**5. Applications**

It is worth noting that in Definition 4 for *μ* = 0, the generalized strongly preinvex functions reduce to the preinvex functions. Moreover, if we put *η* = *v* − *u* in Definition 2, then the preinvex functions reduce to the classical convex functions. Besides, the quantum integral inequalities would lead to the corresponding Riemann integral inequalities by taking the limit *q* → 1<sup>−</sup>. Thus, several new and previously known results can be derived from Theorems 1 and 2 as special cases. Here, we illustrate the applications of our main results by three examples.

**Example 1.** *Recently, Zhang and Du et al. [26] investigated the quantum integral inequalities for convex functions, they established the following inequality:*

$$\begin{aligned} & \left| \frac{1}{3} \left[ \frac{qf(a) + f(b)}{1 + q} + 2f \left( \frac{qa + b}{1 + q} \right) \right] - \frac{1}{b - a} \int\_a^b f(t) \, \_a \mathbf{d}\_q t \right| \\ & \le \min \left\{ \mathcal{H}\_1 \left( \frac{1}{3}, \frac{1}{1 + q}, 1, 1 \right), \mathcal{H}\_2 \left( \frac{1}{3}, \frac{1}{1 + q}, 1, 1 \right) \right\}, \end{aligned} \tag{7}$$

*where f* : [*a*, *b*] → R *is a q-differentiable function and* | *a*D*q f* | *is an integrable and convex function with* 0 < *q* < 1*, the expressions of* H1 *and* H2 *are given by [26] (Theorem 3.2).*

Further, in [26], the authors derived a remarkable inequality from (7), as follows:

$$\left| \frac{1}{3} \left[ \frac{f(a) + f(b)}{2} + 2f\left(\frac{a+b}{2}\right) \right] - \frac{1}{b-a} \int\_a^b f(t) \, dt \right| \le \frac{5(b-a)}{72} \left[ |f'(a)| + |f'(b)| \right],\tag{8}$$

where *f* : [*a*, *b*] → R is a differentiable function, and | *f* | is an integrable and convex function on [*a*, *b*].

In the following, we show a new result analogous to the inequality (7), which can be obtained directly by taking *μ* = 0 in Theorem 1.

**Corollary 1.** *Let f* : *I* = [*a*, *a* + *η*(*b*, *a*)] → R *be a q-differentiable function on I with η*(*b*, *a*) > 0*. If* | *a*D*q f* | *is an integrable and preinvex function,* 0 < *q* < 1*, then*

$$\begin{aligned} & \left| \frac{1}{3} \left[ \frac{f(a) + f(a + \eta(b, a))}{2} + 2f \left( \frac{2a + \eta(b, a)}{2} \right) \right] - \frac{1}{\eta(b, a)} \int\_{a}^{a + \eta(b, a)} f(t) \, \_{a} \mathbf{d}\_{q} t \right| \\ & \le \eta(b, a) \left[ (A\_{1}(q) + A\_{4}(q)) \big| \_{a} \mathbf{D}\_{q} f(a) \big| + (A\_{2}(q) + A\_{5}(q)) \big| \, \_{a} \mathbf{D}\_{q} f(b) \big| \right], \end{aligned} \tag{9}$$

*where <sup>A</sup>*1(*q*), *<sup>A</sup>*2(*q*), *<sup>A</sup>*4(*q*)*, and <sup>A</sup>*5(*q*) *are the coefficients as described in Theorem 1.*

Putting *η*(*b*, *a*) = *b* − *a* in Corollary 1, it follows that

**Corollary 2.** *Let f* : [*a*, *b*] → R *be a q-differentiable function. If* | *a*D*q f* | *is an integrable and convex function,* 0 < *q* < 1*, then*

$$\begin{aligned} & \left| \frac{1}{3} \left[ \frac{f(a) + f(b)}{2} + 2f\left(\frac{a+b}{2}\right) \right] - \frac{1}{b-a} \int\_a^b f(t) \,\_a \mathbf{d}\_q t \right| \\ & \le (b-a) \left[ (A\_1(q) + A\_4(q)) \big| \,\_a \mathbf{D}\_{\overline{q}} f(a) \big| + (A\_2(q) + A\_5(q)) \big| \,\_a \mathbf{D}\_{\overline{q}} f(b) \big| \right], \end{aligned} \tag{10}$$

*where <sup>A</sup>*1(*q*), *<sup>A</sup>*2(*q*), *<sup>A</sup>*4(*q*)*, and <sup>A</sup>*5(*q*) *are the coefficients as described in Theorem 1.*

