Prove that $\lim\limits_{x\to+\infty}\frac{1}{x}\int_0^x|\sin t|\mathrm{d}t=\frac{2}{\pi}$ I came cross the following equation:
$$\lim_{x\to+\infty}\frac{1}{x}\int_0^x|\sin t|\mathrm{d}t=\frac{2}{\pi}$$
I wonder how to prove it.
Using the Mathematica I got the following result:

Could you suggest some ideas how to prove this?  Any hints will be appreciated.
 A: $\textbf{Hint}$: the limmand can be rewritten as
$$ \frac{\int_0^{\left\lfloor \frac{x}{\pi}\right\rfloor \pi} |\sin t|dt + \int_{\left\lfloor \frac{x}{\pi}\right\rfloor \pi}^x |\sin t|dt}{x}$$
$$=2 \frac{\left\lfloor \frac{x}{\pi}\right\rfloor}{x}+\frac{\int_{ \left\lfloor \frac{x}{\pi}\right\rfloor}^x|\sin t|dt}{x}$$
because the integrand is $\pi$-periodic. The equality $z \equiv \lfloor z \rfloor + \{z\}$ (where $0\leq \{ \cdot \} < 1$ denotes the fractional part function) can be used to calculate the limit on the left. Squeeze theorem can be used for the remainder on the right.
A: Let $x=A+\pi C$ where $A$ is between $0$ & $\pi$ , with Integer $C$.
$D(A,C) = \frac{1}{A+\pi C}\int_0^{A+\pi C}|\sin t|\mathrm{d}t$
$D(A,C) = \frac{1}{A+\pi c}\int_0^{\pi C}|\sin t|\mathrm{d}t + \frac{1}{A+\pi c}\int_{0+\pi C}^{A+\pi C}|\sin t|\mathrm{d}t$
$D(A,C) = \frac{1}{A+\pi C} [[ \int_0^{\pi C}|\sin t|\mathrm{d}t ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$
$D(A,C) = \frac{1}{A+\pi C} [[ \int_0^{\pi}|\sin t|\mathrm{d}t + \int_{\pi}^{2\pi}|\sin t|\mathrm{d}t + \int_{2\pi}^{3\pi}|\sin t|\mathrm{d}t +\cdots \int_{\pi (C-1)}^{\pi (C)}|\sin t|\mathrm{d}t ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$
Pictorially :

Here $x$ is given in terms of $A$ & $C$.
Each Purple Area is the Positive Part of a Cycle of the $\sin$ Curve having $Area=2$.
Each Green Area is the Negative Part of a Cycle of the $\sin$ Curve having $Area=2$.
The last Gray Area is the Partial Cycle having $Area$ between $0$ & $2$.
$D(A,C) = \frac{1}{A+\pi C} [[ 2 + 2 + 2 +\cdots 2 ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$
$D(A,C) = \frac{1}{A+\pi C} [[ 2C ]] + \frac{1}{A+\pi c}\int_0^{A}|\sin t|\mathrm{d}t$
$D(A,C) = \frac{2C}{A+\pi C} + \frac{k}{A+\pi C}$ where $k$ is a number (Depending on $A$) between $0$ & $2$
$\large{\displaystyle\lim_{x\to+\infty}\frac{1}{x}\int_0^x|\sin t|\mathrm{d}t = \lim_{C\to+\infty}D(A,C) = \frac{2}{\pi}}$
The Limit will not change with $A$ or $k$ because the Grey Area is too negligible when $C$ is very large.
