The Simpson’s inequality cannot be applied to a function that is twice differentiable but not four times differentiable or have a bounded fourth derivative in the interval under consideration. Loads of articles are bound for twice differentiable convex functions but nothing, to the best of our knowledge, is known yet for twice differentiable exponentially convex and quasi-convex functions. In this paper, we aim to do justice to this query. For this, we prove several Simpson’s type inequalities for exponentially convex and exponentially quasi-convex functions. Our findings refine, generalize and complement existing results in the literature. We regain previously known results by taking \(\alpha=0\). In addition, we also show the importance of our results by applying them to some special means of positive real numbers and to the Simpson’s quadrature rule. The obtained results can be extended for different kinds of convex functions.
Suppose \(g:[a_1,a_2]\to \mathbb{R}\) is a four times continuously differentiable mapping on \((a_1,a_2)\) and \(||g^{(4)}||_{\infty}=\displaystyle\sup_{x\in (a_1,a_2)}|g^{(4)}(x)|< \infty\). Then the following inequality
\begin{equation*} \begin{aligned} &\Big|\frac{1}{3}\Big[\frac{g(a_1)+g(a_2)}{2}+2g\Big(\frac{a_1+a_2}{2}\Big)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\le\frac{1}{2880}||g^{(4)}||_{\infty}(a_2-a_1)^4 \end{aligned} \end{equation*} holds, and it is well known in the literature as Simpson’s inequality, named after the English mathematician Thomas Simpson. If the mapping \(g\) is neither four times differentiable nor is the fourth derivative \(g^{(4)}\) bounded on \((a_1, a_2)\), then we cannot apply the classical Simpson quadrature formula. In [1] Dragomir et al., proved some recent developments on Simpson’s inequality for which the remainder is expressed in terms of lower derivatives than the fourth. For recent refinements, counterparts, generalizations and new Simpson’s type inequalities, see [1,2,3,4].The classical convexity of functions is a fundamental notions in mathematics, they have widely applications in many branches of mathematics and physics. A function \(g:I\subset \mathbb{R}\to \mathbb{R}\) is said to be a convex in the classical sense, if \[g(ta_2+(1-t)a_1)\le tg(a_2)+(1-t)g(a_1)\] for all \(a_1,a_2 \in I\) and \(t \in [0,1]\). We call \(g\) concave if the inequality is reversed. A somewhat generalization of the above definition is given by Awan et al., [5] as follows:
Definition 1([5]). A function \(g:I\subset \mathbb{R}\to \mathbb{R}\) is called exponentially convex, if
For example, the function \(g:\mathbb{R} \to \mathbb{R}\), defined by \(g(x)=-x^2\) is a concave function, thus this function is exponentially convex for all \(\alpha < 0\). An exponentially convex function on a closed interval is bounded, it also satisfies the Lipschitzian condition on any closed interval \([a_1,a_2]\subset \mathring{I}\) (interior of I). Therefore an exponentially convex function is absolutely continuous on \([a_1,a_2]\subset \mathring{I}\) and continuous on \(\mathring{I}\).
Recently, Nie et al., [6] introduced the notion of exponentially quasi-convex as thus:
Definition 2([6]). Let \(\alpha \in \mathbb{R}\). Then a mapping \(g:I\subset \mathbb{R}\to \mathbb{R}\) is said to be exponentially quasi-convex if \begin{eqnarray*}\label{ec 1.4} g(ta_2+(1-t)a_1)\le \max\left\{\frac{g(a_2)}{e^{\alpha a_2}}, \frac{g(a_1)}{e^{\alpha a_1}}\right\} \end{eqnarray*} for all \(a_1, a_2 \in I\), and \(t\in [0,1]\).
Remark 1. Note that if \(\alpha =0\), then the classes of exponentially convex and quasi-convex functions reduce to the classes of classical convex and quasi-convex function.
Inspired by the work of Sarikaya et al., [7] and Vivas et al., [8], we aim to establish new Simpson’s type results for the class of functions whose derivatives in absolute value at certain powers are exponentially convex and exponentially quasi-convex. By taking \(\alpha=0\), we recapture some already established results in the literature. The main results are framed and justified in section symbol §2, followed by applications of our results to some special means in section symbol §3, and Simpson’s quadrature in section symbol §4.Lemma 1([7]). Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). Then the following equality holds: \begin{equation*} \begin{aligned} &\frac{1}{6}\left[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\right]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)\,dx \\ &=(a_2-a_1)^2\int_{0}^{1}k(t)g''(ta_2+(1-t)a_1)\,dt, \end{aligned} \end{equation*} where \begin{equation*} k(t)= \begin{cases} \frac{t}{2}\left(\frac{1}{3}-t\right)\quad \mbox if \quad t\in \left[0,\frac{1}{2}\right)\\[1ex] (1-t)\left(\frac{t}{2}-\frac{1}{3}\right) \quad \mbox if \quad t\in \left[\frac{1}{2},1\right]. \end{cases} \end{equation*}
Theorem 1. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*}\label{4} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\le \frac{(a_2-a_1)^2}{162}\left[\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ \frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\right]. \end{aligned} \end{equation*}
Proof. Using Lemma 1 and since \(|g”|\) is exponentially convex, we have \begin{align*} \Big|\frac{1}{6}\Big[g(a_1)&+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ \leq& (a_2-a_1)^2\int_0^1|k(t)|\left|g”(ta_2+(1-t)a_1)\right|dt \\ \le& (a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\Big(t\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ &+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\Big(t\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ =& (a_2-a_1)^2(I_1+I_2), \end{align*} where \begin{align*} I_1=&\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left(t\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}} \right)\,dt\\ =&\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|tdt+\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|(1-t)\,dt\\ =&\frac{59}{31104}\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+\frac{133}{31104}\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}}, \end{align*} and \begin{align*} I_2=&\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\Big(t\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ =&\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|tdt+\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}}\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|(1-t)dt\\ =&\frac{133}{31104}\frac{\left|g”(a_2)\right|}{e^{\alpha a_2}}+\frac{59}{31104}\frac{\left|g”(a_1)\right|}{e^{\alpha a_1}}, \end{align*} which completes the proof.
