What determines the direction of a path on a line integral (vector case)? Line integrals are very important to use in Physics. For example, we calculate work by: W=∫<F,dr>. But I just got confused about something. What determines the direction of motion? The integral limits, or the vector dr?
Well, when we do the internal product of the Force by the path(dr), we are aligning this force on the path's direction(I mean, the shape of the curve) . But if I want a path from the position B to the position A, I determine this inserting on Integral's limits(B inferior limit and A superior limit), or the vector dr would indicate the direction of motion (from B to A) and the integral limits would be (A inferior, B superior)?
 A: This is concerned with getting the sign right. Signs in physics are notoriously tricky. There is no general rule. You have to understand the physics of what you are doing, and construct your mathematical equation accordingly. For example, if a force $\bf f$ acts on a body $A$, and the body undergoes a displacement by $d\bf r$, then the energy of body $A$ (e.g. kinetic energy) goes up by
$$
{\bf f} \cdot d{\bf r}
$$
But this force was provided by some other body $B$, so the energy of body $B$ must have gone down by this amount. In other words the energy of body $B$ increases by $(-{\bf f} \cdot d{\bf r})$. Hence my claim that there is no alternative but to understand the physics of what you are doing and construct your mathematics accordingly.
When calculating work, we normally want to know the work done by a force, so we integrate
$$
W = \int_{\rm initial\; location}^{\rm final\; location} {\bf f} \cdot d{\bf r}
$$
where the variable $\bf r$ is the location of the body on which the force is acting (or, more precisely, it is the location of the part of the body on which the force acts).
A: $d\mathbf{r}$ is the line element of the whole path on which you're calculating the line integral, so saying that the direction of motion is determind by $d\mathbf{r}$ or by the path is just the same thing.
A: When you do a line integral, you must specify the path. Giving the endpoints is not enough, because there are many paths connecting two endpoints, and in general the value of the integral depends on the path. So the answer to your question is: neither. The direction of motion is specified by the definition of the path on which you do the integral.
A: The correct form of an integral over a path takes the form $\int_{\overrightarrow{r_i}}^{\overrightarrow{r_f}}d\overrightarrow{r}$. The direction of motion was determined when you chose the path and the limits were determined when you chose the path. 
If you know the velocity, the direction of the path is obvious because $\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}$.
A: *

*The line segments $\mathrm d\vec r$ determine - define - the exact path.


*The integral itself decides - via the order of its limits - the direction along this path. Meaning, flipping around the integral will only change the sign of the integration.
Both are thus involved with determining the path direction, but the integral limits to a lesser degree.
A: The parametrization of your path determines the direction of integration/motion. Let's say you have a semi-circular path of radius 1 moving anti-clockwise (i.e. from the positive x-axis to the positive y-axis to the negative x-axis). We can parametrize it as the following:
$r(t) = (\cos(t), \sin(t))$,   $t\epsilon [0, \pi]$
However, imagine you instead parametrize a semi-circular path like this:
$r(t) = (-\cos(t), \sin(t))$,   $t\epsilon [0, \pi]$
While both these paths look the same on a plot, the first one starts at (\cos(0), \sin(0)) = (1, 0), while the second one starts at (-cos(0), sin(0)) = (-1, 0).
This is how the direction of travel can be defined for a simple example. If you need to find the direction of some parametric equation, all you need to do is plug in the endpoints and see which way the path flows. Here's another example.
Consider the parametric equation:
$r(t)=(\cos(2t),\sin(t+5))$ $t\epsilon[0, 2\pi]$
Looking at it, I don't have an intuition for where it starts and where it ends.
But we can alleviate this! Just plug in the endpoints.
$r(0) = (\cos(0), \sin(5)) = (1, -0.96)$
$r(2\pi) = (\cos(4\pi), \sin(2\pi + 5)) = (1, -0.96)$
Oh no! We've gotten the same endpoint. From the plot, we can deduce this because it loops over itself. Not to worry, however, we can plug in a number close to one of the endpoints, which will help us make our determination of path.
$r(0.1) = (0.98, -0.93)$
Now it's clear which direction the path goes in.
