**2. Preliminaries**

Let us recall some preliminary concepts and results.

**Definition 1** ([15])**.** *A set Kη* ⊂ R*n is said to be invex with respect to bifunction η*(., .) : R*n* × R*n* → R*n, if*

$$
\mu + \lambda \eta(\upsilon, \mu) \in \mathcal{K}\_{\eta \prime} \quad \forall \ \mu, \upsilon \in \mathcal{K}\_{\eta \prime} \ \ \ \lambda \in [0, 1] .
$$

**Definition 2** ([15])**.** *A function φ on the invex set Kη* ⊂ R*n is said to be preinvex with respect to bifunction η*(., .) : R*n* × R*n* → R*<sup>n</sup>*, *if*

$$
\phi(u + \lambda \eta(v, u)) \le (1 - \lambda)\phi(u) + \lambda\phi(v), \quad \forall u, v \in K\_{\eta}, \ \lambda \in [0, 1].
$$

**Definition 3** ([16])**.** *A function φ on the invex set Kη* ⊂ R*n is said to be strongly preinvex with respect to bifunction η*(., .) : R*n* × R*n* → R*<sup>n</sup>*, *and modulus μ* > 0*, if*

$$
\phi(u + \lambda \eta(v, u)) \le (1 - \lambda)\phi(u) + \lambda\phi(v) - \mu\lambda(1 - \lambda)\eta^2(v, u), \quad \forall u, v \in K\_{\eta}, \ \lambda \in [0, 1].
$$

Here, we introduce a new definition which combines the preinvex functions and the strongly preinvex functions given above.

**Definition 4.** *A function φ on the invex set Kη* ⊂ R*n is said to be generalized strongly preinvex with respect to bifunction η*(., .) : R*n* × R*n* → R*n and modulus μ* ≥ 0*, if*

$$
\phi(u + \lambda \eta(v, u)) \le (1 - \lambda)\phi(u) + \lambda\phi(v) - \mu\lambda(1 - \lambda)\eta^2(v, u), \quad \forall u, v \in \mathcal{K}\_{\mathbb{H}}, \ \lambda \in [0, 1].
$$

Clearly, if *μ* = 0, then the class of generalized strongly preinvex functions reduces to the class of preinvex functions as defined in Definition 2.

In the following, we recall some basic properties of *q*-calculus.

Let *J* = [*a*, *b*] ⊆ R be an interval and 0 < *q* < 1 be a constant. The *q*-derivative of a function *φ* : *J* → R at a point *u* ∈ *J* on [*a*, *b*] is defined as follows:

**Definition 5** ([25])**.** *Let φ* : *J* = [*a*, *b*] → R *be a continuous function and let u* ∈ *J. Then, the q-derivative of φ on J at u is defined as*

$$\_{d}D\_{q}\phi(u) = \frac{\phi(u) - \phi(qu + (1-q)a)}{(1-q)(u-a)}, \quad u \neq a. \tag{2}$$

**Definition 6** ([25])**.** *Let φ* : *J* = [*a*, *b*] → R *is a continuous function. A second-order q-derivative on J, which is denoted as aD*<sup>2</sup> *qφ, provided that aDqφ is q-differentiable on J with aD*<sup>2</sup> *qφ* <sup>=</sup>*a Dq*(*aDqφ*) : *J* → R*. Similarly higher order q-derivative on J is defined by aDn q φ* : *J* → R*.*

In [25], Tariboon and Ntouyas defined the *q*-integral as follows:

**Definition 7** ([25])**.** *Let φ* : *J* = [*a*, *b*] → R *be a continuous function. Then, the q-integral on J is defined as:*

$$
\int\_{a}^{u} \phi(\nu) \,\_{d} \mathbf{d}\_{q} \nu = (1 - q)(u - a) \sum\_{n=0}^{\infty} q^{n} \phi(q^{n} u + (1 - q^{n}) a), \tag{3}
$$

*for u* ∈ *J.*

> The following results are useful in the computation of *q*-integral in subsequent section.

