Definite or indefinite integral [Beginning calculus question.] From Edwards and Penney (6e) p. 810, worked Example 8:

Suppose that a moving point has given initial position vector $\boldsymbol{r}(0) = 2\boldsymbol{i}$, initial velocity vector $\boldsymbol{v}(0) = i-j$, and acceleration vector $\boldsymbol{a}(t) = \boldsymbol 2i + 6t \boldsymbol j$. Find its position and evolcity at time $t$.
Solution: $$ \boldsymbol{v}(t) = \int \boldsymbol a(t)dt = \int (2\boldsymbol i + 6t\boldsymbol j)dt = 2t\boldsymbol i+3t^2\boldsymbol j + C_1 $$

What I am confused about is how to decide whether to use a definite or indefinite integral here: $ \boldsymbol{v}(t) = \int \boldsymbol a(t)dt $.
I'm imagining taking that as a definite integral from $0$ to $t$. Is that implicitly what is being done here?
Perhaps, also, my question doesn't make sense, in which case an explanation in elementary language would be most helpful.
 A: You can use any of them in your problem as initial conditions are given.
If you use indefinite integral you are going to get a constant of integration which can be found by substituting the initial values. If initial values are not given then you go for indefinite integration.
If you use definite integration and use the limits 0 to t then you will get an answer which depends on t and will be the same answer if you were to find it by using indefinite integration by substituting initial values. 
By substituting different values of t you can get position and velocity at different times. In indefinite integration you require initial values to be given.
If you are asked to find a function (given the initial conditions) then you can use any of definite or indefinite integration.
A: Note that:
$$
\int_{t_0}^t\left(2\mathbf{i}+6x\mathbf{j} \right) dx= 2t\mathbf{i}+3t^2\mathbf{j}-\left(2t_0\mathbf{i}+3t_0^2\mathbf{j}\right)
$$
that is the same as your result with $C_1=-2t_0\mathbf{i}-3t_0^2\mathbf{j}$.
In this form it is more clear that the constant is a vector, that represent the velocity at an ''initial'' time $t_0$ ( that can be $0$ depending on the given initial conditions) , but the result is essentially the same.
