(BTW, I think it should be $r-\Delta r$ in keeping with the axis.)
Let me first re derived the formula as in my book, then ask my question.
Suppose a object is at some distance r away from center of earth, and falls by gravity a distance of $\Delta r$. As in the picture. The work done $\Delta W$ is equal to $\bar{F}=-\frac{GMm}{\bar{r}^2}$. $\bar{r}$ is somewhere between $r$ and $r+\Delta r$.
Then, $\frac{dW}{dr}=lim_{\Delta r} \frac{\Delta W}{\Delta r}= F= -\frac{GMm}{r^2}$
Then, integrating, $W=\frac{GmM}{r}+C$. If we substitute $W=0$ at $r=r_1$ (initial position), then $C=-\frac{GMm}{r_1}$.
My Question is, after integrating why is $r$ the end position, I thought $r+\Delta r$ is the final position. And when they take the limit to arrive at the derivative of W with respect to r, the derivative is equal to F, the force at the initial position, as $\Delta r$, the distance, approaches 0.
As you stated, $F=-\frac{GMm}{r^2}.$ If we proceed very far away from the planet, i.e. take $r\rightarrow \infty$, then $F\rightarrow 0$ (this is what your problem assumes with $r_1$). Suppose we stop at a point "infinitely" far away from the planet and move towards the planet some minor distance $\Delta r$. Because $F=0$, $W=0$. But if we then approach the planet so that $r\rightarrow R,$ we note $F\rightarrow -\frac{GMm}{R^2}$. To calculate the exact amount of work, we use these bounds--namely, $$W=-\int_\infty^{r_2}\frac{GMm}{r^2}dr= \frac{GMm}{r_2},$$ which is the amount of work done on the object by the planet when the object is initially "infinitely" far out in space but then proceeds near the planet to a distance $r_2$ (where $r_2\geq R$).