Denoting the antiderivative of velocity With simple Newtonian laws (and in a specific context), I learned that the speed $\vec{v}$ of an object is the derivative of the corresponding position vector $\vec{OM}$. So that means that 
$$\vec{v}(t)=\frac{\mathrm{d}\vec{OM}(t)}{\mathrm{d}t}.$$
So, if we only have the speed of an object, we need to calculate the primitive of $\vec{v}$ to find the corresponding $\vec{OM}$ vector.
But how can I write it with math symbols? Many people use the integral notation without any domain specified (just the $\int{}$ symbol and I think there isn't any domain specified because it's a notation?) but it's not exactly the same as a primitive right? Or may be I just don't know the correct way for writing primitives.
I heard about the $\sideset{^a_b}{}{\left[\vec{v}\right]}$ notation, but I'm not sure about what it means and how to use it.
 A: The way to write it is
$$\mathbf{r}(t_2)-\mathbf{r}(t_1)=\int_{t_1}^{t_2}\mathbf{v}(t)dt.$$
This is the integrated version of
$$\mathbf{v}(t)=\frac{d\mathbf{r}(t)}{dt}.$$
A: Let $\vec{v}$ be te velocity vector, and $\vec{r}$ the position vector.
Indeed, you are right, it is true that
$$  \vec{v}=\frac{d\vec{r}}{dt}$$
But, you should be aware of one thing: all derivatives imply some loss of information, because you lose track of the constants. See:


*

*The derivative of $x^2$ is $2x$.

*The derivative of $x^2+3$ is also $2x$.

*In general $x^2+\mathrm{constant}$ yields $2x$.


This is a loss of information. Now, if you want to go "backwards", you have many possibilities. IF someone tells you that "the derivative is $2x$, there is no way you know what the constant was. That's why it is called "indefinite integral".
So, when you want to get position starting from its derivative (velocity), you need one more thing.
That extra data is called Boundary conditions. IT means that you need to know an extra equation, so that you can solve all the unknowns. This extra information is usually "at a certain time, the position is this one". That is enough to solve anything.
That said, you've got two options now.
A) DEFINITE INTEGRAL.
As the others have pointed out, if $\vec{v}=d\vec{r}/dt$, then
$$  \Delta\vec{r}=\int_{t_1}^{t_2} \vec{v}\cdot dt $$
which means
$$  \vec{r}_{t_2}-\vec{r}_{t_1}=\int_{t_1}^{t_2} v\cdot dt $$
Check that, if you want the new position, $\vec{r}_{t_2}$, you'd need $\vec{r}_{t_1}$, the boundary condition.
Example: 1D constant velocity
$r_2-r_1=v\cdot \Delta t$. 
It is the usual equation:  $r=r_0+vt$

B) INDEFINITE INTEGRAL + Boundary conditions
You can also solve 
$$  \vec{r}=\int \vec{v}\cdot dt $$
For a 1D constant velocity, you will get
$r=v\cdot t + \mathrm{constant}$
And now you must use the extra data to complete the information.
For $t=0$, $r=r_0$.
Then, substituting in the equation:    $r_0=v\cdot0+\mathrm{constant} \ \Rightarrow  \mathrm{constant}=r_0$
So you get the same result: $r=r_0+vt$.
Both methods are equivalent.
