Conceptual knowledge in area measurement for primary school students: A systematic review

Hafiz Idrus; Suzieleez Syrene Abdul Rahim; Hutkemri Zulnaidi; Hafiz Idrus; Suzieleez Syrene Abdul Rahim; Hutkemri Zulnaidi

doi:10.3934/steme.2022003

STEM Education

2022, Volume 2, Issue 1: 47-58. doi: 10.3934/steme.2022003

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Review

Conceptual knowledge in area measurement for primary school students: A systematic review

Department of Mathematics and Sciences Education, Faculty of Education, Universiti Malaya, 50603 Kuala Lumpur, Malaysia; idrus.hafiz89@gmail.com (H.I.); suzieleez@um.edu.my (S.S.A.R.); hutkemri@um.edu.my (H.Z.)

Academic Editor: Stephen Xu

Received: 01 January 2022 Revised: 01 February 2022

Discussions about teaching area measurement in primary school have been ongoing over some decades. However, investigations that thoroughly examine the current research on conceptual understanding in area measuring in elementary schools are still lacking. The objective of this paper is to review whether conceptual knowledge in area measurement may support students to obtain better results in primary schools. This study is to gain insight into how conceptual knowledge in area measurement has been portrayed for primary school students, and reveal possible omissions and gaps in the synthesized literature on the subject. To gather information, two databases were used: Scopus and Web of Science. Primary searches pulled up many studies on the subject of investigation. After analyzing abstracts and eliminating duplicates, our systematic review indicates that there seems a direct link between conceptual understanding and area measurement in primary school mathematics. Hence, teaching children the principle of area measurement rather than a procedure for solving problems seems to be the most effective way of improving problem-solving skills and conceptual understanding for primary students.

Keywords:

Citation: Hafiz Idrus, Suzieleez Syrene Abdul Rahim, Hutkemri Zulnaidi. Conceptual knowledge in area measurement for primary school students: A systematic review[J]. STEM Education, 2022, 2(1): 47-58. doi: 10.3934/steme.2022003

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Abstract

1. Introduction

For a convex function $\sigma:\mathrm{I}\subseteq \bf{R}\to \bf{R}$ on $\mathrm{I}$ with $\nu_{1}, \nu_{2}\in \mathrm{I}$ and $\nu_{1} < \nu_{2}$ , the Hermite-Hadamard inequality is defined by ^[1]:

$\begin{equation} \sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \leq \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma(\eta)d\eta \leq \frac{\sigma(\nu_{1})+\sigma(\nu_{2})}{2}. \end{equation}$

(1.1)

The Hermite-Hadamard integral inequality (1.1) is one of the most famous and commonly used inequalities. The recently published papers ^[2,3,4] are focused on extending and generalizing the convexity and Hermite-Hadamard inequality.

The situation of the fractional calculus (integrals and derivatives) has won vast popularity and significance throughout the previous five decades or so, due generally to its demonstrated applications in numerous seemingly numerous and great fields of science and engineering ^[5,6,7].

Now, we recall the definitions of Riemann-Liouville fractional integrals.

Definition 1.1 (^[5,6,7]). Let $\sigma\in L_{1}[\nu_{1}, \nu_{2}].$ The Riemann-Liouville integrals $J_{\nu_{1}+}^{\vartheta }\sigma$ and $J_{\nu_{2}-}^{\vartheta }\sigma$ of order $\vartheta > 0$ with $\nu_{1}\geq 0$ are defined by

$\begin{equation} J_{\nu_{1}+}^{\vartheta }\sigma(x) = \frac{1}{\Gamma (\vartheta )}\int_{\nu_{1}}^{x}(x-\eta)^{\vartheta -1}\sigma(\eta)d\eta,\; \ \ \ \nu_{1} < x \end{equation}$

(1.2)

and

$\begin{equation} J_{\nu_{2}-}^{\vartheta }\sigma(x) = \frac{1}{\Gamma (\vartheta )}\int_{x}^{\nu_{2}}(\eta-x)^{\vartheta -1}\sigma(\eta)d\eta,\; \ \ x < \nu_{2}, \end{equation}$

(1.3)

respectively. Here $\Gamma (\vartheta)$ is the well-known Gamma function and $J_{\nu_{1}+}^{0}\sigma(x) = J_{\nu_{2}-}^{0}\sigma(x) = \sigma(x).$

With a huge application of fractional integration and Hermite-Hadamard inequality, many researchers in the field of fractional calculus extended their research to the Hermite-Hadamard inequality, including fractional integration rather than ordinary integration; for example see ^{[8,9,10,11,12,13,14,15,16,17,18,19,20,21]}.

