## Introduction

### Definition 1

The function $$f:[a,b]\subset \mathbb {R}\rightarrow \mathbb {R}$$ is said to be convex if the following inequality holds

\begin{aligned} f(\lambda x+(1-\lambda )y)\le \lambda f(x)+(1-\lambda )f(y) \end{aligned}

for all $$x,y\in [a,b]$$ and $$\lambda \in \left[ 0,1\right] .$$ We say that $$f$$ is concave if $$(-f)$$ is convex.

The following inequality is well known in the literature as the Hermite–Hadamard integral inequality (see, [2, 4]):

\begin{aligned} f\left( \frac{a+b}{2}\right) \le \frac{1}{b-a}\int _{a}^{b}f(x)dx\le \frac{f(a)+f(b)}{2} \end{aligned}
(1.1)

where $$f:I\subset \mathbb {R}\rightarrow \mathbb {R}$$ is a convex function on the interval $$I$$ of real numbers and $$a,b\in I$$ with $$a<b$$.

In [1], Dragomir and Agarwal proved the following results connected with the right part of (1.1).

### Lemma 1

Let $$f:I^{\circ }\subseteq \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$. If $$f^{\prime }\in L[a,b]$$, then the following equality holds:

\begin{aligned} \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a} \int _{a}^{b}f(x)dx=\frac{b-a}{2}\int _{0}^{1}(1-2t)f^{\prime }(ta+(1-t)b)dt. \end{aligned}
(1.2)

### Theorem 1

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

\begin{aligned} \left| \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a} \int _{a}^{b}f(x)dx\right| \le \frac{\left( b-a\right) }{8}\left( \left| f^{\prime }(a)\right| +\left| f^{\prime }(b)\right| \right) . \end{aligned}
(1.3)

### Theorem 2

Let $$f:I^{\circ }\subset \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b,$$ $$f^{\prime }\in L(a,b)$$ and $$p>1.$$ If the mapping $$\left| f^{\prime }\right| ^{p/\left( p-1\right) }$$ is convex on $$\left[ a,b\right]$$, then the following inequality holds:

\begin{aligned} \left| \dfrac{f(a)+f(b)}{2}-\dfrac{1}{b-a}\int _{a}^{b}f(x)dx\right| \le \frac{b-a}{2(p+1)^{1/p}}\left( \frac{\left| f^{\prime }(a)\right| ^{p/(p-1)}+\left| f^{\prime }(b)\right| ^{p/(p-1)}}{2 }\right) ^{(p-1)/p}. \end{aligned}
(1.4)

The most well-known inequalities related to the integral mean of a convex function are the Hermite–Hadamard inequalities or its weighted versions, the so-called Hermite–Hadamard–Fejér inequalities (see [514]). In [3], Fejer gave a weighted generalization of the inequalities (1.1) as the following:

### Theorem 3

$$f:[a,b]\rightarrow \mathbb {R}$$ be a convex function, then the inequality

\begin{aligned} f\left( \frac{a+b}{2}\right) \int _{a}^{b}w(x)dx\le \frac{1}{b-a} \int _{a}^{b}f(x)w(x)dx\le \frac{f(a)+f(b)}{2}\int _{a}^{b}w(x)dx \end{aligned}
(1.5)

holds, where $$w:[a,b]\rightarrow \mathbb {R}$$ is nonnegative, integrable, and symmetric about $$x=\frac{a+b}{2}.$$

In [5], some inequalities of Hermite–Hadamard–Fejer type for differentiable convex mappings were proved using the following lemma.

### Lemma 2

Let $$f:I^{\circ }\subset \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$, and $$w:[a,b]\rightarrow [0,\infty )$$ be a differentiable mapping. If $$f^{\prime }\in L[a,b]$$, then the following equality holds:

\begin{aligned} \frac{f\left( a\right) +f(b)}{2}\int _{a}^{b}w(x)dx-\int _{a}^{b}f(x)w(x)dx= \frac{\left( b-a\right) ^{2}}{2}\int _{0}^{1}p(t)f^{\prime }(ta+(1-t)b)dt \end{aligned}
(1.6)

for each $$t\in [0,1],$$ where

\begin{aligned} p(t)=\int _{t}^{1}w(as+(1-s)b)ds-\int _{0}^{t}w(as+(1-s)b)ds. \end{aligned}

In this article, using functions whose derivatives’ absolute values are convex, we obtained new inequalities of Hermite–Hadamard–Fejer type. The results presented here would provide extensions of those given in earlier works.