**Proposition 1** ([25])**.** *Let f* , *g* : *J* = [*a*, *b*] → R *be continuous functions, c* ∈ R. *Then, for x* ∈ *J*,

$$\begin{aligned} \int\_{a}^{\chi} (f(\nu) + \mathfrak{g}(\nu))\_{\,\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu &= \int\_{a}^{\chi} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu + \int\_{a}^{\chi} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu, \\ \int\_{a}^{\chi} \mathfrak{c} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu &= \mathfrak{c} \int\_{a}^{\chi} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu, \\ \int\_{\tilde{\xi}}^{\chi} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu &= \int\_{a}^{\chi} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu - \int\_{a}^{\tilde{\xi}} f(\nu) \,\_{\,\,\mathrm{d}} \mathbf{d}\_{\eta} \nu, \ \,\_{\,\tilde{\xi}}^{\mathfrak{x}} \in (a, \infty). \end{aligned}$$

**Proposition 2** ([25])**.** *For q-integral, we have the following identities*

$$\begin{aligned} \int\_{a}^{u} \mathbf{1}\_{\ \!\!\!-\!\!d} \upsilon &= u - a, \\ \int\_{a}^{u} \upsilon \, \!\!\!d\_{q} \upsilon &= \frac{(u-a)(u+qa)}{1+q}, \\ \int\_{a}^{u} (\upsilon - a)^{\tau} \, \!\!\!d\_{q} \upsilon &= \left(\frac{1-q}{1-q^{\tau+1}}\right)(u-a)^{\tau+1}, \quad \tau \neq -1, \\ \int\_{\tilde{\xi}}^{u} (\upsilon - \tilde{\xi}) \, \!\!\!d\_{q} \upsilon &= \frac{u^{2} - (1+q)u\sharp + q\sharp^{2}}{1+q} - \frac{a(1-q)(u-\tilde{\xi})}{1+q}, \quad \tilde{\xi} \in (a,u). \end{aligned}$$

## **3. A Key Lemma**

In this section, we present an identity associated with *q*-integral, which plays an important role in establishing our main results.

**Lemma 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 integrable on I and* 0 < *q* < 1*, then*

$$\frac{1}{\delta} \left[ f(a) + 4f\left(\frac{2a + \eta(b, a)}{2}\right) + f(a + \eta(b, a)) \right] - \frac{1}{\eta(b, a)} \int\_{a}^{a + \eta(b, a)} f(t) \, \_a \mathrm{d}\_{\eta} t$$

$$\eta = \eta(b, a) \int\_{0}^{1} \Psi(t, q) \, \_a \mathrm{D}\_{\eta} f(a + t\eta(b, a)) \, \_0 \mathrm{d}\_{\eta} t,\tag{4}$$

*where*

$$\Psi(t,q) = \begin{cases} qt - \frac{1}{6}, & \text{if } 0 \le t < \frac{1}{2}, \\ qt - \frac{5}{6}, & \text{if } \frac{1}{2} \le t \le 1. \end{cases}$$

**Proof.** Let

$$\begin{aligned} Q\_1 &= \int\_0^{\frac{1}{2}} \left(qt - \frac{1}{6}\right) \,\_a\mathrm{D}\_q f(a + t\eta(b, a)) \,\_0\mathrm{d}\_q t, \\ Q\_2 &= \int\_{\frac{1}{2}}^1 \left(qt - \frac{5}{6}\right) \,\_a\mathrm{D}\_q f(a + t\eta(b, a)) \,\_0\mathrm{d}\_q t, \\\\ &\vdots \\\\ &\vdots \end{aligned}$$

then

$$\int\_0^1 \Psi(t, q) \,\_a \mathrm{D}\_\emptyset f(a + t\eta(b, a)) \,\_0 \mathrm{d}\_\emptyset t = Q\_1 + Q\_2.$$