In this paper, we consider the integral inequality of Hermite-Hadamard-Mercer type that relies on the Hermite-Hadamard and Jensen-Mercer inequalities. For this purpose, we recall the Jensen-Mercer inequality: Let $0 < x_{1}\leq x_{2}\leq \cdots\leq x_{n}$ and $\mu = (\mu _{1}, \mu_{2}, \ldots, \mu _{n})$ nonnegative weights such that $\sum_{i = 1}^{n}\mu_{i} = 1$ . Then, the Jensen inequality ^[22,23] is as follows, for a convex function $\sigma$ on the interval $[\nu_{1}, \nu_{2}]$ , we have

$\begin{equation} \sigma\Bigg(\sum\limits_{i = 1}^{n}\mu _{i}x_{i}\Bigg)\leq \sum\limits_{i = 1}^{n}\mu _{i}\sigma(x_{i}), \end{equation}$

(1.4)

where for all $x_{i}\in \lbrack \nu_{1}, \nu_{2}]$ and $\mu_{i}\in \lbrack 0, 1]$ , $(i = \overline{1, n})$ .

Theorem 1.1 (^[2,23]). If $\sigma$ is convex function on $[\nu_{1}, \nu_{2}]$ , then

$\begin{equation} \sigma\left(\nu_{1}+\nu_{2}-\sum\limits^{n}_{i = 1}\mu_{i}x_{i}\right)\leq \sigma(\nu_{1})+\sigma(\nu_{2})-\sum\limits^{n}_{i = 1}\mu_{i}\sigma(x_{i}), \end{equation}$

(1.5)

for each $x_{i}\in[\nu_{1}, \nu_{2}]$ and $\mu_{i}\in[0, 1]$ , $(i = \overline{1, n})$ with $\sum^{n}_{i = 1}\mu_{i} = 1$ . For some results related with Jensen-Mercer inequality, see ^[24,25,26].

In view of above indices, we establish new integral inequalities of Hermite-Hadamard-Mercer type for convex functions via the Riemann-Liouville fractional integrals in the current project. Particularly, we see that our results can cover the previous researches.

2. Main results

Theorem 2.1. For a convex function $\sigma:[\nu_{1}, \nu_{2}]\subseteq \bf{R} \to\bf{R}$ with $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

$\begin{multline} \sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\leq \frac{2^{\vartheta -1} \Gamma\left(\vartheta +1\right)}{\left( y-x\right)^{\vartheta}} \Biggl[ J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \\ +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\Biggr] \leq \sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)-\frac{\sigma\left(x\right)+\sigma\left(y\right)}{2}. \end{multline}$

(2.1)

Proof. By using the convexity of $\sigma$ , we have

$\begin{equation} \sigma\left( \nu_{1}+\nu_{2}-\frac{u+v}{2}\right) \leq \frac{1}{2}\left[ \sigma\left( \nu_{1}+\nu_{2}-u\right) +\sigma\left( \nu_{1}+\nu_{2}-v\right) \right], \end{equation}$

(2.2)

and above with $u = \frac{1-\eta}{2}x+\frac{1+\eta}{2}y$ , $v = \frac{1+\eta}{2}x+\frac{1-\eta}{2}y$ , where $x, y\in[\nu_{1}, \nu_{2}]$ and $\eta\in [0, 1]$ , leads to

$\begin{multline} \sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\leq\frac{1}{2} \Biggl[\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) \\ +\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\Biggr]. \end{multline}$

(2.3)

Multiplying both sides of (2.3) by $\eta^{\vartheta -1}$ and then integrating with respect to $\eta$ over $[0, 1]$ , we get

$\begin{align*} \frac{1}{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) &\leq \frac{1}{2}\left[\int_{0}^{1}\eta^{\vartheta-1} \sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)d\eta\right. \\ &+\left. \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)d\eta\right] \\ & = \frac{1}{2}\left[ \frac{2^{\vartheta }}{\left( y-x\right) ^{\vartheta }}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-\frac{x+y}{2}} \left( \left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)-w\right) ^{\vartheta -1}\sigma\left( w\right) dw\right. \\ &\left. +\frac{2^{\vartheta}}{\left(y-x\right)^{\vartheta}}\int_{\nu_{1}+\nu_{2}-\frac{x+y}{2}}^{\nu_{1}+\nu_{2}-x} \left( w-\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right) ^{\vartheta-1}\sigma\left( w\right) dw\right] \\ & = \frac{2^{\vartheta-1}\Gamma \left(\vartheta\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right], \end{align*}$

and thus the proof of first inequality in (2.1) is completed.