**Remark 1.** *In Corollary 2, if we take the limit q* → 1− *in (10) and use the basic properties of q-derivative and q-integral ([25], see also [1])*

$$\lim\_{q \to 1^{-}} \, \_aD\_q f(t) = f'(t), \quad \lim\_{q \to 1^{-}} \int\_a^b f(t) \, \_a \mathbf{d}\_q t = \int\_a^b f(t)dt.$$

*along with the equalities*

$$\lim\_{q \to 1^{-}} \left( A\_1(q) + A\_4(q) \right) \\ = \lim\_{q \to 1^{-}} \left( \frac{1 + 12q + 12q^2 + 36q^3}{216(1+q)(1+q+q^2)} + \frac{12q^2 + 12q + 5}{216(1+q)(1+q+q^2)} \right) \\ = \frac{5}{72}, \quad \lim\_{q \to 1^{-}} \left( \frac{1}{1+q} \right) = \frac{5}{72}$$

$$\lim\_{q \to 1^{-}} \left( A\_2(q) + A\_5(q) \right) \\ = \lim\_{q \to 1^{-}} \left( \frac{18q^2 + 18q - 7}{216(1+q)(1+q+q^2)} + \frac{18q^2 + 18q + 25}{216(1+q)(1+q+q^2)} \right) \\ = \frac{5}{72}A\_2$$

*then we obtain the inequality*

$$\left| \frac{1}{3} \left[ \frac{f(a) + f(b)}{2} + 2f\left(\frac{a+b}{2}\right) \right] - \frac{1}{b-a} \int\_{a}^{b} f(t) \, dt \right| \le \frac{5(b-a)}{72} \left[ |f'(a)| + |f'(b)| \right]. \tag{11}$$

This is exactly the above-mentioned inequality (8) due to Zhang and Du et al. [26].

**Example 2.** *In a recent paper [27], Tunç, Göv, and Balgeçti established a Simpson-type quantum integral inequality for convex functions ([27] Theorem 1), as follows:*

$$\left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right] - \frac{1}{b-a} \int\_{a}^{b} f(t) \, \_{d}\mathbf{d}\_{q}t \right|$$

$$\leq \frac{(b-a)}{12} \left[ \frac{2q^2 + 2q + 1}{q^3 + 2q^2 + 2q + 1} \, \_{d}\mathbf{D}\_{q}f(b) \, \vert + \frac{1}{3} \cdot \frac{6q^3 + 4q^2 + 4q + 1}{q^3 + 2q^2 + 2q + 1} \vert \, \_{d}\mathbf{D}\_{q}f(a) \vert \right],\tag{12}$$

*where f* : [*a*, *b*] → R *is a continuous function,* | *a*D*q f* | *is a convex and integrable function with* 0 < *q* < 1*.*

**Remark 2.** *Before we describe the related result of inequality (12), we should point out that in (12) there is an error occurring in the coefficients of* |*a*D*q f*(*b*)| *and* |*a*D*q f*(*a*)|*. The mistakes arise from the calculations of quantum integrals in [27] (Lemmas 4 and 5), the details are as follows:*

*As an auxiliary for establishing the inequality (12), in [27] (Lemmas 4 and 5) , the authors gave the following results involving q-integrals (*0 < *q* < 1*):*

$$\int\_0^{\frac{1}{2}} (1-t) \left| qt - \frac{1}{6} \right|\_0^{\cdot} \mathbf{d}\_q t \quad = \frac{36q^3 + 12q^2 + 12q + 1}{216(q^3 + 2q^2 + 2q + 1)} . \tag{13}$$

$$\int\_{\frac{1}{2}}^{1} (1-t) \left| qt - \frac{5}{6} \right|\_{0} \mathrm{d}\_{\mathbb{P}} t \quad = \begin{array}{c} 12q^2 + 12q + 5 \\ \hline 216(q^3 + 2q^2 + 2q + 1) \end{array} . \tag{14}$$

*However, the equality (13) is incorrect for the case of* 0 < *q* < 13 *; and the equality (14) is incorrect for the case of* 0 < *q* < 56 *, which can be observed by direct computation of q-integrals. In fact, by the formulas and algorithms for q-integrals stated in Propositions 1 and 2, when* 0 < *q* < 13*, we have*

