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<h1>Generalized Ostrowski Inequalities and Computational Integration</h1>
<p class="authors">
<span class="author">Nazia Irshad\(^1\), Asif R. Khan\(^2\) Hina Musharraf\(^2\)</span>
</p>
<p class="date">August 25, 2020; accepted: December 1, 2020; published online: February 20, 2021.</p>
</div>
<div class="abstract"><p> We state and prove three generalized results related to Ostrowski inequality by using differentiable functions which are bounded, bounded below only and bounded above only, respectively. From our proposed results we get number of established results as our special cases. Some applications in numerical integration are also given which gives us some standard and nonstandard quadrature rules. </p>
<p><b class="bf">MSC.</b> 26D07, 26D10, 26D20, 45H99 </p>
<p><b class="bf">Keywords.</b> Ostrowski Inequality, Quadrature Rules </p>
</div>
<p>\(1-\)Department of Basic Sciences, Mathematics and Humanities, Dawood University of Engineering and Technology, New M. A. Jinnah Road, Karachi-74800, Pakistan. <span class="tt">nazia.irshad@duet.edu.pk</span> </p>
<p>\(2-\)Department of Mathematics, University of Karachi, University Road, Karachi-75270, Pakistan, <span class="tt">asifrk@uok.edu.pk, hinamusharraf93@gmail.com</span> </p>
<h1 id="a0000000002">1 Introduction.</h1>
<p>Ostrowski inequality has gained supreme position among many types of integral inequalities. This interesting and useful inequality <span class="cite">
	[
	<a href="#H9" >19</a>
	]
</span> was first presented by the Ukrainian mathematician Alexander Markowich Ostrowski in 1938, which is stated as follows: </p>
<p><div class="thm_thmwrapper " id="thm1">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1.1</span>
  </div>
  <div class="thm_thmcontent">
  <p> <i class="it">Let \(g:I\subset \mathbb {R}\to \mathbb {R}\) be a differentiable mappings on \(I^o\), the interior of the interval \(I\), such that \(g\) is differentiable and belongs to \( L[a_0,a_1]\), where \(a_0,a_1 \in I\) with \(a_0{\lt}a_1\). If \(|g'(\eta )|\leq M\), valid for all \(\eta \in [a_0,a_1]\) and \(M\) is positive real constant, then we have the following inequality: <div class="displaymath" id="a0000000003">
  \begin{align*}  \left|g(\eta )-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds \right|\leq M(a_1-a_0)\bigg[\tfrac {1}{4}+\tfrac {\left(\eta -\frac{a_0+a_1}{2}\right)^2}{(a_1-a_0)^2} \bigg], \end{align*}
</div> where \(\frac{1}{4}\) is the best possible constant that it cannot be replaced by any smaller one value. </i></p>

  </div>
</div> </p>
<p>Ostrowski inequality can be used to determine the absolute deviation of functional value from its integral mean. It also approximates area under the curve of a function by a rectangle. It has great importance because of its number of applications in statistics, probability theory, integral operator theory, numerical quadrature rules and special means. Due to its high importance most of the researchers are continuously working on its generalization by using various techniques. Even we can find research work on Ostrowski in 70’s as can be seen in <span class="cite">
	[
	<a href="#mil1" >15</a>
	, 
	<a href="#mil2" >16</a>
	]
</span>. For some of its recent generalizations and different variants we refer the reader to the following articles <span class="cite">
	[
	<a href="#double-sided" >1</a>
	, 
	<a href="#nazo1" >7</a>
	, 
	<a href="#nazo2" >8</a>
	, 
	<a href="#nazo3" >9</a>
	, 
	<a href="#nazo5" >10</a>
	, 
	<a href="#nazo6" >11</a>
	, 
	<a href="#nazo7" >12</a>
	, 
	<a href="#nazo4" >20</a>
	, 
	<a href="#H5" >13</a>
	, 
	<a href="#H6" >14</a>
	, 
	<a href="#3" >22</a>
	, 
	<a href="#4" >23</a>
	]
</span>. </p>
<p>In this paper, we would use our main results to introduce standard quadratures and nonstandard quadratures rules:</p>
<div class="displaymath" id="a0000000004">
  \begin{align*} & A_1(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong \tfrac {g(a_0)+g(a_1)}{2},\\ & A_2(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong g\left(\tfrac {a_0+a_1}{2}\right),\\ & A_3(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong \tfrac {1}{2}\left[g\left(\tfrac {a_0+a_1}{2}\right)+\tfrac {g(a_0)+g(a_1)}{2}\right],\\ & A_4(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong \tfrac {1}{2}\left[-g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right)+g(a_1)\right],\\ & A_5(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong \tfrac {1}{2}\left[g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right)-g(a_1)\right],\\ & A_6(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong g(a_1),\\ & A_7(g):\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(\eta )d\eta \cong g(a_0). \end{align*}
</div>
<p> We need some lemmas that would be helpful in our main result. <div class="lem_thmwrapper " id="a0000000005">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">1.2</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(g\) be as in <a href="#thm1">theorem 1.1</a>. Consider the following kernel on \([a_0,a_1]\) for all \( \delta \in [0,1]\) and <div class="displaymath" id="a0000000006">
  \[ \eta +\delta \tfrac {a_1-a_0}{2} \leq \eta \leq \tfrac {a_0+a_1}{2} \]
</div> <div class="displaymath" id="a0000000007">
  \begin{align}  K(\eta ,s)=\begin{cases}  s-\eta +\delta \frac{a_1-a_0}{2},& \text{if}\quad s\in [a_0,\eta ],\\[2mm] s-\eta -\delta \frac{a_1-a_0}{2},& \text{if}\quad s\in (\eta ,a_1]. \end{cases}\label{2.1} \end{align}
</div> Then the following identity holds <div class="displaymath" id="a0000000008">
  \begin{align}  \int _{a_0}^{a_1} K(\eta ,s)g{’}(\eta )ds =& \delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+\nonumber \\ & +(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)-\int _{a_0}^{a_1} g(s) ds\label{2.2}. \end{align}
</div> </i></p>

  </div>
</div> <div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Applying integration-by-parts for the Riemann-Stieltjes integral in kernel <a href="#2.1" class="eqref">1.3</a>, we get </p>
<div class="displaymath" id="2.3">
  \begin{align} & \int _{a_0}^\eta \left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)dg(s) =\delta \tfrac {a_1-a_0}{2}g(\eta )-\left(\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right)g(\eta )-\! \int _{a_0}^\eta g(s)ds,\label{2.3} \end{align}
</div>
<p> and </p>
<div class="displaymath" id="2.4">
  \begin{align} & \int _\eta ^{a_1}\! \! \left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)dg(s) =\left(a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right)g(a_1)+\delta \tfrac {a_1-a_0}{2}g(\eta )\! -\! \! \int _{a_1}^\eta g(s)ds.\label{2.4} \end{align}
</div>
<p> By adding the equalities <a href="#2.3" class="eqref">1.5</a> and <a href="#2.4" class="eqref">1.6</a>, we get <a href="#2.2" class="eqref">1.4</a>. <div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We use the following lemma from <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> to proceed further. <div class="lem_thmwrapper " id="a0000000011">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">1.3</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">If \(\gamma (\eta )\) \(\leq \) \(g{’}(\eta ) \leq \) \(\tau (\eta )\) for any \(\gamma , \tau \in C[a_0,a_1]\) and \(\eta \) \(\in [a_0,a_1],\) then we have <div class="displaymath" id="2.5">
  \begin{align}  \left|g{’}(\eta )- \tfrac {\gamma (s)+\tau (s)}{2}\right|\leq \tfrac {\tau (s)-\gamma (s)}{2}.\label{2.5} \end{align}
</div> </i></p>

  </div>
</div> With the help of kernel <a href="#2.1" class="eqref">1.3</a>, our concentration is to derive different bounds of Ostrowski type inequality. Also explicit error bounds for numerical quadrature and nonstandard quadrature formulas will be discussed. Also we recaptured many established results from articles <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span> and <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span>. </p>
<h1 id="a0000000012">2 Main results</h1>
<p> <div class="thm_thmwrapper " id="thm2">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.1</span>
  </div>
  <div class="thm_thmcontent">
  <p> Let g: \(I \to \mathbb {R}\), be a differentiable function in the interior \(I^0\) of interval \(I\) and let \([a_0,a_1]\) \(\subset \) \(I^0\). If \(\gamma (\eta )\) \(\leq \) \(g{’}(\eta ) \leq \) \(\tau (\eta )\) for any \(\gamma , \tau \in C[a_0,a_1]\), \(\eta \) \(\in [a_0,a_1]\), then for all \(\delta \in [0,1]\) the following inequality holds </p>
<div class="displaymath" id="a0000000013">
  \begin{align}  m(\eta ,\delta )\leq & \delta \left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+\tfrac {a_1-\eta }{a_1-a_0}g(a_1)\nonumber \\ & +\tfrac {\eta -a_0}{a_1-a_0}g(a_0)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M(\eta ,\delta ),\label{2.6} \end{align}
</div>
<p> or </p>
<div class="displaymath" id="a0000000014">
  \begin{align}  (a_1-a_0)m(\eta ,\delta )\leq & \delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)\nonumber \\ & +(\eta -a_0)g(a_0)-\int _{a_0}^{a_1} g(s)ds\leq (a_1-a_0) M(\eta ,\delta ),\label{2.6a} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000015">