## Main results

We will establish some new results connected with the right-hand side of (1.5) and (1.1). Now, we prove our main theorems:

### Theorem 4

Let $$f:I^{\circ }\subseteq \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$ and let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be continuous on $$\left[ a,b \right]$$. If $$\left| f^{\prime }\right|$$ is convex on $$[a,b]$$, then for all $$x\in \left[ a,b\right] ,$$ the following inequalities hold:

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }\left\{ \frac{\left| f^{^{\prime }}(a)\right| }{b-a}\left[ \frac{\left( x-a\right) ^{\alpha +1}\left( b-x\right) }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\alpha +2}\right] +\frac{\left| f^{^{\prime }}(b)\right| }{b-a}\frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }\right\} \\&\\&+\left\| w\right\| _{\left[ x,b\right] ,\infty }^{\alpha }\left\{ \frac{\left| f^{^{\prime }}(a)\right| }{b-a}\frac{\left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+\frac{ \left| f^{^{\prime }}(b)\right| }{b-a}\left[ \frac{\left( b-x\right) ^{\alpha +1}\left( x-a\right) }{\alpha +1}+\frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2}\right] \right\} \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{\left( b-a\right) }\left\{ \left| f^{^{\prime }}(a)\right| \left[ \frac{\left( x-a\right) ^{\alpha +1}\left( b-x\right) }{\alpha +1}+\frac{ \left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+\frac{\left( x-a\right) ^{\alpha +2}}{\alpha +2}\right] \right. \\&\\&+\left. \left| f^{^{\prime }}(b)\right| \left[ \frac{\left( b-x\right) ^{\alpha +1}\left( x-a\right) }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+ \frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2}\right] \right\} \end{aligned}

where $$\alpha >0$$ and $$\left\| w\right\| _{\infty }=\underset{t\in \left[ a,b\right] }{\sup }\left| w(t)\right| .$$

### Proof

By integration by parts, we have the following equalities:

\begin{aligned}&\int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha }f^{^{\prime }}(t)dt \\&\nonumber \\&= \left. \left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha }f(t)\right| _{a}^{b}-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt \nonumber \\&\nonumber \\&= \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt. \nonumber \end{aligned}
(2.1)

We take absolute value of (2.1) and use convexity of $$\left| f^{^{\prime }}\right|$$, we find that

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\le \int \limits _{a}^{x}\left( \left| \int \limits _{x}^{t}w(s)ds\right| \right) ^{\alpha }\left| f^{^{\prime }}(t)\right| dt+\int \limits _{x}^{b}\left( \left| \int \limits _{x}^{t}w(s)ds\right| \right) ^{\alpha }\left| f^{^{\prime }}(t)\right| dt \\&\le \left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }\int \limits _{a}^{x}\left( x-t\right) ^{\alpha }\left| f^{^{\prime }}(t)\right| dt+\left\| w\right\| _{\left[ x,b\right] ,\infty }\int \limits _{x}^{b}\left( t-x\right) ^{\alpha }\left| f^{^{\prime }}(t)\right| dt \\&= \left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }\left[ \int \limits _{a}^{x}\left( x-t\right) ^{\alpha }\left| f^{^{\prime }}( \frac{b-t}{b-a}a+\frac{t-a}{b-a}b)\right| dt\right] \\&+\left\| w\right\| _{\left[ x,b\right] ,\infty }^{\alpha }\left[ \int \limits _{x}^{b}\left( t-x\right) ^{\alpha }\left| f^{^{\prime }}( \frac{b-t}{b-a}a+\frac{t-a}{b-a}b)\right| dt\right] \\&\le \left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }\left\{ \frac{\left| f^{^{\prime }}(a)\right| }{b-a}\left[ \frac{ \left( x-a\right) ^{\alpha +1}\left( b-x\right) }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\alpha +2}\right] +\frac{\left| f^{^{\prime }}(b)\right| }{b-a}\frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }\right\} \\&+\left\| w\right\| _{\left[ x,b\right] ,\infty }^{\alpha }\left\{ \frac{\left| f^{^{\prime }}(a)\right| }{b-a}\frac{\left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+\frac{ \left| f^{^{\prime }}(b)\right| }{b-a}\left[ \frac{\left( b-x\right) ^{\alpha +1}\left( x-a\right) }{\alpha +1}+\frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2}\right] \right\} \\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{\left( b-a\right) }\left\{ \left| f^{^{\prime }}(a)\right| \left[ \frac{\left( x-a\right) ^{\alpha +1}\left( b-x\right) }{\alpha +1}+\frac{ \left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+\frac{\left( x-a\right) ^{\alpha +2}}{\alpha +2}\right] \right. \\&+\left. \left| f^{^{\prime }}(b)\right| \left[ \frac{\left( b-x\right) ^{\alpha +1}\left( x-a\right) }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+ \frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2}\right] \right\} \end{aligned}

for all $$x\in \left[ a,b\right] .$$ Hence, the proof of theorem is completed.

