Novel Analysis of Hermite–Hadamard Type Integral Inequalities via Generalized Exponential Type m-Convex Functions

Abstract

The theory of convexity has a rich and paramount history and has been the interest of intense research for longer than a century in mathematics. It has not just fascinating and profound outcomes in different branches of engineering and mathematical sciences, it also has plenty of uses because of its geometrical interpretation and definition. It also provides numerical quadrature rules and tools for researchers to tackle and solve a wide class of related and unrelated problems. The main focus of this paper is to introduce and explore the concept of a new family of convex functions namely generalized exponential type $\mathfrak{m}$-convex functions. Further, to upgrade its numerical significance, we present some of its algebraic properties. Using the newly introduced definition, we investigate the novel version of Hermite–Hadamard type integral inequality. Furthermore, we establish some integral identities, and employing these identities, we present several new Hermite–Hadamard H–H type integral inequalities for generalized exponential type $\mathfrak{m}$-convex functions. These new results yield some generalizations of the prior results in the literature.

1. Introduction

The concept of convexity and its generalization is a rapidly growing area of mathematics, with numerous applications in machine (deep) learning, optimization, engineering, management, control theory, economics, and other disciplines. Nowadays, numerous mathematicians, economists, and physicists use this novel concept for their theoretical aspects as well. Mathematicians incorporate it to provide solution to problems that arise in different branches of sciences. This hypothesis gives us fascinating and amazing mathematical strategies to tackle and to take care of numerous issues that emerge in pure and applied sciences. During the last few decades, numerous scientists especially mathematicians have added to the advancement of this beautiful concept in different directions. It is especially important in the study of optimization problems, where it is distinguished by many convenient properties (for example, any minimum of a convex function is a global minimum, or the maximum is attained at a boundary point). This explains why there is a very rich theory of convex functions and convex sets. Optimization of convex functions has many practical applications (circuit design, controller design, modeling, etc.). Due to a lot of importance, it has gathered a lot of attention and has been the focus of several researchers. For the attention of the readers, we encourage the references [1,2,3,4,5,6].

It has a strong connection and also plays a significant role in the development of the theory of inequalities. Both convexity and inequality are closely related to each other. The concept of inequality is one of the main instruments in many parts of designing and science, for example, optimization engineering, probability theory, numerical analysis, finite element method, modeling, etc. At last, the hypothesis of inequalities might be viewed as an autonomous branch of mathematics. During the last twenty years, numerous mathematicians and researchers thought of their incredible commitments and considerations to investigate the Hermite–Hadamard inequality. In this manner, a significant amount of literature can be found for the researchers. For the attention of the readers, we encourage exploring references [7,8,9,10,11,12].

In the year 1929, Bernstein, introduced the term exponential convexity to the literature. After Bernstein, Widder [13] explored these functions as a subclass of convex functions in a given interval $(a,b)$. The sizeable and worthwhile research on big data analysis and extensive learning has recently increased the attentiveness in information theory involving exponentially convex functions. So especially in the last few decades, different mathematicians worked on the idea of exponential type convexity in different directions and contributed to the field of analysis. Due to the earlier works, these functions have proceeded as a remarkable and new class of convex functions, which have worthy benefits in technology, data science, information sciences, data mining, statistics, stochastic optimization, statistical learning, and sequential prediction. The reader may refer to [14,15,16,17,18,19] for the background of exponential convexity.

Motivated by the ongoing research activities, this paper aims to introduce a new class of exponential type convex function, called generalized exponential type $\mathfrak{m}$-convex function. We explore some algebraic properties and examples in the manner of newly introduced definitions. A new version of Hermite–Hadamard inequality and its refinements are investigated employing this new convexity.

2. Preliminaries

In this section we recall some known concepts.

In the year 1905, Jensen, presented the meaning of convex function, which reads as follows:

Definition 1

([1,20]). Let $\mathbb{X}$ be a convex subset of a real vector space $\mathbb{R}$ and let $\mathcal{G}:\mathbb{X}\to \mathbb{R}$ be a function. Then a function $\mathcal{G}$ is said to be convex, if

$$\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\tau +\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\le \tau \mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\left(1-\tau \right)\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right),$$

(1)

holds for all ${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$ and $\tau \in [0,1].$

Any paper on Hermite inequalities seems to be incomplete without mentioning the well-known Hermite–Hadamard inequality. This inequality states that, if $\mathcal{G}:\mathbb{X}\subseteq \mathbb{R}\to \mathbb{R}$ is convex in $\mathbb{X}$ for ${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$ and ${\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$, then

$$\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\le \frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\chi \right)d\chi \le \frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}.$$

(2)

The Hermite–Hadamard inequality plays an amazing and magnificent role in the literature. Several mathematicians have collaborated variant concepts in the subject of inequalities and its applications. Interested readers can refer to [21,22,23,24].

The family of $\mathfrak{m}$-convex functions was first time explored and introduced by G. Toader in [25].

