequality (ii) of Theorem 4.2 gives the following result: If f (x) is
2
increasing, then
Z b
a+b
1
a+b
(4.1)
f
≤
f (x)dx ≤ f (a) + f (b) − f
.
2
b−a a
2
Hillel Gauchman
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The left inequality (2.1) is a half of the famous Hermite-Hadamard’s inequality
[3]: If f (x) is convex, then
Z b
a+b
1
f (a) + f (b)
(4.2)
f
≤
f (x)dx ≤
.
2
b−a a
2
Note that the right inequality (4.1) is weaker than the right inequality (4.2).
Thus, inequality (ii) of Theorem 4.2 can be considered
R b as a generalization of
1
a+b
the Hermite-Hadamard’s inequality f 2 ≤ b−a a f (x)dx, where f (x) is
convex.
Case 4.2. n = 1.
In this case Theorem 4.2 implies the following two results:
Theorem 4.3. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let
f ∈ C 2 ([a, b]). Assume that f 00 (x) is increasing on [a, b]. Then
(b − a)2 0
b−a
0
(i)
f a+
+ f (a)
2
3
Z b
f (a) + f (b)
f 0 (b) − f 0 (a)
≤
f (x)dx −
(b − a) +
(b − a)2
2
4
a
(b − a)2 0
b−a
0
≤
f (b) − f
;
2
3
f (b) − f (a) f 0 (a) + f 0 (b) (ii) −
b−a
2
1 0
b−a
b−a
0
0
0
≤
f (a) − f a +
−f b−
+ f (b) .
2
3
3
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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References
[1] R.P. AGARWAL AND S.S. DRAGOMIR, An application of Hayashi’s
inequality for differentiable functions, Computers Math. Applic., 32(6)
(1996), 95–99.
[2] K.S.K. IYENGAR, Note on an inequality, Math. Student, 6 (1938), 75–76.
[3] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex functions, Partial
Orderings, and Statistical Applications, Acad. Press, 1992.
[4] J.F. STEFFENSEN, On certain inequalities and methods of approximation,
J. Inst. Actuaries, 51 (1919), 274–297.
Some Integral Inequalities
Involving Taylor’s Remainder. I
Hillel Gauchman
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J. Ineq. Pure and Appl. Math. 3(2) Art. 26, 2002
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