On the other hand, we have by using the Jensen-Mercer inequality:

$\begin{eqnarray} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) &\leq &\sigma\left( \nu_{1}\right) +\sigma\left( \nu_{2}\right) -\left( \frac{1-\eta}{2}\sigma\left( x\right) +\frac{1+\eta}{2}\sigma\left( y\right) \right) \end{eqnarray}$

(2.4)

$\begin{eqnarray} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) &\leq &\sigma\left( \nu_{1}\right) +\sigma\left( \nu_{2}\right) -\left( \frac{1+\eta}{2}\sigma\left( x\right) +\frac{1-\eta}{2}\sigma\left( y\right) \right). \end{eqnarray}$

(2.5)

Adding inequalities (2.4) and (2.5) to get

$\begin{multline} \sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) +\sigma\left(\nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) \\ \leq 2\left[\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)\right]-\left[\sigma\left(x\right)+\sigma\left(y\right)\right]. \end{multline}$

(2.6)

Multiplying both sides of (2.6) by $\eta^{\vartheta -1}$ and then integrating with respect to $\eta$ over $[0, 1]$ to obtain

$\begin{multline*} \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right)d\eta +\int_{0}^{1}\eta^{\vartheta-1}\sigma\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)d\eta \\ \leq \frac{2}{\vartheta}\left[\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right)\right] -\frac{1}{\vartheta }\left[ \sigma\left( x\right) +\sigma\left( y\right) \right]. \end{multline*}$

By making use of change of variables and then multiplying by $\frac{\vartheta }{2}$ , we get the second inequality in (2.1).

Remark 2.1. If we choose $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ in Theorem 2.1, then the inequality (2.1) reduces to (1.1).

Corollary 2.1. Theorem 2.1 with

● $\vartheta = 1$ becomes [24, Theorem 2.1].

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} \sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\leq\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)+\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\right] \leq \frac{\sigma\left(\nu_{1}\right)+\sigma\left(\nu_{2}\right) }{2}, \end{align*}$

which is obtained by Mohammed and Brevik in ^[10].

The following Lemma linked with the left inequality of (2.1) is useful to obtain our next results.

Lemma 2.1. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R} \to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\sigma'\in L[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( y-x\right)^{\vartheta}} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right)-}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right] \\ -\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) = \frac{\left(y-x\right)}{4}\int_{0}^{1}\eta^{\vartheta} \Bigl[\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) \\ -\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\Bigr]d\eta. \end{multline}$

(2.7)

Proof. From right hand side of (2.7), we set

$\begin{align} \varpi_{1}-\varpi_{2}&:= \int_{0}^{1}\eta^{\vartheta}\left[\sigma'\left(\nu_{1}+\nu_{2} -\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right) -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right]d\eta \\ & = \int_{0}^{1}\eta^{\vartheta }\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)d\eta \\ &-\int_{0}^{1}\eta^{\vartheta}\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right) d\eta. \end{align}$

(2.8)

By integrating by parts with $w = \nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)$ , we can deduce:

$\begin{align*} \varpi_{1}& = -\frac{2}{\left( y-x\right) }\sigma\left(\nu_{1}+\nu_{2}-y\right)+\frac{2\vartheta }{\left(y-x\right)} \int_{0}^{1}\eta^{\vartheta -1}\sigma\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) d\eta \\ & = -\frac{2}{\left(y-x\right)}\sigma\left(\nu_{1}+\nu_{2}-y\right)+\frac{2^{\vartheta+1}\vartheta}{\left(y-x\right)^{\vartheta+1}} \int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-\frac{x+y}{2}}\sigma\left(\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)-w\right)^{\vartheta -1} \sigma\left(w\right) dw \\ & = -\frac{2}{\left(y-x\right)}\sigma\left( \nu_{1}+\nu_{2}-y\right) +\frac{2^{\vartheta+1}\Gamma\left(\vartheta+1\right)}{\left( y-x\right) ^{\vartheta +1}} J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right). \end{align*}$

Similarly, we can deduce:

$\begin{equation*} \varpi_{2} = \frac{2}{y-x}\sigma\left(\nu_{1}+\nu_{2}-x\right) -\frac{2^{\vartheta+1}\Gamma\left(\vartheta+1\right)}{\left( y-x\right) ^{\vartheta +1}} J_{\left( \nu_{1}+\nu_{2}-x\right)-}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right). \end{equation*}$

By substituting $\varpi_{1}$ and $\varpi_{2}$ in (2.8) and then multiplying by $\frac{\left(y-x\right)}{4}$ , we obtain required identity (2.7).