### Corollary 1

Under the same assumptions of Theorem 4 with $$w(s)=1$$, then the following inequality holds:

\begin{aligned}&\left| \left( b-x\right) ^{\alpha }f(b)-\left( a-x\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( t-x\right) ^{\alpha -1}f(t)dt\right| \nonumber \\&\nonumber \\&\le \frac{1}{\left( b-a\right) }\left\{ \left| f^{^{\prime }}(a)\right| \left[ \frac{\left( x-a\right) ^{\alpha +1}\left( b-x\right) }{\alpha +1}+\frac{\left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+\frac{\left( x-a\right) ^{\alpha +2}}{ \alpha +2}\right] \right. \nonumber \\&\\&+\left. \left| f^{^{\prime }}(b)\right| \left[ \frac{\left( b-x\right) ^{\alpha +1}\left( x-a\right) }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }+ \frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2}\right] \right\} \nonumber \end{aligned}
(2.2)

for all $$x\in \left[ a,b\right] .$$

### Remark 1

If we take $$\alpha =1$$ and $$x=\frac{a+b}{2}$$ in (2.2), the inequality (2.2) reduces to (1.3).

### Corollary 2

(Fejer Type Inequality) Under the same assumptions of Theorem 4 with $$\alpha =1$$, then the following inequalities hold:

\begin{aligned}&\left| f(b)\int \limits _{x}^{b}w(s)ds+f(a)\int \limits _{a}^{x}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\\&\le \left| f^{^{\prime }}(a)\right| \frac{\left( x-a\right) ^{2}\left( 3b-2a-x\right) \left\| w\right\| _{\left[ a,x\right] ,\infty }+\left\| w\right\| _{\left[ x,b\right] ,\infty }\left( b-x\right) ^{3}}{6\left( b-a\right) } \\&\\&+\left| f^{^{\prime }}(b)\right| \frac{\left( b-x\right) ^{2}\left( x-3a-2b\right) \left\| w\right\| _{\left[ x,b\right] ,\infty }+\left( x-a\right) ^{3}\left\| w\right\| _{\left[ a,x\right] ,\infty }}{6\left( b-a\right) } \\&\\&\le \left| f^{^{\prime }}(a)\right| \left[ \frac{\left( x-a\right) ^{2}\left( 3b-2a-x\right) +\left( b-x\right) ^{3}}{6\left( b-a\right) }\right] \left\| w\right\| _{\left[ a,b\right] ,\infty } \\&\\&+\left| f^{^{\prime }}(b)\right| \left[ \frac{\left( b-x\right) ^{2}\left( x-3a-2b\right) +\left( x-a\right) ^{3}}{6\left( b-a\right) } \right] \left\| w\right\| _{\left[ a,b\right] ,\infty } \end{aligned}

which is proved by Tseng et al. in [8].

### Corollary 3

(Weighted Trapezoid Inequality) Let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be symmetric to $$\frac{a+b}{ 2}$$ and $$x=\frac{a+b}{2}$$ in Corollary 2. Then the following inequalities hold:

\begin{aligned}&\left| \frac{f(a)+f(b)}{2}\int \limits _{a}^{b}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\\&\le \frac{\left( b-a\right) ^{2}}{48}\left[ 5\left\| w\right\| _{ \left[ a,\frac{a+b}{2}\right] ,\infty }^{\alpha }+\left\| w\right\| _{ \left[ \frac{a+b}{2},b\right] ,\infty }^{\alpha }\right] \left| f^{^{\prime }}(a)\right| \\&\\&+\left[ \left\| w\right\| _{\left[ a,\frac{a+b}{2}\right] ,\infty }^{\alpha }+5\left\| w\right\| _{\left[ \frac{a+b}{2},b\right] ,\infty }^{\alpha }\right] \left| f^{^{\prime }}(b)\right| \\&\\&\le \left( b-a\right) ^{2}\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }\left( \frac{\left| f^{^{\prime }}(a)\right| +\left| f^{^{\prime }}(b)\right| }{8}\right) \end{aligned}

which is proved by Tseng et al. in [8].

