Int. J. Anal. Appl. (2022), 20:28
A New Ostrowski’s Type Inequality for Quadratic Kernel
M. M. Saleem1, Z. Ullah2, T. Abbas1, M. B. Raza1, A. Qayyum1,∗
1Institute of Southern Punjab Multan, Pakistan
2Department of Mathematics, Division of Science and Technology, University of Education Lahore,
Pakistan
∗Corresponding author: atherqayyum@isp.edu.pk
Abstract. From the past few decades, the integral inequalities have been extensively researched. In-
tegral inequalities are applied in innumerable mathematical problems. In current paper, we obtain new
versions of Ostrowski’s type integral inequalities by implementing proposed general form of 7-step
quadratic kernel. Applications for cumulative distribution are also provided.
1. Introduction
In 1938, a Ukrainian Mathematician A. M. Ostrowski [1] discovered the integral inequality. Many
articles and research books have been dedicated to inequalities and its applications [3]-[7]. In this
paper we present 7-step quadratic kernel that further generalize many earlier results contained in [8]-
[12]. Several authors have recently addressed the generalization of the Ostrowski’s type inequalities.
Qayyum et.al [9]-[10] applied a 5-step kernel to generalize some Ostrowski’s type inequalities.
2. Main Findings
Theorem 2.1. Let f :
[
č, ď
]
→R be differentiable on
(
č, ď
)
, f ′is absolutely continuous on
[
č, ď
]
and
γ ≤ f ′′(ţ) ≤ Γ,∀ ţ∈
[
č, ď
]
,then ∀ ÿ ∈
[
č, č+ď
2
]
, we get
Received: Mar. 29, 2022.
2010 Mathematics Subject Classification. 26D10.
Key words and phrases. Ostrowski inequality; numerical integration; quadratic kernel.
https://doi.org/10.28924/2291-8639-20-2022-28
ISSN: 2291-8639
© 2022 the author(s).
https://doi.org/10.28924/2291-8639-20-2022-28
2 Int. J. Anal. Appl. (2022), 20:28
∣∣∣∣14
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
(2.1)
+
1
2
f
(
č + 2ď − ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
+
f ′(ď) − f ′(č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}
−
1
ď − č
∫ ď
č
f (ţ) dţ
∣∣∣∣∣
≤ ω(ÿ)(ď − č)(S −γ)
and ∣∣∣∣14
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
(2.2)
+
1
2
f
(
č + 2ď − ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
+
f ′(ď) − f ′(č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}
−
1
ď − č
∫ ď
č
f (ţ) dţ
∣∣∣∣∣
≤ ω(ÿ)(ď − č)(Γ −S),
where
S =
f ′(ď) − f ′(č)
ď − č
and
ω(ÿ)
=
1
768
max
{∣∣−65č2 + 231čÿ + 165ďÿ − 101čď
−32ď2 − 198ÿ2
∣∣ ,∣∣−35č2 + 135čÿ + 117ďÿ − 65čď − 26ď2 − 126ÿ2∣∣ ,
Int. J. Anal. Appl. (2022), 20:28 3∣∣−83č2 + 255čÿ + 141čÿ − 89čď − 26ď2 − 198ÿ2∣∣ ,∣∣127č2 − 297čÿ − 27ďÿ + 43čď − 8ď2 + 162ÿ2∣∣ ,∣∣7č2 − 105čÿ − 219ďÿ + 91čď + 64ď2 + 162ÿ2∣∣ ,∣∣−65č2 + 183čÿ + 69ďÿ − 53čď − 8ď2 − 126ÿ2∣∣ ,∣∣89č2 − 279čÿ − 165ďÿ + 101čď + 32ď2 + 222ÿ2∣∣} .