Corollary 2.2. Lemma 2.1 with

● $\vartheta = 1$ becomes:

$\begin{multline*} \frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right) dw-\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) = \frac{\left( y-x\right) }{4}\int_{0}^{1}\eta\left[ \sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) \right. \\ \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right) \right] d\eta. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) = \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\int_{0}^{1}\eta \left[ \sigma'\left(\nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}\nu_{1}+\frac{1+\eta}{2}\nu_{2}\right) \right) \right. \\ \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}\nu_{1}+\frac{1-\eta}{2}\nu_{2}\right)\right) \right] d\eta. \end{multline*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( \nu_{2}-\nu_{1}\right)^{\vartheta }} \left[ J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) +\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right]-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \\ = \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\int_{0}^{1}\eta^{\vartheta } \left[\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}\nu_{1}+\frac{1+\eta}{2}\nu_{2}\right) \right) \right. \left. -\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}\nu_{1}+\frac{1-\eta}{2}\nu_{2}\right) \right) \right] d\eta. \end{multline*}$

Theorem 2.2. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $|\sigma'|$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \Biggl| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \Biggl[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr] \\ -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\Biggr| \leq \frac{\left( y-x\right) }{2\left( 1+\vartheta \right) }\left[ \left| \sigma'\left( \nu_{1}\right) \right| +\left| \sigma'\left(\nu_{2}\right) \right| -\frac{\left| \sigma'\left( x\right) \right| +\left| \sigma'\left( y\right) \right| }{2}\right]. \end{multline}$

(2.9)

Proof. By taking modulus of identity (2.7), we get

$\begin{multline*} \Biggl|\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] \\ -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr|\leq \frac{\left( y-x\right) }{4} \left[\int_{0}^{1}\eta^{\vartheta }\left|\sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right) \right) \right| d\eta\right. \\ \left. +\int_{0}^{1}\eta^{\vartheta }\left| \sigma'\left( \nu_{1}+\nu_{2}-\left( \frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right) \right) \right| d\eta\right]. \end{multline*}$

Then, by applying the convexity of $|\sigma'|$ and the Jensen-Mercer inequality on above inequality, we get

$\begin{align*} &\Biggl|\frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(y-x\right)^{\vartheta}} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] \\ &-\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \Biggr|\leq \frac{\left( y-x\right) }{4}\left[ \int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left( \nu_{1}\right) \right| +\left| \sigma'\left( \nu_{2}\right) \right| -\left( \frac{1+\eta}{2}\left| \sigma^{\prime }\left( x\right) \right| +\frac{1-\eta}{2}\right) \left| \sigma'\left( y\right) \right| \right] d\eta\right. \\ &\left. +\int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left(\nu_{1}\right) \right| +\left| \sigma'\left( \nu_{2}\right) \right| -\left(\frac{1-\eta}{2}\left|\sigma'\left(x\right) \right| +\frac{1+\eta}{2}\right)\left| \sigma'\left( y\right)\right| \right] d\eta\right] \\ & = \frac{\left(y-x\right)}{2\left(1+\vartheta \right)}\left[ \left|\sigma'\left( \nu_{1}\right)\right| +\left|\sigma'(\nu_{2})\right|-\frac{\left|\sigma'\left(x\right)\right|+\left|\sigma'\left(y\right)\right|}{2}\right], \end{align*}$

which completes the proof of Theorem 2.2.

Corollary 2.3. Theorem 2.2 with

● $\vartheta = 1$ becomes:

$\begin{align*} \left|\frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma(w)dw-\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right| \leq \frac{\left(y-x\right)}{4}\left[\left| \sigma'\left(\nu_{1}\right)\right|+\left|\sigma'\left(\nu_{2}\right)\right| -\frac{\left|\sigma'\left(x\right)\right|+\left|\sigma'\left(y\right)\right|}{2}\right]. \end{align*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes [27, Theorem 2.2].

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} \left\vert \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \leq \frac{\left( \nu_{2}-\nu_{1}\right) }{4}\left[ \frac{\left\vert \sigma^{\prime }\left( \nu_{1}\right) \right\vert +\left\vert \sigma^{\prime }\left( \nu_{2}\right) \right\vert }{2}\right] . \end{align*}$

Theorem 2.3. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\left| \sigma'\right|^{q}, q > 1$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left( y-x\right) ^{\vartheta }}\left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}- \frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4\sqrt[p]{\vartheta p+1}} \left[ \left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left| \sigma'\left(\nu_{2}\right)\right| ^{q} -\left(\frac{\left|\sigma'(x) \right|^{q}+3\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left(\nu_{2}\right)\right|^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right], \end{multline}$

(2.10)

where $\frac{1}{p}+\frac{1}{q} = 1$ .

