This question might be similar to MSE Question but I didn't really get the answer.
Suppose I want to evaluate the integral $$W = \int \underline{F}\cdot \underline{dr}$$ which actually has the physical meaning of giving the work done by the force $F$ along $r$. Now, Imagine I have a non conservative vector field (or force), i.e. $F$ is non conservative, it is a resistive force which resists with constant value $R$, suppose it acts for a distance $d$. Then can I do the following? $$W = \int \underline{F}\cdot\underline{dr} =\underline{F}\cdot\int\underline{dr} = \underline{F}\cdot\underline{\triangle r} = -Rd$$
and if so
why can I do that? When can I take out one of the factors of a dot product inside an integral?
I.e. can I take out the force? This happens in the last line of this page of my lecture notes:
If the force is constant, \begin{align*} \int {\bf F}\cdot d{\bf r} &= \int \sum_i F_i dr_i \\ &= \sum_i F_i \int dr_i \\ &= {\bf F}\cdot \int d{\bf r} \\ &= {\bf F}\cdot \Delta{\bf r}. \end{align*} This is allowed due to the definition of the dot product and the fact that for constant $c$, $\int c f(x) dx = c\int f(x) dx$.