Utilizing the Definitions 5 and 7, and the properties of *q*-derivative and *q*-integral described in Propositions 1, a direct computation gives

*Q*1 = 1 2 0 *qt a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*)) <sup>0</sup>d*q<sup>t</sup>* − 1 2 0 1 6 *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*)) <sup>0</sup>d*q<sup>t</sup>* = 1 2 0 *q f*(*a* + *<sup>t</sup>η*(*b*, *a*)) − *f*(*a* + *qtη*(*b*, *a*)) (1 − *q*)*η*(*b*, *a*) <sup>0</sup>d*q<sup>t</sup>* − 1 6 1 2 0 *f*(*a* + *<sup>t</sup>η*(*b*, *a*)) − *f*(*a* + *qtη*(*b*, *a*)) (1 − *q*)*<sup>t</sup>η*(*b*, *a*) <sup>0</sup>d*q<sup>t</sup>* = 1 2 ∞ ∑ *<sup>n</sup>*=0 *q<sup>n</sup>*+<sup>1</sup> *f* <sup>2</sup>*a*+*q<sup>n</sup>η*(*b*,*<sup>a</sup>*) 2 − *f* <sup>2</sup>*a*+*q<sup>n</sup>*+1*η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) − 1 6 ∞ ∑ *<sup>n</sup>*=0 *f* <sup>2</sup>*a*+*q<sup>n</sup>η*(*b*,*<sup>a</sup>*) 2 − *f* <sup>2</sup>*a*+*q<sup>n</sup>*+1*η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) =1 2 *f* <sup>2</sup>*a*+*η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) − ∞∑*<sup>n</sup>*=<sup>0</sup>(<sup>1</sup> − *q*)*q<sup>n</sup> f* <sup>2</sup>*a*+*q<sup>n</sup>η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) − 16 · *f* <sup>2</sup>*a*+*η*(*b*,*<sup>a</sup>*) 2 − *f*(*a*) *η*(*b*, *a*) = 1 3 · *f* <sup>2</sup>*a*+*η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) + 1 6 · *f*(*a*) *η*(*b*, *a*) − 1 2 ∞ ∑ *<sup>n</sup>*=0 (1 − *q*)*q<sup>n</sup> f* <sup>2</sup>*a*+*q<sup>n</sup>η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) = 1 3 · *f* <sup>2</sup>*a*+*η*(*b*,*<sup>a</sup>*) 2 *η*(*b*, *a*) + 1 6 · *f*(*a*) *η*(*b*, *a*) − 1 *η*(*b*, *a*) 1 2 0 *f*(*a* + *<sup>t</sup>η*(*b*, *a*)) <sup>0</sup>d*qt*.

On the other hand, one has

$$\begin{split} Q\_{2} &= \int\_{\frac{1}{2}}^{1} q t\_{\
u} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t - \int\_{\frac{1}{2}}^{1} \frac{5}{6} \,\_{a} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t \\ &= \int\_{0}^{1} q t\_{\
u} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t - \int\_{0}^{1} \frac{5}{6} \,\_{a} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t \\ &- \left( \int\_{0}^{\frac{1}{2}} q t\_{\
u} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t - \int\_{0}^{\frac{1}{2}} \frac{5}{6} \,\_{a} \mathrm{D}\_{q} f(a + t\eta(b,a)) \,\_{0} \mathrm{d}\_{q} t \right). \end{split}$$

