Calculate the integral using probability. I am stuck with the following question:
Let $\operatorname{f}:\left[0,1\right]\to \mathbb{R}$ be a bounded, three times continuously differentiable function. Evaluate the following limit:
$$
\lim_{n \to \infty}\int_{0}^{1}\!\!\!
\int_{0}^{1}\!\!\cdots\!\!\int_{0}^{1}
\!\!\!n\left[\operatorname{f}\left({x_1+x_2+\dots+x_n \over n}\right)-\operatorname{f}\left(\frac{1}{2}\right)\right]
\mathrm{d}x_{1}\,\mathrm{d}x_{2}\ldots \mathrm{d}x_{n}
$$
The hint is to express this integral as an expectation.
I know that I can express $x_{1},x_{2},\ldots,x_{n}$ as i.i.d. random variables $X_{1},X_{2},\ldots,X_{n}$ that has the distribution of $\mathrm{U}\left[0,1\right]$, and the integral can be considered as
$$
E\left[n\left(\operatorname{f}\left(\frac{S_{n}}{n}\right)-\operatorname{f}\left(\mu\right)\right)\right],
$$
where $S_{n} = X_{1} + X_{2} + \cdots + X_{n}$ and $\mu = E\left[X_{1}\right] = 1/2$.
But I do not know how to continue.
Maybe laws of large numbers can be used, but how to use it to evaluate this limit of multiple integral
$?$.
 A: 
Hint Let $\bar{X}_n = S_n/n$.
Taylor's theorem implies
$$f(\bar{X}_n) - f(\mu) = f'(\mu)(\bar{X}_n-\mu) + \frac{1}{2} f''(\mu)(\bar{X}_n - \mu)^2 + \frac{1}{6} f'''(\xi)(\bar{X}_n-\mu)^3$$
where $\xi$ depends only on $\bar{X}_n$ and lies between $\bar{X}_n$ and $\mu$.

Solution sketch:

 Taking the expectation and multiplying by $n$ yields $$nE[f(\bar{X}_n) - f(\mu)] = \frac{n}{2} f''(\mu) \text{Var}(\bar{X}_n) + \frac{n}{6}  E[f'''(\xi)(\bar{X}_n - \mu)^3].$$ Since $\text{Var}(\bar{X}_n) = \text{Var}(X_1)/n$, the first term converges to $\frac{1}{2} f''(\mu) \text{Var}(X_1)$. The second term can be shown to converge to zero because it is of the order $1/n$; use the fact that $f'''$ is continuous on $[0,1]$.

Proof that the last term vanishes:

  Let $M = \sup_{x \in [0,1]} |f'''(x)|$; $M$ exists because $f'''$ is continuous on $[0,1]$. Then \begin{align}|E[f'''(\xi)(\bar{X}_n - \mu)^3]| &\le M E[|\bar{X}_n-\mu|^3] \\&= ME\left[\left|\frac{1}{n}\sum_{k=1}^n (X_k - \mu)\right|^3\right]\\& \le \frac{M}{n^2} E[|X_1 - \mu|^3] \\&\to 0.\end{align}