  \begin{align*} & m(\eta ,\delta )= \\ & =\tfrac {1}{a_1-a_0}\Bigg[\int ^{a_1-\eta -\delta \frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta \! +\! \eta \! +\! \delta \tfrac {a_1-a_0}{2}\right) + \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta \! +\! \eta \! +\! \delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \\ & \quad +\int _{\eta -a_0+\delta \frac{a_1-a_0}{2}}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right) + \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \Bigg] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000016">
  \begin{align*} & M(\eta , \delta )= \\ & =\tfrac {1}{a_1-a_0} \Bigg[\int ^{a_1-\eta -\delta \frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\! \left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta \! +\! \eta \! +\! \delta \tfrac {a_1-a_0}{2}\right) + \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta \! +\! \eta \! +\! \delta \tfrac {a_1-a_0}{2}\right)}\right)\! d\zeta + \end{align*}
</div>
<div class="displaymath" id="a0000000017">
  \begin{align*} & \quad +\int _{\eta -a_0+\delta \frac{a_1-a_0}{2}}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right) + \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \Bigg]. \end{align*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By referring to kernel <a href="#2.1" class="eqref">1.3</a> and identity <a href="#2.2" class="eqref">1.4</a>, we first have </p>
<div class="displaymath" id="a0000000019">
  \begin{align} & \int _{a_0}^{a_1} K(\eta ,s)\left(g{’}(\eta )-\tfrac {\gamma (s)+\tau (s)}{2}\right)ds=\nonumber \\ & =\int _{a_0}^{a_1} K(\eta ,s)g{’}(\eta )ds - \tfrac {1}{2}\left(\int _{a_0}^{a_1} K(\eta ,s)\left(\gamma (s)+\tau (s)\right)ds\right)\nonumber \\ & =(a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)\nonumber \\ & \quad -\int _{a_0}^{a_1}g(s)ds-\tfrac {1}{2}\left[\int _{a_0}^{\eta }\left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)(\gamma (s)+\tau (s))ds\right.\nonumber \\ & \hspace{3.5cm}\left.+\int _{\eta }^{a_1}\left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)(\gamma (s)+\tau (s))ds\right].\label{2.7} \end{align}
</div>
<p> Therefore, we can conclude from <a href="#2.5" class="eqref">1.7</a> and <a href="#2.7" class="eqref">2.3</a> that </p>
<div class="displaymath" id="a0000000020">
  \begin{align} & \bigg|(a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)\nonumber \\ & \quad -\int _{a_0}^{a_1}g(s)ds-\tfrac {1}{2}\left[\int _{a_0}^{\eta }\left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)(\gamma (s)+\tau (s))ds\right.\nonumber \\ & \quad \left.+\int _{\eta }^{a_1}\left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)(\gamma (s)+\tau (s))ds\right]\bigg|=\nonumber \\ &  =\left|\int _{a_0}^{a_1} K(\eta ,s) \left(g{’}(\eta )- \tfrac {\gamma (s)+\tau (s)}{2}\right) ds\right| \nonumber \\ & \leq \int _{a_0}^{a_1} \left|K(\eta ,s)\right| \left| \left(g{’}(\eta )- \tfrac {\gamma (s)+\tau (s)}{2}\right) ds\right| \nonumber \\ & \leq \int _{a_0}^{a_1} \left|K(\eta ,s)\right|\left(\tfrac {\tau (s)-\gamma (s)}{2}\right)ds\nonumber \\ & =\tfrac {1}{2}\left[ \int _{a_0}^\eta \left|s-\eta +\delta \tfrac {a_1-a_0}{2}\right| (\tau (s)-\gamma (s))ds\right.\nonumber \\ & \quad \left.+\int _\eta ^{a_1}\left|s-\eta -\delta \tfrac {a_1-a_0}{2}\right|(\tau (s)-\gamma (s))ds\right]. \hspace{1cm} \label{2.8} \end{align}
</div>
<p> After re-arranging <a href="#2.8" class="eqref">2.4</a>, we obtain </p>
<div class="displaymath" id="a0000000021">
  \begin{align*}  m(\eta , \delta )=& \tfrac {1}{a_1-a_0} \left[\int _{a_0}^\eta \left(\left(s-\eta +\delta \tfrac {a_1-a_0}{2}-\left|s-\eta +\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\tau (s)}{2}\right.\right. \\ & + \left.\left(s-\eta +\delta \tfrac {a_1-a_0}{2}+\left|s-\eta +\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\gamma (s)}{2}\right)ds\\ & +\int _\eta ^{a_1}\left(\left(s-\eta -\delta \tfrac {a_1-a_0}{2}-\left|s-\eta -\delta \tfrac {a_1-a_0}{2}\right|\right)\right.\tfrac {\tau (s)}{2}\\ & +\left.\left(s-\eta -\delta \tfrac {a_1-a_0}{2}+\left|s-\eta -\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\gamma (s)}{2}\right)ds= \end{align*}
</div>
<div class="displaymath" id="a0000000022">
  \begin{align*}  =& \tfrac {1}{a_1-a_0}\left[\int ^{a_1-\eta -\delta \frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\eta +\delta \tfrac {a_1-a_0}{2}\right)\right.\right.\nonumber \\ & +\left.\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\eta +\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \nonumber \\ & +\int _{\eta -a_0+\delta \frac{a_1-a_0}{2}}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right) \right.\nonumber \\ & +\left.\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000023">
  \begin{align*}  M(\eta , \delta )=& \tfrac {1}{a_1-a_0} \left[\int _{a_0}^\eta \left(\left(s-\eta +\delta \tfrac {a_1-a_0}{2}+\left|s-\eta +\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\tau (s)}{2}\right.\right. \\ & + \left.\left(s-\eta +\delta \tfrac {a_1-a_0}{2}-\left|s-\eta +\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\gamma (s)}{2}\right)ds \\ & +\int _\eta ^{a_1}\left(\left(s-\eta -\delta \tfrac {a_1-a_0}{2}+\left|s-\eta -\delta \tfrac {a_1-a_0}{2}\right|\right)\right.\frac{\tau (s)}{2} \\ & +\left.\left(s-\eta -\delta \tfrac {a_1-a_0}{2}-\left|s-\eta -\delta \tfrac {a_1-a_0}{2}\right|\right)\tfrac {\gamma (s)}{2}\right)ds \\ =& \tfrac {1}{a_1-a_0}\left[\int ^{a_1-\eta -\delta \frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\eta +\delta \tfrac {a_1-a_0}{2}\right)\right.\right.\nonumber \\ & +\left.\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\eta +\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \nonumber \\ & +\int _{\eta -a_0+\delta \frac{a_1-a_0}{2}}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right) \right.\nonumber \\ & +\left.\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\eta -\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta . \end{align*}
</div>
<p> This completes the proof of <a href="#thm2">theorem 2.1</a>. <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="cor_thmwrapper " id="a0000000025">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">2.2</span>
  </div>
  <div class="cor_thmcontent">
  <p>Suppose that all the assumptions of <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\eta = \frac{a_0+a_1}{2}\) in <a href="#2.6" class="eqref">2.1</a>, we get the following bounds </p>
<div class="displaymath" id="2.9">
  \begin{align}  m_{1}(\delta ) \leq \delta \left[g\left(\tfrac {a_0+a_1}{2}\right)- \tfrac {g(a_0)+g(a_1)}{2}\right]+\tfrac {g(a_0)+g(a_1)}{2} &  -\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{1}(\delta )\hspace{1cm}\label{2.9} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000026">
  \begin{align*} & m_{1}(\delta )= \\ & \tfrac {1}{a_1-a_0} \Bigg[\int ^{(1-\delta )\frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\! \left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta \! +\! \tfrac {a_0+a_1}{2}\! +\! \delta \tfrac {a_1-a_0}{2}\right)\right. \left. \! + \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta \! +\! \tfrac {a_0+a_1}{2}\! +\! \delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \\ & \left. +\int ^{\delta \frac{a_1-a_0}{2}}_{(\delta -1)\frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+a_1}{2}-\delta \tfrac {a_1-a_0}{2}\right)\right.\right. \left.+ \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+a_1}{2}-\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \Bigg] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000027">
  \begin{align*} & M_{1}(\delta )= \\ & \tfrac {1}{a_1-a_0} \Bigg[\! \int ^{(1-\delta )\frac{a_1-a_0}{2}}_{-\delta \frac{a_1-a_0}{2}}\! \left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta \! +\! \tfrac {a_0+a_1}{2}\! +\! \delta \tfrac {a_1-a_0}{2}\right) \left.+\tfrac {\zeta \! +\! |\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+a_1}{2}\! +\! \delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \right.\\ &  +\int ^{\delta \frac{a_1-a_0}{2}}_{(\delta -1)\frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+a_1}{2}-\delta \tfrac {a_1-a_0}{2}\right)\right. \left.+ \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+a_1}{2}-\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \Bigg]. \end{align*}
</div>

  </div>
</div> Corollary \(3.2\) can be more useful to get different quadrature bounds, which we see it as follows. Throughout the section \(\gamma _0, \gamma _1, \tau _0\) and \(\tau _1\) are real constants. </p>
<p><div class="rem_thmwrapper " id="a0000000028">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.3</span>
  </div>
  <div class="rem_thmcontent">
  <p>If we choose \(\delta =0\) in <a href="#2.9" class="eqref">2.-1</a> then we get the bounds for trapezoidal rule </p>
<div class="displaymath" id="2.10">
  \begin{align} &  m_{2}\leq \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{2},\hspace{1cm}\label{2.10} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  m_{2}=\tfrac {1}{a_1-a_0} \Bigg[\int ^{\frac{a_1-a_0}{2}}_{-\frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+a_1}{2}\right)\right. \left.+\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+a_1}{2}\right)}\right)d\zeta \Bigg] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000030">
  \begin{align*}  M_{2}=\tfrac {1}{a_1-a_0} \Bigg[\int ^{\frac{a_1-a_0}{2}}_{-\frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+a_1}{2}\right)\right. & \left.+\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+a_1}{2}\right)}\right)d\zeta \Bigg]. \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> <b class="bf">Special Case 1.</b> If we choose, \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.10" class="eqref">2.0</a>, then </p>