Proof. By taking modulus of identity (2.7) and using Hölder's inequality, we get

$\begin{multline*} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right)^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left(\nu_{1}+\nu_{2}-x\right)-}^{\vartheta }\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta p}\right)^{\frac{1}{p}} \left\{\left(\int_{0}^{1} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)\right|^{q}d\eta\right) ^{\frac{1}{q}}\right. \\ \left. +\left(\int_{0}^{1} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right|^{q}d\eta\right) ^{\frac{1}{q}}\right\}. \end{multline*}$

Then, by applying the Jensen-Mercer inequality with the convexity of $\left|\sigma'\right|^{q}$ , we can deduce

$\begin{align*} &\left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[J_{\left(\nu_{1}+\nu_{2}-y\right)+}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ &\left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ &\leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta p}\right)^{\frac{1}{p}} \left\{ \left( \int_{0}^{1}\left| \sigma'\left( \nu_{1}\right)\right| ^{q}+\left| \sigma'\left( \nu_{2}\right) \right|^{q} -\left(\frac{1-\eta}{2}\left|\sigma'(x)\right|^{q}+\frac{1+\eta}{2}\left|\sigma'(y)\right|^{q}\right)\right)^{\frac{1}{q}}\right. \\ &\left. +\left( \int_{0}^{1}\left| \sigma'\left( \nu_{1}\right)\right| ^{q}+\left| \sigma'\left( \nu_{2}\right) \right|^{q} -\left(\frac{1+\eta}{2}\left| \sigma'\left( x\right)\right|^{q}+\frac{1-\eta}{2}\left| \sigma'\left( y\right) \right|^{q} \right) \right)^{\frac{1}{q}}\right\} \\ & = \frac{\left(y-x\right)}{4\sqrt[p]{\vartheta p+1}}\left[\left( \left| \sigma'\left( \nu_{1}\right) \right| ^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left|\sigma'\left( x\right) \right|^{q} +3\left| \sigma'\left(y\right) \right|^{q} }{4}\right) \right) ^{\frac{1}{q}}\right. \\ &\left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left| \sigma'\left( \nu_{2}\right) \right| ^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right], \end{align*}$

which completes the proof of Theorem 2.3.

Corollary 2.4. Theorem 2.3 with

● $\vartheta = 1$ becomes:

$\begin{multline*} \left|\frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right)dw -\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right| \\ \leq \frac{\left( y-x\right)}{4\sqrt[p]{p+1}}\left[\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left|\sigma'\left( x\right) \right|^{q} +3\left|\sigma'\left(y\right) \right|^{q}}{4}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left(\left|\sigma'\left( \nu_{1}\right)\right|^{q}+\left| \sigma'\left( \nu_{2}\right)\right|^{q} -\left(\frac{3\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q} }{4}\right)\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} &&\left\vert \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw-\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \leq \frac{\left( \nu_{2}-\nu_{1}\right) }{2^{\frac{2}{p}}}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left[ \left\vert \sigma^{\prime }\left( \nu_{1}\right) \right\vert +\left\vert \sigma^{\prime }\left( \nu_{2}\right) \right\vert \right] . \end{align*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{align*} &\left\vert \frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[ J_{\nu_{1}+}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right. \right. \left. \left. +\; J_{\nu_{2}-}^{\vartheta }\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right] -\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right) \right\vert \\ &\leq \frac{2^{\vartheta-1-\frac{2}{q}}}{\nu_{2}-\nu_{1}}\left(\frac{1}{p+1}\right)^{\frac{1}{p}}\left[ \left\vert \sigma'\left( \nu_{1}\right) \right\vert +\left\vert \sigma'\left( \nu_{2}\right) \right\vert \right]. \end{align*}$

Theorem 2.4. Let $\sigma:[\nu_{1}, \nu_{2}]\subseteq\bf{R}\to\bf{R}$ be a differentiable function on $(\nu_{1}, \nu_{2})$ and $\left|\sigma'\right|^{q}, q\geq 1$ is convex on $[\nu_{1}, \nu_{2}]$ with $\nu_{1}\leq \nu_{2}$ and $x, y\in[\nu_{1}, \nu_{2}]$ . Then, we have:

$\begin{multline} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right.\right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left(y-x\right)}{4\left(\vartheta+1\right)}\left[\left(\left|\sigma'(\nu_{1})\right|^{q} +\left|\sigma'\left(\nu_{2}\right)\right|^{q}-\left( \frac{\left| \sigma'\left(x\right)\right|^{q} +\left( 2\vartheta+3\right)\left|\sigma'(y)\right|^{q}}{2\left(\vartheta+2\right) }\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left|\sigma'\left(\nu_{1}\right)\right|^{q}+\left| \sigma'\left( \nu_{2}\right)\right|^{q} -\left(\frac{\left(2\vartheta+3\right)\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{2\left( \vartheta+2\right) }\right) \right) ^{\frac{1}{q}}\right]. \end{multline}$

(2.11)

Proof. By taking modulus of identity (2.7) with the well-known power mean inequality, we can deduce