### Theorem 5

Let $$f:I^{\circ }\subseteq \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$ and let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be continuous on $$\left[ a,b \right]$$. If $$\left| f^{\prime }\right| ^{q}$$ is convex on $$[a,b]$$, $$q>1,$$ then for all $$x\in \left[ a,b\right] ,$$ the following inequalities hold:

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \frac{\left( x-a\right) ^{\alpha +\frac{1}{p}}\left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }}{\left( b-a\right) ^{\frac{1}{q} }\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+ \frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \\&\\&+\frac{\left( b-x\right) ^{\alpha +\frac{1}{p}}\left\| w\right\| _{ \left[ x,b\right] ,\infty }^{\alpha }}{\left( b-a\right) ^{\frac{1}{q} }\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-x\right) ^{2} }{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{\left( b-a\right) ^{\frac{1}{q}}\left( \alpha p+1\right) ^{\frac{1}{p}}} \left\{ \left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}}\right. \\&\\&+\left. \left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \left[ \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right] \right) ^{\frac{1}{q}}\right\} \end{aligned}

where $$\alpha >0,$$ $$\frac{1}{p}+\frac{1}{q}=1,\$$and $$\left\| w\right\| _{\infty }=\underset{t\in \left[ a,b\right] }{\sup }\left| w(t)\right| .$$

### Proof

We take absolute value of (2.1). Using Holder’s inequality, we find that

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \int \limits _{a}^{x}\left( \left| \int \limits _{x}^{t}w(s)ds\right| \right) ^{\alpha }f^{^{\prime }}(t)dt+\int \limits _{x}^{b}\left( \left| \int \limits _{x}^{t}w(s)ds\right| \right) ^{\alpha }f^{^{\prime }}(t)dt \\&\\&\le \left( \int \limits _{a}^{x}\left| \int \limits _{x}^{t}w(s)ds\right| ^{\alpha p}dt\right) ^{\frac{1}{p} }\left( \int \limits _{a}^{x}\left| f^{^{\prime }}(t)\right| ^{q}dt\right) ^{\frac{1}{q}}+\left( \int \limits _{x}^{b}\left| \int \limits _{x}^{t}w(s)ds\right| ^{\alpha p}dt\right) ^{\frac{1}{p} }\left( \int \limits _{x}^{b}\left| f^{^{\prime }}(t)\right| ^{q}dt\right) ^{\frac{1}{q}}. \end{aligned}

Since $$\left| f^{^{\prime }}(t)\right| ^{q}$$ is convex on $$\left[ a,b \right]$$

\begin{aligned} \left| f^{^{\prime }}\left( \frac{b-t}{b-a}a+\frac{t-a}{b-a}b\right) \right| ^{q}\le \frac{b-t}{b-a}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{t-a}{b-a}\left| f^{^{\prime }}(b)\right| ^{q}. \end{aligned}
(2.3)

From (2.3), it follows that

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }\left( \int \limits _{a}^{x}\left( x-t\right) ^{\alpha p}dt\right) ^{\frac{1}{p} }\left( \int \limits _{a}^{x}\left[ \frac{b-t}{b-a}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{t-a}{b-a}\left| f^{^{\prime }}(b)\right| ^{q}\right] dt\right) ^{\frac{1}{q}} \\&\\&+\left\| w\right\| _{\left[ x,b\right] ,\infty }^{\alpha }\left( \int \limits _{x}^{b}\left( t-x\right) ^{\alpha p}dt\right) ^{\frac{1}{p} }\left( \int \limits _{x}^{b}\left[ \frac{b-t}{b-a}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{t-a}{b-a}\left| f^{^{\prime }}(b)\right| ^{q}\right] dt\right) ^{\frac{1}{q}} \\&\\&\le \frac{\left( x-a\right) ^{\alpha +\frac{1}{p}}\left\| w\right\| _{\left[ a,x\right] ,\infty }^{\alpha }}{\left( b-a\right) ^{\frac{1}{q} }\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+ \frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \\&\\&+\frac{\left( b-x\right) ^{\alpha +\frac{1}{p}}\left\| w\right\| _{ \left[ x,b\right] ,\infty }^{\alpha }}{\left( b-a\right) ^{\frac{1}{q} }\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \left[ \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right] \right) ^{\frac{1}{q}} \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{\left( b-a\right) ^{\frac{1}{q}}\left( \alpha p+1\right) ^{\frac{1}{p}}} \left\{ \left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}}\right. \\&\\&+\left. \left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \left[ \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right] \right) ^{\frac{1}{q}}\right\} \end{aligned}

which this completes the proof.