<div class="displaymath" id="a0000000031">
  \begin{align*}  \tfrac {(a_1-a_0)}{8}(\gamma _0 -\tau _0)\leq \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq \tfrac {(a_1-a_0)}{8}(\tau _0-\gamma _0). \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span>. </p>
<p><b class="bf">Special Case 2.</b> If we choose, \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.10" class="eqref">2.0</a>, then </p>
<div class="displaymath" id="a0000000032">
  \begin{align}  m_{3}\leq \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{3},\nonumber \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000033">
  \begin{align*}  m_{3}=\tfrac {a_1-a_0}{8}\left(\tfrac {a_1-a_0}{3}(\gamma _1 + \tau _1)+\tfrac {a_0+a_1}{2}(\gamma _1 - \tau _1) + \gamma _0 - \tau _0 \right) \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000034">
  \begin{align*}  M_{3}=\tfrac {a_1-a_0}{8}\left(\tfrac {a_1-a_0}{3}(\gamma _1 + \tau _1)+\tfrac {a_0+a_1}{2}(\tau _1 - \gamma _1) + \tau _0 - \gamma _0 \right). \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>. </p>
<p><div class="rem_thmwrapper " id="a0000000035">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.4</span>
  </div>
  <div class="rem_thmcontent">
  <p>If we put \(\delta =1\) in <a href="#2.9" class="eqref">2.-1</a>, then we get the bounds for midpoint rule </p>
<div class="displaymath" id="2.11">
  \begin{align}  m_{4}\leq g\left(\tfrac {a_0+a_1}{2}\right)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1}g(s)ds\leq M_{4},\label{2.11} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000036">
  \begin{align*}  m_{4}=& \tfrac {1}{a_1-a_0} \Bigg[\int _{-\frac{a_1-a_0}{2}}^{0}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +a_1\right) +\tfrac {\zeta -|\zeta |}{2}\tau \left(\zeta +a_1\right)\right)d\zeta \nonumber \\ &  +\int ^{\frac{a_1-a_0}{2}}_{0}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +a_0\right) +\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +a_0\right)}\right)d\zeta \Bigg] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000037">
  \begin{align*}  M_4=& \tfrac {1}{a_1-a_0} \Bigg[\int _{-\frac{a_1-a_0}{2}}^{0}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +a_1\right) + \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +a_1\right)}\right)d\zeta \nonumber \\ & +\int ^{\frac{a_1-a_0}{2}}_{0}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +a_0\right) + \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +a_0\right)}\right)d\zeta \Bigg]. \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> <b class="bf">Special Case 3.</b> If we choose, \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.11" class="eqref">2.1</a>, then </p>
<div class="displaymath" id="a0000000038">
  \begin{align*}  \tfrac {(a_1-a_0)}{8}(\gamma _0 -\tau _0)& \leq g\left(\tfrac {a_0+a_1}{2}\right)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds \leq \tfrac {(a_1-a_0)}{8}(\tau _0-\gamma _0). \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span>. </p>
<p><b class="bf">Special Case 4.</b> If we choose, \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.11" class="eqref">2.1</a>, then </p>
<div class="displaymath" id="a0000000039">
  \begin{align*}  m_5\leq g\left(\tfrac {a_0+a_1}{2}\right)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_3, \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000040">
  \begin{align*}  m_5=\tfrac {(a_1-a_0)}{8}\Big(\tfrac {a_1-a_0}{3}\left(\gamma _1 +\tau _1\right)+a_0\gamma _1 - a_1\tau _1+\gamma _0-\tau _0 \Big) \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000041">
  \begin{align*}  M_5=\tfrac {(a_1-a_0)}{8}\Big(\tfrac {a_1-a_0}{3}\left(\gamma _1 +\tau _1\right)+ a_0\tau _1 - a_1\gamma _1+\tau _0 - \gamma _0 \Big). \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span>. </p>
<p><div class="rem_thmwrapper " id="a0000000042">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.5</span>
  </div>
  <div class="rem_thmcontent">
  <p>If we choose \(\delta =\frac{1}{2}\) in <a href="#2.9" class="eqref">2.-1</a> we get the bounds for the average of midpoint and trapezoidal rule </p>
<div class="displaymath" id="2.12">
  \begin{align}  m_{6}\leq \tfrac {1}{2}\left[\tfrac {g(a_0)+g(a_1)}{2}+ g\left(\tfrac {a_0+a_1}{2}\right)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{6},\label{2.12} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000043">
  \begin{align*}  m_{6}=& \tfrac {1}{a_1-a_0} \Bigg[\int ^{\frac{a_1-a_0}{4}}_{-\frac{a_1-a_0}{4}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\tfrac {3a_0+a_1}{4}\right) \right.\nonumber + \left.\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\tfrac {3a_0+a_1}{4}\right)}\right)d\zeta \nonumber \\ & +\int _{-\frac{a_1-a_0}{4}}^{\frac{a_1-a_0}{4}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+3a_1}{4}\right) + \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+3a_1}{4}\right)}\right)d\zeta \Bigg] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000044">
  \begin{align*}  M_{6}=& \tfrac {1}{a_1-a_0} \Bigg[\int ^{\frac{a_1-a_0}{4}}_{-\frac{a_1-a_0}{4}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\tfrac {3a_0+a_1}{4}\right) \right.+ \left.\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\tfrac {3a_0+a_1}{4}\right)}\right)d\zeta \nonumber \\ & +\int _{-\frac{a_1-a_0}{4}}^{\frac{a_1-a_0}{4}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +\tfrac {a_0+3a_1}{4}\right) \right.+ \left.\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +\tfrac {a_0+3a_1}{4}\right)}\right)d\zeta \Bigg]. \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <b class="bf">Special Case 5.</b> If we choose, \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.12" class="eqref">2.0</a> then </p>
<div class="displaymath" id="a0000000045">
  \begin{align*} & \tfrac {(a_1-a_0)}{16}(\gamma _0-\tau _0)\leq \\ & \leq \tfrac {1}{2}\left[\tfrac {g(a_0)+g(a_1)}{2}+ g\left(\tfrac {a_0+a_1}{2}\right)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds \leq \tfrac {(a_1-a_0)}{16}(\tau _0-\gamma _0). \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span>. </p>
<p><b class="bf">Special Case 6.</b> If we choose, \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.12" class="eqref">2.0</a>, then </p>
<div class="displaymath" id="a0000000046">
  \begin{align*}  m_{7}(\eta )\leq \tfrac {1}{2}\left[\tfrac {g(a_0)+g(a_1)}{2}+ g\left(\tfrac {a_0+a_1}{2}\right)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{7}, \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000047">
  \begin{align*}  m_{7}=\tfrac {(a_1-a_0)}{16}\left[\tfrac {(a_1-a_0)}{6}(\gamma _1+\tau _1)+\tfrac {a_0}{2}(\gamma _1-\tau _1)+\tfrac {a_1}{2}(\gamma _1-\tau _1)+\gamma _0-\tau _0\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000048">
  \begin{align*}  M_{7}=\tfrac {(a_1-a_0)}{16}\left[\tfrac {(a_1-a_0)}{6}(\gamma _1+\tau _1)+\tfrac {a_0}{2}(\tau _1-\gamma _1)+\tfrac {a_1}{2}(\tau _1-\gamma _1)+\tau _0-\gamma _0\right]. \end{align*}
</div>
<p> The above result is obtained in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>.<div class="cor_thmwrapper " id="a0000000049">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">2.6</span>
  </div>
  <div class="cor_thmcontent">
  <p>Suppose that all the assumptions of the <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\eta =a_0\) in <a href="#2.6" class="eqref">2.1</a>, then for any value of \(\delta \in [0,1]\), we get the following bounds </p>
<div class="displaymath" id="2.13">
  \begin{align}  m_{8}\leq \delta \left[\tfrac {g(a_0)-g(a_1)}{2}\right]+g(a_1) -\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{8},\label{2.13} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000050">
  \begin{align*} & m_{8}= \\ & \tfrac {1}{a_1-a_0} \left[ \int _{-\delta \frac{a_1-a_0}{2}}^{\left(1-\frac{\delta }{2}\right)(a_1-a_0)}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +a_0+\delta \tfrac {a_1-a_0}{2}\right) \! +\!  \tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +a_0+\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000051">
  \begin{align*} & M_{8}= \\ & \tfrac {1}{a_1-a_0} \left[ \int _{-\delta \frac{a_1-a_0}{2}}^{\left(1-\frac{\delta }{2}\right)(a_1-a_0)}\! \left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta \! +\! a_0\! +\! \delta \tfrac {a_1-a_0}{2}\right) \! +\!  \tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta \! +\! a_0\! +\! \delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \right]. \end{align*}
</div>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000052">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.7</span>
  </div>
  <div class="rem_thmcontent">
  <p>Suppose that all the assumptions of the <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\delta = 0\) in <a href="#2.13" class="eqref">2.-2</a>, then we get the bounds for nonstandard quadrature rule as follows </p>
<div class="displaymath" id="2.14">
  \begin{align}  m_{9}\leq g(a_1) - \tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds\leq M_{9},\label{2.14} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000053">
  \begin{align*}  m_{9} =\tfrac {1}{a_1-a_0} \left[\int _{0}^{a_1-a_0}\left(\tfrac {\zeta +|\zeta |}{2}\gamma (\zeta +a_0)+\tfrac {\zeta -|\zeta |}{2}\tau (\zeta +a_0)\right)d\zeta \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000054">
