Why Is Zero the Upper Limit in Escape Velocity Derivation?

[shaan_aragorn]

escape velocity derivation

The problem is thus:
The acceleration of a satellite is given by -gR^2/r^2 where R= radius of Earth and r = distance of satellite from center of the earth. Find escape velocity.

Now I have read the solution to this problem and in it the author at one point has integrated v dv = -gR^2/r^2 with the upper limit for vdv being 0 and the lower limit being Ve(Escape velocity).

Now my problem is, why is 0 the upper limit and Ve the lower limit if in magnitude Ve>0?

 

[dicerandom]

Think of it this way: the escape velocity is the bare minimum velocity needed to always move away from the planet, so you could say that "at infinity" (which is a strange concept, let's just say "very far away") the satelite's velocity is "zero" (really close to zero). Thus the satelite starts out near the Earth at the escape velocity Ve and ends up out at infinity with zero velocity.