I am reading http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S4

In the beginning of equation 13.18 it says: $$W=\frac{Gm'2 \pi a \mu}{R} \int \limits_{R+a}^{R-a} \, dr$$

The result of the integral should be: $$W=\frac{Gm'2 \pi a \mu}{R} \cdot ((R-a)-(R+a))$$

$(R-a)-(R+a)$ is $-2a$ and so $$W=-\frac{Gm'4 \pi a^2 \mu}{R}$$ $$W=-\frac{Gm'm}{R}$$

Now my question is: How does Mr. Feynman know, how to choose the limits of the integral? If I chose the limits from $R-a$ to $R+a$, I would get in the end: $W=\frac{Gm'm}{R}$. The only difference is the algebraic sign at the beginning. Wje

I don't know, why you have to choose the beginning of the integral further away from P ($R+a$) and end closer to P ($R-a$). Is there any rule that states that, or why did Mr. Feynman choose the limits of the integral the way he did? Did he think: I've got to get a negative result, so I choose the limits from $R+a$ to $R-a$???

P.S.: Feel free to edit, since I'm not a native English speaker.

I am reading http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S4

In the beginning of equation 13.18, in which Mr. Feynman calculates the potential energy of an object outside a spherical shell, it says: $$W=\frac{Gm'2 \pi a \mu}{R} \int \limits_{R+a}^{R-a} \, dr$$

The result of the integral should be: $$W=\frac{Gm'2 \pi a \mu}{R} \bigl((R-a)-(R+a)\bigr)$$

$(R-a)-(R+a)$ is $-2a$ and so $$W=-\frac{Gm'4 \pi a^2 \mu}{R}$$ $$W=-\frac{Gm'm}{R}$$

How does Mr. Feynman know how to choose the limits of the integral? If I chose the limits from $R-a$ to $R+a$, I would get in the end: $W=\frac{Gm'm}{R}$. The only difference is the algebraic sign at the beginning.

I don't know why you have to choose the beginning of the integral further away from P ($R+a$) and end closer to P ($R-a$). Is there any rule that states that, or why did Mr. Feynman choose the limits of the integral the way he did? Did he think: I've got to get a negative result, so I choose the limits from $R+a$ to $R-a$?