# Integral inequalities of Hermite-Hadamard type for functions whose derivatives are α-preinvex

## Abstract

In the article, the authors introduce a new notion, ‘α-preinvex function’, establish an integral identity for the newly introduced function, and find some Hermite-Hadamard type integral inequalities for a function of which the power of the absolute value of the first derivative is α-preinvex.

MSC: 26D15, 26A51, 26B12, 41A55.

## 1 Introduction

Let us recall some definitions of various convex functions.

Definition 1 A function $f:I\subseteq \mathbb{R}=\left(-\mathrm{\infty },\mathrm{\infty }\right)\to \mathbb{R}$ is said to be convex if

$f\left(\lambda x+\left(1-\lambda \right)y\right)\le \lambda f\left(x\right)+\left(1-\lambda \right)f\left(y\right)$
(1)

holds for all $x,y\in I$ and $\lambda \in \left[0,1\right]$.

Definition 2 ()

For $f:\left[0,b\right]\to \mathbb{R}$ and $m\in \left(0,1\right]$, if

$f\left(tx+m\left(1-t\right)y\right)\le tf\left(x\right)+m\left(1-t\right)f\left(y\right)$
(2)

is valid for all $x,y\in \left[0,b\right]$ and $t\in \left[0,1\right]$, then we say that $f\left(x\right)$ is an m-convex function on $\left[0,b\right]$.

Definition 3 ()

For $f:\left[0,b\right]\to \mathbb{R}$ and $\left(\alpha ,m\right)\in \left(0,1\right]×\left(0,1\right]$, if

$f\left(tx+m\left(1-t\right)y\right)\le {t}^{\alpha }f\left(x\right)+m\left(1-{t}^{\alpha }\right)f\left(y\right)$
(3)

is valid for all $x,y\in \left[0,b\right]$ and $t\in \left[0,1\right]$, then we say that $f\left(x\right)$ is an $\left(\alpha ,m\right)$-convex function on $\left[0,b\right]$.

Definition 4 ()

A set $S\subseteq {\mathbb{R}}^{n}$ is said to be invex with respect to the map $\eta :S×S\to {\mathbb{R}}^{n}$ if for every $x,y\in S$ and $t\in \left[0,1\right]$

$y+t\eta \left(x,y\right)\in S.$
(4)

Definition 5 ()

Let $S\subseteq {\mathbb{R}}^{n}$ be an invex set with respect to $\eta :S×S\to {\mathbb{R}}^{n}$. For every $x,y\in S$, the η-path ${P}_{xv}$ joining the points x and $v=x+\eta \left(y,x\right)$ is defined by

${P}_{xv}=\left\{z\mid z=x+t\eta \left(y,x\right),t\in \left[0,1\right]\right\}.$
(5)

Definition 6 ()

Let $S\subseteq {\mathbb{R}}^{n}$ be an invex set with respect to $\eta :S×S\to {\mathbb{R}}^{n}$. A function $f:S\to \mathbb{R}$ is said to be preinvex with respect to η, if for every $x,y\in S$ and $t\in \left[0,1\right]$,

$f\left(y+t\eta \left(x,y\right)\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right).$
(6)

Let us reformulate some inequalities of Hermite-Hadamard type for the above mentioned convex functions.

Theorem 1 ([, Theorem 2.2])

Let $f:{I}^{\circ }\subseteq \mathbb{R}\to \mathbb{R}$ be a differentiable mapping and $a,b\in {I}^{\circ }$ with $a. If $|{f}^{\prime }|$ is convex on $\left[a,b\right]$, then

$|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\le \frac{b-a}{8}\left(|{f}^{\prime }\left(a\right)|+|{f}^{\prime }\left(b\right)|\right).$
(7)

Theorem 2 ([8, 9])

Let $f:{\mathbb{R}}_{0}=\left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be m-convex and $m\in \left(0,1\right]$. If $f\in L\left(\left[a,b\right]\right)$ for $0\le a, then

$\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le min\left\{\frac{f\left(a\right)+mf\left(b/m\right)}{2},\frac{mf\left(a/m\right)+f\left(b\right)}{2}\right\}.$
(8)

