Given a continuous function $f$ prove $\lim\limits \frac 1n \sum_{k=0}^{n-1} k \int_{\tfrac kn}^{\tfrac{k+1}{n}} f(t)dt = \int_0^1 xf(x)dx$ Given a continuous function $f$ on $[0,1]$, it is asked to prove the follwoing 
$$\lim\limits \frac 1n \displaystyle\sum_{k=0}^{n-1} k \displaystyle\int_{\tfrac kn}^{\tfrac{k+1}{n}} f(t)dt = \displaystyle\int_0^1 xf(x)dx$$
I tried to use Riemann sums this way :
We know that  
$$\displaystyle\int_0^1 xf(x)dx = \lim\limits \frac 1n \displaystyle\sum_{k=0}^{n-1} \frac kn f\left(\frac kn \right)$$
Setting 
$$u_n := \frac 1n \displaystyle\sum_{k=0}^{n-1} \frac kn f\left(\frac kn \right) $$
 and 
$$ v_n :=\frac 1n \displaystyle\sum_{k=0}^{n-1} k \displaystyle\int_{\tfrac kn}^{\tfrac{k+1}{n}} f(t)dt $$ 
My idea is to prove that $(v_n - u_n) \to 0 $
Then, i wrote 
$$v_n - u_n = \dfrac 1n \displaystyle\sum_{k=0}^{n-1} k \displaystyle\int_{\tfrac kn}^{\tfrac{k+1}{n}} \left( f(t) - f\left(\frac kn \right) \right) dt $$
So I bounded above (in absolute values) all integrals in that sum by $2M$ where $M = \max \mid f \mid$ on $[0,1]$
Unfortunately the upper bound I got tends to $M$ (not to $0$)
Anu ideas on how to continue are welcome
thanks for your time
 A: Let $n$ be a positive integer greater than $ 1 $, we have the following : \begin{aligned}\frac{1}{n}\sum_{k=0}^{n-1}{k\int_{\frac{k}{n}}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}}&=\frac{1}{n}\sum_{k=0}^{n-1}{\left(k\int_{0}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}-k\int_{0}^{\frac{k}{n}}{f\left(t\right)\mathrm{d}t}\right)}\\ &=-\frac{1}{n}\sum_{k=0}^{n-1}{\int_{0}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}}+\frac{1}{n}\sum_{k=0}^{n-1}{\left(\left(k+1\right)\int_{0}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}-k\int_{0}^{\frac{k}{n}}{f\left(t\right)\mathrm{d}t}\right)}\\ \frac{1}{n}\sum_{k=0}^{n-1}{k\int_{\frac{k}{n}}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}}&=-\frac{1}{n}\sum_{k=1}^{n}{\int_{0}^{\frac{k}{n}}{f\left(t\right)\mathrm{d}t}}+\int_{0}^{1}{f\left(t\right)\mathrm{d}t}\end{aligned}
Since $ \frac{1}{n}\sum\limits_{k=1}^{n}{\int\limits_{0}^{\frac{k}{n}}{f\left(t\right)\mathrm{d}t}}\underset{n\to +\infty}{\longrightarrow}\int\limits_{0}^{1}{\int\limits_{0}^{x}{f\left(t\right)\mathrm{d}t}\,\mathrm{d}x}=\left[\left(x-1\right)\int\limits_{0}^{x}{f\left(t\right)\mathrm{d}t}\right]_{0}^{1}-\int\limits_{0}^{1}{\left(x-1\right)f\left(x\right)\mathrm{d}x} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int\limits_{0}^{1}{\left(1-x\right)f\left(x\right)\mathrm{d}x} $
We get $$ \frac{1}{n}\sum_{k=0}^{n-1}{k\int_{\frac{k}{n}}^{\frac{k+1}{n}}{f\left(t\right)\mathrm{d}t}}\underset{n\to +\infty}{\longrightarrow}\int_{0}^{1}{f\left(t\right)\mathrm{d}t}-\int_{0}^{1}{\left(1-x\right)f\left(x\right)\mathrm{d}x}=\int_{0}^{1}{xf\left(x\right)\mathrm{d}x}$$
A: Since $f(x)$ is continuous in $[0,1]$, so $f(x)$ is uniformly continuous; namely, for $\forall \epsilon>0$, there is $\delta>0$ such that, when $x_1,x_2\in[0,1]$ with $|x_1-x_2|<\delta$, 
$$ |f(x_1)-f(x_2)|<\epsilon. $$
So for big $n$ with $\frac{1}{n}<\delta$, one has $|t-\frac{k}{n}|<\delta$ for $t\in[\frac{k}{n}-\frac{k+1}{n}]$ and hence
$$ |f(t)-\frac{k}{n}|<\epsilon. $$
Thus for big $n$,
\begin{eqnarray}
|v_n - u_n| &=& \bigg|\dfrac 1n\sum_{k=0}^{n-1} k \displaystyle\int_{\tfrac kn}^{\tfrac{k+1}{n}} \left( f(t) - f\left(\frac kn \right) \right) dt\bigg|\\
&\le& \dfrac 1n \sum_{k=0}^{n-1} k\int_{\tfrac kn}^{\tfrac{k+1}{n}} \bigg|f(t) - f\left(\frac kn \right) \bigg|dt\\
&\le&\frac{1}{n}\sum_{k=0}^{n-1}\frac{\epsilon}{n}\\
&=&\frac{n(n-1)}{2n^2}\epsilon\le \frac{1}{2}\epsilon
\end{eqnarray}
and hence
$$ \lim_{n\to\infty}u_n=\lim_{n\to\infty}v_n=\int_0^1xf(x)dx. $$
