Question

why is the following true?

Answer 1

\[\int\limits_{r _{A}}^{r _{B}}\frac{ dR }{ R^2 } = \left( \frac{ 1 }{ r _{B} } - \frac{ 1 }{ r _{A} }\right)\] is it because they reformed the integral to \[\int\limits_{r _{A}}^{r _{B}} R ^{-2}dR \]

Answer 2

@ivancsc1996 @mashy

Answer 3

Actually thes integral should be:\[\int\limits\limits\limits_{r _{A}}^{r _{B}}\frac{ dR }{ R^2 } =\int\limits\limits\limits_{r _{A}}^{r _{B}}R ^{-2}dR= \left(- \frac{ 1 }{ r _{B} } -(- \frac{ 1 }{ r _{A} })\right)=\left( \frac{ 1 }{ r _{A} } - \frac{ 1 }{ r _{B} }\right)\]

Answer 4

maybe I should show the whole thing:
so what I don't understand is how the scalar product of the vector unit and the displacement vector is dR, the red circles on the diagram.

Answer 5

What language?