$$\begin{split} &\int\_{0}^{\frac{1}{2}} (1-t) \left| qt - \frac{1}{6} \right| \,\_{0} \mathbf{d}\_{q} t \\ &= \quad \int\_{0}^{\frac{1}{2}} (1-t) (\frac{1}{6} - qt) \,\_{0} \mathbf{d}\_{q} t \\ &= \quad \int\_{0}^{\frac{1}{2}} (qt^{2} - \frac{1}{6}t - qt + \frac{1}{6}) \,\_{0} \mathbf{d}\_{q} t \\ &= \quad \,\_{0} \int\_{0}^{\frac{1}{2}} t^{2} \,\_{0} \mathbf{d}\_{q} t - (\frac{1}{6} + q) \int\_{0}^{\frac{1}{2}} t \,\_{0} \mathbf{d}\_{q} t + \frac{1}{6} \int\_{0}^{\frac{1}{2}} 1 \,\_{0} \mathbf{d}\_{q} t \\ &= \quad \,\_{0} \frac{1}{8(q + q^{2} + 1)} - (\frac{1}{6} + q) \frac{1}{4(1 + q)} + \frac{1}{12} \\ &= \quad \frac{1 - 4q^{3}}{24(q^{3} + 2q^{2} + 2q + 1)}. \end{split}$$

*When* 0 < *q* < 56 *, we have*

$$\begin{split} &\quad \int\_{\frac{1}{2}}^{1} (1-t) \left| qt - \frac{5}{6} \right|\_{0} \mathrm{d}\_{q}t \\ &= \quad \int\_{\frac{1}{2}}^{1} (1-t) (\frac{5}{6} - qt) \, \_{0} \mathrm{d}\_{q}t \\ &= \quad \int\_{\frac{1}{2}}^{1} (qt^{2} - \frac{1}{6}t - qt + \frac{1}{6}) \, \_{0} \mathrm{d}\_{q}t \\ &= \quad q \int\_{\frac{1}{2}}^{1} t^{2} \, \_{0} \mathrm{d}\_{q}t - (\frac{5}{6} + q) \int\_{\frac{1}{2}}^{1} t \, \_{0} \mathrm{d}\_{q}t + \frac{5}{6} \int\_{\frac{1}{2}}^{1} 1 \, \_{0} \mathrm{d}\_{q}t \\ &= \quad q (\int\_{0}^{1} t^{2} \, \_{0} \mathrm{d}\_{q}t - \int\_{0}^{\frac{1}{2}} t^{2} \, \_{0} \mathrm{d}\_{q}t) - (\frac{5}{6} + q) (\int\_{0}^{1} t \, \_{0} \mathrm{d}\_{q}t - \int\_{0}^{\frac{1}{2}} t \, \_{0} \mathrm{d}\_{q}t) + \frac{5}{12} \\ &= \quad q (\frac{1}{q + q^{2} + 1} - \frac{1}{8(q + q^{2} + 1)}) - (\frac{5}{6} + q)(\frac{1}{1 + q} - \frac{1}{4(1 + q)}) + \frac{5}{12} \\ &= \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{split}$$

In the same way, one can verify that the equality (13) is valid for 13 ≤ *q* < 1, the equality (14) is valid for 56 ≤ *q* < 1.

In the following we provide a modified version of inequality (12).

**Corollary 3.** *Let f* : [*a*, *b*] → R *be a q-differentiable function. If* | *a*D*q f* | *is an integrable and convex function,* 0 < *q* < 1*, then*

$$\begin{aligned} &\left|\frac{1}{3}\left[\frac{f(a)+f(b)}{2}+2f\left(\frac{a+b}{2}\right)\right]-\frac{1}{b-a}\int\_{a}^{b}f(t)\,\_{d}\mathbf{d}\_{\eta}t\right|\\ &\leq \left(b-a\right)\left[\mathbf{C}\_{1}(q)\,\big|\,\_{d}\mathbf{D}\_{\eta}f(a)\big|+\mathbf{C}\_{2}(q)\,\_{d}\mathbf{D}\_{\eta}f(b)\big|\,\_{\eta}\end{aligned} \tag{15}$$

*where <sup>C</sup>*1(*q*) *and <sup>C</sup>*2(*q*) *are given by*

$$\mathcal{C}\_{1}(q) = \begin{cases} \frac{-3q^{3} + 2q^{2} + 2q - 1}{6(q^{3} + 2q^{2} + 2q + 1)}, & 0 < q < \frac{5}{3}, \\\frac{-9q^{3} + 21q^{2} + 21q - 11}{54(q^{3} + 2q^{2} + 2q + 1)}, & \frac{1}{3} \le q < \frac{5}{5}, \\\frac{6q^{3} + 4q^{2} + 4q + 1}{36(q^{3} + 2q^{2} + 2q + 1)}, & \frac{5}{8} \le q < 1. \end{cases}$$

$$\mathcal{C}\_{2}(q) = \begin{cases} \frac{-q^{2} - q + 2}{3(q^{3} + 2q^{2} + 2q + 1)}, & 0 < q < \frac{1}{3}, \\\frac{-9q^{2} - 9q + 32}{54(q^{3} + 2q^{2} + 2q + 1)}, & \frac{1}{3} \le q < \frac{5}{6}, \\\frac{2q^{2} + 2q + 1}{12(q^{3} + 2q^{2} + 2q + 1)}, & \frac{5}{6} \le q < 1. \end{cases}$$

