What represent ito integral ? What does means that $\int_0^t B_sdB_s=\frac{B_t^2}{2}-\frac{t}{2}$ I still have difficulties to understand Itô integral, and unfortunately, I don't really understand what represent the Itô integral. 
Just to inform, I asked, but it has been erased without explanation. 
I already ask this question previously here but it has been put on Hold without any explanation. Also, I found similar question on MSE here, here but I'm not really convinced by the answers. Let take an example $$\int_0^t B_sdB_s=\lim_{n\to \infty }\sum_{i=0}^{n-1} B_{t_i}(B_{t_{i+1}}-B_{t_i}),$$
where $\{t_0,...,t_{n+1}\}$ is a subdivision of $[0,t]$ where $$\max_{i}|t_{i+1}-t_i|\to 0.$$
By calculation, we get $$\int_0^t B_sdB_s=\frac{B_t^2}{2}-\frac{t}{2}.$$
But how can we interpret this result ? What does it mean exactly ? First I get crazy to understand what represent $\int_0^t B_sdB_s$ (instead of the fact that it's continuous martingale). There is a picture of this integral on wikipedia, but I can't really understand what it represent. 


*

*For example, $\int_0^af(x)dx$ is the area between $y=0$, $x=a$, $x=b$ and the curve $y=f(x)$.

*If $g$ is a positive function, then $\int_a^b fdg$ can be seen as the density, of the the surface delimited by $x=a$, $x=b$, $x=0$ and $y=f(x)$.

*Now, what could represent $\int_0^t B_sdB_s$ ?
 A: I have the same question but I can not find an answer. I understand the technique but I can not understand the meaning of Itô integral. Your example suggests me a way to explain it.
First I try to find some physical meaning as you, but I can not. Then I think about the Brownian motion $B_s$ as a random walk, then a random walk as a game.
Think about the path of $B$ as a game. If the path goes up, you win. If it goes down, you lose. Suppose that you gets the amount of money $c$ if you win, and you lost $c$ (or get $-c$) if you lose. Denote the amount of money at time $t$ by $M_t$ and let $M_0=0$. Note that $B_t$ can be thought as the number of win minus the number of loss. Hence at time $t$, the amount of money is $cB_t$ which is the same as Itô integral
$\int_0^t c d B_s = cB_t$. Note that we have the constant $c$ in the integral because $d B_s <0$ when you lose.
Now, if you replace the amount of money you get/lost at time $t$ by $B_t$, then the amount of money at time $t$ is $\int_0^t B_s dB_s=\frac{B_t^2}{2}-\frac{t}{2}$. This is not a good example because it is weird for $B_t<0$ (you lost money even you win).
To make it more intuitive, I consider a discrete version. Consider a random walk on $\mathbb{Z}$. It starts at $0$. At time $t \in \mathbb{N}$, it can only go one step to left or right. Denote by $S_n$ the position at time $n$. Denote $\Delta S_t=S_{t+1}-S_t$. You get (lost, resp.) the amount of money $S_t$ at time $t$ if you win (lose, resp.). The amount of money at time $n$ is $M_n=\sum_1^n S_t \Delta S_t$. We can show that $2M_{n-1}=S_n^2 -n$ (I did prove it myself).
In conclusion, it can be thought as the following. You have an object which moves randomly. At each time $t$, the object has some kind of reward/penalty such as money or energy which only depends on the past and present (measurable with respect to the $\sigma$-algebra $\mathcal{F}_t$).