Proof. To prove our required results, first of all we introduce a mapping. Let f :
[
č, ď
]
→R be such
that f ′ is absolutely continuous on
[
č, ď
]
. Define the kernel K (ÿ,ţ) as:
K (ÿ,ţ) =
1
2
(ţ− č)2 ţ ∈
(
č,
3č+ÿ
4
]
1
2
(
ţ− 7č+ď
8
)2
ţ ∈
(
3č+ÿ
4
,
č+ÿ
2
]
1
2
(
ţ− 3č+ď
4
)2
ţ ∈
(
č+ÿ
2
, ÿ
]
1
2
(
ţ− č+ď
2
)2
ţ ∈
(
ÿ, č + ď − ÿ
]
1
2
(
ţ− č+3ď
4
)2
ţ ∈
(
č + ď − ÿ, č+2ď−ÿ
2
]
1
2
(
ţ− č+7ď
8
)2
ţ ∈
(
č+2ď−ÿ
2
,
č+4ď−ÿ
4
]
1
2
(
ţ− ď
)2
ţ ∈
(
č+4ď−ÿ
4
, ď
]
(2.3)
∀ ÿ ∈
[
č, č+ď
2
]
. Then the following identity
∫ ď
č
K (ÿ,ţ) df ′ (ţ) (2.4)
=
∫ ď
č
f (ţ) dţ−
ď − č
4
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
+
1
2
f
(
č + 2ď − ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3a + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
holds. We know that
1
ď − č
∫ ď
č
f ′′ (ţ) dţ =
f ′(ď) − f ′(č)
ď − č
(2.5)
4 Int. J. Anal. Appl. (2022), 20:28
1
ď − č
∫ ď
č
K (ÿ,ţ) dţ =
1
ď − č
[
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
(2.6)
−
73
192
(
ÿ −
č + ď
2
)3]
and implies that
1
ď − č
∫ ď
č
K (ÿ,ţ) f ′′ (ţ) dţ−
1(
ď − č
)2
∫ ď
č
K (ÿ,ţ) dţ
∫ ď
č
f ′′ (ţ) dţ (2.7)
=
1
ď − č
∫ ď
č
f (ţ) dţ−
1
4
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
+
1
2
f
(
č + 2ď − ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
−
f ′(ď) − f ′(č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}
.
We suppose that
Rn(ÿ) (2.8)
=
1
ď − č
∫ ď
č
K (ÿ,ţ) f ′′ (ţ) dt −
1(
ď − č
)2
∫ ď
č
K (ÿ,ţ) dt
∫ ď
č
f ′′ (ţ) dt.
If C ∈ R is an arbitray constant, then we have
Rn(ÿ) (2.9)
=
1
ď − č
∫ ď
č
(
f ′′ (ţ) −C
)[
K (ÿ,ţ) −
1
ď − č
∫ ď
č
K (ÿ, s) ds
]
dţ.
Furthermore, we have
|Rn(ÿ)| (2.10)
≤
1
ď − č
max
ţ∈[č,ď]
∣∣∣∣∣K (ÿ,ţ) − 1ď − č
∫ ď
č
K (ÿ, s) ds
∣∣∣∣∣
∫ ď
č
∣∣f ′′ (ţ) −C∣∣dţ.
Int. J. Anal. Appl. (2022), 20:28 5
Now
max
∣∣∣∣∣K (ÿ,ţ) − 1ď − č
∫ ď
č
K (ÿ, s) ds
∣∣∣∣∣ (2.11)
= max
{∣∣∣∣∣12
(
ÿ − č
4
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ ,
∣∣∣∣∣18
(
ÿ −
3č + ď
4
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ ,∣∣∣∣∣ 132
(
ÿ −
č + ď
2
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ ,
∣∣∣∣∣12
(
ÿ −
3č + ď
4
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ ,∣∣∣∣∣12
(
ÿ −
č + ď
2
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ ,
∣∣∣∣∣18
(
ÿ −
č + ď
2
)2
−
β(ÿ)
ď − č
∣∣∣∣∣ , β(ÿ)ď − č
}
where
β(ÿ) =
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3
.
and
ω(ÿ) (2.12)
=
1
768
max
{∣∣−65č2 + 231čÿ + 165ďÿ − 101čď
−32ď2 − 198ÿ2
∣∣ ,∣∣−35č2 + 135čÿ + 117ďÿ − 65čď − 26ď2 − 126ÿ2∣∣ ,∣∣−83č2 + 255čÿ + 141čÿ − 89čď − 26ď2 − 198ÿ2∣∣ ,∣∣127č2 − 297čÿ − 27ďÿ + 43čď − 8ď2 + 162ÿ2∣∣ ,∣∣7č2 − 105čÿ − 219ďÿ + 91čď + 64ď2 + 162ÿ2∣∣ ,∣∣−65č2 + 183čÿ + 69ďÿ − 53čď − 8ď2 − 126ÿ2∣∣ ,∣∣89č2 − 279čÿ − 165ďÿ + 101čď + 32ď2 + 222ÿ2∣∣} .