$\begin{multline*} \left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right) ^{\vartheta }} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right) +}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ \left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{4}\left( \int_{0}^{1}\eta^{\vartheta }\right) ^{1-\frac{1}{q}} \left\{\left(\int_{0}^{1}\eta^{\vartheta} \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1-\eta}{2}x+\frac{1+\eta}{2}y\right)\right)\right|^{q}d\eta\right)^{\frac{1}{q}}\right. \\ \left. +\left( \int_{0}^{1}\eta^{\vartheta } \left|\sigma'\left(\nu_{1}+\nu_{2}-\left(\frac{1+\eta}{2}x+\frac{1-\eta}{2}y\right)\right)\right|^{q}d\eta\right)^{\frac{1}{q}}\right\} . \end{multline*}$

By applying the Jensen-Mercer inequality with the convexity of $\left|\sigma'\right|^{q}$ , we can deduce

$\begin{align*} &\left| \frac{2^{\vartheta -1}\Gamma \left( \vartheta +1\right) }{\left(y-x\right)^{\vartheta}} \left[ J_{\left( \nu_{1}+\nu_{2}-y\right)+}^{\vartheta}\sigma\left(\nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right. \right. \\ &\left. \left. +\; J_{\left( \nu_{1}+\nu_{2}-x\right) -}^{\vartheta }\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right)\right] -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ &\leq \frac{\left( y-x\right)}{4}\left(\int_{0}^{1}\eta^{\vartheta }\right)^{1-\frac{1}{q}} \left\{ \left( \int_{0}^{1}\eta^{\vartheta }\left[ \left|\sigma'\left( \nu_{1}\right) \right|^{q} +\left| \sigma'\left(\nu_{2}\right) \right| ^{q}-\left( \frac{1-\eta}{2}\left| \sigma'\left(x\right)\right|^{q} +\frac{1+\eta}{2}\left| \sigma'\left( y\right)\right| ^{q}\right) \right] \right) ^{\frac{1}{q}}\right. \\ &\left. +\left( \int_{0}^{1}\eta^{\vartheta }\left[ \left| \sigma'\left(\nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right) \right|^{q}-\left( \frac{1+\eta}{2}\left| \sigma'\left( x\right) \right|^{q} +\frac{1-\eta}{2}\left| \sigma'\left( y\right) \right|^{q}\right) \right] \right) ^{\frac{1}{q}}\right\} \\ & = \frac{\left( y-x\right) }{4\left( \vartheta +1\right) }\left[\left(\left|\sigma'\left(\nu_{1}\right)\right|^{q} +\left|\sigma'\left( \nu_{2}\right) \right| ^{q}-\left( \frac{\left| \sigma'\left(x\right) \right|^{q} +\left(2\vartheta +3\right)\left|\sigma'(y)\right|^{q}}{2\left(\vartheta+2\right)}\right)\right)^{\frac{1}{q}}\right. \\ &\left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left( \nu_{2}\right)\right|^{q} -\left( \frac{\left( 2\vartheta +3\right) \left| \sigma'\left( x\right) \right|^{q} +\left|\sigma'\left( y\right)\right|^{q}}{2\left(\vartheta+2\right)}\right)\right)^{\frac{1}{q}}\right], \end{align*}$

which completes the proof of Theorem 2.4.

Corollary 5. Theorem 2.4 with

● $q = 1$ becomes Theorem 2.2.

● $\vartheta = 1$ becomes:

$\begin{multline*} \left| \frac{1}{y-x}\int_{\nu_{1}+\nu_{2}-y}^{\nu_{1}+\nu_{2}-x}\sigma\left(w\right) dw -\sigma\left( \nu_{1}+\nu_{2}-\frac{x+y}{2}\right) \right| \\ \leq \frac{\left( y-x\right) }{8}\left[ \left( \left|\sigma'\left( \nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right)\right|^{q}-\left( \frac{\left| \sigma'\left( x\right)\right|^{q} +5\left|\sigma'\left(y\right)\right|^{q}}{6}\right)\right)^{\frac{1}{q}}\right. \\ \left. +\left( \left| \sigma'\left( \nu_{1}\right) \right|^{q}+\left|\sigma'\left(\nu_{2}\right)\right|^{q} -\left(\frac{5\left|\sigma'(x)\right|^{q}+\left|\sigma'(y)\right|^{q}}{6}\right)\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $\vartheta = 1$ , $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \left| \frac{1}{\nu_{2}-\nu_{1}}\int_{\nu_{1}}^{\nu_{2}}\sigma\left( w\right) dw -\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)\right| \\ \leq \frac{\left(y-x\right)}{8}\left[\left(\frac{5\left|\sigma'\left(\nu_{1}\right)\right|^{q} +\left| \sigma'\left( \nu_{2}\right)\right| ^{q}}{6}\right)^{\frac{1}{q}} +\left(\frac{\left|\sigma'(\nu_{1})\right|^{q}+5\left| \sigma'(\nu_{2})\right|^{q}}{6}\right)^{\frac{1}{q}}\right]. \end{multline*}$