  \begin{align*}  M_{9} =\tfrac {1}{a_1-a_0} \left[\int _{0}^{a_1-a_0}\left(\tfrac {\zeta -|\zeta |}{2}\gamma (\zeta +a_0)+\tfrac {\zeta +|\zeta |}{2}\tau (\zeta +a_0)\right)d\zeta \right]. \quad \end{align*}
</div>

  </div>
</div> <b class="bf">Special Case 7.</b> If we choose, \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.14" class="eqref">2.-1</a> then </p>
<div class="displaymath" id="a0000000055">
  \begin{align*}  \tfrac {a_1-a_0}{2}\gamma _0\leq g(a_1) - \tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq \tfrac {a_1-a_0}{2}\tau _0. \end{align*}
</div>
<p> <b class="bf">Special Case 8.</b> If we choose, \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.14" class="eqref">2.-1</a>, then </p>
<div class="displaymath" id="a0000000056">
  \begin{align*}  m_{10} \leq g(a_1)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq M_{10}, \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000057">
  \begin{align*}  m_{10}=\tfrac {a_1-a_0}{6}\left[\gamma _1(2a_1+a_0)+3\gamma _0 \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000058">
  \begin{align*}  M_{10}=\tfrac {a_1-a_0}{6}\left[\tau _1(2a_1+a_0)+3\tau _0\right]. \end{align*}
</div>
<p> <div class="cor_thmwrapper " id="a0000000059">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">2.8</span>
  </div>
  <div class="cor_thmcontent">
  <p>Suppose that all the assumptions of the <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\eta =a_1\) in <a href="#2.6" class="eqref">2.1</a>, then for any value of \(\delta \in [0,1]\), we get the following bound </p>
<div class="displaymath" id="2.15">
  \begin{align}  m_{11}\leq \delta \left[\tfrac {g(a_1)-g(a_0)}{2}\right]+g(a_0)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{11},\label{2.15} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000060">
  \begin{align*} & m_{11}= \\ &  \tfrac {1}{a_1-a_0} \left[ \int _{\left(\frac{\delta }{2}-1\right)(a_1-a_0)}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2}\gamma \left(\zeta +a_1-\delta \tfrac {a_1-a_0}{2}\right) +\tfrac {\zeta -|\zeta |}{2}{\tau \left(\zeta +a_1-\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000061">
  \begin{align*} & M_{11}= \\ & \tfrac {1}{a_1-a_0} \left[ \int _{\left(\frac{\delta }{2}-1\right)(a_1-a_0)}^{\delta \frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2}\gamma \left(\zeta +a_1-\delta \tfrac {a_1-a_0}{2}\right) +\tfrac {\zeta +|\zeta |}{2}{\tau \left(\zeta +a_1-\delta \tfrac {a_1-a_0}{2}\right)}\right)d\zeta \right]\! \! . \end{align*}
</div>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000062">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.9</span>
  </div>
  <div class="rem_thmcontent">
  <p>Suppose that all the assumptions of the <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\delta = 0\) in <a href="#2.15" class="eqref">2.0</a>, then we get the following bound for nonstandard quadrature </p>
<div class="displaymath" id="2.16">
  \begin{align}  m_{12}\leq g(a_0) - \tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds\leq M_{12},\label{2.16} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000063">
  \begin{align*}  m_{12} =\tfrac {1}{a_1-a_0} \left[\int _{-(a_1-a_0)}^{0}\left(\tfrac {\zeta +|\zeta |}{2}\gamma (\zeta +a_0)+\tfrac {\zeta -|\zeta |}{2}\tau (\zeta +a_0)\right)d\zeta \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000064">
  \begin{align*}  M_{12} =\tfrac {1}{a_1-a_0} \left[\int _{-(a_1-a_0)}^{0}\left(\tfrac {\zeta -|\zeta |}{2}\gamma (\zeta +a_0)+\tfrac {\zeta +|\zeta |}{2}\tau (\zeta +a_0)\right)d\zeta \right].\quad \end{align*}
</div>

  </div>
</div> <b class="bf">Special Case 9.</b> If we choose \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.16" class="eqref">2.1</a>, gives the following inequality as </p>
<div class="displaymath" id="a0000000065">
  \begin{align*}  -\tfrac {a_1-a_0}{2}\tau _0\leq g(a_0) - \tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq -\tfrac {a_1-a_0}{2}\gamma _0. \end{align*}
</div>
<p> <b class="bf">Special Case 10.</b> If we choose \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.16" class="eqref">2.1</a>, gives </p>
<div class="displaymath" id="a0000000066">
  \begin{align*}  m_{13} \leq g(a_0)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq M_{13}, \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000067">
  \begin{align*}  m_{13}=\tfrac {a_1-a_0}{6}\left[\tau _1(2a_1 - 5a_0) - 3\tau _0 \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000068">
  \begin{align*}  M_{13}=\tfrac {a_1-a_0}{6}\left[\gamma _1(2a_1-5a_0)-3\gamma _1\right]. \end{align*}
</div>
<p> <div class="rem_thmwrapper " id="a0000000069">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.10</span>
  </div>
  <div class="rem_thmcontent">
  <p>Suppose that all the assumptions of the <a href="#thm2">theorem 2.1</a> hold, also if we substitute \(\delta = 1\) in both <a href="#2.13" class="eqref">2.-2</a> and <a href="#2.15" class="eqref">2.0</a>, gives the bound for trapezoidal rule </p>
<div class="displaymath" id="2.17">
  \begin{align}  m_{14}\leq \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq M_{14}, \label{2.17} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000070">
  \begin{align*}  m_{14}=& \tfrac {1}{a_1-a_0}\left[\int _{-\frac{a_1-a_0}{2}}^{\frac{a_1-a_0}{2}}\left(\tfrac {\zeta +|\zeta |}{2} \gamma \left(\zeta +\tfrac {a_0+a_1}{2}\right) +\tfrac {\zeta -|\zeta |}{2}\tau \left(\zeta +\tfrac {a_0+a_1}{2}\right)\right)d\zeta \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000071">
  \begin{align*}  M_{14}=& \tfrac {1}{a_1-a_0}\left[\int _{-\frac{a_1-a_0}{2}}^{\frac{a_1-a_0}{2}}\left(\tfrac {\zeta -|\zeta |}{2} \gamma \left(\zeta +\tfrac {a_0+a_1}{2}\right)+\tfrac {\zeta +|\zeta |}{2}\tau \left(\zeta +\tfrac {a_0+a_1}{2}\right)\right)d\zeta \right]. \end{align*}
</div>

  </div>
</div> <b class="bf">Special Case 11.</b> If we choose, \(\gamma (\eta )=\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _0\neq 0\) in <a href="#2.17" class="eqref">2.2</a> then </p>
<div class="displaymath" id="a0000000072">
  \begin{align*}  \tfrac {a_1-a_0}{8}(\gamma _0 - \tau _0)\leq \tfrac {g(a_0)+g(a_1)}{2} - \tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq -\tfrac {a_1-a_0}{8}(\tau _0 - \gamma _0). \end{align*}
</div>
<p> which is Corollary 2 of <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span> </p>
<p><b class="bf">Special Case 12.</b> If we choose, \(\gamma (\eta )=\gamma _1 \eta +\gamma _0 \neq 0\) and \(\tau (\eta )=\tau _1 \eta +\tau _0\neq 0\) in <a href="#2.17" class="eqref">2.2</a>, then </p>
<div class="displaymath" id="a0000000073">
  \begin{align*}  m_{15} \leq \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s) ds \leq M_{15}, \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000074">
  \begin{align*}  m_{15}=\tfrac {(a_1-a_0)}{8}\left[\tfrac {a_1-a_0}{3}(\gamma _1+\tau _1)+\tfrac {a_0+a_1}{2}(\gamma _1 - \tau _1) + \gamma _0 - \tau _0 \right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000075">
  \begin{align*}  M_{15}=\tfrac {(a_1-a_0)}{8}\left[\tfrac {a_1-a_0}{3}(\gamma _1+\tau _1)+\tfrac {a_0+a_1}{2}(\tau _1 - \gamma _1) + \tau _0 - \gamma _0 \right]. \end{align*}
</div>
<p> Although we have discussed bounded condition in <a href="#thm2">theorem 2.1</a>, however sometimes we are not able to find both the bounds. In order to be more effective, we need to obtain the theorems only for the cases of bounded below and bounded above. So the first theorem would be useful when \(a_0(\eta )\leq g{’}(\eta )\) and the second one would be useful when \(g{’}(\eta )\leq a_1(\eta )\). <div class="thm_thmwrapper " id="thm3">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.11</span>
  </div>
  <div class="thm_thmcontent">
  <p> Let \(g:I\to \mathbb {R}\), where \(I\) is an interval, be differentiable function in the interior \(I^0\) of \(I\), and let \([a_0,a_1]\) \(\subset I^0\). If \(g'\) is unbounded from above then \(\gamma (\eta )\leq g{’}(\eta )\) for any \(\gamma \in C[a_0,a_1]\), \(\eta \in [a_0,a_1]\), then for all \(\delta \in [0,1]\) the following inequality holds </p>
<div class="displaymath" id="a0000000076">
  \begin{align}  m_{16}(\eta ,\delta )\leq &  \delta \left[g(\eta )- \tfrac {g(a_0)+g(a_1)}{2}\right]+ \tfrac {a_1-\eta }{a_1-a_0}g(a_1)\nonumber \\ & +\tfrac {\eta -a_0}{a_1-a_0}g(a_0)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{16}(\eta ,\delta ), \label{2.18} \end{align}
</div>
<p> or </p>
<div class="displaymath" id="a0000000077">
  \begin{align}  (a_1-a_0)m_{16}(\eta ,\delta )& \leq (a_1-a_0)\delta \left[g(\eta )- \tfrac {g(a_0)+g(a_1)}{2}\right]+ (a_1-\eta )g(a_1)\nonumber \\ & \quad +(\eta -a_0)g(a_0)-\int _{a_0}^{a_1} g(s)ds\leq (a_1-a_0)M_{16}(\eta ,\delta ), \label{2.18a} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000078">
  \begin{align*} & m_{16}(\eta ,\delta )=\\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}(s-\eta )\gamma (s)ds\right.+\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^\eta \gamma (s)ds \right. - \left.\int _{\eta }^{a_1}\gamma (s)ds\right) \nonumber \\ & \quad -\max \Big\{ \delta \tfrac {a_1-a_0}{2},\left| \left(\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right)\right|, \left| \left(a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right)\right| \Big\}  \\ & \quad \times \left.\left(g(a_1)- g(a_0)-\int _{a_0}^{a_1}\gamma (s)ds \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000079">