**Proof.** Using Corollary 2 and performing a simple calculation in the expressions *<sup>C</sup>*1(*q*) = *<sup>A</sup>*1(*q*) + *<sup>A</sup>*4(*q*) and *<sup>C</sup>*2(*q*) = *<sup>A</sup>*2(*q*) + *<sup>A</sup>*5(*q*), where *<sup>A</sup>*1(*q*), *<sup>A</sup>*2(*q*), *<sup>A</sup>*4(*q*), and *<sup>A</sup>*5(*q*) are the coefficients from Theorem 1, we obtain the inequality (15).

**Example 3.** *We provide an estimation of upper bound for the q-integral* . *<sup>a</sup>*+*η*(*b*,*<sup>a</sup>*) *a f*(*t*) *a*d*qt.*

**Corollary 4.** *Let f* : *I* = [*a*, *a* + *η*(*b*, *a*)] → R *be a q-differentiable function on I with η*(*b*, *a*) > 0*. If* | *a*D*q f* | *is an integrable and generalized strongly preinvex function with modulus μ* ≥ 0 *and* 0 < *q* < 1*, then*

$$\begin{aligned} \left| \int\_{a}^{a+\eta(b,a)} f(t) \, \_{a} \mathrm{d}\_{q} t \right| &\leq \eta(b,a) \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{2a + \eta(b,a)}{2}\right) + f(a + \eta(b,a)) \right] \right| \\\ &+ \eta^{2}(b,a) \left[ \left( A\_{1}(q) + A\_{4}(q) \right) \left| \, \_{a} \mathrm{D}\_{\emptyset} f(a) \right| + \left( A\_{2}(q) + A\_{5}(q) \right) \left| \, \_{a} \mathrm{D}\_{\emptyset} f(b) \right| - \mu(A\_{3}(q) + A\_{6}(q)) \eta^{2}(b,a) \right], \end{aligned} \tag{16}$$

*where <sup>A</sup>*1(*q*), *<sup>A</sup>*2(*q*), *<sup>A</sup>*3(*q*), *<sup>A</sup>*4(*q*), *<sup>A</sup>*5(*q*) *and <sup>A</sup>*6(*q*) *are the coefficients as described in Theorem 1.*

## **Proof.** Note that

$$\left|\frac{1}{\eta(b,a)}\int\_{a}^{a+\eta(b,a)}f(t)\,\_{a}\mathbf{d}\_{\eta}t\right| \leq \left|\frac{1}{6}\left[f(a)+4f\left(\frac{2a+\eta(b,a)}{2}\right)+f(a+\eta(b,a))\right]\right|.$$

$$+\left|\frac{1}{\eta(b,a)}\int\_{a}^{a+\eta(b,a)}f(t)\,\_{a}\mathbf{d}\_{\eta}t-\frac{1}{6}\left[f(a)+4f\left(\frac{2a+\eta(b,a)}{2}\right)+f(a+\eta(b,a))\right]\right|.$$

Utilizing Theorem 1, one has

$$\left| \frac{1}{\eta(b,a)} \int\_{a}^{a+\eta(b,a)} f(t) \, \_{a} \mathbf{d}\_{\eta} t \right| \leq \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{2a + \eta(b,a)}{2}\right) + f(a + \eta(b,a)) \right] \right| $$

$$+\eta(b,a)\left[\left(A\_1(q) + A\_4(q)\right)|\_{\
u} \mathcal{D}\_{\mathfrak{h}}f(a)| + \left(A\_2(q) + A\_5(q)\right)|\_{\
u} \mathcal{D}\_{\mathfrak{h}}f(b)| - \mu(A\_3(q) + A\_6(q))\eta^2(b,a)\right].$$

Multiplying both sides of the above inequality by *η*(*b*, *a*) leads to the desired inequality (16).

**Author Contributions:** Y.D., M.U.A., and S.W. finished the proofs of the main results and the writing work. Both authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

**Funding:** This work was supported by the Teaching Reform Project of Longyan University (Grant No. 2017JZ02) and the Teaching Reform Project of Fujian Provincial Education Department (Grant No. FBJG20180120).

**Acknowledgments:** The authors would like to express sincere appreciation to the editors and the anonymous reviewers for their valuable comments and suggestions which helped to improve the manuscript.

**Conflicts of Interest:** The authors declare no conflict of interest.