Definition 2

([25]). A function $\mathcal{G}:\mathbb{X}=[0,\mathfrak{b}]\to \mathbb{R},$$\mathfrak{b}>0,$ is said to be $\mathfrak{m}$-convex, where $\mathfrak{m}\in [0,1],$ if

$$\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\le \tau \mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathfrak{m}\left(1-\tau \right)\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)$$

(3)

holds ∀${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$ and $\tau \in [0,1].$

If the above inequality (3) holds in the reversed sense, then $\mathcal{G}$ is said to be $\mathfrak{m}$-concave.

Definition 3

([26]). Let $\mathcal{G}$ be a nonnegative function. Then $\mathcal{G}:\mathbb{X}\to \mathbb{R},$ is said exponential type convex, if

$$\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\le \left({\mathsf{e}}^{\tau }-1\right)\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)$$

(4)

holds ∀${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$ and $\tau \in [0,1].$

Definition 4

([27]). A nonnegative real-valued function $\mathcal{G}:\mathbb{X}\to \mathbb{R}$ is known as an n-polynomial convex function, if

$$\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}[1-{(1-\tau )}^{\zeta }]\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}[1-{\tau }^{\zeta }]\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right),$$

(5)

holds for every ${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$, $\tau \in [0,1]$, and $\mathfrak{n}\in \mathbb{N}$.

Inspired by the above results and literature of inequality theory, we organize the paper as follow: In Section 3, we elaborate the concept and fundamental properties of the newly introduced definition, namely the generalized exponential type $\mathfrak{m}$-convex function. In Section 4, we deduce new generalization of Hermite–Hadamard type inequality in the manner of a newly introduced concept. Next, in Section 5, we establish some modifications and estimations of the Hermite–Hadamard type inequality in mode of newly introduced idea. Finally, in Section 6, we give a brief conclusion.

3. Algebraic Properties of Generalized Exponential Type $\mathfrak{m}$–Convex Functions

The principal focus of this section, we present our main definition of generalized exponential type $\mathfrak{m}$-convex function and its associated properties.

Definition 5.

Let $\mathcal{G}$ be a nonnegative function, then $\mathcal{G}:\mathbb{X}\to \mathbb{R},$ is said to be a generalized exponential type $\mathfrak{m}$-convex, if

$$\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)$$

(6)

holds ∀${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$, $\mathfrak{m}\in [0,1],$$\tau \in [0,1]$ and $\mathfrak{n}\in \mathbb{N}$.

Remark 1.

For $\mathfrak{n}=\mathfrak{m}=1,$ we attain exponential type convexity, explored by Kadakal and İşcan in [26].

Remark 2.

The range of the exponential type $\mathfrak{m}$-convex functions for $\mathfrak{m}\in [0,1]$ is $[0,+\infty ).$

Proof. 

Let ${\mathfrak{b}}_{\mathfrak{2}}\in \mathbb{X}$ be arbitrary. Using the definition of exponential type $\mathfrak{m}$-convex function, for $\tau =0$, we have

$$\begin{array}{cc}\hfill \mathcal{G}\left({\mathfrak{mb}}_{\mathfrak{2}}\right)& \le \mathfrak{m}[\mathsf{e}-1]\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ \hfill \Rightarrow 0& \le \mathfrak{m}\mathsf{e}\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)-2\mathfrak{m}\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ \hfill \Rightarrow 0& \le \mathfrak{m}(\mathsf{e}-2)\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ \hfill \Rightarrow 0& \le \mathfrak{m}\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right).\hfill \end{array}$$

□

We explore some relations between the class generalized exponential type $\mathfrak{m}$-convex functions and other class of generalized convex functions.

Lemma 1.

The following inequalities $({\mathsf{e}}^{\tau }-1)\ge \tau$ and $({\mathsf{e}}^{1-\tau }-1)\ge (1-\tau )$ hold, $\forall \tau \in [0,1]$.

Proof. 

The proof is clearly seen and hence omitted. □

Proposition 1.

If $\mathfrak{m}\in [0,1]$, then every nonnegative $\mathfrak{m}$-convex function is generalized exponential type $\mathfrak{m}$-convex function.

Proof. 

Since $\mathfrak{m}\in [0,1]$, and by using Lemma 1, we have

$$\begin{array}{cc}\hfill & \mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & \le \tau \mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathfrak{m}(1-\tau )\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right).\hfill \end{array}$$

□

Theorem 1.

Let $\mathcal{G},\mathsf{P}:[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}.$ If $\mathcal{G}$ and $\mathsf{P}$ are generalized exponential type $\mathfrak{m}$-convex functions for $\mathfrak{m}\in [0,1],$ then $\mathcal{G}+\mathsf{P}$ is generalized exponential type $\mathfrak{m}$-convex function.

Proof. 

