Show $\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)=\frac{f(1)-f(0)}{2}$ 
Show that if $f$ is continuously differentiable on $[0,1]$, then $$\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)=\frac{f(1)-f(0)}{2}$$


Observe that 
\begin{align*}
\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)&=\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f(x)dx\right)\\
 &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}\left[f\left(\frac{i}{n}\right)-f(x)\right]dx\right)\\
 &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\right)
 \end{align*}
    where the last equality follows from the Mean Value Theorem.
Let $m_i=\inf\{f'(x):x\in[(i-1)/n,i/n]\}$ and $M_i=\sup\{f'(x):x\in[(i-1)/n,i/n]\}$, then we have the follow inequality: $$m_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx\leq\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq M_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx$$
Consequently $$\frac{1}{2n}\sum_{i=1}^{n}m_i\leq n\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq\frac{1}{2n}\sum_{i=1}^{n} M_i$$ where $\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx=\frac{1}{2n^2}$. 

I stuck at this step. And it seems not right because when I take the limit both sides, I have $0$. Can someone give me a hint or suggestion. Thanks in advanced.
 A: Your bounds
$$ \frac{1}{2n} \sum_{i=1}^{n} m_i
\quad \text{and} \quad
\frac{1}{2n} \sum_{i=1}^{n} M_i $$
converge to the same quantity, namely $\frac{1}{2}(f(1) - f(0))$. This is essentially because they are Riemann sums for $\frac{1}{2}f'(x)$.

Proof using Taylor Theorem. Let $x_i = i/n$ for brevity and consider $F(x) = \int_{0}^{x} f(t) \, dt$. Then we may write
$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right)
= n \sum_{i=1}^{n} (F(x_{i-1}) - F(x_i) + \tfrac{1}{n}F'(x_i)). $$
By Taylor Theorem, we can pick $c_i \in [x_{i-1}, x_i]$ such that
$$ F(x_{i-1}) = F(x_i - \tfrac{1}{n}) = F(x_i) - \tfrac{1}{n}F'(x_i) + \tfrac{1}{2n^2}F''(c_i). $$
Plugging this back, we have
$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right)
= \frac{1}{2n} \sum_{i=1}^{n} f'(c_i). $$
Taking $n \to \infty$, this converges to $\frac{1}{2}(f(1) - f(0))$ as desired.
A: Hint: Use Taylor's theorem and write the following:
$$f(x) = f(\frac{i}{n}) +f'(\frac{i}{n})(x-\frac{i}{n})+h_i(x)(x-\frac{i}{n}) $$, where $h_i(x)$ is a continuous function that has a limit 0 at $x = \frac{i}{n}$. The rest should be a matter of computation. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Indeed, it involves the application of Bernoulli Numbers $\braces{B_{i}}$:
\begin{align}
\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} & =
\int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x +
\sum_{i = 1}^{m}{B_{i} \over i!\,n^{i - 1}}\bracks{%
\mathrm{f}^{\pars{i - 1}}\pars{1} -
\mathrm{f}^{\pars{i - 1}}\pars{0}} + R_{+}\pars{\mathrm{f},m}
\\[3mm] & =
n\int_{0}^{1}\mathrm{f}\pars{x}\,\dd x\ +\
\overbrace{B_{1}}^{\ds{\half}}\
\bracks{\mathrm{f}\pars{1} - \mathrm{f}\pars{0}} +
{1 \over 2n}\
\overbrace{B_{2}}^{\ds{{1 \over 6}}}\bracks{\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0}} + R_{+}\pars{\mathrm{f},2}
\end{align}
which yields
\begin{align}
&n\bracks{{1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} -
\int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x} =
{\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2} +
{\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0} \over 12n} + R_{+}\pars{\mathrm{f},2}
\end{align}

Then,
\begin{align}
\color{#f00}{\lim_{n \to \infty}n\bracks{%
{1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} - \int_{0}^{1}\mathrm{f}\pars{x}\,\dd x}} & =
\color{#f00}{{\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2}}
\end{align}

whenever the remaining term $R_{+}\pars{\mathrm{f},2}$ vanishes out ( see the above cited link ) in the $n \to \infty$ limit.
