If $ x \in \mathbb{R}^n $ fixed, then $ \int_{Q_k}f(x,2^k)dx = f(x,2^k)\mu(Q_k) $. [closed]

Question

I have a question in the following.

$Q_k$ is a open cubes such that $Q_k = \{ x = (x_1,x_2,\cdots, x_n)\in \mathbb{R}^n : a_{k-1} < x_i < a_k\,, i = 1,2, \cdots,n\},$ where $a_0 \in \mathbb{R}$.

If we take a fixed $x \in \mathbb{R}^n$, then can i write $\int_{Q_k}f(x,2^k)dx =f(x,2^k)\mu(Q_k)$ where $f: \mathbb{R}^n \times [0,\infty) \rightarrow [0,\infty]$. Is it true or not?