Corollary 1. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{eqnarray*} \Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \le \frac{(a_2-a_1)^2}{162}\left[\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ \frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\right]. \end{eqnarray*}
Remark 2. In Theorem 1, by letting \(\alpha=0\) we get [7,Theorem 2.2].
Theorem 2. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*}\label{5} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right.\\ &\quad\quad\left.+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right]. \end{aligned} \end{equation*}
Proof. Suppose that \(q\ge 1\). From Lemma 1, we have \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+f(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\le (a_2-a_1)^2\int_0^1|k(t)|\left|g”(ta_2+(1-t)a_1)\right|dt \\ &=(a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g”(ta_2+(1-t)a_1)\right|dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g”(ta_2+(1-t)a_1)\right|dt. \end{aligned} \end{equation*} Using the Hölder’s inequality for functions \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{1-\frac{1}{q}}\] and \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{\frac{1}{q}}\left|g”(ta_2+(1-t)a_1)\right|,\] for the first integral, and the functions \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{1-\frac{1}{q}}\] and \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{\frac{1}{q}}\left|g”(ta_2+(1-t)a_1)\right|,\] for the second integral, from the above relation we get the inequality: \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\leq (a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\left(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g”(ta_2+(1-t)a_1)\right|^qdt\right)^{\frac{1}{q}}\\ &\quad+ (a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\,dt\Big)^{1-\frac{1}{q}}\\ &\quad\times \Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g”(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}. \end{aligned} \end{equation*} Using the fact that \[\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt=\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt=\frac{1}{162}\] and the exponentially convexity of \(|g”|^q\), we have
Corollary 2. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right.\\ &\left.\quad+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right]. \end{aligned} \end{equation*}
Remark 3. By setting \(\alpha =0\) in Theorem 2, we recapture [7,Theorem 2.5].
Corollary 3. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1, a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^2\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)\,dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{\frac{1}{2}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right)^{\frac{1}{2}}\right.\\ &\left.\quad+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right)^{\frac{1}{2}}\right]. \end{aligned} \end{equation*}
Theorem 3. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right| \\ &\le \frac{(a_2-a_1)^2}{81}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}. \end{aligned} \end{equation*}
Proof. From Lemma 1 and by using the exponentially quasi-convexity of \(|g”|\), we get \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\leq (a_2-a_1)^2\int_0^1|k(t)|\left|g”(ta_2+(1-t)a_1)\right|dt \\ &\le (a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=(a_2-a_1)^2(I_1+I_2), \end{aligned} \end{equation*} where \begin{equation*} \begin{aligned} I_1&=\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\\ &=\frac{1}{162}\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}, \end{aligned} \end{equation*} and \begin{equation*} \begin{aligned} I_2&=\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\\ &=\frac{1}{162}\max\left\{\frac{|g”(a_2)|}{e^{\alpha a_2}},\frac{|g”(a_1)|}{e^{\alpha a_1}}\right\}, \end{aligned} \end{equation*} which completes the proof.
