On the convexity of a particular form of integrals EDIT: I made some critical corrections below.
Let $\mathcal{H}\colon\mathbf{w}\cdot\mathbf{x}+c=0$ be a hypeplane in $\mathbb{R}^n$. Also, let $g\colon\mathbb{R}^n\to\mathbb{R}_+$, be a non-negative, real-valued function. I would like to decide on the convexity of the function $f\colon\mathbb{R}^n\times\mathbb{R}\to\mathbb{R}$, given by
$$
f(\mathbf{a},b)=\int_{\Omega}\! (\mathbf{a}\cdot\mathbf{x}+b)g(\mathbf{x}) \,\mathrm{d}\mathbf{x},
$$
where $\mathbf{x}\in\mathbb{R}^n$, $b\in\mathbb{R}$, and $\Omega$ is the half-space defined by $\mathcal{H}$ as $\Omega=\{\mathbf{x}\in\mathbb{R}^n\rvert\mathbf{w}\cdot\mathbf{x}+c\geq0\}$.
What I have thought so far is to express the integral as a sum (if such a thing is feasible) and use the property of the affine (and consequently convex) quantity $(\mathbf{a}\cdot\mathbf{x}+b)\cdot k$, where $k\in\mathbb{R}$. Albeit, I am not sure at all this is going to work. In addition, I am not sure in what direction should I work to...
Is there any appropriate argument I could use instead?

Not sure if you can use this, but a basic property of convex functions is that nonnegative combinations of them are convex. This extends to nonnegative integrals as well. If $h(x,t)$ is convex in $x$ and $g$ is nonnegative over the domain $\Omega$, and
$$
f(x) = \int_\Omega h(x,t)g(t)dt
$$
then $f$ is convex.
This is a consequence of the definition of convexity and some basic properties of integrals:
$$
\begin{aligned}
f(\theta x + (1-\theta) y) &= \int_\Omega h(\theta x + (1-\theta) y,t)g(t)dt \\
&\le \int_\Omega (\theta h(x,t) + (1-\theta) h(y,t))g(t)dt \\
&= \theta \int_\Omega h(x,t) g(t)dt + (1-\theta) \int_\Omega h(y,t)g(t)dt \\
&= \theta f(x) + (1-\theta) f(y)
\end{aligned}
$$
Conversely, if you can express $f$ in this form with a nonconvex $h$, then you might be able to use it to find a example where $f$ is nonconvex.
In your specific case, it seems, the function is not only convex, but affine, which you can see by factoring:
$$
f(\mathbf{a},b)=\mathbf{a} \cdot \left(\int_{\Omega}\! \mathbf{x}g(\mathbf{x}) \,\mathrm{d}\mathbf{x}\right) + b \left(\int_{\Omega}\! g(\mathbf{x}) \,\mathrm{d}\mathbf{x}\right)
$$
