Why should I find value of constant of integration instead of using definite integration?

Question

$v = 2t$, find $s$ (displacement) at 3 seconds if at $t=0$ body is 2m behind origin.

My teacher said that I cannot use definite integration as we do not know the boundaries for time, so I have to first use indefinite integration and find the value of c. Apparently, I have to do the following steps:

$$ s = \int v \space dt $$ $$ s = \int (2t) \space dt$$ $$ s= {t^2} + c$$

Now, since we are given that initially (at $t=0$), the body is 2m behind origin (the displacement is apparently -2m), we plug in $t=0$ and $s=-2$ into the above equation, getting $c=-2$. Next, I have to plug in $c=-2$ and $t=3$ into the above equation, to find the required answer.

1. Just because the body is starting from 2m behind the origin, does that mean the displacement at $t=0$ is -2m? What does that even mean? I'm confused about the difference between position and displacement
2. Shouldn't the boundaries be $t=0$ and $t=3$ ? But when I use those as boundaries and try to solve using the definite integral, vs. when I use the indefinite integral method shown above, my answer differs by $c$. What does this mean?
3. What does it mean to "find displacement at 3 seconds"?
4. What exactly is the difference between asking to "find displacement at 3 seconds" and "find displacement from 0 to 3 seconds"? In the latter I can use definite integration by putting $t=0$ and $t=3$ as the boundaries, and can't do so in the former, so there must be some difference.

Answer 1

1. Position is a vector, and displacement is a vector. They are equal when the initial position of the displacement is 0. Displacement is just the difference between two positions.

2. True you find the displacement by integrating, but that's not the final position. The answer you got is the change in position. To find the final position, add the displacement to the initial position.

3. Doesn't quite make sense in our normal language. You should say the displacement between 2 time intervals. I believe by the displacement at ($t=0$) you mean the final position. So similar to 2., you add the displacement to the initial position to find the final position.

4. Answered in 2. and 3.