● $x = \nu_{1}$ and $y = \nu_{2}$ becomes:

$\begin{multline*} \left| \frac{2^{\vartheta-1}\Gamma\left(\vartheta+1\right)}{\left(\nu_{2}-\nu_{1}\right)^{\vartheta}} \left[J_{\nu_{1} +}^{\vartheta}\sigma\left(\frac{\nu_{1}+\nu_{2}}{2}\right)+\; J_{\nu_{2} -}^{\vartheta }\sigma\left( \frac{\nu_{1}+\nu_{2}}{2}\right)\right] -\sigma\left( \frac{\nu_{1}+\nu_{2}}{2}\right)\right| \\ \leq \frac{\left(\nu_{2}-\nu_{1}\right)}{4\left(\vartheta+1\right)} \left[\left(\frac{\left(2\vartheta+3\right)\left|\sigma'(\nu_{1})\right|^{q}+\left|\sigma'(\nu_{2})\right|^{q}}{2\left(\vartheta+2\right)}\right)^{\frac{1}{q}} +\left(\frac{\left|\sigma'(\nu_{1})\right|^{q}+\left(2\vartheta+3\right)\left|\sigma'(\nu_{2})\right|^{q}}{2\left(\vartheta+2\right)}\right) ^{\frac{1}{q}}\right]. \end{multline*}$

3. Applications to special means

Here, we consider the following special means:

● The arithmetic mean:

$\mathcal{A}(\nu_{1},\nu_{2}) = \frac{\nu_{1}+\nu_{2}}{2}, \quad \nu_{1}, \nu_{2}\geq 0.$

● The harmonic mean:

$\mathcal{H}(\nu_{1},\nu_{2}) = \frac{2\nu_{1}\nu_{2}}{\nu_{1}+\nu_{2}}, \quad \nu_{1}, \nu_{2} > 0.$

● The logarithmic mean:

$\begin{equation*} \mathcal{L}(\nu_{1},\nu_{2}) = \begin{cases} \frac{\nu_{2}-\nu_{1}}{\ln \nu_{2}-\ln \nu_{1}}, & \text{if}\; \nu_{1}\neq \nu_{2}, \\ \nu_{1}, & \text{if}\; \nu_{1} = \nu_{2}, \end{cases} \quad \nu_{1}, \nu_{2} > 0. \end{equation*}$

● The generalized logarithmic mean:

$\begin{equation*} \mathcal{L}_{n}(\nu_{1},\nu_{2}) = \begin{cases} \left[\frac{\nu_{2}^{n+1}-\nu_{1}^{n+1}}{\left(n+1\right)(\nu_{2}-\nu_{1})}\right]^{\frac{1}{n}}, & \text{if}\; \nu_{1}\neq \nu_{2} \\ \nu_{1}, & \text{if}\; \nu_{1} = \nu_{2}, \end{cases} \quad \nu_{1},\nu_{2} > 0;\; n\in\bf{Z}\setminus \{-1,0\}. \end{equation*}$

Proposition 3.1. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

(3.1)

Proof. By applying Corollary 2.3 (first item) for the convex function $\sigma(x) = x^{n}, \, x > 0$ , one can obtain the result directly.

Proposition 3.2. Let $0 < \nu_{1} < \nu_{2}$ . Then, for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

(3.2)

Proof. By applying Corollary 2.3 (first item) for the convex function $\sigma(x) = \frac{1}{x}, \, x > 0$ , one can obtain the result directly.

Proposition 3.3. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, we have:

$\begin{equation} \left| \mathcal{L}_{n}^{n}(\nu_{1},\nu_{2}) -\mathcal{A}^{n}(\nu_{1},\nu_{2}) \right| \leq \frac{n(\nu_{2}-\nu_{1})}{4} \left[ \mathcal{A}\left(\nu_{1}^{n-1}, \nu_{2}^{n-1}\right) \right], \end{equation}$

(3.3)

and

$\begin{equation} \left| \mathcal{L}^{-1}(\nu_{1},\nu_{2}) -\mathcal{A}^{-1}(\nu_{1},\nu_{2}) \right| \leq \frac{(\nu_{2}-\nu_{1})}{4}\mathcal{H}^{-1}\left( \nu_{1}^{2},\nu_{2}^{2}\right) . \end{equation}$

(3.4)

Proof. By setting $x = \nu_{1}$ and $y = \nu_{2}$ in results of Proposition 3.1 and Proposition 3.2, one can obtain the Proposition 3.3.