Let $\mathcal{G}$ and $\mathsf{P}$ be generalized exponential type $\mathfrak{m}$-convex functions, then

$$\begin{array}{cc}\hfill & \left(\mathcal{G}+\mathsf{P}\right)\left[\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & =\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)+\mathsf{P}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ \hfill & +\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathsf{P}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathsf{P}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & =\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\left[\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathsf{P}\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right]+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\left[\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+\mathsf{P}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & =\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\left(\mathcal{G}+\mathsf{P}\right)\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\left(\mathcal{G}+\mathsf{P}\right)\left({\mathfrak{b}}_{\mathfrak{2}}\right).\hfill \end{array}$$

□

Theorem 2.

Let $\mathcal{G}:[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}.$ If $\mathcal{G}$ is generalized exponential type $\mathfrak{m}$-convex functions for $\mathfrak{m}\in [0,1],$ then nonnegative real number $c,$ $c\mathcal{G}$ is generalized exponential type $\mathfrak{m}$-convex function.

Proof. 

Let $\mathcal{G}$ be generalized exponential type $\mathfrak{m}$-convex, then

$$\begin{array}{cc}\hfill & \left(c\mathcal{G}\right)\left[\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & =c\left[\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & \le c\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & =\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }c\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }c\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & =\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\left(c\mathcal{G}\right)\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\left(c\mathcal{G}\right)\left({\mathfrak{b}}_{\mathfrak{2}}\right).\hfill \end{array}$$

□

Theorem 3.

Let $\mathcal{G}:[0,\mathfrak{b}]\to J$ be $\mathfrak{m}$-convex function for $\mathfrak{b}>0$ and $\mathfrak{m}\in [0,1]$ and $\mathcal{G}:\mathbb{X}\to \mathbb{R}$ is nondecreasing and generalized exponential type $\mathfrak{m}$-convex function. Then for the same fixed numbers $\mathfrak{m}\in [0,1],$ the function $\mathsf{P}\circ \mathcal{G}:[0,\mathfrak{b}]\to \mathbb{R}$ is generalized exponential type $\mathfrak{m}$-convex.

Proof. 

$\forall {\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in [0,\mathfrak{b}]$, $\mathfrak{m}\in [0,1]$, and $\tau \in [0,1],$ we have

$$\begin{array}{cc}\hfill & (\mathsf{P}\circ \mathcal{G})\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+m\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & =\mathsf{P}\left(\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\right)\hfill \\ & \le \mathsf{P}(\tau \mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathfrak{m}(1-\tau )\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right))\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\zeta \tau }-1\right)}^{\zeta }(\mathsf{P}\circ \mathcal{G})\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }(\mathsf{P}\circ \mathcal{G})\left({\mathfrak{b}}_{\mathfrak{2}}\right).\hfill \end{array}$$

□

Theorem 4.

Let ${\mathcal{G}}_i:[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ be a class of generalized exponential type $\mathfrak{m}$-convex functions for $\mathfrak{m}\in [0,1]$ and let $\mathcal{G}(♭)={sup}_i{\mathcal{G}}_i(♭).$ If $E=\{♭\in [{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]:\mathcal{G}(♭)<+\infty \}\ne \varnothing ,$ then E is an interval and $\mathcal{G}$ is generalized exponential type $\mathfrak{m}$-convex function on $E.$

Proof. 

For all ${\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\in E$, $\mathfrak{m}\in [0,1],$ and $\tau \in [0,1],$ we have

$$\begin{array}{cc}\hfill & \mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+m\left(1-\mathcal{G}\right){\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & =\underset{i}{sup}{\mathcal{G}}_i\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & \le \underset{i}{sup}\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\mathcal{G}}_i\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\mathcal{G}}_i\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right]\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\underset{i}{sup}{\mathcal{G}}_i\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\underset{i}{sup}{\mathcal{G}}_i\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & =\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)<+\infty .\hfill \end{array}$$

□

Theorem 5.

If the function $\mathcal{G}:[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is generalized exponential type $\mathfrak{m}$-convex for $\mathfrak{m}\in [0,1],$ then $\mathcal{G}$ is bounded on $[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{mb}}_{\mathfrak{2}}].$

Proof. 

Suppose $x\in [{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]$ be a point and $\mathfrak{m}\in [0,1]$ and $L=max\left\{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right),\mathfrak{m}\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right\}$ and then ∃$\tau \in [0,1]$ such that $x=\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}}.$ Thus, since ${\mathsf{e}}^{\tau }\le \mathsf{e}$ and ${\mathsf{e}}^{1-\tau }\le \mathsf{e}$, we have

$$\begin{array}{cc}\hfill & \mathcal{G}\left(x\right)=\mathcal{G}(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}})\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\hfill \\ & \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-1\right)}^{\zeta }L+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left(\mathsf{e}-1\right)}^{\zeta }L\hfill \\ & =L({\mathfrak{m}}^{\zeta }+1)\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{(\mathsf{e}-1)}^{\zeta }=M.\hfill \end{array}$$

□

4. New Generalization of H–H Type Inequality Using Generalized Exponential Type $\mathfrak{m}$-Convex Function

The subject of this section is to deduce new generalizations of H–H type integral inequality involving generalized exponential type $\mathfrak{m}$-convex function $\mathcal{G}$.

