Newton's Law of Gravitation with a different unit vector my question concerns what the correct form of newton's law of universal law of gravity is when my vector r goes from the center of the body acted on towards the center of the body acting on it. In images:
I'm interested in the force F of the bigger body on the smaller one.
Since it's in the direction of my unit vector r, I can write
$$ \mathbf F = \frac{Gm_1m_2}{R^2}\mathbf r  $$
Now, here is why this whole business seems awfully fishy to me. Let's admit that work is the integral of the force with respect to distance, ie:
$$ W = \int \mathbf F \cdot d\mathbf s $$
Since we're moving along the radius, we can write that as:
$$ W = \int \mathbf F \cdot d\mathbf r $$
And if we go ahead and compute the integral:
$$ W = -Gm_1m_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right) $$
The reason why I feel like this has to be wrong is because this implies that if we move from a distance a to a distance b, where a > b, the work done by the force is negative, we gain potential energy, and we lose kinetic energy. Which is all... backwards really!
So what's wrong with the reasoning above?
EDIT: Rob's answer made me think about the consequences of choosing my unit vector r going downwards and I have come to the following diagram (sorry about the quality of the drawing):

The seemingly fixes everything, but is it the right way to think about it?
 A: It's really impossible to answer this without seeing how you did the integral, but I'll take a guess.
To get the correct final sign we need to make sure we get the sign of the force correct, and the sign of the displacement correct.  You have chosen a coordinate system that makes that tricky.  If $r$ increases going down, then $r_f>r_i$ when the object moves down.
It's almost always easier to 1.) lay it out horizontally and 2.) choose $x$ to increase going to the right.  Our brains seem to work better with that.  Work the problem that way, and you'll have no problems.
A: You're using $\hat{\mathbf r}$, a downward-pointing unit vector, and $d\mathbf r = dx\hat{\mathbf x} + dy\hat{\mathbf y} + dz\hat{\mathbf z}$, an infinitesimal displacement, in two different ways, and becoming confused because you have given them similar symbols.  The coordinate radius, $r = \sqrt{x^2 + y^2 + z^2}$, acquires a larger value as you move away from the origin of your coordinate system.  So your scalar radius becomes larger as you move in the $(-\hat{\mathbf r})$ direction, according to your definition, and your potential energy is off by a sign.
Most people choose $\hat{\mathbf r}$ to point in the outward direction so that the scalar $r$ becomes larger in the $+\hat{\mathbf r}$ direction.  This is like choosing your $+\hat{\mathbf x}$-axis in the direction that the coordinate $x$ increases, which is how your calculus intuition was developed.
A: The $\vec{r}$ points from the center of the large object outward.  But the gravitational force is in the opposite direction.  Therefore $\vec{F}=-\frac{Gm_1m_2}{R^2}\vec{r}$.  When you carry the negative sign through, it works out properly.
