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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?

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You're essentially integrating distance, which gives you a quantity that has units of (distance)$\times$(time), say "meter-seconds" (NOT meters-per-second).

Think of an integral of a function $r(t)$ as the average value $r_{av}$ of that function on an interval $[a,b]$:


Then if your $r(t)$ represents displacement, your integral is essentially the average displacement (which becomes a vector in higher dimensions), times the amount of time spent moving. Or, $r_{av}$ is exactly the average displacement over a time period $[a,b]$.

As far as physical units of meter-seconds, I haven't seen this used anywhere in physics as a fundamental quantity like velocity, force, momentum, etc.

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One thing you do see are volt-seconds in the context of electromagnetism. But that's a unit of magnetic flux i.e. the integral $\int_S \mathbf{B}\cdot d\mathbf{A}$ over a closed surface $S$, and thus is not a vector quantity; so units alone aren't enough. On the other hand, the impulse $\Delta \mathbf{p}=\int_{t_0}^{t_1} \mathbf{F}\,dt$ is a genuinely vector-based quantity; however, since momentum is mass times velocity, this rus into the 'not a derivative' requirement... – Semiclassical Sep 6 '14 at 1:30
Physical interpretation: the average of $\mathbf r$ is the center of mass of uniformly thick wire parametrized by $\mathbf r$. When $[a,b]=[0,1]$, the integral gives exactly that. – user147263 Sep 6 '14 at 18:36

If you want an intuitive understanding of an integral $\int_a^b{\bf r}(t)\>dt$ you first have to give some intuitive sense to products of the form $${\bf r}\>\Delta t\ ,\tag{1}$$ where $t$ is a one-dimensional variable (not necessarily time), $\Delta t$ an interval-length in this variable, and ${\bf r}$ is a constant vector.

Given such a physical interpretation of $(1)$ we then can envisage situations of the following kind: We are given a vector-valued function of this variable $t$ (not necessarily a parametrical representation of a curve) $$t\mapsto{\bf r}(t)\qquad(a\leq t\leq b)\ ,\tag{2}$$ and we want to know the total impact in the sense of $(1)$ this variable vector has along the interval $[a,b]$. This total impact is called the integral of ${\bf r}(\cdot)$ over $[a,b]$, and is denoted by $$\int_{[a,b]}{\bf r}(t)\>{\rm d}t\ .$$ The integral should have the following properties: It is linear in ${\bf r}(\cdot)$, additive with respect to the concatenation of intervals, and for a constant ${\bf r}(\cdot)$ one has $$\int_{[a,b]}{\bf r}\>{\rm d}t={\bf r}\cdot(b-a)\ .$$ Using these axioms it is easy to see that for continuous functions $(2)$ the integral $\int_{[a,b]}{\bf r}(t)\>{\rm d}t$ has to be a limit of Riemann sums: $$\int_{[a,b]}{\bf r}(t)\>{\rm d}t=\lim_\ldots\>\sum_{k=1}^N {\bf r}(\tau_k)\>(t_k-t_{k-1})\ ,$$ where one has partitions $$a=t_0<t_1<\ldots<t_N=b$$ and sampling points $\tau_k\in[t_{k-1},t_k]$ in mind.

Note that these explanations did not mention the FTC, and they make sense as well for multivariate functions defined on domains $B\subset {\mathbb R}^d$. The FTC and Fubini's theorem come into play when we actually want to compute such limits of Riemann sums in a systematic way.

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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.

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