Proposition 3.4. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for $q > 1, \, \frac{1}{p}+\frac{1}{q} = 1$ and for all $x, y\in[\nu_{1}, \nu_{2}],$ we have:

$\begin{multline} \left| \mathcal{L}_{n}^{n}\left( \nu_{1}+\nu_{2}-y,\nu_{1}+\nu_{2}-x\right) -\left( 2\mathcal{A}(\nu_{1},\nu_{2}) -\mathcal{A}\left( x,y\right) \right) ^{n}\right| \\ \leq \frac{n(y-x)}{4\sqrt[p]{p+1}}\Bigg\{\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(x^{q(n-1)}, 3y^{q(n-1)}\right) \right]^{\frac{1}{q}} \\ +\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(3x^{q(n-1)}, y^{q(n-1)}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.5)

Proof. By applying Corollary 2.4 (first item) for convex function $\sigma\left(x\right) = x^{n}, \, x > 0$ , one can obtain the result directly.

Proposition 3.5. Let $0 < \nu_{1} < \nu_{2}$ . Then, for $q > 1, \, \frac{1}{p}+\frac{1}{q} = 1$ and for all $x, y\in[\nu_{1}, \nu_{2}]$ , we have:

$\begin{multline} \left| \mathcal{L}^{-1}\left( \nu_{1}+\nu_{2}-y,\nu_{1}+\nu_{2}-x\right) -\left( 2\mathcal{A}(\nu_{1},\nu_{2}) -\mathcal{A}\left(x,y\right)\right)^{-1}\right| \\ \leq \frac{\sqrt[q]{2}(y-x)}{4\sqrt[p]{p+1}}\Bigg\{\left[ \mathcal{H}^{-1}\left(\nu_{1}^{2q},\nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(x^{2q}, 3y^{2q}\right) \right]^{\frac{1}{q}} \\ +\left[\mathcal{H}^{-1}\left(\nu_{1}^{2q},\nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(3x^{2q}, y^{2q}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.6)

Proof. By applying Corollary 2.4 (first item) for the convex function $\sigma\left(x\right) = \frac{1}{x}, \, x > 0$ , one can obtain the result directly.

Proposition 3.6. Let $0 < \nu_{1} < \nu_{2}$ and $n\in\bf{N}$ , $n \geq 2$ . Then, for $q > 1$ and $\frac{1}{p}+\frac{1}{q} = 1$ , we have:

$\begin{multline} \left| \mathcal{L}_{n}^{n}(\nu_{1},\nu_{2}) -\mathcal{A}^{n}(\nu_{1},\nu_{2}) \right| \leq \frac{n(\nu_{2}-\nu_{1})}{4\sqrt[p]{p+1}} \Bigg\{\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left( \nu_{1}^{q(n-1)}, 3\nu_{2}^{q(n-1)}\right) \right]^{\frac{1}{q}} \\ +\left[ 2\mathcal{A}\left( \nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) -\frac{1}{2}\mathcal{A}\left(3\nu_{1}^{q(n-1)}, \nu_{2}^{q(n-1)}\right) \right]^{\frac{1}{q}}\Bigg\}, \end{multline}$

(3.7)

and

$\begin{multline} \left| \mathcal{L}^{-1}(\nu_{1},\nu_{2}) -\mathcal{A}^{-1}(\nu_{1},\nu_{2}) \right| \leq \frac{\sqrt[q]{2}(\nu_{2}-\nu_{1})}{4\sqrt[p]{p+1}} \Bigg\{\left[ \mathcal{H}^{-1}\left( \nu_{1}^{2q}, \nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(\nu_{1}^{2q}, 3\nu_{2}^{2q}\right) \right]^{\frac{1}{q}} \\ +\left[ \mathcal{H}^{-1}\left( \nu_{1}^{2q}, \nu_{2}^{2q}\right) -\frac{3}{4}\mathcal{H}^{-1}\left(3\nu_{1}^{2q}, \nu_{2}^{2q}\right) \right]^{\frac{1}{q}}\Bigg\}. \end{multline}$

(3.8)

Proof. By setting $x = \nu_{1}$ and $y = \nu_{2}$ in results of Proposition 3.4 and Proposition 3.5, one can obtain the Proposition 3.6.

4. Conclusions

As we emphasized in the introduction, integral inequality is the most important field of mathematical analysis and fractional calculus. By using the well-known Jensen-Mercer and power mean inequalities, we have proved new inequalities of Hermite-Hadamard-Mercer type involving Riemann-Liouville fractional operators. In the last section, we have considered some propositions in the context of special functions; these confirm the efficiency of our results.

Acknowledgements

We would like to express our special thanks to the editor and referees. Also, the first author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.

Conflict of interest

The authors declare no conflict of interest.

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