To what random variable does this converge in distribution at n→∞$n\to \mathrm{\infty }$?

Let Wi(i=0,1,2,...)$W_i(i=0,1,2,...)$ be i.i.d. random variables following a standard normal distribution N(0,1)$\mathcal{N}(0,1)$. For a continuously differentiable function f:[0,1]→R$f:[0,1]\to \mathbb{R}$, consider the following random variable:

Xn=∑n−1i=0n−−√(f(i+1n)−f(in))Wi$X_n=\sum _{i=0}^{n-1}\sqrt{n}(f\left(\frac{i+1}{n}\right)-f\left(\frac{i}{n}\right))W_i$

Well, the equation itself looks a lot like gaussian quadrature, so I thought maybe this has something to do with it...

EDIT1:

I just realized that this is essentially a sum of random variables following a standard normal distribution, so the sum also follows a standard normal distribution.

Then, we have E[Xn]=0,Var[Xn]=∑n−1i=0n(f(i+1n)−f(in))2$E[X_n]=0,Var[X_n]=\sum _{i=0}^{n-1}n{(f\left(\frac{i+1}{n}\right)-f\left(\frac{i}{n}\right))}^2$

So,Xn$X_n$ converges in distribution to a normal distribution, due to the central limit theorem?

EDIT 2: Thank you to those who have commented. I now have for n→∞$n\to \mathrm{\infty }$ E[Xn]=0,Var[Xn]→∫10f′(x)2dx$E[X_n]=0,Var[X_n]\to {\int }_0^1{f}^{\prime }(x{)}^{2}dx$.

Then Xn$X_n$ would intuitively converge to a normal distribution with the above average and variance, but how can I show this more rigorously?