All functions such that $\lim_{n\to \infty} \int_n^{n+\frac 1n} f(x)dx$ is a non-zero finite value

If we assume that $f$ is continuous we have, by the mean value theorem, that $$\int_{n}^{n+1/n}f\left(x\right)dx=\frac{f\left(c_{n}\right)}{n}$$ where $c_{n}\in\left[n,n+1/n\right]$. Hence $$\lim_{n\rightarrow+\infty}\frac{f\left(c_{n}\right)}{n}=\lim_{n\rightarrow+\infty}\frac{f\left(n\right)}{n}$$ and this limit is finite if and only if $f\left(x\right)\sim ax,\,a\in\mathbb{R}$, as $x\rightarrow+\infty$.

We want to consider the integral $$I=\int\limits ^{n+1/n}_{n} f( x) dx$$ Where $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous. Suppose $f$ has antiderivative $F$. Then $$I=F(n+\frac{1}{n})-F(n)$$ Using a first order Taylor approximation, $$F\left( n+\frac{1}{n}\right) \approx F( n) +F'( n)\left(\frac{1}{n}\right)$$ And since $F'=f$, $$I \approx \frac{f(n)}{n}$$ And this approximation will clearly get better and better for larger and larger $n$. Therefore for $\lim _{n\rightarrow \infty }\frac{f( n)}{n}$ to be finite and nonzero, $\text{deg}(f)=1$. So linear equations are the only types of equations that have this property. EDIT: functions with linear-like growth will also have this property. So functions such as $\frac{x}{\log(x)}$, $x+e^{-x}$, and so on.