We also have ∫ ď
č
∣∣f ′′ (ţ) −γ∣∣dt = (S −γ) (ď − č) (2.13)
and ∫ ď
č
∣∣f ′′ (ţ) − Γ∣∣dţ = (Γ −S) (ď − č) . (2.14)
So, we attain (2.1) and (2.2) by using (2.5) to (2.14) and taking C = γ and C = Γ in (2.10)
respectively.
Corollary 2.1. By replacing ÿ = č in (2.1) and (2.2) , we get∣∣∣∣∣f (č) + f (ď)2 −(ď − č) f
′(ď) − f ′(č)
12
−
1
ď − č
∫ ď
č
f (ţ)dţ
∣∣∣∣∣
≤
1
96
(
ď − č
)3
(S −γ) ,
6 Int. J. Anal. Appl. (2022), 20:28∣∣∣∣∣f (č) + f (ď)2 −(ď − č) f
′(ď) − f ′(č)
12
−
1
ď − č
∫ ď
č
f (ţ)dţ
∣∣∣∣∣
≤
1
96
(
ď − č
)3
(Γ −S) .
�
Now some new perturbed Ostrowski type inequalities are presented by working with differentiable
mapping whose first derivatives are absolutely continuous and the second derivatives belong to f ′′′ ∈ L2
the usual Lebesgue spaces which refine and generalize some previous inequalities of this domain.
Theorem 2.2. Let f :
[
č, ď
]
→ R be three times differentiable function on
(
č, ď
)
. If f ′′′ ∈
L2
[
č, ď
]
,then for all ÿ ∈
[
č, č+ď
2
]
, we have∣∣∣∣14
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
(2.15)
+
1
2
f
(
č + 2b− ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
+
f ′(ď) − f ′(č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}
−
1
ď − č
∫ ď
č
f (ţ) dţ
∣∣∣∣∣
≤
1
π
∥∥f ′′′∥∥
2
[
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
×
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
1
2
.
Proof. Let Rn(x) be defined by (2.7) and (2.8), If we take C = f ′′
(
č+ď
2
)
in (2.9) by applying the
Cauchy Inequality, then
|Rn(ÿ)| ≤
1
ď − č
∫ ď
č
∣∣∣∣
(
f ′′ (ţ) − f ′′
(
č + ď
2
))∣∣∣∣ (2.16)
×
∣∣∣∣∣K (ÿ,ţ) − 1ď − č
∫ ď
č
K (ÿ, s) ds
∣∣∣∣∣dţ.
Int. J. Anal. Appl. (2022), 20:28 7
≤
1
ď − č
[∫ ď
č
(
f ′′ (ţ) − f ′′
(
č + ď
2
))2
dţ
]1
2
×
∫ ď
č
(
K (ÿ,ţ) −
1
ď − č
∫ ď
č
K (ÿ, s) ds
)2
dţ
1
2
.
We apply the Diaze-Metcalf inequality from [16] to get∫ ď
č
(
f ′′ (ţ) − f ′′
(
č + ď
2
))2
dţ ≤
(
ď − č
)2
π2
∥∥f ′′′∥∥2
2
and ∫ ď
č
(
K (ÿ,ţ) −
1
ď − č
∫ ď
č
K (ÿ, s) ds
)2
dţ (2.17)
=
∫ ď
č
K (ÿ,ţ)2 dţ−
1
ď − č
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
=
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
.
So, by using the above relations (2.18)-(2.19) , we attain (2.17) .
Corollary 2.2. By replacing ÿ = č in (2.17) , we get∣∣∣∣∣f (č) + f (ď)2 −(ď − č) f
′(č) + f ′(ď)
12
−
1
ď − č
∫ ď
č
f (ţ)dţ
∣∣∣∣∣
≤
1
π
∥∥f ′′′∥∥
2
(
ď − č
)5
2
1
12
1
√
5
.
