Interpreting the integral of a vector-valued function Say we have some vector-valued function that, when feed a $t$-value, gives a particle's position. So, this vector function measures the distance from the origin to a point on the curve that the vector function traces. Now, if we were to integrate this function from a point "$a$" to a point "$b$", where $a > b$, would this act correspond to me adding up all the vectors that define the curve between "$a$" and "$b$"? And would this tell me how far the particle has traveled relative to the origin?
If this isn't the case, how should I be viewing the act of integrating a vector function? 
 A: Let $ \mathbf{v}: [0,\infty) \to \mathbb{R}^{3} $ denote the velocity function of a particle. As rlgordonma has mentioned,
$$
\int_{a}^{b} \mathbf{v}(t) ~ d{t} = \mathbf{s}(b) - \mathbf{s}(a),
$$
which is the particle’s change in displacement between $ t = a $ and $ t = b $. If, instead, you wish to calculate the distance that the particle traveled in the time interval $ [a,b] $, then use the formula
$$
\int_{a}^{b} |\mathbf{v}(t)| ~ d{t}.
$$
The difference here is that you are integrating the speed function $ |\mathbf{v}| $ instead of the velocity function $ \mathbf{v} $.
A: No.  I think you mean integrating a velocity with respect to time to get a net displacement.  In that case, yes, you may integrate component by component to get displacements for each component to form a displacement vector.
In slightly more complicated cases, we have an integral of a force over a path $\Gamma$ to compute work done.  In that case, we are dealing with a component of the force parallel to the tangent to $\Gamma$ at each point, so you get for the work done
$$W = \int_{\Gamma} d\vec{s} \cdot \vec{F}$$
where $d\vec{s}$ is an element of the path $\Gamma$ in the direction of the tangent to $\Gamma$.