### Corollary 4

Under the same assumptions of Theorem 5 with $$w(s)=1$$, then the following inequalities hold:

\begin{aligned}&\left| \left( b-x\right) ^{\alpha }f(b)-\left( a-x\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( t-x\right) ^{\alpha -1}f(t)dt\right| \nonumber \\&\nonumber \\&\le \frac{\left( x-a\right) ^{\alpha +\frac{1}{p}}}{\left( b-a\right) ^{ \frac{1}{q}}\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \nonumber \\&\nonumber \\&+\frac{\left( b-x\right) ^{\alpha +\frac{1}{p}}}{\left( b-a\right) ^{\frac{ 1}{q}}\left( \alpha p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \nonumber \\&\,\,\,\,\,\\&\le \frac{1}{\left( b-a\right) ^{\frac{1}{q}}\left( \alpha p+1\right) ^{ \frac{1}{p}}}\left\{ \left( x-a\right) ^{\alpha +\frac{1}{p}}\left( \frac{ \left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}}\right. \nonumber \\&\nonumber \\&+\left. \left( b-x\right) ^{\alpha +\frac{1}{p}}\left( \left[ \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right] \right) ^{\frac{1}{q}}\right\} \nonumber \end{aligned}
(2.4)

### Corollary 5

Let the conditions of Corollary 4 hold. If we take $$\alpha =1$$ and $$x=\frac{a+b}{2}$$ in (2.4), then the following inequality holds:

\begin{aligned}&\left| \frac{f(a)+f(b)}{2}-\frac{1}{b-a}\int \limits _{a}^{b}f(t)dt \right| \\&\\&\le \frac{\left( b-a\right) }{4\left( p+1\right) ^{\frac{1}{p}}}\left[ \left( \frac{3\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{4}\right) ^{\frac{1}{q}}+\left( \frac{ \left| f^{^{\prime }}(a)\right| ^{q}+3\left| f^{^{\prime }}(b)\right| ^{q}}{4}\right) ^{\frac{1}{q}}\right] . \end{aligned}

### Corollary 6

(Fejer Type Inequality) Under the same assumptions of Theorem 5 with $$\alpha =1$$, then the following inequalities hold:

\begin{aligned}&\left| f(b)\int \limits _{x}^{b}w(s)ds+f(a)\int \limits _{a}^{x}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\\&\le \frac{\left( x-a\right) ^{1+\frac{1}{p}}\left\| w\right\| _{ \left[ a,x\right] ,\infty }}{\left( b-a\right) ^{\frac{1}{q}}\left( p+1\right) ^{\frac{1}{p}}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{ \frac{1}{q}} \\&\\&+\frac{\left( b-x\right) ^{1+\frac{1}{p}}\left\| w\right\| _{\left[ x,b\right] ,\infty }}{\left( b-a\right) ^{\frac{1}{q}}\left( p+1\right) ^{ \frac{1}{p}}}\left( \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2} \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }}{\left( b-a\right) ^{\frac{1}{q}}\left( p+1\right) ^{\frac{1}{p}}}\left\{ \left( x-a\right) ^{1+\frac{1}{p}}\left( \frac{\left( b-a\right) ^{2}-\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right) ^{ \frac{1}{q}}\right. \\&\\&+\left. \left( b-x\right) ^{1+\frac{1}{p}}\left( \left[ \frac{\left( b-x\right) ^{2}}{2}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{\left( b-a\right) ^{2}-\left( x-a\right) ^{2}}{2}\left| f^{^{\prime }}(b)\right| ^{q}\right] \right) ^{\frac{1}{q}}\right\} . \end{aligned}

