The constant is a vector BC only

For $\mathbf{v}(t) = \langle f(t),\, g(t)\rangle$,

$$\int \mathbf{v}(t)\,dt = \left\langle \int f(t)\,dt,\; \int g(t)\,dt \right\rangle + \mathbf{C}, \qquad \mathbf{C} = \langle C_1,\, C_2\rangle.$$

The bold $\mathbf{C}$ is the whole lesson. Each component was antidifferentiated separately, so each carries its own unknown constant, and there is no reason for the two to be equal. Writing a single scalar $C$ and adding it to both entries asserts that they are, which is an assumption the problem never made.

The count is worth stating plainly: two unknowns, two equations. The initial condition $\mathbf{r}(t_0) = \langle a,\, b\rangle$ is not one equation about vectors; it is two equations about numbers, one per slot, and each determines its own constant.

Take $\mathbf{v}(t) = \langle 2t,\, 3t^{2}\rangle$ with $\mathbf{r}(0) = \langle 1,\, -4\rangle$. Antidifferentiating gives $\langle t^{2},\, t^{3}\rangle + \mathbf{C}$. Substituting $t = 0$ gives $\langle 0,\, 0\rangle + \mathbf{C} = \langle 1,\, -4\rangle$, so $\mathbf{C} = \langle 1,\, -4\rangle$ and

$$\mathbf{r}(t) = \langle t^{2} + 1,\; t^{3} - 4\rangle.$$

Against $\mathbf{r}(2) = \langle 5,\, 4\rangle$ for the example above:

1. Dropped entirely: $\langle 4,\, 8\rangle$. This is the antiderivative evaluated at $t = 2$ with no constant, so it describes a particle that started at the origin.
2. Applied to the first slot only: $\langle 5,\, 8\rangle$. Half right, which is what makes it durable: the $x$-coordinate checks out against the initial condition and the $y$-coordinate silently does not.
3. Applied to the second slot only: $\langle 4,\, 4\rangle$. The same failure mirrored.

The verification that catches all three costs one substitution: put $t_0$ back into your answer and confirm you get the given initial position, in both slots. Here $\mathbf{r}(0) = \langle 0 + 1,\, 0 - 4\rangle = \langle 1,\, -4\rangle$, which matches. Any answer failing that check is wrong, and the check runs in seconds.

Geometrically the constant is a translation. The antiderivative without $\mathbf{C}$ traces a curve of exactly the right shape starting at the origin; adding $\mathbf{C}$ slides that whole curve into position. The shape is never wrong, only its location, which is why the error survives a glance at the picture.

Rather than finding $\mathbf{C}$ explicitly, the position can be built directly:

$$\mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^{t} \mathbf{v}(u)\,du.$$

This is 8.2's identity with a second component, and it has the same shape for the same reason: an integral of a rate is a change, and a change plus a starting value is a position. Written this way the initial condition is structurally impossible to omit, because it is the first term rather than a constant to be solved for afterwards.

The same construction runs one rung up the chain. Given acceleration and both initial values,

$$\mathbf{v}(t) = \mathbf{v}(t_0) + \int_{t_0}^{t}\mathbf{a}(u)\,du, \qquad \mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^{t}\mathbf{v}(u)\,du,$$

and a problem that supplies $\mathbf{a}$ needs two initial conditions, one for each integration. Four unknown constants in total, four equations. Being handed only $\mathbf{r}(0)$ when the acceleration is given is a signal that something is missing from the problem, not a licence to proceed.

Without an initial position, $\int_{t_1}^{t_2}\mathbf{v}(t)\,dt$ is still meaningful. It is the displacement vector: the change in position over that interval, entry by entry.

For the running example over $[0, 2]$, $\int_0^2 \langle 2t,\, 3t^{2}\rangle\,dt = \langle 4,\, 8\rangle$. So the particle finished $4$ to the right and $8$ up from wherever it began, regardless of where that was. Adding the starting point $\langle 1,\, -4\rangle$ gives the final position $\langle 5,\, 4\rangle$.

Three quantities to keep apart, and 9.6 is about telling them apart under pressure:

1. Displacement: $\int \mathbf{v}\,dt$, a vector.
2. Position: initial position plus that displacement, a vector.
3. Distance travelled: $\int |\mathbf{v}|\,dt$, a scalar, and generally larger than the length of the displacement.

Note that the third one is not the magnitude of the first. $|\int \mathbf{v}\,dt|$ and $\int |\mathbf{v}|\,dt$ differ whenever the direction of travel changes, for the same reason $\left|\int v\right|$ and $\int |v|$ differed in 8.2 and 8.6.