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Say we have some vector-valued function that, when feed a t-value, gives a particles 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?

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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} $.

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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 $\Gamma$.

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