Theorem 3 ([, Theorem 3.1])

Let $I\supset {\mathbb{R}}_{0}$ be an open real interval and let $f:I\to \mathbb{R}$ be a differentiable function on I such that ${f}^{\prime }\in L\left(\left[a,b\right]\right)$ for $0\le a. If ${|{f}^{\prime }|}^{q}$ is $\left(\alpha ,m\right)$-convex on $\left[a,b\right]$ for some given numbers $m,\alpha \in \left(0,1\right]$ and $q\ge 1$, then

$\begin{array}{c}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{b-a}{2}{\left(\frac{1}{2}\right)}^{1-1/q}min\left\{{\left[{v}_{1}|{f}^{\prime }\left(a\right){|}^{q}+{v}_{2}m{|{f}^{\prime }\left(\frac{b}{m}\right)|}^{q}\right]}^{1/q},\hfill \\ \phantom{\rule{2em}{0ex}}{\left[{v}_{2}m{|{f}^{\prime }\left(\frac{a}{m}\right)|}^{q}+{v}_{1}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\},\hfill \end{array}$

where

${v}_{1}=\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right)}\left(\alpha +\frac{1}{{2}^{\alpha }}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{v}_{2}=\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right)}\left(\frac{{\alpha }^{2}+\alpha +2}{2}-\frac{1}{{2}^{\alpha }}\right).$

Theorem 4 ([, Theorem 2.1])

Let $A\subseteq \mathbb{R}$ be an open invex subset with respect to $\theta :A×A\to \mathbb{R}$ and let $f:A\to \mathbb{R}$ be a differentiable function. If $|{f}^{\prime }|$ is preinvex on A, then for every $a,b\in A$ with $\theta \left(a,b\right)\ne 0$ we have

$|\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\le \frac{|\theta \left(a,b\right)|}{8}\left[|{f}^{\prime }\left(a\right)|+|{f}^{\prime }\left(b\right)|\right].$
(9)

For more information on Hermite-Hadamard type inequalities for various convex functions, please refer to recently published articles  and closely related references therein.

In this article, we will introduce a new notion ‘α-preinvex function’, establish an integral identity for such a kind of functions, and find some Hermite-Hadamard type integral inequalities for a function that the power of the absolute value of its first derivative is α-preinvex.

## 2 A new definition and a lemma

The so-called ‘α-preinvex function’ may be introduced as follows.

Definition 7 Let $S\subseteq {\mathbb{R}}^{n}$ be an invex set with respect to $\eta :S×S\to {\mathbb{R}}^{n}$. A function $f:S\to \mathbb{R}$ is said to be α-preinvex with respect to η for $\alpha \in \left(0,1\right]$, if for every $x,y\in S$ and $t\in \left[0,1\right]$,

$f\left(y+t\eta \left(x,y\right)\right)\le {t}^{\alpha }f\left(x\right)+\left(1-{t}^{\alpha }\right)f\left(y\right).$
(10)

Remark 1 If $\alpha =1$ and $f\left(x\right)$ is an α-preinvex function, then $f\left(x\right)$ is a preinvex function.

For establishing our new integral inequalities of Hermite-Hadamard type for α-preinvex functions, we need the following integral identity.

Lemma 1 Let $A\subseteq \mathbb{R}$ be an open invex subset with respect to $\theta :A×A\to \mathbb{R}$ and let $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. If $f:A\to \mathbb{R}$ is a differentiable function and ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$, then

$\begin{array}{c}\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{\theta \left(a,b\right)}{4}{\int }_{0}^{1}\left(\frac{1}{2}-t\right)\left[{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)+{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t.\hfill \end{array}$

Proof Since $a,b\in A$ and A is an invex set with respect to θ, for every $t\in \left[0,1\right]$, we have $b+t\theta \left(a,b\right)\in A$. Integrating by parts gives