### Corollary 7

(Weighted Trapezoid Inequality) Let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be symmetric to $$\frac{a+b}{ 2}$$ and $$x=\frac{a+b}{2}$$ in Corollary 6. Then the following inequalities hold:

\begin{aligned}&\left| \frac{f(a)+f(b)}{2}\int \limits _{a}^{b}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\le \frac{\left( b-a\right) ^{2}}{4\left( p+1\right) ^{\frac{1}{p}}}\left[ \left\| w\right\| _{\left[ a,\frac{a+b}{2}\right] ,\infty }\left( \frac{3\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{4}\right) ^{\frac{1}{q}}+\left\| w\right\| _{ \left[ \frac{a+b}{2},b\right] ,\infty }\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+3\left| f^{^{\prime }}(b)\right| ^{q}}{4} \right) ^{\frac{1}{q}}\right] \\&\le \frac{\left( b-a\right) ^{2}\left\| w\right\| _{\left[ a,b \right] ,\infty }}{4\left( p+1\right) ^{\frac{1}{p}}}\left[ \left( \frac{ 3\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{4}\right) ^{\frac{1}{q}}+\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+3\left| f^{^{\prime }}(b)\right| ^{q}}{4}\right) ^{\frac{1}{q}}\right] . \end{aligned}

### Theorem 6

Let $$f:I^{\circ }\subseteq \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$ and let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be continuous on $$\left[ a,b \right]$$. If $$\left| f^{\prime }\right| ^{q}$$ is convex on $$[a,b]$$, $$q>1,$$ then for all $$x\in \left[ a,b\right] ,$$ the following inequality holds:

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \frac{\left( b-a\right) ^{\frac{1}{q}}\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }}{\left( \alpha p+1\right) ^{\frac{1}{p}}} \left[ \left( x-a\right) ^{\alpha p+1}+\left( b-x\right) ^{\alpha p+1}\right] ^{\frac{1}{p}}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2}\right) ^{\frac{1}{q}} \end{aligned}

where $$\alpha >0,$$ $$\frac{1}{p}+\frac{1}{q}=1,\$$and $$\left\| w\right\| _{\infty }=\underset{t\in \left[ a,b\right] }{\sup }\left| w(t)\right| .$$

### Proof

We take absolute value of (2.1). Using Holder’s inequality and the convexity of $$\left| f^{^{\prime }}\right| ^{q},$$ we find that

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \left( \int \limits _{a}^{b}\left| \int \limits _{x}^{t}w(s)ds\right| ^{\alpha p}dt\right) ^{\frac{1}{p} }\left( \int \limits _{a}^{b}\left| f^{^{\prime }}(t)\right| ^{q}dt\right) ^{\frac{1}{q}} \\&\\&\le \left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }\left( \int \limits _{a}^{b}\left| t-x\right| ^{\alpha p}dt\right) ^{\frac{1}{ p}}\left( \int \limits _{a}^{b}\left[ \frac{b-t}{b-a}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{t-a}{b-a}\left| f^{^{\prime }}(b)\right| ^{q}\right] dt\right) ^{\frac{1}{q}} \\&\\&= \frac{\left( b-a\right) ^{\frac{1}{q}}\left\| w\right\| _{\left[ a,b \right] ,\infty }^{\alpha }}{\left( \alpha p+1\right) ^{\frac{1}{p}}}\left[ \left( x-a\right) ^{\alpha p+1}+\left( b-x\right) ^{\alpha p+1}\right] ^{ \frac{1}{p}}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2}\right) ^{\frac{1}{q}} \end{aligned}

which this completes the proof.

### Corollary 8

Under the same assumptions of Theorem 6 with $$w(s)=1$$, then the following inequality holds:

\begin{aligned}&\left| \left( b-x\right) ^{\alpha }f(b)-\left( a-x\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( t-x\right) ^{\alpha -1}f(t)dt\right| \\&\nonumber \\&\le \frac{\left( b-a\right) ^{\frac{1}{q}}}{\left( \alpha p+1\right) ^{ \frac{1}{p}}}\left[ \left( x-a\right) ^{\alpha p+1}+\left( b-x\right) ^{\alpha p+1}\right] ^{\frac{1}{p}}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2} \right) ^{\frac{1}{q}}. \nonumber \end{aligned}
(2.5)

### Remark 2

Let the conditions of Corollary 8 hold. If we take $$\alpha =1$$ and $$x=\frac{a+b}{2}$$ in (2.5), then the inequality becomes the inequality ( 1.4).

