TITLE: Intuitive understanding of integral of vector valued functions
QUESTION [16 upvotes]: Today in class we were introducing complex line integrals. And that got me thinking, I don't know of a good interpretation for integrals of functions from $\mathbb{R}$ to $\mathbb{R}^2$ or $\mathbb{R}^3$. By a good interpretation I mean some (possibly geometric) way of intuitively understanding what such an integral means and why it's calculated like it is.
It's easy to understand why, if $\mathbf{r}(t)$ describes the position of a particle at time $t$, then $\int_a^b \mathbf{r}'(t)\ dt = \mathbf{r}(b)-\mathbf{r}(a)$, by imagining lots of small tangents to the curve $\mathbf{r}(t)$ being added up. It's harder to picture what $\int_a^b \mathbf{r}(t)\ dt$ means without thinking of $\mathbf{r}(t)$ as a derivative. Does anyone know of an intuitive interpretation for this kind of integrals?

REPLY [3 votes]: Since you're asking for intuition, I won't give the proofs of the statements below, although they're not hard.
I'll assume that you already have an intuitive understanding of what the integral of a single-variable function means. Although it's not technically necessary, things will be easier to picture if we imagine that the vectors belong to Euclidean space, with a concept of length, angles, etc., rather than just a vector space.
Let an integrable (say, continuous) vector function $r \colon [a,b] \to \mathbf{R}^n$ be given, and let $e \in \mathbf{R}^n$ be a vector. Although it's not technically necessary for what follows, things will be easier to picture if we take $e$ to have length $1$, and think of it as indicating a direction.
Everything I'm going to say is based on the following (true) equation.
$$\int_a^b r(t) \cdot e \, dt = \left( \int_a^b r(t) \, dt \right) \cdot e$$
Now let's interpret this equation. The dot product $r(t) \cdot e$ tells us how far the vector $r(t)$ goes in the direction of $e$. More precisely, if we project the vector $r(t)$ orthogonally onto the line determined by $e$, the number $r(t) \cdot e$ is the length of the projection, together with a $\pm$ sign depending on whether the projection is in the same direction as $e$ or the opposite one. We'll call this the $e$-component of $r(t)$. It's as though we've cut out of $r(t)$ everything except what was in the $e$-direction.
With this terminology in place, we can now make the following statement. Let $I$ be the vector integral $\int_a^b r(t) \, dt$ whose interpretation we're interested in. Now select any direction $e$. Then the $e$-component of $I$ is the integral of the single-variable function consisting, at all times $t$, of the $e$-component of $r(t)$. 
In other words, if you cut everything out of $r(t)$ except what's in the direction of $e$, and then integrate the resulting single-variable function, the number you get is the same one you would obtain by cutting everything out of $I$ except for what's in the $e$-direction.
This characterizes the vector $I$ completely, because if you know what the $e$-component of a vector is in every possible direction $e$, then you automatically know what that vector is.
It's not clear whether you wanted an interpretation in general, or specifically in the case where $r(t)$ is a radius vector. In this case, the difficulty of interpretation is not really related to the fact that it's a vector function you're integrating. In fact, it's not that easy to interpret $\int_a^b x(t) \, dt$ in the case in which the particle is moving only along the $x$-axis. In the single variable case, as others have said, the integral represents $b - a$ times the average displacement. In the vector case, it is also $b-a$ times the average displacement vector, where by "average", in light of the foregoing, I mean that vector $m$ for which, for every direction $e$, the number $m \cdot e$ is the average of the particle's (single-variable) displacement in the $e$-direction.