Corollary 4. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} \left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \le \frac{(a_2-a_1)^2}{81}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}. \end{aligned} \end{equation*}
Theorem 4. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1, a_2 \in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right| \\ &\le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{aligned} \end{equation*}
Proof. Suppose that \(q\ge 1\). From Lemma 1, we have \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le(a_2-a_1)^2\int_0^1|k(t)|\left|g”(ta_2+(1-t)a_1)\right|dt \\ &=(a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g”(ta_2+(1-t)a_1)\right|dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g”(ta_2+(1-t)a_1)\right|dt. \end{aligned} \end{equation*} Using the Hölder’s inequality for functions \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{1-\frac{1}{q}}\] and \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{\frac{1}{q}}\left|g”(ta_2+(1-t)a_1)\right|\] for the first integral and the functions \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{1-\frac{1}{q}}\] and \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{\frac{1}{q}}\left|g”(ta_2+(1-t)a_1)\right|,\] for the second integral, from the above relation we get the inequalities: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le (a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g”(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}\\ &\quad+(a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g”(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}. \end{aligned} \end{equation*} Since \(|g”|^q\) is exponentially quasi-convex, therefore we have
Corollary 5. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{aligned} \end{equation*}
Corollary 6. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^2\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right\}\right)^{\frac{1}{2}}. \end{aligned} \end{equation*}
Proposition 1. Let \(a_1, a_2 \in \mathbb{R}\), \(0< a_1< a_2\). Then, we have \begin{eqnarray*}\label{eq 12} \left|\frac{1}{3}\mathcal{A}(a_1^4,a_2^4)+\frac{2}{3}\mathcal{A}^4(a_1,a_2)-\mathcal{L}_4^4(a_1,a_2)\right|\le \frac{2(a_2-a_1)^2}{27}\left[\frac{a_2^2}{e^{\alpha a_2}}+\frac{a_1^2}{e^{\alpha a_1}}\right]. \end{eqnarray*}
Proof. The assertion follows from Theorem 1 and a simple computation applied to \(g(x)=\frac{x^4}{12}, \ \ x\in [a_1, a_2]\), where \(|g”|\) is exponentially convex mapping.
Proposition 2. Let \(a_1, a_2 \in \mathbb{R}\), \(0< a_1< a_2\). Then, we have \begin{eqnarray*} \left|\frac{1}{3}\mathcal{A}(a_1^{r+1},a_2^{r+1})+\frac{2}{3}\mathcal{A}^{r+1}(a_1, a_2)-\mathcal{L}_{r+1}^{r+1}(a_1, a_2)\right|\le \frac{r(r+1)(a_2-a_1)^2}{81}\max\left\{\frac{a_2^{r-1}}{e^{\alpha a_2}},\frac{a_1^{r-1}}{e^{\alpha a_1}}\right\}. \end{eqnarray*}
Proof. This time we use Theorem 3 and a simple computation applied to \(g(x)=\frac{x^{r+1}}{r+1}, \ r\ge 1, \ \ x\in [a_1,a_2]\). Here, the function \(|g”(x)|=rx^{r-1}\) is increasing and exponentially quasi-convex.
Theorem 5. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially convex on \([a_1,a_2]\), then in (6) for every division \(\mathbf{P}\) of \([a_1,a_2]\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{162}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left[\frac{\left|g''(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g''(s_{i})\right|}{e^{\alpha s_i}}\right]. \end{eqnarray*}
Proof. Applying Theorem 1 on the subintervals \([s_i,s_{i+1}],\ (i=0,1,2,\cdots, n-1)\) of the division \(\mathbf{P}\), we get \begin{eqnarray*} &&\left|\frac{(s_{i+1}-s_i)}{6}\left[g(s_i)+4g\left(\frac{s_{i+1}-s_i}{2}\right)+g(s_{i+1})\right]-\int_{s_i}^{s_{i+1}}g(s)ds\right|\\ &&\;\;\;\;\le \frac{(s_{i+1}-s_i)^3}{162}\left[\frac{\left|g”(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g”(s_{i})\right|}{e^{\alpha s_i}}\right]. \end{eqnarray*} Adding over \(i\) for \(0\) to \(n-1\) and taking into account that \(|g”|\) is exponentially convex, we have: \begin{eqnarray*} \left|S(g,\mathbf{P})-\int_{a_1}^{a_2}g(s)ds\right|\le \sum_{i=0}^{n-1}\frac{(s_{i+1}-s_i)^3}{162}\left[\frac{\left|g”(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g”(s_{i})\right|}{e^{\alpha s_i}}\right], \end{eqnarray*} which completes the proof.
Corollary 7. If \(\alpha =0\), we get \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{162}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left[g”(s_i)+ g”(s_{i+1})\right]. \end{eqnarray*}
Theorem 6. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1, a_2\in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then in (6) for every division \(\mathbf{P}\) of \([a_1,a_2]\), the following holds: \begin{eqnarray*} \left|E_s(g,\mathbf{P})\right|\le \frac{1}{81}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\max\left\{\frac{|g''(s_{i+1})|}{e^{\alpha s_{i+1}}},\frac{|g''(s_i)|}{e^{\alpha s_i}}\right\}. \end{eqnarray*}
Proof. Applying Theorem 3 and proceeding as in the proof of Theorem 5, we obtain the desired result.
Proposition 3. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \left(\frac{1}{162}\right)^{1-\frac{1}{q}}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\Big[M_{\alpha}^q(g''(s_i), g''(s_{i+1}))\Big], \end{eqnarray*} where \begin{align*} M_{\alpha}^q\left(g''(s_i), g''(s_{i+1})\right)&=\left(\frac{59}{31104}\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q+\frac{133}{31104}\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right)^{\frac{1}{q}}\\ &\quad+ \left(\frac{133}{31104}\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q+\frac{59}{31104}\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right)^{\frac{1}{q}}. \end{align*}
Proof. The proof is immediate by using Theorem 2.
Proposition 4. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g”\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{81}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left(\max\left\{\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q,\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{eqnarray*}
Proof. The proof follows by applying Theorem 4.