  \begin{align*} & M_{16}(\eta ,\delta )=\\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}(s-\eta )\gamma (s)ds\right.+\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^\eta \gamma (s)ds \right. - \left.\int _{\eta }^{a_1}\gamma (s)ds\right) \nonumber \\ & \quad +\max \Big\{ \delta \tfrac {a_1-a_0}{2},\left| \left(\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right)\right|,\left| \left(a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right)\right| \Big\}  \\ & \quad \times \left.\left(g(a_1)- g(a_0)-\int _{a_0}^{a_1}\gamma (s)ds \right)\right]. \end{align*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000080">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Since </p>
<div class="displaymath" id="a0000000081">
  \begin{align*} & \int _{a_0}^{a_1} K(\eta ,s)\left(g{’}(s)-\gamma (s)\right)ds=\nonumber \\ & =\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0) \\ & \quad -\int _{a_0}^{a_1} g(s)ds -\int _{a_0}^{a_1} K(\eta ,s)\gamma (s)ds. \\ & =\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)\nonumber \\ & \quad -\int _{a_0}^{a_1}\! \!  g(s)ds-\left[\int _{a_0}^\eta \!  \left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)\gamma (s)ds +\left.\int _{\eta }^{a_1}\! \left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)\gamma (s)ds\right.\right], \end{align*}
</div>
<p> using modulus property, we have </p>
<div class="displaymath" id="a0000000082">
  \begin{align*} & \bigg|\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1) +(\eta -a_0)g(a_0) \\ & \quad -\int _{a_0}^{a_1}\! \! \!  g(s)ds-\left[\int _{a_0}^\eta \! \!  \left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)\gamma (s)ds \nonumber +\left. \! \! \int _{\eta }^{a_1}\! \left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)\gamma (s)ds\right]\right|=\nonumber \\ & =\left|\int _{a_0}^{a_1} K(\eta ,s)\left(g{’}(\eta )-\gamma (s)\right)ds\right|\nonumber \\ & \leq \int _{a_0}^{a_1} \left|K(\eta ,s)\right|\left(g{’}(s)-\gamma (s)\right)ds\nonumber \\ & \leq \max _{s\in [a_0,a_1]}\left|K(\eta ,s)\right|\int _{a_0}^{a_1}\left(g{’}(s)-\gamma (s)\right)ds.\nonumber \end{align*}
</div>
<div class="displaymath" id="a0000000083">
  \begin{align} & =\max \Big\{ \delta \tfrac {a_1-a_0}{2},\left(\eta -a_0+ \delta \tfrac {a_1-a_0}{2}\right),\left(a_1-\eta - \delta \tfrac {a_1-a_0}{2}\right)\Big\} \nonumber \\ & \quad \times \left(g(a_1)- g(a_0)-\int _{a_0}^{a_1}\gamma (s)ds \right).\label{2.19} \end{align}
</div>
<p> Arrangement of <a href="#2.19" class="eqref">2.-4</a> gives inequality <a href="#2.18" class="eqref">2.3</a>. <div class="proof_wrapper" id="a0000000084">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="rem_thmwrapper " id="a0000000085">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.12</span>
  </div>
  <div class="rem_thmcontent">
  <p>The inequality <a href="#2.18" class="eqref">2.3</a> represents the generalized case of Theorem 2 presented in <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <div class="cor_thmwrapper " id="cor:2.13">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">2.13</span>
  </div>
  <div class="cor_thmcontent">
  <p> Suppose that all the assumptions of <a href="#thm3">theorem 2.11</a> holds and if we substitute \(\eta =\frac{a_0+a_1}{2}\) in <a href="#2.18" class="eqref">2.3</a>, then we obtain </p>
<div class="displaymath" id="2.20">
  \begin{align}  m_{17}(\delta )& \leq \delta \left[g \left(\tfrac {a_0+a_1}{2}\right)- \tfrac {g(a_0)+g(a_1)}{2}\right] + \tfrac {g(a_0)+g(a_1)}{2}-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{17}(\delta ), \label{2.20} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000086">
  \begin{align*} & m_{17}(\delta )= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-\tfrac {a_0+a_1}{2}\right)\gamma (s)ds\right.+\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^{\eta }\gamma (s)ds - \int _{\eta }^{a_1}\gamma (s)ds\right) \\ & \quad -\max \Big\{ \delta \tfrac {a_1-a_0}{2},(1-\delta )\tfrac {a_1-a_0}{2},(\delta -1)\tfrac {a_1-a_0}{2}\Big\}  \\ & \quad \times \left.\left(g(a_1)- g(a_0)-\int _{a_0}^{a_1}\gamma (s)ds \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000087">
  \begin{align*} & M_{17}(\delta )= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-\tfrac {a_0+a_1}{2}\right)\gamma (s)ds\right. +\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^{\eta }\gamma (s)ds - \int _{\eta }^{a_1}\gamma (s)ds\right) \\ & \quad \left.+\max \Big\{ \delta \tfrac {a_1-a_0}{2},(1-\delta )\tfrac {a_1-a_0}{2},(\delta -1)\tfrac {a_1-a_0}{2}\Big\} \right.\\ & \quad \times \left.\left(g(a_1)- g(a_0)-\int _{a_0}^{a_1}\gamma (s)ds \right)\right]. \end{align*}
</div>

  </div>
</div> <div class="thm_thmwrapper " id="thm4">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.14</span>
  </div>
  <div class="thm_thmcontent">
  <p> Let \(g:I\to \mathbb {R}\), where \(I\) is an interval, be a function differentiable in the interior \(I^0\) of \(I\), and let \([a_0,a_1]\) \(\subset I^0\). If \(g'\) is unbounded from below then \(g{’}(\eta )\leq \tau (\eta )\) for any \(\tau \in C[a_0,a_1]\), \(\eta \in [a_0,a_1]\), then for all \(\delta \in [0,1]\) the following inequality holds </p>
<div class="displaymath" id="2.21">
  \begin{align} \label{2.21} m_{18}(\eta ,\delta )\leq &  \delta \left[g(\eta )- \tfrac {g(a_0)+g(a_1)}{2}\right]+ \tfrac {a_1-\eta }{a_1-a_0}g(a_1) \\ &  +\tfrac {\eta -a_0}{a_1-a_0}g(a_0)-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds \leq M_{18}(\eta ,\delta ), \nonumber \end{align}
</div>
<p> or </p>
<div class="displaymath" id="a0000000088">
  \begin{align}  (a_1-a_0)m_{18}(\eta ,\delta )& \leq (a_1-a_0)\delta \left[g(\eta )- \tfrac {g(a_0)+g(a_1)}{2}\right]+ (a_1-\eta )g(a_1)\nonumber \\ & \quad +(\eta -a_0)g(a_0)-\int _{a_0}^{a_1} g(s)ds \leq M_{18}(\eta ,\delta ), \label{2.21a} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000089">
  \begin{align*} & m_{18}(\eta ,\delta )= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}(s-\eta )\tau (s) ds\right.+\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^\eta \tau (s)ds \right. - \left.\int _{\eta }^{a_1}\tau (s)ds\right) \nonumber \\ & \quad -\max \Big\{ \delta \tfrac {a_1-a_0}{2},\left| \left(\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right)\right|, \left| \left(a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right)\right| \Big\}  \\ & \quad \times \left.\left(\int _{a_0}^{a_1}\tau (s)-g(a_1)-g(a_0)ds \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000090">
  \begin{align*}  M_{18}(\eta ,\delta )=& \tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}(s-\eta )\tau (s) ds\right.+\delta \tfrac {a_1-a_0}{2}\left(\int _{a_0}^\eta \tau (s)ds \right. - \left.\int _{\eta }^{a_1}\tau (s)ds\right) \nonumber \\ & +\max \Big\{ \delta \tfrac {a_1-a_0}{2},\left| \left(\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right)\right|, \left| \left(a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right)\right| \Big\}  \\ & \times \left.\left(\int _{a_0}^{a_1}\tau (s)-g(a_1)-g(a_0)ds \right)\right]. \end{align*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000091">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Since </p>
<div class="displaymath" id="a0000000092">
  \begin{align*} & \int _{a_0}^{a_1} K(\eta ,s)\left(g{’}(s)-\tau (s)\right)ds=\nonumber \\ & =\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0) \\ & \quad -\int _{a_0}^{a_1} g(s)ds-\int _{a_0}^{a_1} K(\eta ,s)\tau (s)ds\nonumber \\ & =\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)\nonumber \\ & \quad -\int _{a_0}^{a_1}\!  g(s)ds-\left[\int _{a_0}^\eta \! \left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)\tau (s)ds\right. \left. \! +\! \int _{\eta }^{a_1}\! \left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)\tau (s)ds \right]\! , \end{align*}
</div>
<p> so we have </p>
<div class="displaymath" id="a0000000093">
  \begin{align} & \left|\delta (a_1-a_0)\left[g(\eta )-\tfrac {g(a_0)+g(a_1)}{2}\right]+(a_1-\eta )g(a_1)+(\eta -a_0)g(a_0)\right.\nonumber \\ & -\! \int _{a_0}^{a_1}\!  g(s)ds\left.-\left[\int _{a_0}^\eta \! \left(s-\eta +\delta \tfrac {a_1-a_0}{2}\right)\tau (s)ds\right.\nonumber \left.\! +\! \int _{\eta }^{a_1}\! \! \left(s-\eta -\delta \tfrac {a_1-a_0}{2}\right)\tau (s)ds \right]\right|= \\ & =\left|\int _{a_0}^{a_1} K(\eta ,s)\left(g’(t)-\tau (s)\right)ds\right|\nonumber \\ & \leq \int _{a_0}^{a_1} \left|K(\eta ,s)\right|\left(\tau (s)-g{’}(s)\right)ds\nonumber \end{align}
</div>
<div class="displaymath" id="a0000000094">
  \begin{align}  \leq &  \max _{s\in [a_0,a_1]}\left|K(\eta ,s)\right|\int _{a_0}^{a_1}\left(\tau (s)-g{’}(s)\right)ds\nonumber \\ =& \max \Big\{ \left|\eta -a_0+\delta \tfrac {a_1-a_0}{2}\right|,\delta \tfrac {a_1-a_0}{2},\left|a_1-\eta -\delta \tfrac {a_1-a_0}{2}\right|\Big\}  \nonumber \\ & \quad \times \left(\int _{a_0}^{a_1}\tau (s)ds-g(a_1)+ g(a_0) \right).\label{2.22} \end{align}
</div>
<p> Rearrangement of <a href="#2.22" class="eqref">2.-5</a> gives the inequality <a href="#2.21" class="eqref">2.-2</a>. <div class="proof_wrapper" id="a0000000095">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="rem_thmwrapper " id="a0000000096">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.15</span>