$\begin{array}{c}{\int }_{0}^{1}\left(\frac{1}{2}-t\right)\left[{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)+{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{2}{\theta \left(a,b\right)}\left[\left(\frac{1}{2}-t\right)f\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right){|}_{0}^{1}+{\int }_{0}^{1}f\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{2em}{0ex}}+\left(\frac{1}{2}-t\right)f\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right){|}_{0}^{1}+{\int }_{0}^{1}f\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{2}{\theta \left(a,b\right)}\left[\frac{1}{2}f\left(b\right)+\frac{1}{2}f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)-{\int }_{0}^{1}f\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{2}{\theta \left(a,b\right)}\left[\frac{1}{2}f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)+\frac{1}{2}f\left(b+\theta \left(a,b\right)\right)-{\int }_{0}^{1}f\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{2}{\theta \left(a,b\right)}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{4}{{\theta }^{2}\left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.\hfill \end{array}$

The proof of Lemma 1 is completed. □

## 3 Some new integral inequalities of Hermite-Hadamard type

We are now in a position to establish some Hermite-Hadamard type integral inequalities for a function that the power of the absolute value of its first derivative is α-preinvex.

Theorem 5 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A×A\to \mathbb{R}$ and $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$, and $\alpha \in \left(0,1\right]$. If ${|{f}^{\prime }|}^{q}$ is α-preinvex on A for $q\ge 1$, then

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{16}{\left[\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right){2}^{2\alpha }}\right]}^{1/q}\left\{\left[2\left(1+\alpha {2}^{\alpha }\right)|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}{+\left({2}^{2\alpha +1}-2+\alpha \left(3×{2}^{\alpha }-2\right){2}^{\alpha }+{\alpha }^{2}{2}^{2\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+\left[\left(\alpha \left({2}^{\alpha +1}-1\right){2}^{\alpha +1}+2×{3}^{\alpha +2}-\left({2}^{\alpha }+1\right){2}^{\alpha +3}\right)|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}{+\left({\alpha }^{2}{2}^{2\alpha }+\left(\alpha \left(1-{2}^{\alpha -1}\right)+4+5×{2}^{\alpha }\right){2}^{\alpha +1}-2×{3}^{\alpha +2}\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$

Proof Since $b+t\theta \left(a,b\right)\in A$ for every $t\in \left[0,1\right]$, by Lemma 1 and Hölder’s inequality, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\int }_{0}^{1}|\frac{1}{2}-t|\left[|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|+|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}|\frac{1}{2}-t|\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\left\{{\left[{\int }_{0}^{1}|\frac{1}{2}-t|{|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}|\frac{1}{2}-t|{|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}.\hfill \end{array}$
(11)

Using the α-preinvexity of ${|{f}^{\prime }|}^{q}$, we have

$\begin{array}{c}{\int }_{0}^{1}|\frac{1}{2}-t|{|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{1}|\frac{1}{2}-t|\left[{\left(\frac{1-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{1-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right){2}^{2\left(\alpha +1\right)}}\left[2\left(1+\alpha {2}^{\alpha }\right)|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}+\left({2}^{2\alpha +1}-2+\alpha \left(3×{2}^{\alpha }-2\right){2}^{\alpha }+{\alpha }^{2}{2}^{2\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]\hfill \end{array}$

and

$\begin{array}{c}{\int }_{0}^{1}|\frac{1}{2}-t|{|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{1}|\frac{1}{2}-t|\left[{\left(\frac{2-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{2-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right){2}^{2\left(\alpha +1\right)}}\left[\left(\alpha \left({2}^{\alpha +1}-1\right){2}^{\alpha +1}+2×{3}^{\alpha +2}-\left({2}^{\alpha }+1\right){2}^{\alpha +3}\right)\hfill \\ \phantom{\rule{2em}{0ex}}×|{f}^{\prime }\left(a\right){|}^{q}+\left({\alpha }^{2}{2}^{2\alpha }+\left(\alpha \left(1-{2}^{\alpha -1}\right)+4+5×{2}^{\alpha }\right){2}^{\alpha +1}-2×{3}^{\alpha +2}\right)|{f}^{\prime }\left(b\right){|}^{q}\right].\hfill \end{array}$

