What does instantaneous displacement really mean? I've read and (I think) understood that instantaneous velocity is the velocity of an object for an "infinitesimally" small time interval: $$ \lim_{\Delta t\to0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt} $$
This makes sense in my mind.
But what does instantaneous displacement mean? Let's say we have an expression for velocity, $v$ in terms of time $t$: $$v=t +1$$
If we find the antiderivative of this, we get an  expression for displacement, $s$:
$$s = \int{(t+1)dt}$$
$$ s = \frac{t^2}{2} + t + c $$
Let us say we somehow find out the value of $c$ to be $0$. So:
$$s = \frac{t^2}{2} + t $$
If we plug in, let's say, $t = 2$ into this expression, we get s = 4. What does this 4 mean/represent?
Also, is it different to take the definite integral from $t=0$ to $t=2$:
$$ \int_{0}^{2}(t+1)\space dt$$
and using this to find displacement at $t=2$, compared to what we did using the indefinite integral?
Is there a difference between saying "displacement at $t=2$" and "displacement from $t=0$ to $t=2$"?
 A: Answer in response to the original question and your following chat message:
"The answer on the link I've sent says that displacement is not the position, and it is only the difference between final and initial position (change in position). You have said "you can say $\int v dt = s(t)$ (absolute position)". In my mind, there are two possibilities here: either the s here refers to position, or the s refers to displacement and you are saying that the displacement is the absolute position. If it is the latter, that is what the contradiction.
And yes, the specific problem is: "v = t+1, find s (displacement) at the end of 4 seconds if initially body is at origin." My teacher said that I cannot use definite integration as we do not know the boundaries for time, so I have to first use indefinite integration and find the value of c. Aren't the boundaries t=0 and t=4 (they used the word initially). When I tried both approaches, my answer differed by the value of c, as I said."
Response:
Your teacher is creating unnecessary distinctions, but possibly for a good reason to create the least confusion for students new to the topic. In my personal experience, I would call a function s(t) that describes a particles trajectory "position," and I would call a subtraction of two positions $Δs = (s_2–s_1) = s(t_2)–s(t_1)$ "displacement."  I would say the majority of users in Western Industry and Academia would agree with those terms. I would not call s(t) "displacement" in normal usage, but you should recognize that the concept of "position" simply means "displacement from an origin which is agreed to be zero." 
As for your specific problem, you can use either indefinite or definite integral to achieve your answer, as long as you use them correctly. This is because mathematically they are related. If there is a function $F(t)$ whose derivative is $f(t)$, which is to say
$$\frac{dF}{dt} = f(t)$$
then the following are true:
$$ \int f(t) dt = F(t) + c$$
$$ \int_a^t f(t) dt = F(t)-F(a)$$
So we get that $F(a)=-c$.  If you want to obtain $F(t)$ in isolation, choose $a$ such that $F(a)=0$.
When learning these topics, it is best to learn to understand the physical concepts involved rather than be married to strict definitions of terms. You can find "how far the object with velocity $v$ has traveled," and "where it's current location with respect to an origin is," and call those whatever you want.
One last thing, you do know the boundaries for time because the problem says "at the end of 4 seconds." Therefore you are starting your clock at $t=0$ and stopping it at $t=4$.  You also know the initial position because it says "if initially [$t=0$] the body is at origin [$s=0$]."
A: The instantaneous displacement is just the displacement of the system evaluated at some instant $t$, $x(t)$.
As we intuitively think the speed of a body in rest should be zero, it is not so intuitive to think the displacement of a body from a reference point should be zero too. Since our intuitive reference for speed is zero, we should also expect our intuitive reference for displacement to be zero too.
If you think in the mass+spring+damper system:
$$
\ddot x+2\zeta \omega\dot x+\omega^2x=0, x(0)=1
$$
Its zero energy state is at $x=0$ and $\dot x=0$, when both the instantaneous velocity and the instantaneous displacement are zero. In this case, the intuition is somehow more explicit.
In your specific example, if $x(t)=\frac{t^2}2+t$, the instantaneous displacement at $t=2$ is $x(2)=4$, meaning the instantaneous displacement measured from the reference is 4.
In the case of the mass spring damper, for example, if we have an underdamped system $x(t)=e^{-\zeta\omega t}\cos(\omega\sqrt{1-\zeta^2}t)$, the same concept applies, the instantaneous displacement at $t=0$ is $x(0)=1$ but in here it is more evident that the displacement is a deviation from the reference.
In contrast, check the following: Suppose we are measuring a displacement with respect to the last measured position, i.e. suppose $dx'(t)=dt'$ represents the change of displacement from the last time with a resolution of $dt'$, that is, after each $dt'$ interval, the displacement increases in that same measure. In this case, we should, again, integrate to obtain the displacement:
$$
x(t)=\int_0^{t} dx'=\int_0^t dt'=t
$$
That is, since the position is constantly changing, we expect the position to be linearly increasing.
Again, $x(t)$ represent a measurement wrt to a reference, and $\delta x(t)$ is a mathematical object to depict what you could understand by change of displacement. Note, how this last concept is by far less used, hence the use of that notation, less used too.