Theorem 6.

Let $\mathcal{G}:[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{mb}}_{\mathfrak{2}}]\to \mathbb{R}$ be generalized exponential type $\mathfrak{m}$-convex function for $\mathfrak{m}\in (0,1]$ and ${\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{mb}}_{\mathfrak{2}}.$ If $\mathcal{G}\in L_1[{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{mb}}_{\mathfrak{2}}],$ then

$$\begin{array}{c}\frac{1}{\frac{1}{\mathfrak{n}}{\displaystyle \sum _{\zeta =1}^n}{\left(\sqrt{\mathsf{e}}-1\right)}^{\zeta }}\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2}\right)\le \frac{1}{({\mathfrak{mb}}_{\mathfrak{2}}-{\mathfrak{b}}_{\mathfrak{1}})}\left\{{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{mb}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx+\mathfrak{m}{\int }_{\frac{{\mathfrak{b}}_{\mathfrak{1}}}{m}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx\right\}\hfill \\ \hfill \le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left[\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+{\mathfrak{m}}^{\zeta }\left(\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}}{{\mathfrak{m}}^{\zeta +1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right)\right]\end{array}$$

(7)

Proof. 

Let us denote,

$$a_1=\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}},a_2=(1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{1}}}{\mathfrak{m}}+\tau {\mathfrak{b}}_{\mathfrak{2}},\forall \tau \in [0,1].$$

Using the definition of exponential type $\mathfrak{m}$-convexity of $\mathcal{G}$, we have

$$\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2}\right)=\mathcal{G}\left(\frac{a_1+\mathfrak{m}{a}_2}{2}\right)$$

$$=\mathcal{G}\left(\frac{[\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}}]+[(1-\tau ){\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\tau {\mathfrak{b}}_{\mathfrak{2}}]}{2}\right)$$

$$\le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\sqrt{\mathsf{e}}-1\right)}^{\zeta }\left[\mathcal{G}(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}})+\mathcal{G}((1-\tau ){\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\tau {\mathfrak{b}}_{\mathfrak{2}})\right].$$

Now, integrating on both sides in the last inequality with respect to $\tau$ over $[0,1],$ we obtain

$$\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2}\right)\le \frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\sqrt{\mathsf{e}}-1\right)}^{\zeta }$$

$$\times \left[{\int }_0^1\mathcal{G}(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}})d\tau +{\int }_0^1\mathcal{G}\left((1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{1}}}{\mathfrak{m}}+\tau {\mathfrak{b}}_{\mathfrak{2}}\right)d\tau \right]$$

$$=\frac{\frac{1}{\mathfrak{n}}{\sum }_{\zeta =1}^{\mathfrak{n}}{\left(\sqrt{\mathsf{e}}-1\right)}^{\zeta }}{({\mathfrak{mb}}_{\mathfrak{2}}-{\mathfrak{b}}_{\mathfrak{1}})}\left\{{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{mb}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx+\mathfrak{m}{\int }_{\frac{{\mathfrak{b}}_{\mathfrak{1}}}{\mathfrak{m}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx\right\}.$$

This completes the left-side inequality. For the right-side inequality, using exponential type $\mathfrak{m}$-convexity of $\mathcal{G}$, we obtain

$$\frac{1}{({\mathfrak{mb}}_{\mathfrak{2}}-{\mathfrak{b}}_{\mathfrak{1}})}\left\{{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{mb}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx+\mathfrak{m}{\int }_{\frac{{\mathfrak{b}}_{\mathfrak{1}}}{\mathfrak{m}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(x\right)dx\right\}$$

$$={\int }_0^1\mathcal{G}(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau ){\mathfrak{b}}_{\mathfrak{2}})d\tau +{\int }_0^1\mathcal{G}\left((1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{1}}}{\mathfrak{m}}+\tau {\mathfrak{b}}_{\mathfrak{2}}\right)d\tau$$

$$\le {\int }_0^1\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right]d\tau$$

$$+{\int }_0^1\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}}{{\mathfrak{m}}^2}\right)\right]d\tau$$

$$=\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left[\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+{\mathfrak{m}}^{\zeta }\left(\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}}{{\mathfrak{m}}^{\zeta +1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right)\right].$$

The proof is completed. □

Corollary 1.

If $\mathfrak{n}=\mathfrak{m}=1$ in Theorem 6, we obtain (Theorem 3.1, [26]).

5. Refinements of H–H Type Inequality via Generalized Exponential Type $\mathfrak{m}$-Convex Function

Let us establish some refinements of the H–H inequality for functions whose first derivative in absolute value at certain power is generalized exponential type $\mathfrak{m}$-convex. First we need some new useful lemmas.

Lemma 2.