Substituting the above two inequalities into (11) yields

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}|\frac{1}{2}-t|\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\left\{\left[{\int }_{0}^{1}|\frac{1}{2}-t|\left({\left(\frac{1-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}{+\left(1-{\left(\frac{1-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}|\frac{1}{2}-t|\left({\left(\frac{2-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{2-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{|\theta \left(a,b\right)|}{4}{\left(\frac{1}{4}\right)}^{1-1/q}{\left[\frac{1}{\left(\alpha +1\right)\left(\alpha +2\right){2}^{2\left(\alpha +1\right)}}\right]}^{1/q}\left\{\left[2\left(1+\alpha {2}^{\alpha }\right)|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}{+\left({2}^{2\alpha +1}-2+\alpha \left(3×{2}^{\alpha }-2\right){2}^{\alpha }+{\alpha }^{2}{2}^{2\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+\left[\left(\alpha \left({2}^{\alpha +1}-1\right){2}^{\alpha +1}+2×{3}^{\alpha +2}-\left({2}^{\alpha }+1\right){2}^{\alpha +3}\right)|{f}^{\prime }\left(a\right){|}^{q}\hfill \\ \phantom{\rule{2em}{0ex}}{+\left({\alpha }^{2}{2}^{2\alpha }+\left(\alpha \left(1-{2}^{\alpha -1}\right)+4+5×{2}^{\alpha }\right){2}^{\alpha +1}-2×{3}^{\alpha +2}\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$

The proof of Theorem 5 is completed. □

Corollary 1 Under the conditions of Theorem  5, if $\alpha =1$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{16}\left\{{\left[\frac{|{f}^{\prime }\left(a\right){|}^{q}+3|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}+{\left[\frac{3|{f}^{\prime }\left(a\right){|}^{q}+|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}\right\}.\hfill \end{array}$

Theorem 6 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A×A\to \mathbb{R}$ and $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$, and $\alpha \in \left(0,1\right]$. If ${|{f}^{\prime }|}^{q}$ is α-preinvex on A for $q>1$, then

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}{\left[\frac{1}{\left(\alpha +1\right){2}^{\alpha }}\right]}^{1/q}\left\{{\left[|{f}^{\prime }\left(a\right){|}^{q}+\left(\left(\alpha +1\right){2}^{\alpha }-1\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[\left({2}^{\alpha +1}-1\right)|{f}^{\prime }\left(a\right){|}^{q}+\left(1-\left(1-\alpha \right){2}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$

Proof Since $b+t\theta \left(a,b\right)\in A$ for every $t\in \left[0,1\right]$, by Lemma 1, Hölder’s inequality, and the α-preinvexity of ${|{f}^{\prime }|}^{q}$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\int }_{0}^{1}|\frac{1}{2}-t|\left[|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|+|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}{|\frac{1}{2}-t|}^{q/\left(q-1\right)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\left\{{\left[{\int }_{0}^{1}{|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}{|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}{|\frac{1}{2}-t|}^{q/\left(q-1\right)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[{\int }_{0}^{1}\left({\left(\frac{1-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{1-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}\left({\left(\frac{2-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{2-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{|\theta \left(a,b\right)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}{\left[\frac{1}{\left(\alpha +1\right){2}^{\alpha }}\right]}^{1/q}\left\{{\left[|{f}^{\prime }\left(a\right){|}^{q}+\left(\left(\alpha +1\right){2}^{\alpha }-1\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[\left({2}^{\alpha +1}-1\right)|{f}^{\prime }\left(a\right){|}^{q}+\left(1-\left(1-\alpha \right){2}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$

The proof of Theorem 6 is complete. □

Corollary 2 Under the conditions of Theorem  6, if $\alpha =1$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}\left\{{\left[\frac{|{f}^{\prime }\left(a\right){|}^{q}+3|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}+{\left[\frac{3|{f}^{\prime }\left(a\right){|}^{q}+|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}\right\}.\hfill \end{array}$

Theorem 7 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A×A\to \mathbb{R}$ and $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$, and $\alpha \in \left(0,1\right]$. If ${|{f}^{\prime }|}^{q}$ is α-preinvex on A for $q>1$ and $q\ge r>0$, then

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{\left(q-r\right)/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[\left(\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{1}{\left(2\alpha +1\right){2}^{2\alpha +1}}\right)|{f}^{\prime }\left(a\right){|}^{q}+\frac{3}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[\left(\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{{2}^{2\alpha +1}-1}{\left(2\alpha +1\right){2}^{2\alpha +1}}\right)|{f}^{\prime }\left(a\right){|}^{q}+\frac{1}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$
(12)