�
Theorem 2.3. Let f :
[
č, ď
]
→R be an absolutely continuous function on
(
č, ď
)
, with f ′′ ∈ L2
[
č, ď
]
.
Then ∣∣∣∣14
[
f (ÿ) + f
(
č + ď − ÿ
)
+
1
2
f
(
č + ÿ
2
)
+
1
2
f
(
3č + ÿ
4
)
(2.18)
+
1
2
f
(
č + 2ď − ÿ
2
)
+
1
2
f
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f ′
(
č + ď − ÿ
)
− f ′ (ÿ) +
1
4
f ′
(
č + 2ď − ÿ
2
)
−
1
4
f ′
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f ′
(
č + 4ď − ÿ
4
)
− f ′
(
3č + ÿ
4
)}]
8 Int. J. Anal. Appl. (2022), 20:28
+
f ′(ď) − f ′(č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}
−
1
ď − č
∫ ď
č
f (ţ) dţ
∣∣∣∣∣
≤
√
σ(f ′′)
ď − č
[
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3
+
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
1
2
∀ÿ ∈
[
č, č+ď
2
]
, where
σ(f ′′) =
∥∥f ′′∥∥2
2
−
(
f ′(ď) − f ′(č)
)2
ď − č
=
∥∥f ′′∥∥2
2
−S2
(
ď − č
)
(2.19)
where
S =
f ′(ď) − f ′(č)
ď − č
.
Proof. Let Rn(x) be defined by (2.7) and (2.8) .If we take C =
1
ď−č
∫ ď
č
f
′′
(s) ds in (2.9) and applying
the Cauchy Inequality, then we have
|Rn(ÿ)|
≤
1
ď − č
∫ ď
č
∣∣∣∣∣
(
f ′′ (ţ) −
1
ď − č
∫ ď
č
f
′′
(s) ds
)∣∣∣∣∣
×
∣∣∣∣∣K (ÿ,ţ) − 1ď − č
∫ ď
č
K (ÿ, s) ds
∣∣∣∣∣dţ.
≤
1
ď − č
∫ ď
č
(
f ′′ (ţ) −
1
ď − č
∫ ď
č
f
′′
(s) ds
)2
dţ
1
2
×
∫ ď
č
(
K (ÿ,ţ) −
1
ď − č
∫ ď
č
K (ÿ, s) ds
)2
dţ
1
2
=
√
σ(f ′′)
ď − č
[
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3
Int. J. Anal. Appl. (2022), 20:28 9
+
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
1
2
.
Hence proved (2.21) �
Corollary 2.3. By replacing ÿ = č in (2.21) we get∣∣∣∣∣f (č) + f (ď)2 −(ď − č) f
′(č) + f ′(ď)
12
−
1
ď − č
∫ ď
č
f (ţ)dţ
∣∣∣∣∣
≤
√
σ(f ′′)
(
ď − č
)3
2
1
12
1
√
5
.
3. An Application to Cumulative Distribution Function
Consider X is a random variable taking values in the finite interval
[
č, ď
]
with the probability density
function f :
[
č, ď
]
→ [0, 1] and cumulative distributive function
F (ÿ) = Pr (X ≤ ÿ) =
∫ ÿ
č
f (ţ)dţ, (3.1)
F
(
ď
)
= Pr
(
X ≤ ď
)
=
∫ ď
č
f (u)du = 1. (3.2)
In this section, we may use the inequalities (2.1)-(2.2), (2.17) and (2.21) to get useful application
for cumulative distribution function by using probability density function with smaller error than that
which may be obtained by the classical results.