Let $0<k\le 1$ and a mapping $\mathcal{G}:\left[{\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right]\to \mathbb{R}$ is differentiable on $\left({\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $\mathfrak{m}\in [0,1]$. If ${\mathcal{G}}^{\prime }\in L_1\left[{\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right],$ then

$$\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right)$$

(8)

$$\times {\int }_0^1\left(1-2\tau \right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau .$$

Proof. 

Using the integrating by parts, we have

$$\begin{array}{cc}\hfill & \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\int }_0^1\left(1-2\tau \right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \hfill \\ \hfill & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right)\left\{\frac{\left(1-2\tau \right)\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{{\mathfrak{b}}_{\mathfrak{1}}-\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}{|}_0^1-{\int }_0^1\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{{\mathfrak{b}}_{\mathfrak{1}}-\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\left(-2\right)d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right)\left\{\frac{-\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)-\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{\frac{k{\mathfrak{b}}_{\mathfrak{1}}-{\mathfrak{mb}}_{\mathfrak{2}}}{k}}+\frac{2k}{k{\mathfrak{b}}_{\mathfrak{1}}-{\mathfrak{mb}}_{\mathfrak{2}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right)\left\{\frac{k\left(\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)\right)}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}-\frac{2k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \right\}\hfill \\ & =\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(d\right)d\tau .\hfill \end{array}$$

This completes the proof. □

Lemma 3.

Let $0<k\le 1$ and $\mathcal{G}:[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\left(k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right)$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $\mathfrak{m}\in [0,1]$. If ${\mathcal{G}}^{\prime }\in L_1\left[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right],$ then

$$\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right)$$

(9)

$$\times {\int }_0^1\left(2\tau -1\right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau .$$

Proof. 

Using the integrating by parts, we have

$$\begin{array}{cc}\hfill & \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\int }_0^1\left(2\tau -1\right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\hfill \\ \hfill & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right)\hfill \\ \hfill & \times \left\{\frac{\left(2\tau -1\right)\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{|}_0^1-{\int }_0^1\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}\left(2\right)d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right)\left\{\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}-\frac{2}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right)\left\{\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)+\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}-\frac{2}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \right\}\hfill \\ & =\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau ,\hfill \end{array}$$

which completes the proof. □

Lemma 4.

Let $0<k\le 1$ and a function $\mathcal{G}:[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\left(k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right)$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $\mathfrak{m}\in [0,1]$. If ${\mathcal{G}}^{\prime }\in L_1\left[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right],$ then

$$\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)$$

(10)

$$\times {\int }_0^1\left(2\tau -1\right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau .$$

Proof. 

Applying the by parts integration, we have

$$\begin{array}{cc}\hfill & \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\int }_0^1\left(2\tau -1\right){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\hfill \\ \hfill & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)\hfill \\ \hfill & \times \left\{\frac{\left(2\tau -1\right)\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{|}_0^1-2{\int }_0^1\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)\left\{\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}-\frac{2}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \right\}\hfill \\ & =\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{k+1}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \hfill \\ & =\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau .\hfill \end{array}$$

□

Lemma 5.

Let $0<k\le 1$ and a function $\mathcal{G}:\left[{\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right]\to \mathbb{R}$ is differentiable on $\left({\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $\mathfrak{m}\in [0,1]$. If ${\mathcal{G}}^{\prime }\in L_1\left[{\mathfrak{b}}_{\mathfrak{1}},\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right],$ then

$$\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)=\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)$$

(11)

$$\times \left\{{\int }_0^1\tau {\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau -{\int }_{\frac{1}{2}}^1{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \right\}.$$

Proof. 

Applying the by parts integration, we have

$$\begin{array}{cc}\hfill & \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\left\{{\int }_0^1\tau {\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau -{\int }_{\frac{1}{2}}^1{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \right\}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ \hfill & \times \left\{\frac{\tau \mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{{\mathfrak{b}}_{\mathfrak{1}}-\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}{|}_0^1-{\int }_0^1\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{{\mathfrak{b}}_{\mathfrak{1}}-\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}d\tau -\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{{\mathfrak{b}}_{\mathfrak{1}}-\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}{|}_{\frac{1}{2}}^1\right\}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\{\frac{k\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)}{k{\mathfrak{b}}_{\mathfrak{1}}-{\mathfrak{mb}}_{\mathfrak{2}}}-\frac{k}{k{\mathfrak{b}}_{\mathfrak{1}}-{\mathfrak{mb}}_{\mathfrak{2}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}(1-\tau )\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)d\tau \hfill \\ \hfill & -\frac{k}{k{\mathfrak{b}}_{\mathfrak{1}}-{\mathfrak{mb}}_{\mathfrak{2}}}\left(\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)-\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)\right)\}\hfill \\ & =\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right).\hfill \end{array}$$

□

Lemma 6.

Let $0<k\le 1$ and a function $\mathcal{G}:[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\left(k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right)$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $\mathfrak{m}\in [0,1]$. If ${\mathcal{G}}^{\prime }\in L_1\left[k{\mathfrak{b}}_{\mathfrak{1}},{\mathfrak{b}}_{\mathfrak{2}}\right],$ then

$$\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)=\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)$$

(12)

$$\times \left\{{\int }_0^1(-\tau ){\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau +{\int }_{\frac{1}{2}}^1{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \right\}.$$

Proof. 