Proof Since $b+t\theta \left(a,b\right)\in A$ for every $t\in \left[0,1\right]$, by Lemma 1, Hölder’s inequality, and the α-preinvexity of ${|{f}^{\prime }|}^{q}$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\int }_{0}^{1}|\frac{1}{2}-t|\left[|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|+|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}{|\frac{1}{2}-t|}^{\left(q-r\right)/\left(q-1\right)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\left\{{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}{|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}{|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{\left(q-r\right)/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left({\left(\frac{1-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{1-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left({\left(\frac{2-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{2-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}.\hfill \end{array}$
(13)

Since ${x}^{r}{y}^{\alpha }\le \frac{{x}^{2r}+{y}^{2\alpha }}{2}$ and $z\le {z}^{\alpha }$ for $x,y\ge 0$ and $0\le z\le 1$, we obtain

$\begin{array}{c}{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left[{\left(\frac{1-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{1-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{1}\left[\frac{1}{2}\left({|\frac{1}{2}-t|}^{2r}+{\left(\frac{1-t}{2}\right)}^{2\alpha }\right)|{f}^{\prime }\left(a\right){|}^{q}+\left(\frac{1+t}{2}\right){|\frac{1}{2}-t|}^{r}|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}=\left[\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{1}{\left(2\alpha +1\right){2}^{2\alpha +1}}\right]|{f}^{\prime }\left(a\right){|}^{q}+\frac{3}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\hfill \end{array}$
(14)

and

$\begin{array}{c}{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left[{\left(\frac{2-t}{2}\right)}^{\alpha }|{f}^{\prime }\left(a\right){|}^{q}+\left(1-{\left(\frac{2-t}{2}\right)}^{\alpha }\right)|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{1}\left[\frac{1}{2}\left({|\frac{1}{2}-t|}^{2r}+{\left(\frac{2-t}{2}\right)}^{2\alpha }\right)|{f}^{\prime }\left(a\right){|}^{q}+\frac{t}{2}{|\frac{1}{2}-t|}^{r}|{f}^{\prime }\left(b\right){|}^{q}\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}=\left[\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{{2}^{2\alpha +1}-1}{\left(2\alpha +1\right){2}^{2\alpha +1}}\right]|{f}^{\prime }\left(a\right){|}^{q}+\frac{1}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}.\hfill \end{array}$
(15)

Substituting (14) and (15) into (13) results in (12). The proof of Theorem 7 is complete. □

Corollary 3 Under the conditions of Theorem  7, if $\alpha =1$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{\left(q-r\right)/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[\left(\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{1}{24}\right)|{f}^{\prime }\left(a\right){|}^{q}+\frac{3}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[\left(\frac{1}{\left(2r+1\right){2}^{2r+1}}+\frac{7}{24}\right)|{f}^{\prime }\left(a\right){|}^{q}+\frac{|{f}^{\prime }\left(b\right){|}^{q}}{\left(r+1\right){2}^{r+2}}\right]}^{1/q}\right\}.\hfill \end{array}$

Theorem 8 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A×A\to \mathbb{R}$ and $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function and ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$. If ${|{f}^{\prime }|}^{q}$ is preinvex on A for $q>1$ and $q\ge r>0$, then

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{8}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{r+1}\right)}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[\frac{|{f}^{\prime }\left(a\right){|}^{q}+3|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}+{\left[\frac{3|{f}^{\prime }\left(a\right){|}^{q}+|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}\right\}.\hfill \end{array}$

