Evaluate $\lim_{n \to \infty}\int^n_1 \frac{\left |\sin x \right |}{n}dx$ 
Evaluate $$\lim_{n \to \infty}\int^n_1 \frac{|\sin x|}{n}dx$$

I think that I should deal with $\int|\sin x|dx$, but I don't know how to go on. Please help. Thank you.
 A: More generally:

For every (locally integrable) function $f$ with period $T$ and every $t_0$,
  $$
\lim\limits_{t\to+\infty}\frac1t\int_{t_0}^tf(x)\,\mathrm dx=\frac1T\int_0^Tf(x)\,\mathrm dx.
$$

To prove this, call $J=\int\limits_0^T|f(x)|\,\mathrm dx$ and $I(t)=\int\limits_0^tf(x)\,\mathrm dx$. Then, for every integer $n\geqslant0$ and every  $nT\leqslant t\lt(n+1)T$,
$|I(t)-I(nT)|\leqslant J$ and $I(nT)=nI(T)$ hence
$$
\left|\frac1t(I(t)-I(t_0))+\frac1tI(t_0)-\frac{nT}t\cdot\frac1{T}I(T)\right|\leqslant\frac1tJ,
$$
Since $\left|\frac{nT}t-1\right|\leqslant\frac1{n+1}\to0$, $\frac1tI(t_0)\to0$ and $\frac1tJ\to0$ when $t\to+\infty$, the proof is complete.
In your case, $T=\pi$ and $I(\pi)=\int\limits_0^\pi\sin(x)\,\mathrm dx=2$ hence the limit is $\frac2\pi$.
A: I guess you know the result $$\int_0^\pi\!dx\,\sin x = \int_\pi^{2\pi}\!dx\,|\sin x|=2.$$
Using that we find (with $n_- = \lfloor n/\pi \rfloor$)
$$\int_1^n\!dx\,|\sin x| = \underbrace{\int_1^\pi\!dx\,\sin x}_{1+\cos 1} +  \underbrace{\int_\pi^{n_-\pi}\!dx\,|\sin x|}_{2(n_- - 1)} + 
\underbrace{\int_{n_- \pi}^n\!dx |\sin(x)|}_{1-\cos (n -n_- \pi)} .$$
Now it should be easy to perform the limit.
A: We have $\int_{k\pi+c}^{(k+1)\pi+c}|\sin x|\,dx=\int_0^\pi\sin x\,dx=2$, therefore
$$2k\le \int_1^n|\sin x|\, dx\le 2(k+1)$$
if $1+k\pi \le n< 1+(k+1)\pi$, i.e $k=\lfloor\frac{n-1}\pi\rfloor$.