Applying the by parts integration, we have

$$\begin{array}{cc}\hfill & \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\left\{{\int }_0^1-\tau {\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau +{\int }_{\frac{1}{2}}^1{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau \right\}\hfill \\ & =\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{\frac{-\tau \mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{|}_0^1-{\int }_0^1\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}\left(-1\right)d\tau \hfill \\ & +\frac{\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{|}_{\frac{1}{2}}^1\}\hfill \\ & =\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\hfill \\ \hfill & \times \left\{\frac{-\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}+\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_0^1\mathcal{G}\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)d\tau +\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)-\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}\right\}\hfill \\ & =\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right).\hfill \end{array}$$

□

Theorem 7.

Let $\mathcal{G}:\mathbb{X}=(0,\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}]\to \mathbb{R}$ be differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q>1$ and $\frac{1}{q}+\frac{1}{p}=1,$ then for $m\in [0,1],$ the following inequality holds:

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(13)

Proof. 

From Lemma 2, Hölder’s inequality and generalized exponential type $\mathfrak{m}$-convexity of ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left({\int }_0^1{\left|1-2\tau \right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left\{{\int }_0^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left({\int }_0^1{\left|1-2\tau \right|}^{p}d\tau \right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{{\int }_0^1\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right]d\tau \right\}}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

This completes the proof. □

Remark 3.

If $n=m=1$ in Theorem 7, we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{\left(\mathsf{e}-2\right)\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 8.

Let $\mathcal{G}:\mathbb{X}=(0,\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}]\to \mathbb{R}$ be differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q\ge 1,$ and $m\in [0,1],$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(14)

Proof. 

From Lemma 2, power mean inequality and generalized exponential type $\mathfrak{m}$-convexity of ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right)\left\{{\int }_0^1\left|1-2\tau \right|\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|d\tau \right\}\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left({\int }_0^1\left|1-2\tau \right|d\tau \right)}^{1-\frac{1}{q}}{\left\{{\int }_0^1\left|1-2\tau \right|{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left({\int }_0^1\left|1-2\tau \right|d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{{\int }_0^1\left|1-2\tau \right|\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right]d\tau \right\}}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{{\left(\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

This completes the proof. □

Remark 4.

If $\mathfrak{n}=\mathfrak{m}=1$ in Theorem 8, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)}{2}-\frac{k}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2k}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 9.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q>1$ and $\frac{1}{q}+\frac{1}{p}=1,$ then for $m\in [0,1]$, we have:

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(15)

Proof. 

From Lemma 3, Hölder’s inequality and generalized exponential type $\mathfrak{m}$-convexity of ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left({\int }_0^1{\left|2\tau -1\right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left\{{\int }_0^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left({\int }_0^1{\left|2\tau -1\right|}^{p}d\tau \right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{{\int }_0^1\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^{{}^q}+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}\right]d\tau \right\}}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

This completes the proof. □

Remark 5.

If $n=m=1$ in Theorem 9, we obtain

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{\left(\mathsf{e}-2\right)\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 10.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q\ge 1,$ and $m\in [0,1],$ then we have:

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(16)

Proof. 

From Lemma 3, power mean inequality and generalized exponential type $\mathfrak{m}$-convexity of ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right)\left\{{\int }_0^1\left|2\tau -1\right|\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|d\tau \right\}\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left({\int }_0^1\left|2\tau -1\right|d\tau \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left\{{\int }_0^1\left|2\tau -1\right|{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left({\int }_0^1\left|2\tau -1\right|d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left[{\int }_0^1\left|2\tau -1\right|\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^{{}^q}\right\}d\tau \right]}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

This completes the proof. □

Remark 6.

If $n=m=1$ in Theorem 10, we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{2}-\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{2}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 11.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q>1$ and $q^{-1}+p^{-1}=1,$ then for $m\in [0,1]$, we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(17)

Proof. 

Using Lemma 4, property of modulus, Hölder’s inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left({\int }_0^1{\left|\left(2\tau -1\right)\right|}^{p}d\tau \right)}^{\frac{1}{p}}{\left({\int }_0^1|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right){|}^{q}d\tau \right)}^{\frac{1}{q}}.\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left({\int }_0^1{\left|2\tau -1\right|}^{p}d\tau \right)}^{\frac{1}{p}}\hfill \\ & \times {\left\{{\int }_0^1\left[\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^{{}^q}\right]d\tau \right\}}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left({\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 7.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\left(\mathsf{e}-2\right)\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 12.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q\ge 1,$ then for $\mathfrak{m}\in [0,1]$, we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)\hfill \\ & \times {\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}{\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(18)

Proof. 