  </div>
  <div class="rem_thmcontent">
  <p>The inequality established in <a href="#2.21" class="eqref">2.-2</a> is the special case of Theorem 3 presented in <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <div class="cor_thmwrapper " id="cor:2.16">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">2.16</span>
  </div>
  <div class="cor_thmcontent">
  <p> Suppose that all the assumptions of <a href="#thm4">theorem 2.14</a> hold and if we substitute \(\eta =\frac{a_0+a_1}{2}\) in <a href="#2.21" class="eqref">2.-2</a>, then we will get </p>
<div class="displaymath" id="2.23">
  \begin{align}  m_{19}(\delta )& \leq \delta \left[g\left(\tfrac {a_0+a_1}{2}\right)-\tfrac {g(a_0)+g(a_1)}{2}\right] +\tfrac {g(a_0)+g(a_1)}{2}-\int _{a_0}^{a_1} g(s)ds\leq M_{19}(\delta ),\label{2.23} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000097">
  \begin{align*} & m_{19}(\delta )= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-\tfrac {a_0+a_1}{2}\right)\tau (s)ds\right. +\delta \tfrac {a_1-a_0}{2}\Big(\int _{a_0}^{\frac{a_0+a_1}{2}} \tau (s)ds - \int _{\frac{a_0+a_1}{2}}^{a_1} \tau (s)ds\Big) \nonumber \\ & \quad -\max \Big\{ \delta \tfrac {a_1-a_0}{2},(1-\delta )\tfrac {a_1-a_0}{2},(\delta -1)\tfrac {a_1-a_0}{2}\Big\}  \left.\left(\int _{a_0}^{a_1}\tau (s)ds-g(a_1)+ g(a_0) \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000098">
  \begin{align*} & M_{19}(\delta )= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-\tfrac {a_0+a_1}{2}\right)\tau (s)ds\right. +\delta \tfrac {a_1-a_0}{2}\Big(\int _{a_0}^{\frac{a_0+a_1}{2}} \tau (s)ds - \int _{\frac{a_0+a_1}{2}}^{a_1} \tau (s)ds\Big)\\ & \quad +\max \Big\{ \delta \tfrac {a_1-a_0}{2},(1-\delta )\tfrac {a_1-a_0}{2},(\delta -1)\tfrac {a_1-a_0}{2}\Big\}  \left.\left(\int _{a_0}^{a_1}\tau (s)ds-g(a_1)+ g(a_0) \right)\right]. \end{align*}
</div>

  </div>
</div> <div class="rem_thmwrapper " id="rem:2.17">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.17</span>
  </div>
  <div class="rem_thmcontent">
  <p> If \(\gamma (\eta )\) \(\leq g'(\eta )\leq \tau (\eta )\) for any \(\eta \in [a_0,a_1]\) and \(\gamma ,\tau \in C[a_0,a_1]\), and if we choose \(\delta =1\), then the error of nonstandard quadrature can be bounded as </p>
<div class="displaymath" id="2.24">
  \begin{align}  m_{20}\leq \tfrac {1}{2}\left[-g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right) +g(a_1)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{20},\label{2.24} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000099">
  \begin{align*}  m_{20}=\tfrac {1}{a_1-a_0}\left[\! \int _{a_0}^{\frac{a_0+a_1}{2}}\! \! \left(s-a_0\right)\gamma (s)ds\right. \left.\! +\!  \int _{\frac{a_0+a_1}{2}}^{a_1}\! (s-a_1)\gamma (s)ds \! +\!  \tfrac {(a_1-a_0)}{2}\! \int _{a_0}^{a_1}\! \! \gamma (s)ds\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000100">
  \begin{align*}  M_{20}=& \tfrac {1}{a_1-a_0}\! \left[\int _{a_0}^{\frac{a_0+a_1}{2}}\! \! \left(s-a_0\right)\tau (s)ds\right. \left.\! +\!  \int _{\frac{a_0+a_1}{2}}^{a_1}\! (s-a_1)\tau (s)ds \! +\!  \tfrac {(a_1-a_0)}{2}\! \int _{a_0}^{a_1}\! \tau (s)ds\right] \end{align*}
</div>
<p> which is the Corollary \(3\) obtained in <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and the Corollary \(4\) obtained in <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <div class="proof_wrapper" id="a0000000101">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> To prove <a href="#2.24" class="eqref">2.-4</a> we should use the results of both <a href="#cor:2.13">corollary 2.13</a> and <a href="#cor:2.16">corollary 2.16</a> simultaneously. First by substituting \(\delta =1\) in <a href="#2.20" class="eqref">2.-3</a>, we have </p>
<div class="displaymath" id="a0000000102">
  \begin{align} & \tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-a_0\right)\gamma (s)ds+\tfrac {a_1-a_0}{2}\left(\int _{a_0}^{\frac{a_0+a_1}{2}} \gamma (s)ds + \int _{\frac{a_0+a_1}{2}}^{a_1} \gamma (s)ds\right)\right] \leq \nonumber \\ & \leq \tfrac {1}{2}\left[-g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right) +g(a_1)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\label{2.25} \end{align}
</div>
<p> provided that \(\gamma (\eta )\leq g'(\eta )\), \(\forall \eta \in [a_0,a_1]\). </p>
<p>On the other hand, by assuming \(\delta =1\) in <a href="#2.23" class="eqref">2.-4</a>, we obtain </p>
<div class="displaymath" id="2.26">
  \begin{align} & \tfrac {1}{2}\left[-g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right) +g(a_1)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq \label{2.26} \\ & \leq \tfrac {1}{a_1-a_0}\left[\int _{a_0}^{a_1}\left(s-a_0\right)\tau (s)ds+\tfrac {a_1-a_0}{2}\left(\int _{a_0}^{\frac{a_0+a_1}{2}} \tau (s)ds + \int _{\frac{a_0+a_1}{2}}^{a_1} \tau (s)ds\right)\right] \nonumber \end{align}
</div>
<p> provided that \(g'(\eta )\leq \tau (\eta )\) \(\forall \eta \in [a_0,a_1]\). </p>
<p>Now by combining the above two results <a href="#2.25" class="eqref">2.-3</a> and <a href="#2.26" class="eqref">2.-2</a>, the inequality <a href="#2.24" class="eqref">2.-4</a> is derived. <div class="proof_wrapper" id="a0000000103">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="rem_thmwrapper " id="a0000000104">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.18</span>
  </div>
  <div class="rem_thmcontent">
  <p>If \(\gamma (\eta )\) \(\leq g'(\eta )\leq \tau (\eta )\) for any \(\eta \in [a_0,a_1]\) and \(\gamma ,\tau \in C[a_0,a_1]\), and if we put \(\delta =1\), then the error of nonstandard quadrature can be bounded as </p>
<div class="displaymath" id="2.27">
  \begin{align}  m_{21}\leq \tfrac {1}{2}\left[g(a_0)+2g\left(\tfrac {a_0+a_1}{2}\right) -g(a_1)\right]-\tfrac {1}{a_1-a_0}\int _{a_0}^{a_1} g(s)ds\leq M_{21} \label{2.27} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000105">
  \begin{align*} & m_{21}= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{\frac{a_0+a_1}{2}}\left(s-a_0\right)\tau (s)ds+ \int _{\frac{a_0+a_1}{2}}^{a_1}(s-a_1)\tau (s)ds - \tfrac {(a_1-a_0)}{2}\int _{a_0}^{a_1}\tau (s)ds\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000106">
  \begin{align*} & M_{21}= \\ & =\tfrac {1}{a_1-a_0}\left[\int _{a_0}^{\frac{a_0+a_1}{2}}\left(s-a_0\right)\gamma (s)ds\right. \left.+ \int _{\frac{a_0+a_1}{2}}^{a_1}(s-a_1)\gamma (s)ds - \tfrac {(a_1-a_0)}{2}\int _{a_0}^{a_1}\gamma (s)ds\right] \end{align*}
</div>
<p> which is the Corollary \(4\) of <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and Corollary \(5\) of <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> <div class="proof_wrapper" id="a0000000107">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Proof of <a href="#2.27" class="eqref">2.-1</a> is similar to that of <a href="#rem:2.17">remark 2.17</a>, if one replaces \(\delta =1\) in <a href="#2.20" class="eqref">2.-3</a> and <a href="#2.23" class="eqref">2.-4</a>, respectively, and combines them together. <div class="proof_wrapper" id="a0000000108">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000109">3 Applications to Numerical Quadrature Rules</h1>
<p>Let \(A_n:a_0=z_0{\lt}z_1{\lt}\cdots {\lt}z_n=a_1\) be a partition of the interval \([a_0,a_1]\) and let \(\Delta z_k=z_{k+1}-z_k, k\in \{ 0,1,2,\cdots ,n-1\} .\) Then </p>
<div class="displaymath" id="3.1">
  \begin{align}  \int _{a_0}^{a_{1}}g(s)ds=Q_n(A_n,g)+R_n(A_n,g)\label{3.1} \end{align}
</div>
<p> Consider a general quadrature formula </p>
<div class="displaymath" id="3.2">
  \begin{align} & Q_n(A_n,g):= \label{3.2} \\ & :=\sum _{k=0}^{n-1} \Big[{\Delta z_k}\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\}  +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big] \nonumber \end{align}
</div>
<p> for all \(\delta \in [0,1]\).<div class="thm_thmwrapper " id="a0000000110">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3.1</span>
  </div>
  <div class="thm_thmcontent">
  <p>Suppose that all the assumptions of <a href="#thm2">theorem 2.1</a> hold. Considering <a href="#3.1" class="eqref">3.1</a>, where \(Q_n(.,.)\) is given by formula <a href="#3.2" class="eqref">3.2</a> and the remainder \(R_n (., .)\) satisfies the estimates </p>
<div class="displaymath" id="3.3">
  \begin{align}  \left|R_n(A_n,g)\right| \leq \sup \left\{ \left|R_1 \right|, \left|R_2\right|\right\} \label{3.3} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000111">
  \begin{align*} & R_1= \\ & =\int ^{z_{k+1}-\eta _k-\delta \frac{\Delta z_k}{2}}_{-\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k+|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right) +\tfrac {\zeta _k-|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k\\ & \quad +\int _{\eta -z_{k+1}+\delta \frac{\Delta z_k}{2}}^{\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k+|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right) +\tfrac {\zeta _k-|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000112">
  \begin{align*} & R_2= \\ & =\int ^{z_{k+1}-\eta _k-\delta \frac{\Delta z_k}{2}}_{-\delta \tfrac {\Delta z_k}{2}}\left(\tfrac {\zeta _k-|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right) +\tfrac {\zeta _k+|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k\\ & \quad +\int _{\eta -z_{k+1}+\delta \tfrac {\Delta z_k}{2}}^{\delta \tfrac {\Delta z_k}{2}}\left(\tfrac {\zeta _k-|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right) +\tfrac {\zeta _k+|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k \end{align*}
