Integrating Vector-Valued Functions BC only

Antidifferentiating a vector is componentwise, and the constant of integration is a vector: one entry per component, with no reason for the two entries to be equal. An initial condition is therefore two equations rather than one, and it determines each constant separately. For $\mathbf{v} = \langle 2t, 3t^{2}\rangle$ with $\mathbf{r}(0) = \langle 1, -4\rangle$, the result is $\mathbf{r}(t) = \langle t^{2}+1,\, t^{3}-4\rangle$. The three failures give $\langle 4, 8\rangle$ with the constant dropped and $\langle 5, 8\rangle$ or $\langle 4, 4\rangle$ with it applied to one slot only, and all three are caught by substituting $t_0$ back and checking both entries.

Geometrically the constant is a translation: the antiderivative without it traces the right shape from the origin, so only the location is wrong, which is why a glance at the picture does not catch it. The definite form $\mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^{t}\mathbf{v}$ makes omission structurally impossible, and a problem supplying acceleration needs two initial conditions rather than one. On its own, $\int_{t_1}^{t_2}\mathbf{v}\,dt$ is the displacement vector; the distance travelled is $\int|\mathbf{v}|\,dt$, which is not the magnitude of the displacement whenever the direction of travel changes.