### Corollary 9

(Fejer Type Inequality) Under the same assumptions of Theorem 6 with $$\alpha =1$$, then the following inequality holds:

\begin{aligned}&\left| f(b)\int \limits _{x}^{b}w(s)ds+f(a)\int \limits _{a}^{x}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\\&\le \frac{\left( b-a\right) ^{\frac{1}{q}}\left\| w\right\| _{\left[ a,b\right] ,\infty }}{\left( p+1\right) ^{\frac{1}{p}}}\left[ \left( x-a\right) ^{p+1}+\left( b-x\right) ^{p+1}\right] ^{\frac{1}{p}}\left( \frac{ \left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2}\right) ^{\frac{1}{q}}. \end{aligned}

### Corollary 10

(Weighted Trapezoid Inequality) Let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be symmetric to $$\frac{a+b}{ 2}$$ and $$x=\frac{a+b}{2}$$ in Corollary 9. Then the following inequality holds:

\begin{aligned} \left| \frac{f(a)+f(b)}{2}\int \limits _{a}^{b}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \le \frac{\left( b-a\right) ^{2}\left\| w\right\| _{\left[ a,b\right] ,\infty }}{2\left( p+1\right) ^{\frac{1}{p}}}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2} \right) ^{\frac{1}{q}}. \end{aligned}

### Theorem 7

Let $$f:I^{\circ }\subseteq \mathbb {R}\rightarrow \mathbb {R}$$ be a differentiable mapping on $$I^{\circ }$$, $$a,b\in I^{\circ }$$ with $$a<b$$ and let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be continuous on $$\left[ a,b \right]$$. If $$\left| f^{\prime }\right| ^{q}$$ is convex on $$[a,b]$$, $$q>1,$$ then for all $$x\in \left[ a,b\right] ,$$ the following inequality holds:

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{\left( \alpha +1\right) \left( \alpha +2\right) ^{\frac{1}{q}}\left( b-a\right) ^{\frac{1}{q}}}\left( \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right) ^{\frac{1}{p}} \\&\times \left( \left( \left( \alpha +1\right) \left( b-a\right) \left( x-a\right) ^{\alpha +1}+\left( b-x\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(a)\right| ^{q}\right. \\&\left. +\left( \left( \alpha +1\right) \left( b-a\right) \left( b-x\right) ^{\alpha +1}+\left( x-a\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \end{aligned}

where $$\alpha >0,$$ $$\frac{1}{p}+\frac{1}{q}=1,\$$and $$\left\| w\right\| _{\infty }=\underset{t\in \left[ a,b\right] }{\sup }\left| w(t)\right| .$$

### Proof

We take absolute value of (2.1). Using Holder’s inequality and the convexity of $$\left| f^{^{\prime }}\right| ^{q},$$ we find that

\begin{aligned}&\left| \left( \int \limits _{x}^{b}w(s)ds\right) ^{\alpha }f(b)-\left( \int \limits _{x}^{a}w(s)ds\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( \int \limits _{x}^{t}w(s)ds\right) ^{\alpha -1}w(t)f(t)dt\right| \\&\\&\le \left( \int \limits _{a}^{b}\left| \int \limits _{x}^{t}w(s)ds\right| ^{\alpha }dt\right) ^{\frac{1}{p} }\left( \int \limits _{a}^{b}\left| \int \limits _{x}^{t}w(s)ds\right| ^{\alpha }\left| f^{^{\prime }}(t)\right| ^{q}dt\right) ^{\frac{1}{q} } \\&\\&\le \left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }\left( \int \limits _{a}^{b}\left| t-x\right| ^{\alpha }dt\right) ^{\frac{1}{p }}\left( \int \limits _{a}^{b}\left| t-x\right| ^{\alpha }\left[ \frac{ b-t}{b-a}\left| f^{^{\prime }}(a)\right| ^{q}+\frac{t-a}{b-a} \left| f^{^{\prime }}(b)\right| ^{q}\right] dt\right) ^{\frac{1}{q}} \\&\\&= \left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }\left( \frac{\left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}}{\alpha +1}\right) ^{\frac{1}{p}} \\&\times \left( \left( \frac{\left( b-x\right) \left( x-a\right) ^{\alpha +1} }{\alpha +1}+\frac{\left( x-a\right) ^{\alpha +2}}{\alpha +2}\right) \left| f^{^{\prime }}(a)\right| ^{q}+\left( \frac{\left( x-a\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \left| f^{^{\prime }}(b)\right| ^{q}\right. \\&+\left. \left( \frac{\left( b-x\right) ^{\alpha +2}}{\left( \alpha +1\right) \left( \alpha +2\right) }\right) \left| f^{^{\prime }}(a)\right| ^{q}+\left( \frac{\left( x-a\right) \left( b-x\right) ^{\alpha +1}}{\alpha +1}+\frac{\left( b-x\right) ^{\alpha +2}}{\alpha +2} \right) \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \\&\\&= \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha }}{ \left( \alpha +1\right) \left( \alpha +2\right) ^{\frac{1}{q}}\left( b-a\right) ^{\frac{1}{q}}}\left( \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right) ^{\frac{1}{p}} \\&\times \left( \left( \left( \alpha +1\right) \left( b-a\right) \left( x-a\right) ^{\alpha +1}+\left( b-x\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(a)\right| ^{q}\right. \\&\left. +\left( \left( \alpha +1\right) \left( b-a\right) \left( b-x\right) ^{\alpha +1}+\left( x-a\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}} \end{aligned}

