117 Let the function $f(x)=x^{2}+x+\frac{1}{2}$ have the domain $[n, n+1]$ (where $n$ is a natural number), then the range of $f(x)$ contains $\qquad$ integers.
$1172(n+1)$. Let $y=f(x)=x^{2}+x+\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{1}{4}$. When $x>-\frac{1}{2}$, $f(x)$ is an increasing function, so
$$
n^{2}+n+\frac{1}{2} \leqslant f(x) \leqslant(n+1)^{2}+n+1+\frac{1}{2}
$$

If $f(x) \in \mathbf{Z}$, then
$$
n^{2}+n \leqslant f(x) \leqslant(n+1)^{2}+n+1
$$

Therefore, the number of integers in the range of $f(x)$ is
$$
(n+1)^{2}+(n+1)-\left(n^{2}+n\right)+1=2(n+1) \text { (integers). }
$$