Since

$$\int\_0^1 qt \,\_a\mathrm{D}\_q f(a+t\eta(b,a)) \,\_0\mathrm{d}\_q t - \int\_0^1 \frac{5}{6} \,\_a\mathrm{D}\_q f(a+t\eta(b,a)) \,\_0\mathrm{d}\_q t$$

$$=\int\_0^1 q \frac{f(a+t\eta(b,a)) - f(a+qt\eta(b,a))}{(1-q)\eta(b,a)} \,\_0\mathrm{d}\_q t$$

$$-\frac{5}{6}\int\_0^1 \frac{f(a+t\eta(b,a))-f(a+qt\eta(b,a))}{(1-q)\eta(b,a)} \,\_0\mathrm{d}\_q t$$

$$=\sum\_{n=0}^\infty q^{n+1} \frac{f\left(a+q^n\eta(b,a)\right)-f\left(a+q^{n+1}\eta(b,a)\right)}{\eta(b,a)}$$

$$-\frac{5}{6}\sum\_{n=0}^\infty \frac{f\left(a+q^n\eta(b,a)\right)-f\left(a+q^{n+1}\eta(b,a)\right)}{\eta(b,a)}$$

$$=\frac{f\left(a+\eta(b,a)\right)}{\eta(b,a)} - \sum\_{n=0}^\infty (1-q)q^n \frac{f\left(a+q^n\eta(b,a)\right)}{\eta(b,a)} - \frac{5}{6} \cdot \frac{f\left(a+\eta(b,a)\right)-f(a)}{\eta(b,a)}$$

$$=\frac{1}{6} \cdot \frac{f\left(a+\eta(b,a)\right)}{\eta(b,a)} + \frac{5}{6} \cdot \frac{f(a)}{\eta(b,a)} - \sum\_{n=0}^\infty (1-q)q^n \frac{f\left(a+q^n\eta(b,a)\right)}{\eta(b,a)}$$

$$=\frac{1}{6} \cdot \frac{f\left(a+\eta(b,a)\right)}{\eta(b,a)} + \frac{5}{6} \cdot \frac{f(a)}{\eta(b,a)} - \frac{1}{\eta(b,a)} \int\_0^1 f(a+t\eta(b,a)) \,\_0\mathrm{d}\_qt$$

and

$$\int\_0^{\frac{1}{2}} q t \, \_a \mathrm{D}\_q f(a + t\eta(b, a)) \, \_0 \mathrm{d}\_q t - \int\_0^{\frac{1}{2}} \frac{5}{6} \, \_a \mathrm{D}\_q f(a + t\eta(b, a)) \, \_0 \mathrm{d}\_q t$$

$$\frac{1}{2} = \frac{1}{2} \left[ \frac{f\left(\frac{2a + \eta(b, a)}{2}\right)}{\eta(b, a)} - \sum\_{n=0}^\infty (1 - q) q^n \frac{f\left(\frac{2a + q^n \eta(b, a)}{2}\right)}{\eta(b, a)} \right] - \frac{5}{6} \cdot \frac{f\left(\frac{2a + \eta(b, a)}{2}\right) - f(a)}{\eta(b, a)}$$

$$= -\frac{1}{3} \cdot \frac{f\left(\frac{2a + \eta(b, a)}{2}\right)}{\eta(b, a)} + \frac{5}{6} \cdot \frac{f(a)}{\eta(b, a)} - \frac{1}{\eta(b, a)} \int\_0^{\frac{1}{2}} f(a + t\eta(b, a)) \, \_0 \mathrm{d}\_q t,$$

we obtain

$$Q\_2 = \frac{1}{6} \cdot \frac{f\left(a + \eta(b, a)\right)}{\eta(b, a)} + \frac{1}{3} \cdot \frac{f\left(\frac{2a + \eta(b, a)}{2}\right)}{\eta(b, a)} - \frac{1}{\eta(b, a)} \int\_0^1 f(a + t\eta(b, a)) \,\_0\mathrm{d}\_q t$$

$$+ \frac{1}{\eta(b, a)} \int\_0^{\frac{1}{2}} f(a + t\eta(b, a)) \,\_0\mathrm{d}\_q t.$$

