What does the integral of position with respect to time mean? The integral of acceleration with respect to time is velocity.
The integral of velocity with respect to time is position.
What is the integral of position with respect to time, and what does it mean?
Please explain so that your answer is understandable by someone who took calculus I.
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$tl; dr: It's true that "velocity is the derivative of position", but "acceleration is the derivative of velocity" is not true in the same sense: The notion of velocity is independent of arbitrary coordinate changes, but acceleration isn't; you have to equip space with "extra structure" before you can make sense of acceleration (which becomes the "covariant derivative" of velocity). In this framework, "integral of position" doesn't even have mathematical meaning; there's no way to add positions. 

Caveat: I don't know how to express these ideas without going beyond the normal high school curriculum. However, I've tried to factor out the technicalities and deeper material as web links.
Let's first take a closer look at the implicit premises:


*

*Velocity is the derivative of position.

*Acceleration is the derivative of velocity.
Despite what we teach in elementary calculus, these statements are not on an equal footing.
In elementary calculus and physics, our model of space is $\Reals^{n}$, the Cartesian space whose points are labeled by ordered $n$-tuples of real numbers. Our model of time is $\Reals$, and an interval $I$ of real numbers represents "an interval of time". The position of a point particle during an interval $I$ is modeled by a continuous (often smooth) mapping $\Vec{x}:I \to \Reals^{n}$, which we unconsciously "decompose into component functions":
$$
\Vec{x}(t) = \bigl(x_{1}(t), x_{2}(t), \dots, x_{n}(t)\bigr),\quad t \in I.
\tag{1}
$$
If the position of our particle is continuously-differentiable, we define the velocity to be
$$
\Vec{x}'(t) = \bigl(x_{1}'(t), x_{2}'(t), \dots, x_{n}'(t)\bigr),\quad t \in I.
\tag{2a}
$$
If the position is twice continuously-differentiable, we define the acceleration to be
$$
\Vec{x}''(t) = \bigl(x_{1}''(t), x_{2}''(t), \dots, x_{n}''(t)\bigr),\quad t \in I.
\tag{3a}
$$
Closer inspection leads us to a more cautious viewpoint: The Cartesian coordinates we've taken for granted are not intrinsic to space; they're extra structure we imposed. In that spirit, we should ask whether the preceding definitions depend on the choice of coordinates.
Remarkably, velocity "transforms linearly (i.e., like a tensor) under change of coordinates". Acceleration does not.
To see why, let $\phi$ represent a coordinate transformation, and write $\Vec{y} = \phi(\Vec{x})$ for the coordinate representation of our particle's position in the "new" coordinates. By the (multivariable) chain rule,
$$
\Vec{y}'(t) = D\phi(\Vec{x})\, \Vec{x}'(t).
\tag{2b}
$$
The coordinate representation of the velocity of our particle in the new system is a linear function of the coordinate representation in the old system.
By contrast, differentiating (2b) and using the product rule gives
$$
y''(t) = D\phi(\Vec{x})\, \Vec{x}''(t) + \bigl[D\bigl(D\phi(\Vec{x})\bigr) \Vec{x}'(t)\bigr] \Vec{x}'(t).
\tag{3b}
$$
The first term on the right is the "pleasant" part, which transforms like a tensor; the second term involves second derivatives of the coordinate change, and is not linear in $\Vec{x}'$. If acceleration of a particle is to transform like a tensor, we must either


*

*Restrict the set of "allowable" changes of coordinate, or

*Modify our notion of differentiation to cancel the second term.
The approach of elementary calculus and physics may be viewed as fixing the Euclidean metric and permitting only changes of coordinate that preserve this extra structure. If $\phi$ is a rigid (Euclidean) motion, then the first derivative $D\phi$ is a constant field of linear transformations, and the second derivative vanishes, so (3b) becomes
$$
\Vec{y}''(t) = D\phi(\Vec{x})\Vec{x}''(t).
$$
The approach of coordinate-free mechanics, and of general relativity, is to fix a Riemannian metric, and to replace the componentwise derivative with covariant differentiation. (Compare the second term on the right in (3b) with the second partials of $\Psi$ appearing in the Wikipedia entry on Christoffel symbols.)
To summarize the preceding discussion:


*

*In elementary calculus and physics, position, velocity, and acceleration are all modeled by ordered $n$-tuples of real numbers, i.e., by points/vectors in $\Reals^{n}$.

*When one looks more closely, a position of a point particle is modeled by a point in a smooth $n$-manifold $M$; a velocity is an element of the tangent bundle of $M$; an acceleration is either


*

*An element of the second tangent bundle $T(TM)$ (if we impose no additional structure on $M$), or

*An element of $TM$ (if we use a connection to identify the horizontal subbundle of $T(TM)$ with $TM$).

With all this understood, it's difficult to understand what is even meant by "the integral of position" in a coordinate-invariant sense. Loosely, integration is a process of summing, but positions—points of a manifold—can't be added in any obvious natural way. (In order to subtract points in a coordinate-invariant manner, we had to construct an entirely new space, the tangent bundle $TM$.)
Further, one would naively expect that "the derivative of 'the integral of position with respect to time' is position (up to an additive constant)". If "the integral of position" could be interpreted as a path in some manifold $P$, the derivative of this path would then "live" both in $TP$ and in $M$; that's impossible, since "most" manifolds $M$ are not the total space of the tangent bundle of another manifold.
While these observations aren't definitive (I may be getting unimaginative with age), they do strongly suggest that


*

*Within the framework of differential geometry, "the integral of position with respect to time" has no mathematical (much less physical) meaning.

*Any useful definition of "the integral of position with respect to time" will require a fundamental reformulation of the notion of position.

*Aside from interpretations within Euclidean geometry (which, as a matter of opinion, I suspect are "not particularly interesting"), the expression
$$
\int \Vec{x}(t)\, dt
  = \left(\int x_{1}(t)\, dt, \int x_{2}(t)\, dt, \dots, \int x_{3}(t)\, dt\right)
$$
is not meaningful. (Contrast with "the integral of position with respect to position", from which one can extract, e.g., the theory and applications of line integrals.)
A: Suppose there's a lever that you can move from $0$ to $1$. Letting $f(t)$ be the position of that lever over time as you move it, you can think of the lever's velocity $f'$, acceleration $f''$, etc. Now imagine that lever controls a floodgate. The floodgate is closed at the $0$ position and opens as you move the lever towards $1$. The integral $\int_0^t f(x)\,\mathrm{d}x$ measures the accumulation of water that has spilled from the floodgate over time. If you leave the lever at some fixed position, the water will be flowing out at a constant rate. The absement in this situation measures the accumulated water, equal to $\int_0^t f(x)\,\mathrm{d}x$, and is a measure of the sustained position of the lever.
A: It's called Absement. From the Wikipedia page,

... absement (or absition) is a measure of sustained displacement of an object from its initial position, i.e. a measure of how far away and for how long.

There are also names for more derivatives/integrals of position:
-4 Abserk
-3 Abseleration
-2 Absity
-1 Absement [Absition]
 0 Displacement [Position]
 1 Velocity
 2 Acceleration
 3 Jerk
 4 Jounce
 etc

A: I would say that integral is the sum ($\int$)of the position ($x$) weighted by the duration ($dt$) stayed.
