## 2 Preliminaries  
Throughout this paper, let \([a,b]\subseteq\mathbb{R}\) be an interval and \(0<q<p\leq 1\) be a constant. The following definitions for the \((p,q)\)-derivative and \((p,q)\)-integral were given in [10; 11]. Definition 1: Let \(f:[a,b]\rightarrow\mathbb{R}\) be a continuous function, and let \(x\in[a,b]\). Then, the \((p,q)\)-derivative of \(f\) on \([a,b]\) at \(x\) is defined as:  
\[{}_{a}D_{p,q}f(x) =\frac{f(px+(1-p)a)-f(qx+(1-q)a)}{(p-q)(x-a)},\quad x\
eq a \tag{3}\] \[{}_{a}D_{p,q}f(a) =\lim_{x\to a}D_{p,q}f(x).\]  
Obviously, a function \(f\) is \((p,q)\)-differentiable on \([a,b]\) if \({}_{a}D_{p,q}f(x)\) exists for all \(x\in[a,b]\). In Definition 1, if \(a=0\), then \({}_{0}D_{p,q}f=D_{p,q}f\), where \(D_{p,q}f\) is defined by:  
\[D_{p,q}f(x)=\frac{f(px)-f(qx)}{(p-q)x},\quad x\
eq 0. \tag{4}\]  
Furthermore, if \(p=1\) in (4), then it reduces to \(D_{q}f\), which is the \(q\)-derivative of the function \(f\); see [5]. Example 1: Define function \(f:[a,b]\rightarrow\mathbb{R}\) by \(f(x)=x^{2}+1\). Let \(0<q<p\leq 1\). Then, for \(x\
eq a\), we have:  
\[{}_{a}D_{p,q}(x^{2}+1) =\frac{\big{[}(px+(1-p)a)^{2}+1\big{]}-\big{[}(qx+(1-q)a)^{2}+1 \big{]}}{(p-q)(x-a)} \tag{5}\] \[=\frac{(p+q)x^{2}+2 ax[1-(p+q)]+x^{2}[(p+q)-2]}{(x-a)}\] \[=\frac{x(p+q)(x-a)-a\rho(x-a)+2a(x-a)}{(x-a)}\] \[=(p+q)(x-a)+2a.\]  
Definition 2: Let \(f:[a,b]\rightarrow\mathbb{R}\) be a continuous function. Then, the \((p,q)\)-integral on \([a,b]\) is defined by:  
\[\int_{a}^{x}f(t)\ {}_{a}d{p}_{a}t=(p-q)(x-a)\sum_{n=0}^{\infty}\frac{q^{n}}{p^{n+1}}f \bigg{(}\frac{q^{n}}{p^{n+1}}x+\bigg{(}1-\frac{q^{n}}{p^{n+1}}\bigg{)}a\bigg{)}, \tag{6}\]  
for \(x\in[a,b]\). If \(a=0\) and \(p=1\) in (6), then we have the classical \(q\)-integral [5]. **Example 2**. : _Define function \(f:[a,b]\rightarrow\mathbb{R}\) by \(f(x)=2x\). Let \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}\int_{a}^{b}f(x)\ _{a}d_{p,q}x&=\int_{a}^{b}2x_{a}d_{p,q}x\\ &=2(p-q)(b-a)\sum_{n=0}^{\infty}\frac{q^{n}}{p^{n+1}}\bigg{(}\frac{q^{n}}{p^{n +1}}b+\bigg{(}1-\frac{q^{n}}{p^{n+1}}\bigg{)}a\bigg{)}\\ &=\frac{2(b-a)(b-a(1-p-q))}{p+q}.\end{array}  \tag{7}\]  
**Theorem 1**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a continuous function. Then, we have the following:_  
1. \(aD_{p,q}\int_{a}^{x}f(t)\ _{a}d_{p,q}t=f(x)\)_;_
2. \(\int_{c}^{x}aD_{p,q}f(t)\ _{a}d_{p,q}t=f(x)-f(c)\) _for_ \(c\in(a,x)\)_._  
**Theorem 2**. : _Let \(f,g:[a,b]\rightarrow\mathbb{R}\) be continuous functions and \(\alpha\in\mathbb{R}\). Then, we have the following:_  
1. \(\int_{a}^{x}[f(t)+g(t)]\ _{a}d_{p,q}t=\int_{a}^{x}f(t)\ _{a}d_{p,q}t+\int_{a}^{x}g(t)\ _{a}d_{p,q}t\)_;_
2. \(\int_{a}^{x}(\alpha f)(t)_{a}d_{p,q}t=\alpha\int_{a}^{x}f(t)_{a}d_{p,q}t\)_;_
3. \(\int_{c}^{x}f(pt+(1-p)a)_{a}D_{p,q}g(t)\ _{a}d_{q}t=(fg)|_{c}^{x}-\int_{c}^{x}g( qt+(1-q)a)_{a}D_{p,q}f(t)\ _{a}d_{p,q}t\) _for_ \(c\in(a,x)\)_._  
For the proof properties of Theorems 1 and 2, we refer to [10; 11]. The proofs of the following theorems were given in [22]. **Theorem 3**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex differentiable function on \((a,b)\) and \(0<q<p\leq 1\). Then, we have:_  
\[f\bigg{(}\frac{qa+pb}{p+q}\bigg{)}\leq\frac{1}{p(b-a)}\int_{a}^{pb+(1-p)a}f(x )\ _{a}d_{p,q}x\leq\frac{qf(a)+pf(b)}{p+q}. \tag{8}\]  
**Theorem 4**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex differentiable function on \((a,b)\) and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}f\bigg{(}\frac{pa+qb}{p+q}\bigg{)}+\frac{(p-q)(b-a)}{p+q}f ^{\prime}\bigg{(}\frac{pa+qb}{p+q}\bigg{)}&\leq\frac{1}{p(b-a)}\int_{a}^{pb+(1- p)a}f(x)\ _{a}d_{p,q}x\\ &\leq\frac{qf(a)+pf(b)}{p+q}.\end{array}  \tag{9}\]  
**Theorem 5**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex differentiable function on \((a,b)\) and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}f\bigg{(}\frac{a+b}{2}\bigg{)}+\frac{(p-q)(b-a)}{2(p+q)}f ^{\prime}\bigg{(}\frac{a+b}{2}\bigg{)}&\leq\frac{1}{p(b-a)}\int_{a}^{pb+(1-p)a }f(x)\ _{a}d_{p,q}x\\ &\leq\frac{qf(a)+pf(b)}{p+q}.\end{array}  \tag{10}\]  
**Lemma 1**.