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