Distance in 2D movement Let the position of a particle be given by the vector:
$$\vec{r}=(2t-t^2) \vec{e_{x}} +(8+6t)\vec{e_{y}}$$
where $t$ is the time, starting from $0$.
When solving for the distance the particle has traveled from $t=[0.0 ; 2.0] s$, my teacher said it was equal to the displacement since:
$$s= \parallel\vec{r}\parallel = \sqrt{(\Delta x^2+\Delta y^2)}$$
Upon solving that would give: $$\vec{r_{0}}=(0) \vec{e_{x}}
+(8)\vec{e_{y}}=8\vec{e_{y}}$$ $$\vec{r_{2}}=(2\times2-2^2) \vec{e_{x}} +(8+6\times 2)\vec{e_{y}}=(0) \vec{e_{x}} +(20)\vec{e_{y}}=20\vec{e_{y}}$$ $$\Delta\vec{r}=(0-0) \vec{e_{x}} +(20-8)\vec{e_{y}}=0 \vec{e_{x}}
+12\vec{e_{y}}$$ $$\mathbf{s=\sqrt{0^2+12^2}=12} meters$$
However, using this method, aren't we forgetting to have in count the distance the particle has traveled in the x-axis during this time?
I also drew the function of the distance traveled in the x-axis by the particle according to time :
$$\Delta x=-t^2-2t$$
And concluded that it is a parabola with a vertice on the point of coordinates (1,1) and intersections on the points (0,0) and (2,0).
I then assumed that the distance the particle traveled in the xx-axis was 2 meters (1 to the left + 1 to the right).

That would give us a total distance of 14 meters instead of 12
meters.

My question is whether any of these solutions are right or whether we need integrals to solve for the distance?
My apologies if this is not a relevant question, but I am still new to movements in 2D.
 A: There is a homework rule that means I cannot give you a worked answer.  However as you ask a conceptual question and are asking for clarification this is how it works.
You have some path the object travels on.
You work out the position at each time you are given.  Hint : have you done this correctly ?
The distance between these two positions is what you are looking for.
Your confusion is :

However, using this method, aren't we forgetting to have in count the distance the particle has traveled in the xx-axis during this time?

This is already accounted for in the equation for position.  It has a time dependency.
As a maybe useful exercise, try and work out what initial velocity, position and acceleration that function represents.  It may help clarify what is going on.  You can find these using differentiation.
A: The displacement of the particle from $t=0$ to $t=2$ is indeed $12$ units because the particle starts at $(0,8)$ and ends at $(0,20)$.
But, as you say, the distance travelled by the particle between $t=0$ and $t=2$ is greater than $12$ units because it does not travel in a straight line between the start and end points, but follows a parabola with its vertex at $(1,14)$.
The find the distance travelled you would have to find the arc length of the parabola between $(0,8)$ and $(0,20)$. This will be more than $12$ units but less than $14$ units.