Thus,

$$\int\_0^1 \Psi(t, q) \,\_a \mathcal{D}\_q f(a + t\eta(b, a)) \,\_0 \mathbf{d}\_q t = Q\_1 + Q\_2$$

$$\frac{1}{6} = \frac{1}{6} \cdot \frac{f\left(a + \eta(b, a)\right) + f(a)}{\eta(b, a)} + \frac{2}{3} \cdot \frac{f\left(\frac{2a + \eta(b, a)}{2}\right)}{\eta(b, a)} - \frac{1}{\eta(b, a)} \int\_0^1 f(a + t\eta(b, a)) \,\_0 \mathbf{d}\_q t$$

$$= \frac{1}{6} \cdot \frac{f\left(a + \eta(b, a)\right) + f(a)}{\eta(b, a)} + \frac{2}{3} \cdot \frac{f\left(\frac{2a + \eta(b, a)}{2}\right)}{\eta(b, a)} - \frac{1}{\eta^2(b, a)} \int\_a^{a + \eta(b, a)} f(t) \, \_a \mathbf{d}\_q t,$$

which leads to the desired identity (4). The proof of Lemma 1 is complete.

## **4. Main Results**

We are in a position to establish the *q*-integral inequalities of Simpson-type for strongly preinvex functions.

**Theorem 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 a generalized strongly preinvex function with modulus μ* ≥ 0 *and* 0 < *q* < 1*, then*

$$\frac{1}{6}\left[f(a) + 4f\left(\frac{2a + \eta(b,a)}{2}\right) + f(a + \eta(b,a))\right] - \frac{1}{\eta(b,a)}\int\_a^{a + \eta(b,a)} f(t) \, \_a \mathrm{d}\_q t \right] \tag{5}$$

$$\leq \eta(b,a) \left[ (A\_1(q) + A\_4(q))|\, \_a \mathrm{D}\_\emptyset f(a)| + (A\_2(q) + A\_5(q))|\, \_a \mathrm{D}\_\emptyset f(b)| - \mu(A\_3(q) + A\_6(q))\eta^2(b,a) \right],$$

*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 given by*

$$A\_1(q) = \begin{cases} \frac{1 - 4q^3}{24(1+q)(1+q+q^2)}, & 0 < q < \frac{1}{3}, \\\\ \frac{1 + 12q + 12q^2 + 36q^3}{216(1+q)(1+q+q^2)}, & \frac{1}{3} \le q < 1, \end{cases}$$

$$A\_2(q) = \begin{cases} \frac{1 - 2q - 2q^2}{24(1 + q)(1 + q + q^2)}, & 0 < q < \frac{1}{3}, \\\\ \frac{18q^2 + 18q - 7}{216(1 + q)(1 + q + q^2)}, & \frac{1}{3} \le q < 1, \end{cases}$$

$$A\_3(q) = \begin{cases} \frac{1 - 2q - 2q^3 - 4q^4}{48(1 + q)(1 + q^2)(1 + q + q^2)}, & 0 < q < \frac{1}{3}, \\\\ \frac{108q^4 + 54q^3 + 12q^2 + 54q - 17}{1296(1 + q)(1 + q^2)(1 + q + q^2)}, & \frac{1}{3} \le q < 1, \end{cases}$$

$$A\_4(q) = \begin{cases} \frac{-5 + 8q + 8q^2 - 8q^3}{24(1+q)(1+q+q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{12q^2 + 12q + 5}{216(1+q)(1+q+q^2)}, & \frac{5}{6} \le q < 1, \end{cases}$$

$$A\mathfrak{s}(q) = \begin{cases} \frac{5 - 2q - 2q^2}{8(1+q)(1+q+q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{18q^2 + 18q + 25}{216(1+q)(1+q+q^2)}, & \frac{5}{6} \le q < 1, \end{cases}$$

$$A\_6(q) = \begin{cases} \frac{5 - 2q + 28q^2 - 2q^3 - 12q^4}{48(1 + q)(1 + q^2)(1 + q + q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{108q^4 - 54q^3 + 96q^2 - 54q + 115}{1296(1 + q)(1 + q^2)(1 + q + q^2)}, & \frac{5}{6} \le q < 1. \end{cases}$$