Proof Since $b+t\theta \left(a,b\right)\in A$ for every $t\in \left[0,1\right]$, by Lemma 1, Hölder’s inequality, and the preinvexity of ${|{f}^{\prime }|}^{q}$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\int }_{0}^{1}|\frac{1}{2}-t|\left[|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|+|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left({\int }_{0}^{1}{|\frac{1}{2}-t|}^{\left(q-r\right)/\left(q-1\right)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right)}^{1-1/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}{|{f}^{\prime }\left(b+\frac{1-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}{|{f}^{\prime }\left(b+\frac{2-t}{2}\theta \left(a,b\right)\right)|}^{q}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{\left(q-r\right)/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left(\frac{1-t}{2}|{f}^{\prime }\left(a\right){|}^{q}+\frac{1+t}{2}|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[{\int }_{0}^{1}{|\frac{1}{2}-t|}^{r}\left(\frac{2-t}{2}|{f}^{\prime }\left(a\right){|}^{q}+\frac{t}{2}|{f}^{\prime }\left(b\right){|}^{q}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\right]}^{1/q}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{|\theta \left(a,b\right)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{\left(q-r\right)/q}\hfill \\ \phantom{\rule{2em}{0ex}}×\left\{{\left[\frac{1}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(a\right){|}^{q}+\frac{3}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\hfill \\ \phantom{\rule{2em}{0ex}}+{\left[\frac{3}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(a\right){|}^{q}+\frac{1}{\left(r+1\right){2}^{r+2}}|{f}^{\prime }\left(b\right){|}^{q}\right]}^{1/q}\right\}.\hfill \end{array}$

The proof of Theorem 8 is complete. □

Corollary 4 Under the conditions of Theorem  8, if $r=q$, we have

$\begin{array}{c}|\frac{1}{2}\left[\frac{f\left(b\right)+f\left(b+\theta \left(a,b\right)\right)}{2}+f\left(\frac{2b+\theta \left(a,b\right)}{2}\right)\right]-\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|\theta \left(a,b\right)|}{8}{\left(\frac{1}{q+1}\right)}^{1/q}\left\{{\left[\frac{|{f}^{\prime }\left(a\right){|}^{q}+3|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}+{\left[\frac{3|{f}^{\prime }\left(a\right){|}^{q}+|{f}^{\prime }\left(b\right){|}^{q}}{4}\right]}^{1/q}\right\}.\hfill \end{array}$

Theorem 9 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A×A\to \mathbb{R}$ and $a,b\in A$ with $\theta \left(a,b\right)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, ${f}^{\prime }$ is integrable on the θ-path ${P}_{bc}$: $c=b+\theta \left(a,b\right)$, and $\alpha \in \left(0,1\right]$. If f is α-preinvex on A, then

$\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le min\left\{\frac{f\left(a\right)+\alpha f\left(b\right)}{\alpha +1},\frac{\alpha f\left(a\right)+f\left(b\right)}{\alpha +1}\right\}.$
(16)

Proof Since $b+t\theta \left(a,b\right)\in A$ for $0\le t\le 1$, letting $x=\left(1-t\right)b+t\left(b+\theta \left(a,b\right)\right)=b+t\theta \left(a,b\right)$ for $0\le t\le 1$ and using the α-preinvexity of f, we have

$\begin{array}{rcl}\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x& =& {\int }_{0}^{1}f\left(b+t\theta \left(a,b\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \le & {\int }_{0}^{1}\left[{t}^{\alpha }f\left(a\right)+\left(1-{t}^{\alpha }\right)f\left(b\right)\right]\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=\frac{f\left(a\right)+\alpha f\left(b\right)}{\alpha +1}.\end{array}$

The proof of Theorem 9 is complete. □

Corollary 5 Under the conditions of Theorem  9, if $\alpha =1$, we have

$\frac{1}{\theta \left(a,b\right)}{\int }_{b}^{b+\theta \left(a,b\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le \frac{f\left(a\right)+f\left(b\right)}{2}.$

## References

1. Toader G: Some generalizations of the convexity. In Proceedings of the Colloquium on Approximation and Optimization. Univ. Cluj-Napoca, Cluj-Napoca; 1985:329-338.

2. Miheşan VG: A generalization of the convexity. Seminar on Functional Equations, Approximation and Convexity Cluj-Napoca, Romania 1993.

