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I just wanted to know why is this possible:

$$\int_{i}^{f}-\frac{dU}{dr}\overrightarrow{e_{r}}\cdot (dr\overrightarrow{e_{r}})=\int_{i}^{f}-dU$$

Could someone help me figure out what is happening with that dot product?

P.S.: Integrating vectors tips are welcome (I'm a first year)

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Since you have two vectors in the same direction with the length one ($\vec{e}_r$) they yield 1 when multiplied with each other. $$ \vec{e}_{r}\cdot (dr\vec{e}_{r})= (\vec{e}_r\cdot \vec{e}_r)dr=dr $$

So you are left with

$$ \int_{i}^{f}-\frac{dU}{dr}\cdot dr=-(U(f)-U(i)) $$

Here you integrate the derivative of U with respect to r (that is $dU/dr$) with respect to r. So you get the function $U$ itself back. This is a direct conclusion from the Fundamental theorem of calculus.

Another way to look at it is by using substitution (which of course uses the fundamental thereom of calculus itself). The formula for this is $$ \int_a^b f(\phi(t))\cdot\phi'(t) dt=\int_{\phi(a)}^{\phi(b)} f(x) dx $$

With $f(x)=1$ you get what you want.

In general when calculating integrals with vectors there is not so much of a difference. When the integrand consists of vectors, the result is either a vector or a "normal" function (in the case above the vectors cancel each other in such a way, that you are left with a function). If this is not a case, i.e. your integrand is a vector, then you can integrate each component separately. This is not always the best approach, but in principle it works.

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  • $\begingroup$ It makes total sense, and in matter of fact you helped me realize one propertie that I never had thought before $\alpha \vec{v_{1}}\cdot \beta \vec{v_{2}}=\alpha \beta (\vec{v_{1}}\cdot \vec{v_{2}})$. Sounds obvious but it wasn't for me, thanks! $\endgroup$
    – arpg
    Nov 6, 2020 at 13:57
  • $\begingroup$ You are welcome! I think I will edit my answer accordingly to make it absolutely clear. $\endgroup$
    – daveh
    Nov 6, 2020 at 14:04

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