**Proof.** Using Lemma 1 and the assumption condition that | *a*D*q f* | is a generalized strongly preinvex function, we have

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

$$= \left|\eta(b,a) \int\_0^1 \Psi(t,q) \, \_a \mathbf{D}\_q f(a + t\eta(b,a)) \, \_0 \mathbf{d}\_q t\right| $$

= *η*(*b*, *a*) 12 0 *qt* − 16 *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*)) <sup>0</sup>d*q<sup>t</sup>* + 112 *qt* − 56 *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*)) 0d*qt* ≤ *η*(*b*, *a*) 12 0 *qt* − 16 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))| <sup>0</sup>d*q<sup>t</sup>* + 112 *qt* − 56 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))| 0d*qt* ≤ *η*(*b*, *a*) 12 0 *qt* − 16 (1 − *t*)| *a*D*q f*(*a*)| + *t*| *a*D*q f*(*b*)| − *μ<sup>t</sup>*(<sup>1</sup> − *<sup>t</sup>*)*η*<sup>2</sup>(*b*, *a*) <sup>0</sup>d*q<sup>t</sup>* + 112 *qt* − 56 (1 − *t*)| *a*D*q f*(*a*)| + *t*| *a*D*q f*(*b*)| − *μ<sup>t</sup>*(<sup>1</sup> − *<sup>t</sup>*)*η*<sup>2</sup>(*b*, *a*) 0d*qt* = *η*(*b*, *a*)| *a*D*q <sup>f</sup>*(*a*)| 12 0 (1 − *t*)*qt* − 16 <sup>0</sup>d*q<sup>t</sup>* + 112 (1 − *t*)*qt* − 56 0d*qt* +| *a*D*q f*(*b*)| 12 0 *t qt* − 16 <sup>0</sup>d*q<sup>t</sup>* + 112 *t qt* − 56 0d*qt* <sup>−</sup>*μη*<sup>2</sup>(*b*, *a*) 12 0 *t*(1 − *t*)*qt* − 16 <sup>0</sup>d*q<sup>t</sup>* + 112 *t*(1 − *t*)*qt* − 56 <sup>0</sup>d*q<sup>t</sup>*.

In view of the Definitions 5 and 7, and Propositions 1 and 2, a direct calculation gives

$$\begin{aligned} A\_{1}(q) &= \int\_{0}^{\frac{1}{2}} (1-t) \left| qt - \frac{1}{6} \right|\_{0} \text{d} \mathbf{d}\_{q} t = \begin{cases} \frac{1-4q^{3}}{24(1+q)(1+q+q^{2})}, & 0 < q < \frac{1}{3}, \\\\ \frac{1+12q+12q^{2}+36q^{3}}{216(1+q)(1+q+q^{2})}, & \frac{1}{3} \le q < 1, \end{cases} \\\\ A\_{2}(q) &= \int\_{0}^{\frac{1}{2}} t \left| qt - \frac{1}{6} \right|\_{0} \text{d} \mathbf{d}\_{q} t = \begin{cases} \frac{1-2q-2q^{2}}{24(1+q)(1+q+q^{2})}, & 0 < q < \frac{1}{3}, \\\\ \frac{18q^{2}+18q-7}{216(1+q)(1+q+q^{2})}, & \frac{1}{3} \le q < 1, \end{cases} \\\\ A\_{3}(q) &= \int\_{0}^{\frac{1}{2}} t (1-t) \left| qt - \frac{1}{6} \right|\_{0} \text{d} \mathbf{d}\_{q} t = \begin{cases} \frac{1-2q-2q^{3}-4q^{4}}{4(1+q)(1+q+q^{2})}, & 0 < q < \frac{1}{3}, \\\\ \frac{108q^{4}+54q^{3}+12q^{2}+54q-17}{1296(1+q)(1+q^{2})(1+q+q^{2})}, & \frac{1}{3} \le q < 1, \end{cases} \\\\ A\_{4}(q) &= \frac{5}{12}, \qquad \frac{5}{12}, \end{aligned}$$