</div>
<p> for all \(\eta _k \in [z_k, z_{k+1}].\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000113">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Applying inequality <a href="#2.6" class="eqref">2.1</a> on the intervals, \([z_k, z_{k+1}]\), we can state that </p>
<div class="displaymath" id="a0000000114">
  \begin{align*} & R_k(A_k,g)= \\ & \! =\! \int _{z_k}^{z_{k+1}}\! \! \! g(s)ds\! -\! {\Delta z_k}\delta \left\{ g(\eta _k)\! -\! \tfrac {g(z_k) \! +\!  g(z_{k+1})}{2}\right\}  \! +\! (z_{k+1}\! -\! \eta _k)g(z_{k+1})\! +\! (\eta _k \! -\!  z_k)g(z_k) \end{align*}
</div>
<p> we sum the inequalities presented above over \(k\) from \(0\) to \(n-1\). This gives </p>
<div class="displaymath" id="a0000000115">
  \begin{align*}  R_n(A_n,g)=\int _{a_0}^{a_1}g(s)ds-\sum _{k=0}^{n-1}& \bigg[{\Delta z_k}\delta \Big\{ g(\eta _k)-\tfrac {g(z_k) + g(z_{k+1})}{2}\Big\}  \\ & +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\bigg] \end{align*}
</div>
<p> It follows from <a href="#2.6a" class="eqref">2.2</a> that </p>
<div class="displaymath" id="a0000000116">
  \begin{align*} & \left|R_n(A_n,g)\right|= \\ =& \left|\int _{a_0}^{a_1}g(s)ds-\sum _{k=0}^{n-1}\left[{\Delta z_k}\delta \left\{ g(\eta _k)-\tfrac {g(z_k) + g(z_{k+1})}{2}\right\} \right.\right.\\ & \left.+(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big]\right| \\ \leq &  \sup \left\{ \left| \left.\int ^{z_{k+1}-\eta _k-\delta \frac{\Delta z_k}{2}}_{-\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k+|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)\right.\right.\right.\right. \nonumber \\ &  +\left.\tfrac {\zeta _k-|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k \\ & +\int _{\eta -z_{k+1}+\delta \frac{\Delta z_k}{2}}^{\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k+|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right) \right.\nonumber \\ & +\left.\left.\left.\tfrac {\zeta _k-|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k\right.\right|, \\ & \left.\left| \left.\int ^{z_{k+1}-\eta _k-\delta \frac{\Delta z_k}{2}}_{-\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k-|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)\right.\right.\right.\right.\nonumber \\ & +\left.\tfrac {\zeta _k+|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k+\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k\\ & +\int _{\eta -z_{k+1}+\delta \frac{\Delta z_k}{2}}^{\delta \frac{\Delta z_k}{2}}\left(\tfrac {\zeta _k-|\zeta _k|}{2}\gamma \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right) \right.\nonumber \\ & +\left.\left.\left.\left.\tfrac {\zeta _k+|\zeta _k|}{2}{\tau \left(\zeta _k+\eta _k-\delta \tfrac {\Delta z_k}{2}\right)}\right)d\zeta _k\right.\right|\right\}  \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000117">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="thm_thmwrapper " id="a0000000118">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3.2</span>
  </div>
  <div class="thm_thmcontent">
  <p>Let \(\phi \) be defined as in <a href="#thm3">theorem 2.11</a>. Then <a href="#3.1" class="eqref">3.1</a> holds where \(Q_n(A_n,g)\) is given by formula <a href="#3.2" class="eqref">3.2</a> and the remainder \(R_n (A_n, g)\) satisfies the estimates </p>
<div class="displaymath" id="a0000000119">
  \begin{eqnarray} & & \left|R_n(A_n,\phi )\right| \leq \sup \left\{ \left|R_3 \right|,\left|R_4\right|\right\}  \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000120">
  \begin{align*}  R_{3}=& \left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\gamma (s)ds\right.+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \gamma (s)ds \right. - \left.\int _{\eta _k}^{z_{k+1}}\gamma (s)ds\right) \nonumber \\ & -\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|, \left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\} \times \end{align*}
</div>
<div class="displaymath" id="a0000000121">
  \begin{align*} & \times \left.\left(g(z_{k+1})- g(z_k)-\int _{z_k}^{z_{k+1}}\gamma (s)ds \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000122">
  \begin{multline*}  R_{4}=\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\gamma (s)ds\right.+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \gamma (s)ds \right. - \left.\int _{\eta _k}^{z_{k+1}}\gamma (s)ds\right) \\ +\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|, \left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\}  \\ \times \left.\left(g(z_{k+1})- g(z_k)-\int _{z_k}^{z_{k+1}}\gamma (s)ds \right)\right] \end{multline*}
</div>
<p> for all \(\eta _k \in [z_k, z_{k+1}].\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000123">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Applying inequality <a href="#2.18a" class="eqref">2.4</a> on the intervals, \([z_k, z_{k+1}]\), we can state that </p>
<div class="displaymath" id="a0000000124">
  \begin{multline*}  R_k(A_k,g)=\int _{z_k}^{z_{k+1}}g(s)ds-\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\} \\ +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k) \end{multline*}
</div>
<p> we sum the inequalities presented above over \(k\) from \(0\) to \(n-1\). This gives </p>
<div class="displaymath" id="a0000000125">
  \begin{multline*}  \left.R_n(A_n,g)\right. =\left.\int _{a_0}^{a_{1}}g(s)ds -\sum _{k=0}^{n-1}\left[\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\} \right.\right.\\ \left.+(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big]\right.\\ \end{multline*}
</div>
<p> It follows from <a href="#2.7" class="eqref">2.3</a> that </p>
<div class="displaymath" id="a0000000126">
  \begin{align*} & \left|R_n(A_n,g)\right|= \\ =& \left|\int _{a_0}^{a_{1}}g(s)ds -\sum _{k=0}^{n-1}\left[\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\} \right.\right.\\ & +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big]\Big| \\ \leq &  \sup \left\{ \left|\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\gamma (s)ds\right.\right.\right. +\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \gamma (s)ds \right. - \left.\int _{\eta _k}^{z_{k+1}}\gamma (s)ds\right) \nonumber \\ & -\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|, \left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\}  \\ & \times \left.\left.\left(g(z_{k+1})- g(z_k)-\int _{z_k}^{z_{k+1}}\gamma (s)ds \right)\right]\right|, \\ & \left|\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\gamma (s)ds\right.+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \gamma (s)ds \right. - \left.\int _{\eta _k}^{z_{k+1}}\gamma (s)ds\right)\right. \nonumber \\ & +\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|, \left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\}  \end{align*}
</div>
<div class="displaymath" id="a0000000127">
  \begin{align*} & \times \left.\left.\left.\left(g(z_{k+1})- g(z_k)-\int _{z_k}^{z_{k+1}}\gamma (s)ds \right)\right]\right|\right\}  \end{align*}
</div>
<p><div class="thm_thmwrapper " id="a0000000128">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3.3</span>
  </div>
  <div class="thm_thmcontent">
  <p>Suppose that all the assumptions of <a href="#thm4">theorem 2.14</a> hold. Considering <a href="#3.1" class="eqref">3.1</a> where \(Q_n(A_n,g)\) is given by formula <a href="#3.2" class="eqref">3.2</a> and the remainder \(R_n (A_n, g)\) satisfies the estimates </p>
<div class="displaymath" id="3.5">
  \begin{align} & \left|R_n(A_n,g)\right| \leq \sup \left\{ \left|R_5 \right|, \left|R_6\right|\right\} \label{3.5} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000129">
  \begin{align*}  R_{5}=& \left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\tau (s)ds+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \tau (s)ds \right.\right. - \left.\int _{\eta _k}^{z_{k+1}}\tau (s)ds\right) \nonumber \\ -& \max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|,\left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\} \\ & \times \left.\left(\int _{z_k}^{z_{k+1}}\tau (s)-g(z_{k+1})-g(z_k)ds \right)\right] \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000130">
  \begin{align*}  R_{6}& =\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\tau (s)ds+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \tau (s)ds \right.\right. - \left.\int _{\eta _k}^{z_{k+1}}\tau (s)ds\right) \nonumber \\ & \quad +\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|,\left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\} \\ & \quad \times \left.\left(\int _{z_k}^{z_{k+1}}\tau (s)-g(z_{k+1})-g(z_k)ds \right)\right] \end{align*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000131">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Applying inequality <a href="#2.21a" class="eqref">2.-1</a> on the intervals, \([z_k, z_{k+1}]\), we can state that </p>
<div class="displaymath" id="a0000000132">
  \begin{align*}  R_k(A_k,g)=& \int _{z_k}^{z_{k+1}}g(s)ds-\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\}  \\ & +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k) \end{align*}
</div>
<p> we sum the inequalities presented above over \(k\) from \(0\) to \(n-1\). This gives </p>
<div class="displaymath" id="a0000000133">
  \begin{align*}  \left.R_n(A_n,g)\right.=& \left.\int _{a_0}^{a_1}g(s)ds -\sum _{k=0}^{n-1}\left[\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\} \right.\right. \\ & \left.+(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big]\right. \end{align*}