which this completes the proof.

### Corollary 11

Under the same assumptions of Theorem 7 with $$w(s)=1$$, then the following inequality holds:

\begin{aligned}&\left| \left( b-x\right) ^{\alpha }f(b)-\left( a-x\right) ^{\alpha }f(a)-\alpha \int \limits _{a}^{b}\left( t-x\right) ^{\alpha -1}f(t)dt\right| \\&\nonumber \\&\le \frac{1}{\left( \alpha +1\right) \left( \alpha +2\right) ^{\frac{1}{q} }\left( b-a\right) ^{\frac{1}{q}}}\left( \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right) ^{\frac{1}{p}} \nonumber \\&\times \left( \left( \left( \alpha +1\right) \left( b-a\right) \left( x-a\right) ^{\alpha +1}+\left( b-x\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(a)\right| ^{q}\right. \nonumber \\&\left. +\left( \left( \alpha +1\right) \left( b-a\right) \left( b-x\right) ^{\alpha +1}+\left( x-a\right) \left[ \left( x-a\right) ^{\alpha +1}+\left( b-x\right) ^{\alpha +1}\right] \right) \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}}. \nonumber \end{aligned}
(2.6)

### Corollary 12

Let the conditions of Corollary 11 hold. If we take $$\alpha =1$$ and $$x=\frac{a+b}{2}$$ in (2.6), then the following inequality holds:

\begin{aligned} \left| \frac{f(a)+f(b)}{2}-\frac{1}{b-a}\int \limits _{a}^{b}f(t)dt\right| \le \frac{\left( b-a\right) }{4}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2} \right) ^{\frac{1}{q}}. \end{aligned}

### Corollary 13

(Fejer Type Inequality) Under the same assumptions of Theorem 7 with $$\alpha =1$$, then the following inequality holds:

\begin{aligned}&\left| f(b)\int \limits _{x}^{b}w(s)ds+f(a)\int \limits _{a}^{x}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\\&\le \frac{\left\| w\right\| _{\left[ a,b\right] ,\infty }^{\alpha } }{2\cdot 3^{\frac{1}{q}}\left( b-a\right) ^{\frac{1}{q}}}\left( \left( x-a\right) ^{2}+\left( b-x\right) ^{2}\right) ^{\frac{1}{p}} \\&\times \left( \left( 2\left( b-a\right) \left( x-a\right) ^{2}+\left( b-x\right) \left[ \left( x-a\right) ^{2}+\left( b-x\right) ^{2}\right] \right) \left| f^{^{\prime }}(a)\right| ^{q}\right. \\&\left. +\left( 2\left( b-a\right) \left( b-x\right) ^{2}+\left( x-a\right) \left[ \left( x-a\right) ^{2}+\left( b-x\right) ^{2}\right] \right) \left| f^{^{\prime }}(b)\right| ^{q}\right) ^{\frac{1}{q}}. \end{aligned}

### Corollary 14

(Weighted Trapezoid Inequality) Let $$w:\left[ a,b\right] \rightarrow \mathbb {R}$$ be symmetric to $$\frac{a+b}{ 2}$$ and $$x=\frac{a+b}{2}$$ in Corollary 13. Then the following inequality holds:

\begin{aligned}&\left| \frac{f(a)+f(b)}{2}\int \limits _{a}^{b}w(s)ds-\int \limits _{a}^{b}w(t)f(t)dt\right| \\&\le \frac{\left( b-a\right) ^{2}\left\| w\right\| _{\left[ a,b \right] ,\infty }^{\alpha }}{4}\left( \frac{\left| f^{^{\prime }}(a)\right| ^{q}+\left| f^{^{\prime }}(b)\right| ^{q}}{2} \right) ^{\frac{1}{q}}. \end{aligned}