Confusion in equation of integration and differentiation for motion? 
It says displacement time equation here but x is position of particle here I think. It is not displacement of particle here.
But when it is given velocity of particle to us and then we integrate it. Then, do we get displacement or position?
Does that make any difference?
I think it will make a difference in cases when body accelerates for some time and then retards again. In that case, s will have a different value but it will also be equal to x as well.
 A: Generally speaking, if you fix the orientation of your coordinate system (which, in your 1D case, just means being consistent about which way is "forward") then position $x$ and displacement $d$ (from any point $x_0$) will be related by a simple equation like $x = d + x_0$.
Here I am interpreting "position" as referring to the coordinates of an object in a certain fixed reference and "displacement from a point $x_0$" as referring to the vector which moves you from $x_0$ to the point of interest, $x$, so $d = x - x_0$.
The thing is, as long as the displacement is considered from a fixed point $x_0$ then the velocity can be equivalently calculated by differentiating $d$ or $x$ since the derivative of $x_0$ will just be zero.
The problem does not explicitly tell you that this is the case, but it is quite reasonable to assume it.
Even if it did not hold, by the principle of Galileian relativity you could argue that if the reference point is moving then what you calculate by differentiating the displacement will be the velocity of $x$ relative to $x_0$, and this is all you can compute with the data you are given.
Edit, regarding what you get when you integrate velocity.
If all you know is the expression for $v(t)$, then you can integrate it with respect to time starting from a certain $t_0$ to find the displacement relative to the point $x_0 = x(t_0)$:
$$
\int_{t_0}^t v(\tau) \mathrm{d} \tau = x(t) - x( t_0)
\,.
$$
Displacement from the starting point is all you can get because an object could perform motion with the same velocity $v(t)$ starting from any point in space.
In a stationary reference frame, displacement from the starting point and position are equal up to a constant shift, so you can use either one in calculating the velocity.
However, you can only recover the first from knowing the velocity: in this sense, the velocity profile contains less information than the position profile.
A concrete example: say there is a car on the highway. I do not tell you where exactly it starts, but you know that it starts off stationary, accelerates for a bit, then decelerates and comes to a stop after 1 unit of time.
Say the velocity profile is known, it might be something like $v = 6t(1-t)$ (disregarding units, as your book does). Then, you can calculate how far it got:
$$
\int _0^1 6 t(1-t) \mathrm{d}t = 1 \text{ unit of distance} = x(1) - x(0)
\,.
$$
Now, you know how far the car has travelled (the displacement from the starting point) but if I were to ask how far the car is from the start of the highway (position, in a reference in which we set $x=0$ there) then you wouldn't have enough information to answer.
By the way the question is phrased in your book, though, I believe they when they write "displacement" they mean "displacement from the starting point of the coordinate system", so in this context position ($x$) and displacement are equivalent.
However, you must be careful since this is not the same "displacement from the starting point of the motion" ($d$) which you get when you integrate the velocity from $t=0$ to another time.
These two are precisely related by $x = d + x(t=0) = d + 20$ in the example there.
A: In this question x is used for displacement of the body. it is not always that x is for position and s for displacement, they are variable and we an use anything. And when we calculate the displacement it will give the position only as we are taking it with origin.
Hope it clears your doubt.
A: Well I don't think there is a difference.
Let's assume a 1 - dimensional motion. A particle is at the origin O and it moves $x$ metres to the right.
What is its displacement and what is its position ?
Both equals $x$ since both are measured from the origin.
What if the particle comes back $\frac{x}{2}$ metres, what is the displacement and corresponding position now ?
Both are still the same : $\frac{x}{2}$ metres. This clearly means that position is the same as displacement.
In case if the initial position of the particle is d (let's assume) and it displaces by x metres then the final position is $(d + x)$.
If you integrate velocity, you get :
$$\int_{x_1} ^{x_2} dx = \int_{t_1}^{t_2} v(t) dt$$
In the above equation, the terms $x_1$ and $x_2$ represents the position of the body at time $t_1$ and $t_2$ and when you do the operation : upper limit - lower limit you get what you need : displacement.
In your linked figure , x is showing position and you can get velocity by differentiating displacement and not position. So you must differentiate (x - 20) and not x to get velocity but mathematically both gives same result and hence it doesn't matter.