$$A\_4(q) = \int\_{\frac{1}{2}}^1 (1-t) \left| qt - \frac{5}{6} \right|\_0 \mathbf{d}\_q t = \begin{cases} \frac{-5 + 8q + 8q^2 - 8q^3}{24(1+q)(1+q+q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{12q^2 + 12q + 5}{216(1+q)(1+q+q^2)}, & \frac{5}{6} \le q < 1, \end{cases}$$

$$\begin{aligned} A\_5(q) &= \int\_{\frac{1}{2}}^1 t \left| qt - \frac{5}{6} \right|\_{0} \mathbf{d}\_q t = \begin{cases} \frac{5 - 2q - 2q^2}{8(1+q)(1+q+q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{18q^2 + 18q + 25}{216(1+q)(1+q+q^2)}, & \frac{5}{6} \le q < 1, \\\\ \frac{1}{2} \end{cases} \\\\ A\_6(q) &= \int\_{\frac{1}{2}}^1 t(1-t) \left| qt - \frac{5}{6} \right|\_{0} \mathbf{d}\_q t = \begin{cases} \frac{5 - 2q + 28q^2 - 2q^3 - 12q^4}{48(1+q)(1+q^2)(1+q+q^2)}, & 0 < q < \frac{5}{6}, \\\\ \frac{108q^4 - 54q^3 + 96q^2 - 54q + 115}{1296(1+q)(1+q+q^2)(1+q+q^2)}, & \frac{5}{6} \le q < 1. \end{cases} \end{aligned}$$

,

Hence, we deduce the required inequality (5). This completes the proof of Theorem 1.

**Theorem 2.** *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*|*<sup>r</sup> is an integrable and a generalized strongly preinvex function with modulus μ* ≥ 0*, r* > 1 *and* 0 < *q* < 1*, then*

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

$$\leq \eta(b,a)\left[\left(B\_{1}(q)\right)^{1-\frac{1}{\tau}}\left(A\_{1}(q)|\,\_{a}\mathrm{D}\_{q}f(a)|^{r}+A\_{2}(q)|\,\_{a}\mathrm{D}\_{q}f(b)|^{r}-\mu A\_{3}(q)\eta^{2}(b,a)\right)^{\frac{1}{\tau}}\right]$$

$$+\left(B\_{2}(q)\right)^{1-\frac{1}{\tau}}\left(A\_{4}(q)|\,\_{a}\mathrm{D}\_{q}f(a)|^{r}+A\_{5}(q)|\,\_{a}\mathrm{D}\_{q}f(b)|^{r}-\mu A\_{6}(q)\eta^{2}(b,a)\right)^{\frac{1}{\tau}}\right]\_{q}$$

*where*

$$\begin{aligned} B\_1(q) &= \begin{cases} \frac{1-2q}{12(1+q)}, & 0 < q < \frac{1}{5}, \\\\ \frac{6q-1}{36(1+q)}, & \frac{1}{3} \le q < 1, \\\\ B\_2(q) &= \begin{cases} \frac{5-4q}{12(1+q)}, & 0 < q < \frac{5}{6}, \\\\ \frac{4q-5}{12(1+q)}, & \frac{5}{6} \le q < 1, \end{cases} \end{aligned}$$

*<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 given by the same expressions as described in Theorem 1.*

