Work is a scalar, not a vector. The dot product $\vec F \cdot \vec dx$ is a scalar, so the integral is over a scalar, not a vector. $\vec F \cdot \vec dx$ is the work done by the force moving the body a differential distance. The total work is the integral over the path travelled by the body, starting at $x_1$ and ending at $x_2$. The limits on the integral merely specify the coordinates of the initial point and the final point; you cannot evaluate the integral until you know the path taken between the initial and final points (similar such integrals dependent on the path occur in thermodynamics).
The work can also be expressed as $\int_{t_1}^{t_2} \vec F \cdot \vec v \enspace dt$, since $d \vec x = \vec v \enspace dt$, where $t$ is time and $\vec v$ is velocity.
All of this is true whether or not the force is conservative. If the force is conservative, the work is independent of the path taken between the initial and final end points, and a potential energy can be defined and used in place of work, as discussed in the Wikipedia article you reference. An advantage of using potential energy is no matter how complex the path from the initial to the final position, the work is simply the negative of the change in potential energy.
See a good basic physics text for the definition of work in mechanics, such as one of the Halliday and Resnick texts.
Here we are using the mechanics concept of work, defined as the change in kinetic energy. Work as used in thermodynamics is a much broader concept; see my answer to Positive and negative Work, question on this exchange.