Theorem 3.1. Under the assumption of Theorem 2.1, we get the following inequality which holds∣∣∣∣ď −E(X)ď − č − 14
[
F (ÿ) + F
(
č + ď − ÿ
)
+
1
2
F
(
č + ÿ
2
)
(3.3)
+
1
2
F
(
3č + ÿ
4
)
+
1
2
F
(
č + 2ď − ÿ
2
)
+
1
2
F
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f
(
č + ď − ÿ
)
− f (ÿ) +
1
4
f
(
č + 2ď − ÿ
2
)
−
1
4
f
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f
(
č + 4ď − ÿ
4
)
− f
(
3č + ÿ
4
)}]
−
f (ď) − f (č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
≤ ω(ÿ)(ď − č)(S −γ)
10 Int. J. Anal. Appl. (2022), 20:28
and
∣∣∣∣ď −E(X)ď − č − 14
[
F (ÿ) + F
(
č + ď − ÿ
)
+
1
2
F
(
č + ÿ
2
)
(3.4)
+
1
2
F
(
3č + ÿ
4
)
+
1
2
F
(
č + 2ď − ÿ
2
)
+
1
2
F
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f
(
č + ď − ÿ
)
− f (ÿ) +
1
4
f
(
č + 2ď − ÿ
2
)
−
1
4
f
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
){
f
(
č + 4ď − ÿ
4
)
− f
(
3č + ÿ
4
)}]
−
f (ď) − f (č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}∣∣∣∣∣ ≤ ω(ÿ)(ď − č)(Γ −S)
for all ÿ ∈
[
č, č+ď
2
]
.
Proof. By (3.1) and (3.2) on taking f = F and by applying the fact
E(X) =
∫ ď
č
ţdF (ţ) = ď −
∫ ď
č
F (ţ)dţ (3.5)
we attain (3.3) and (3.4) . �
Theorem 3.2. By using Theorem 2.1, we get the following inequality which holds
∣∣∣∣ď −E(X)ď − č − 14
[
F (ÿ) + F
(
č + ď − ÿ
)
+
1
2
F
(
č + ÿ
2
)
(3.6)
+
1
2
F
(
3č + ÿ
4
)
+
1
2
F
(
č + 2ď − ÿ
2
)
+
1
2
F
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f
(
č + ď − ÿ
)
− f (ÿ) +
1
4
f
(
č + 2ď − ÿ
2
)
−
1
4
f
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
)
×
{
f
(
č + 4ď − ÿ
4
)
− f
(
3č + ÿ
4
)}]
Int. J. Anal. Appl. (2022), 20:28 11
−
f (ď) − f (č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}∣∣∣∣∣
≤
1
π
∥∥f ′′′∥∥
2
[
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3
+
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
1
2
.
Proof. Using (4.36) and by the conditions that we used in above Theorem , we get (3.6). �
Theorem 3.3. With the statement of Theorem 2.1, we have the following inequality which holds∣∣∣∣ď −E(X)ď − č − 14
[
F (ÿ) + F
(
č + ď − ÿ
)
+
1
2
F
(
č + ÿ
2
)
(3.7)
+
1
2
F
(
3č + ÿ
4
)
+
1
2
F
(
č + 2ď − ÿ
2
)
+
1
2
F
(
č + 4ď − ÿ
4
)
+
(
ÿ −
5č + 3ď
8
)
×
{
f
(
č + ď − ÿ
)
− f (ÿ) +
1
4
f
(
č + 2ď − ÿ
2
)
−
1
4
f
(
č + ÿ
2
)}
+
1
8
(
ÿ −
3č + ď
4
)
×
{
f
(
č + 4ď − ÿ
4
)
− f
(
3č + ÿ
4
)}]
−
f (ď) − f (č)(
ď − č
)2
{
1
192
(ÿ − č)3 +
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}∣∣∣∣∣
≤
√
σ(f ′′)
ď − č
[
1
10240
(ÿ − č)5 +
33
320
(
ÿ −
3č + ď
4
)5
−
1057
10240
(
ÿ −
č + ď
2
)5
−
1
ď − č
{
1
192
(ÿ − č)3
+
3
8
(
ÿ −
3č + ď
4
)3
−
73
192
(
ÿ −
č + ď
2
)3}2
1
2
.
Proof. Applying (4.43) and by set of conditions that we used in Theorem 3.1, we get (3.7). �
Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the
publication of this paper.
12 Int. J. Anal. Appl. (2022), 20:28
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https://doi.org/10.1007/978-94-011-3562-7
1. Introduction
2. Main Findings
3. An Application to Cumulative Distribution Function
References