3. Antczak T: Mean value in invexity analysis. Nonlinear Anal. 2005,60(8):1473-1484. 10.1016/j.na.2004.11.005

4. Barani A, Ghazanfari AG, Dragomir SS: Hermite-Hadamard inequality for functions whose derivatives absolute values are preinvex. J. Inequal. Appl. 2012., 2012: Article ID 247 10.1186/1029-242X-2012-247

5. Yang XM, Li D: On properties of preinvex functions. J. Math. Anal. Appl. 2001,256(1):229-241. 10.1006/jmaa.2000.7310

6. Ion DA: Some estimates on the Hermite-Hadamard inequality through quasi-convex functions. An. Univ. Craiova, Ser. Mat. Inform. 2007, 34: 83-88.

7. Dragomir SS, Agarwal RP: Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula. Appl. Math. Lett. 1998,11(5):91-95. 10.1016/S0893-9659(98)00086-X

8. Dragomir SS: On some new inequalities of Hermite-Hadamard type for m -convex functions. Tamkang J. Math. 2002, 33: 45-55.

9. Dragomir SS, Toader G: Some inequalities for m -convex functions. Stud. Univ. Babeş-Bolyai, Math. 1993,38(1):21-28.

10. Bakula MK, Özdemir ME, Pečarić J: Hadamard type inequalities for m -convex and $\left(\alpha ,m\right)$ -convex functions. J. Inequal. Pure Appl. Math. 2008.,9(4): Article ID 96 http://www.emis.de/journals/JIPAM/article1032.html

11. Bai R-F, Qi F, Xi B-Y: Hermite-Hadamard type inequalities for the m - and$\left(\alpha ,m\right)$ -logarithmically convex functions. Filomat 2013,27(1):1-7. 10.2298/FIL1301001B

12. Bai S-P, Qi F:Some inequalities for $\left({s}_{1},{m}_{1}\right)$-$\left({s}_{2},{m}_{2}\right)$-convex functions on the co-ordinates. Glob. J. Math. Anal. 2013,1(1):22-28.

13. Bai S-P, Wang S-H, Qi F: Some Hermite-Hadamard type inequalities for n -time differentiable $\left(\alpha ,m\right)$ -convex functions. J. Inequal. Appl. 2012., 2012: Article ID 267 10.1186/1029-242X-2012-267

14. Dragomir SS: On Hadamard’s inequalities for convex functions. Math. Balk. 1992,6(3):215-222.

15. Dragomir SS: Two mappings on connection to Hadamard’s inequality. J. Math. Anal. Appl. 1992,167(1):49-56. 10.1016/0022-247X(92)90233-4

16. Dragomir SS, Pečarić J, Persson LE: Some inequalities of Hadamard type. Soochow J. Math. 1995,21(3):335-341.

17. Dragomir SS, Pečarić JE, Sándor J: A note on the Jensen-Hadamard inequality. Anal. Numér. Théor. Approx. 1990,19(1):29-34.

18. Qi F, Wei Z-L, Yang Q: Generalizations and refinements of Hermite-Hadamard’s inequality. Rocky Mt. J. Math. 2005,35(1):235-251. 10.1216/rmjm/1181069779

19. Wang Y, Wang S-H, Qi F: Simpson type integral inequalities in which the power of the absolute value of the first derivative of the integrand is s -preinvex. Facta Univ., Ser. Math. Inform. 2013,28(2):151-159.

20. Xi B-Y, Bai R-F, Qi F: Hermite-Hadamard type inequalities for the m - and$\left(\alpha ,m\right)$-geometrically convex functions. Aequ. Math. 2012,184(3):261-269. 10.1007/s00010-011-0114-x

21. Xi B-Y, Qi F: Some integral inequalities of Hermite-Hadamard type for convex functions with applications to means. J. Funct. Spaces Appl. 2012., 2012: Article ID 980438 10.1155/2012/980438

## Acknowledgements

The authors appreciate the anonymous referees for their helpful corrections to and valuable comments on the original version of this paper. This work was partially supported by the NNSF under Grant No. 11361038 of China and by the Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region under Grant No. NJZY13159, China.

## Author information

Authors

### Corresponding author

Correspondence to Yan Wang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Wang, Y., Zheng, MM. & Qi, F. Integral inequalities of Hermite-Hadamard type for functions whose derivatives are α-preinvex. J Inequal Appl 2014, 97 (2014). https://doi.org/10.1186/1029-242X-2014-97 