Using Lemma 4, property of modulus, power mean inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)\left[{\int }_0^1\left|2\tau -1\right|\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|d\tau \right]\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left({\int }_0^1\left|2\tau -1\right|d\tau \right)}^{1-\frac{1}{q}}{\left\{{\int }_0^1\left|2\tau -1\right|{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ & \le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left({\int }_0^1\left|2\tau -1\right|d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left[{\int }_0^1\left|2\tau -1\right|\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\mathfrak{m}}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^{{}^q}\right\}d\tau \right]}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right){\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\right)}^{\zeta }\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 8.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{\mathcal{G}\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)+\mathcal{G}\left({\mathfrak{b}}_{\mathfrak{2}}\right)}{k+1}-\frac{2}{\left(k+1\right)\left({\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}\right)}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau \right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k+1}\right)\hfill \\ & \times {\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}{\left\{\frac{8\sqrt{\mathsf{e}}-2\mathsf{e}-7}{2}\left({\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

Theorem 13.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q>1$ and $q^{-1}+p^{-1}=1,$ then for $\mathfrak{m}\in [0,1]$, we have

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ & \times [{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

(19)

Proof. 

Using Lemma 5, property of modulus, Hölder’s inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ & \times [{\left({\int }_0^1{\tau }^{p}d\tau \right)}^{\frac{1}{p}}{\left\{{\int }_0^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left({\int }_{\frac{1}{2}}^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}]\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)[{\left({\int }_0^1{\tau }^{p}d\tau \right)}^{\frac{1}{p}}\hfill \\ & \times {\left({\int }_0^1\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\sum _{s=1}^n{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right\}d\tau \right)}^{\frac{1}{q}}\hfill \\ & +{\left({\int }_{\frac{1}{2}}^1\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right\}d\tau \right)}^{\frac{1}{q}}]\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ & \times [{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 9.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ & \times [{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\left(\mathsf{e}-2\right)\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{2\sqrt{\mathsf{e}}-3}{2}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^q\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Theorem 14.

Let $\mathcal{G}:\mathbb{X}=(0,{\mathfrak{b}}_{\mathfrak{2}}]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function $\mathbb{X}$ for $q\ge 1,$ then for $\mathfrak{m}\in [0,1]$, we have

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)[{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-5}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

(20)

Proof. 

Using Lemma 5, property of modulus, power mean inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{mb}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{mb}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\hfill \\ & \times \left\{{\int }_0^1\tau \left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|d\tau +{\int }_{\frac{1}{2}}^1\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|d\tau \right\}\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)\{{\left({\int }_0^1\tau d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left({\int }_0^1\tau {\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}+{\left({\int }_{\frac{1}{2}}^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{1}}+\mathfrak{m}\left(1-\tau \right)\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\}\hfill \\ & \le \left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)[{\left({\int }_0^1\tau d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left({\int }_0^1\tau \left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right\}d\tau \right)}^{\frac{1}{q}}\hfill \\ & +{\left({\int }_{\frac{1}{2}}^1\left\{\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^s{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{\left(1-\tau \right)}-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right\}d\tau \right)}^{\frac{1}{q}}\hfill \\ & =\left(\frac{{\mathfrak{mb}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)[{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-5}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 10.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{k}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{{\mathfrak{b}}_{\mathfrak{1}}}^{\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2k}\right)\right|\le \left(\frac{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{k}\right)[{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left(\frac{-1}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{2\mathsf{e}-5}{2}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q+\frac{2\sqrt{\mathsf{e}}-3}{2}{\left|{\mathcal{G}}^{\prime }\left(\frac{{\mathfrak{b}}_{\mathfrak{2}}}{k}\right)\right|}^{{}^q}\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Theorem 15.

Let $\mathcal{G}:\mathbb{X}=\left(0,{\mathfrak{b}}_{\mathfrak{2}}\right]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q>1$ and $q^{-1}+p^{-1}=1,$ then for $\mathfrak{m}\in [0,1]$ and we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\hfill \\ \hfill & \le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}[{\left\{\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

(21)

Proof. 

Using Lemma 6, property of modulus, Hölder’s inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\hfill \\ & \times \{{\left({\int }_0^1{\tau }^{p}d\tau \right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\hfill \\ \hfill & +{\left({\int }_{\frac{1}{2}}^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\}\hfill \\ & \le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left({\int }_0^1{\tau }^{p}d\tau \right)}^{\frac{1}{p}}\hfill \\ & \times {\left({\int }_0^1\left(\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{1-\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}\right)d\tau \right)}^{\frac{1}{q}}\hfill \\ & +{\left({\int }_{\frac{1}{2}}^1\left(\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{1-\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}\right)d\tau \right)}^{\frac{1}{q}}\}\hfill \\ & =\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)[{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left\{\sum _{\zeta =1}^{\mathfrak{n}}{\left(\mathsf{e}-2\right)}^{\zeta }\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)\right\}}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }+\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 11.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\hfill \\ & \le \left({\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}\right)[{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\left(\mathsf{e}-2\right)\left({\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}].\hfill \end{array}$$

Theorem 16.