**Proof.** Using Lemma 1 and the Hölder inequality, one has

$$\begin{aligned} \left| \frac{1}{6} \left[ f(a) + 4f\left(\frac{2a + \eta(b,a)}{2}\right) + f(a + \eta(b,a)) \right] - \frac{1}{\eta(b,a)} \int\_a^{a + \eta(b,a)} f(x) \, \_a \mathbf{d}\_q x \right| \\ &= \left| \eta(b,a) \int\_0^1 \Psi(t, q) \, \_a \mathbf{D}\_q f(a + t\eta(b,a)) \, \_0 \mathbf{d}\_q t \right| \\ &= \eta(b,a) \left| \int\_0^{\frac{1}{2}} \left( qt - \frac{1}{6} \right) \, \_a \mathbf{D}\_q f(a + t\eta(b,a)) \, \_0 \mathbf{d}\_q t \right| \\ &\quad + \int\_{\frac{1}{2}}^1 \left( qt - \frac{5}{6} \right) \, \_a \mathbf{D}\_q f(a + t\eta(b,a)) \, \_0 \mathbf{d}\_q t \right| \end{aligned}$$

≤ *η*(*b*, *a*) 12 0 *qt* − 16 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))| <sup>0</sup>d*q<sup>t</sup>* + 112 *qt* − 56 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))| 0d*qt* ≤ *η*(*b*, *a*) 12 0 *qt* − 16 <sup>0</sup>d*q<sup>t</sup>*<sup>1</sup><sup>−</sup> 1*r* 12 0 *qt* − 16 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))|*<sup>r</sup>* 0d*qt*1*r* + 112 *qt* − 56 <sup>0</sup>d*q<sup>t</sup>*<sup>1</sup><sup>−</sup> 1*r* 112 *qt* − 56 | *a*D*q f*(*a* + *<sup>t</sup>η*(*b*, *a*))|*<sup>r</sup>* <sup>0</sup>d*q<sup>t</sup>*<sup>1</sup>*<sup>r</sup>* ≤ *η*(*b*, *a*) 12 0 *qt* − 16 <sup>0</sup>d*q<sup>t</sup>*<sup>1</sup><sup>−</sup> 1*r* × 12 0 *qt* − 16 \*(1 − *t*)| *a*D*q f*(*a*)|*<sup>r</sup>* + *t*| *a*D*q f*(*b*)|*r* − *μ<sup>t</sup>*(<sup>1</sup> − *<sup>t</sup>*)*η*<sup>2</sup>(*b*, *a*)+ 0d*qt* 1 *r* + 112 *qt* − 56 <sup>0</sup>d*q<sup>t</sup>*<sup>1</sup><sup>−</sup> 1*r* × 112 *qt* − 56 \*(1 − *t*)| *a*D*q f*(*a*)|*<sup>r</sup>* + *t*| *a*D*q f*(*b*)|*r* − *μ<sup>t</sup>*(<sup>1</sup> − *<sup>t</sup>*)*η*<sup>2</sup>(*b*, *a*)+ 0d*qt* 1 *r* 

By direct computation, we find

$$\begin{aligned} B\_1(q) &= \int\_0^{\frac{1}{2}} \left| qt - \frac{1}{6} \right|\_0 \mathbf{d}\_q t = \begin{cases} \frac{1 - 2q}{12(1 + q)}, & 0 < q < \frac{1}{5}, \\\\ \frac{6q - 1}{36(1 + q)}, & \frac{1}{3} \le q < 1, \end{cases} \\\\ B\_2(q) &= \int\_{\frac{1}{2}}^1 \left| qt - \frac{5}{6} \right|\_0 \mathbf{d}\_q t = \begin{cases} \frac{5 - 4q}{12(1 + q)}, & 0 < q < \frac{5}{6}, \\\\ \frac{4q - 5}{12(1 + q)}, & \frac{5}{6} \le q < 1, \end{cases} \end{aligned}$$

and obtain the integral expressions of *<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*), which have the same formulas as those given in Theorem 1. This completes the proof of Theorem 2.

 <sup>12</sup>(<sup>1</sup>+*q*),  ≤ *q* < 1,