</div>
<p> It follows from <a href="#2.7" class="eqref">2.3</a> that </p>
<div class="displaymath" id="a0000000134">
  \begin{align*} & \left|R_n(A_n,g)\right|= \\ =& \left|\int _{a_0}^{a_1}g(s)ds -\sum _{k=0}^{n-1}\left[\Delta z_k\delta \left\{ g(\eta _k)-\tfrac {g(z_k)+ g(z_{k+1})}{2}\right\} \right.\right.\\ & +(z_{k+1}-\eta _k)g(z_{k+1})+(\eta _k-z_k)g(z_k)\Big]\Big|\leq \end{align*}
</div>
<div class="displaymath" id="a0000000135">
  \begin{align*}  \leq &  \sup \left\{ \left|\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\tau (s)ds+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \tau (s)ds - \left.\int _{\eta _k}^{z_{k+1}}\tau (s)ds\right) \right.\right.\right.\right.\nonumber \\ & \left.-\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \tfrac {\Delta z_k}{2}\right)\right|,\left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\} \right. \\ & \times \left.\left.\left(\int _{z_k}^{z_{k+1}}\tau (s)-g(z_{k+1})-g(z_k)ds \right)\right]\right|, \\ & \left|\left[\int _{z_k}^{z_{k+1}}(s-\eta _k)\tau (s)ds+\delta \tfrac {\Delta z_k}{2}\left(\int _{z_k}^{\eta _k} \tau (s)ds\right.\right.\right. \left. - \int _{\eta _k}^{z_{k+1}}\tau (s)ds\right) \nonumber \\ & \left.+\max \left\{ \delta \tfrac {\Delta z_k}{2},\left| \left(\eta _k-z_k+\delta \frac{\Delta z_k}{2}\right)\right|,\left| \left(z_{k+1}-\eta _k-\delta \tfrac {\Delta z_k}{2}\right)\right| \right\} \right. \\ & \times \left.\left.\left.\left(\int _{z_k}^{z_{k+1}}\tau (s)-g(z_{k+1})-g(z_k)ds \right)\right]\right|\right\}  \end{align*}
</div>
<h1 id="a0000000136">4 conclusion</h1>
<p> Inspired by the work of <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span> and <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span>, we generalized Ostrowski inequality to obtained explicit error bounds for standard and nonstandard numerical quadrature formulae. By using appropriate substitution in our main results we get various established results given in <span class="cite">
	[
	<a href="#nazo7" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#H5" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#H6" >14</a>
	]
</span> and <span class="cite">
	[
	<a href="#H10" >21</a>
	]
</span> as our special cases. In the last section we have discussed its applications in Numerical quadrature rules. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="double-sided">1</a></dt>
  <dd><p><i class="sc">W.G. Alshanti, G.V. Milovanović</i>, <i class="it">Double-sided inequalities of Ostrowski’s type and some applications</i>, J. Comput. Anal. Appl., <b class="bf">28</b> (2020) (4), pp.&#160;724–736. </p>
</dd>
  <dt><a name="H1">2</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/978-3-642-64971-4"> <i class="sc">E.F. Beckenbach, R. Bellman</i>, Inequalities, Springer-Verlag, Berlin, 1961. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="H2">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/s0893-9659(99)00139-1"> <i class="sc">S.S. Dragomir, P. Cerone, J. Roumeliotis</i>, <i class="it">A new generalization of Ostrowski integral inequality for mappings whose derivatives are bounded and applications in numerical integration and for special means</i>, Appl. Math. Lett., <b class="bf">13</b> (2000) no. 1, pp.&#160;19–25. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="H3">4</a></dt>
  <dd><p><i class="sc">W. Gautschi</i>, <i class="it">Numerical Analysis: An Introduction</i>, Birkahauser, Boston 1997 </p>
</dd>
  <dt><a name="H4">5</a></dt>
  <dd><p><i class="sc">G.H. Hardy, J.E. Littlewood, G. Polya</i>, <i class="it">Inequalities</i>, Cambridge University Press, 1934. </p>
</dd>
  <dt><a name="nazo8">6</a></dt>
  <dd><p><i class="sc">M. Imtiaz, N. Irshad, A.R. Khan</i>, <i class="it">Generalization of weighted Ostrowski integral inequality for twice differentiable mappings</i>, Adv. Inequal. Appl., <b class="bf">2016</b> (2016), pp.&#160;1–17, art. no. 20. </p>
</dd>
  <dt><a name="nazo1">7</a></dt>
  <dd><p><a href ="https://doi.org/10.28919/aia/3600 "> <i class="sc">N. Irshad, A.R. Khan, A. Nazir</i>, <i class="it">Extension of Ostrowki type inequality via moment generating function</i>, Adv. Inequal. Appl., <b class="bf">2020</b> (2020) no. 2, pp.&#160;1–15. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="nazo2">8</a></dt>
  <dd><p><a href ="https://doi.org/10.28919/aia/3998"> <i class="sc">N. Irshad, A.R. Khan, M.A. Shaikh</i>, <i class="it">Generalized weighted Ostrowski-Grüss type inequality with applications</i>, Global J. Pure Appl. Math., <b class="bf">15</b> (2019) (5), pp.&#160;675–692. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="nazo3">9</a></dt>
  <dd><p><i class="sc">N. Irshad, A.R. Khan, M.A. Shaikh</i>, <i class="it">Generalization of weighted Ostrowski inequality with applications in numerical integration</i>, Adv. Ineq. Appl., <b class="bf">2019</b> (2019) (7), pp.&#160;1–14. </p>
</dd>
  <dt><a name="nazo5">10</a></dt>
  <dd><p><i class="sc">N. Irshad, A.R. Khan</i>, <i class="it">On Weighted Ostrowski Gruss inequality with applications</i>, Transylvanian J. Math. Mec., <b class="bf">10</b> (2018) (1), pp.&#160;15–22. </p>
</dd>
  <dt><a name="nazo6">11</a></dt>
  <dd><p><a href ="https://ictp.acad.ro/jnaat/journal/article/view/1107"> <i class="sc">N. Irshad, A.R. Khan</i>, <i class="it">Some applications of quadrature rules for mappings on \(L_p[u,v]\) space via Ostrowski-type inequality</i>, J. Numer. Anal. Approx. Theory, <b class="bf">46</b> (2017) no. 2, pp.&#160;141–149. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="nazo7">12</a></dt>
  <dd><p><i class="sc">N. Irshad, A.R. Khan</i>, <i class="it">Generalization of Ostrowski inequality for differentiable functions and its applications to numerical quadrature rules</i>, J. Math. Anal., <b class="bf">8</b> (2017) (1), pp.&#160;79–102. </p>
</dd>
  <dt><a name="H5">13</a></dt>
  <dd><p><a href ="https://doi.org/10.1142/s0219530513500309"> <i class="sc">M. Masjed-Jamei, S.S. Dragomir</i>, <i class="it">Generalization of Ostrowski Gruss inequality</i>, Analysis Appl., <b class="bf">12</b> (2014) (2), pp.&#160;117–130. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="H6">14</a></dt>
  <dd><p><a href ="https://doi.org/10.2298/fil1402373m"> <i class="sc">M. Masjed-Jamei, S.S. Dragomir</i>, <i class="it">An analogue of the Ostrowski inequality and applications</i>, Filomat, <b class="bf">28</b> (2014) (2), pp.&#160;373–381. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="mil1">15</a></dt>
  <dd><p><i class="sc">G.V. Milovanović</i>, <i class="it">On some Integral Inequalities</i>, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., <b class="bf">498</b>–<b class="bf">541</b> (1975), pp.&#160;119–124. </p>
</dd>
  <dt><a name="mil2">16</a></dt>
  <dd><p><i class="sc">G.V. Milovanović, J.E. Pečarić</i>, <i class="it">On Generalization of the inequality of A. Ostrowski and some related applications</i>, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., <b class="bf">544</b>–<b class="bf">576</b> (1976), pp.&#160;155–158. </p>
</dd>
  <dt><a name="H7">17</a></dt>
  <dd><p><i class="sc">D.S. Mitrinović, J.E. Pecarić, A.M. Fink</i>, <i class="it">Classical and New Inequalities in Analysis</i>, Kluwer Academic Publishers, Dordrecht, 1991. </p>
</dd>
  <dt><a name="H8">18</a></dt>
  <dd><p><i class="sc">D.S. Mitrinović, J.E. Pecarić, A.M. Fink</i>, <i class="it">Inequalities Involving Functions and their Integrals and Derivatives</i>, Kluwer Academic Publishers, Dordrecht, 1991. </p>
</dd>
  <dt><a name="H9">19</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/bf01214290"> <i class="sc">A.M. Ostrowski</i>, <i class="it">\(\ddot{U}\)ber die absolutabweichung einer differentiebaren funktion von ihren integralmittelwert</i>, Comment. Math. Helv., <b class="bf">10</b> (1938), pp.&#160;226–227. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="nazo4">20</a></dt>
  <dd><p><a href ="https://doi.org/10.28919/aia/3553"> <i class="sc">M.A. Shaikh, A.R. Khan, N. Irshad</i>, <i class="it">Generalized Ostrowski inequality with applications in numerical integration and special means</i>, Adv. Inequal. Appl., <b class="bf">2018</b> (2018) no. 7, pp.&#160;1–22. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="H10">21</a></dt>
  <dd><p><a href ="https://doi.org"> <i class="sc">N. Ujević</i>, <i class="it">A generalization of Ostrowski’s inequality and applications in numerical integration</i>, Appl. Math. Lett., <b class="bf">17</b> (2004), pp.&#160;133–137. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="3">22</a></dt>
  <dd><p><a href ="https://doi.org/10.3390/sym11121513"> <i class="sc">M.J. Vivas-Cortez, A. Kashuri, R. Liko, J.E. Hernández Hernández</i>, <i class="it">Quantum estimates of Ostrowski inequalities for generalized \(\varphi \)-convex functions</i>, Symmetry, <b class="bf">11</b> (2019) no. 12, pp.&#160;1–16. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="4">23</a></dt>
  <dd><p><a href ="https://doi.org/10.3390/sym11111402"> <i class="sc">M.J. Vivas-Cortez, A. Kashuri, R. Liko, J.E. Hernández Hernández</i>, <i class="it">Some inequalities using generalized convex functions in quantum analysis</i>, Symmetry, <b class="bf">11</b> (2019) no. 11, pp.&#160;1–14. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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