Let $\mathcal{G}:\mathbb{X}=\left(0,{\mathfrak{b}}_{\mathfrak{2}}\right]\to \mathbb{R}$ is differentiable on $\mathbb{X}$ with $0<{\mathfrak{b}}_{\mathfrak{1}}<{\mathfrak{b}}_{\mathfrak{2}}$ and $0<k\le 1$. If ${\left|{\mathcal{G}}^{\prime }\right|}^q$ is generalized exponential type $\mathfrak{m}$-convex function on $\mathbb{X}$ for $q\ge 1,$ then for $\mathfrak{m}\in [0,1]$, we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left(\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-5}{2}\right)}^{\zeta }\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{1}{2}\right)}^{\zeta }\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\}.\hfill \end{array}$$

(22)

Proof. 

Using Lemma 6, property of modulus, power mean inequality and generalized exponential type $\mathfrak{m}$-convex function ${\left|{\mathcal{G}}^{\prime }\right|}^q,$ we have

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\le \left({\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}\right)\hfill \\ & \times \left\{{\int }_0^1\left|-\tau \right|\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|d\tau +{\int }_{\frac{1}{2}}^1\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|d\tau \right\}\hfill \\ & \le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left({\int }_0^1\tau d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left({\int }_0^1\tau {\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}+{\left({\int }_{\frac{1}{2}}^1{\left|{\mathcal{G}}^{\prime }\left(\tau {\mathfrak{b}}_{\mathfrak{2}}+\mathfrak{m}k\left(1-\tau \right){\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{q}d\tau \right)}^{\frac{1}{q}}\}\hfill \\ & \le \left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left({\int }_0^1\tau d\tau \right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left({\int }_0^1\tau \left(\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{1-\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}\right)d\tau \right)}^{\frac{1}{q}}\hfill \\ & +{\left({\int }_{\frac{1}{2}}^1\left(\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\left({\mathsf{e}}^{\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{1}{\mathfrak{n}}\sum _{\zeta =1}^{\mathfrak{n}}{\mathfrak{m}}^{\zeta }{\left({\mathsf{e}}^{1-\tau }-1\right)}^{\zeta }{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^{{}^q}\right)d\tau \right)}^{\frac{1}{q}}\}\hfill \\ & =\left({\mathfrak{b}}_{\mathfrak{2}}-\mathfrak{m}k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}\hfill \\ & \times {\left(\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-5}{2}\right)}^{\zeta }\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{1}{2}\right)}^{\zeta }\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}\right)}^{\zeta }\frac{1}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\sum _{\zeta =1}^{\mathfrak{n}}{\left(\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}\right)}^{\zeta }\frac{{\mathfrak{m}}^{\zeta }}{\mathfrak{n}}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\}.\hfill \end{array}$$

Consequently, the proof is completed. □

Remark 12.

If we put $\mathfrak{m}=\mathfrak{n}=1$ in above theorem, then

$$\begin{array}{cc}\hfill & \left|\frac{1}{{\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}}{\int }_{k{\mathfrak{b}}_{\mathfrak{1}}}^{{\mathfrak{b}}_{\mathfrak{2}}}\mathcal{G}\left(\tau \right)d\tau -\mathcal{G}\left(\frac{k{\mathfrak{b}}_{\mathfrak{1}}+{\mathfrak{b}}_{\mathfrak{2}}}{2}\right)\right|\hfill \\ & \le \left({\mathfrak{b}}_{\mathfrak{2}}-k{\mathfrak{b}}_{\mathfrak{1}}\right)\{{\left(\frac{1}{2}\right)}^{1-\frac{1}{q}}{\left(\frac{2\mathsf{e}-5}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{1}{2}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\hfill \\ & +{\left(\frac{2\sqrt{\mathsf{e}}-3}{2}{\left|{\mathcal{G}}^{\prime }\left({\mathfrak{b}}_{\mathfrak{2}}\right)\right|}^q+\frac{2\mathsf{e}-2\sqrt{\mathsf{e}}-1}{2}{\left|{\mathcal{G}}^{\prime }\left(k{\mathfrak{b}}_{\mathfrak{1}}\right)\right|}^q\right)}^{\frac{1}{q}}\}.\hfill \end{array}$$

6. Conclusions

In this paper, the authors have introduced a new concept namely generalized exponential type $\mathfrak{m}$-convex function. A new evaluation of H–H type inequality for the new class of convexity is introduced. Further, the authors have introduced five new integral identities concerning $\mathfrak{m}$-convexity, then employing these identities some H–H type integral inequalities for generalized exponential type $\mathfrak{m}$-convex functions are presented. Special cases of our established results are also presented for some particular values of parameters of $\mathfrak{m}$ and $\mathfrak{n}$. Since convex functions have enormous applications in numerous mathematical fields, we trust that our results and techniques can be applied in convex analysis, deep learning, quantum calculus, fractional calculus, economics, etc.