Deriving Voltage from Electric Field 
I'm trying to derive the point charge equation for voltage by integrating the point charge equation for an electric field over distance ($dr$) traversed:$ \int (KQ/r^2)\cdot dr$

This is my reasoning:
1) Take a source charge at $r = 0$ and a point charge at $r=-\infty$
2) Assume that both charges are positive. Therefore, as the point charge moves from 
$-\infty$
 to $r$, an electric field acts against the charge, decreasing its kinetic energy.
3) Account for the force and distance being in opposite directions (hence the dot product $(KQ/r^2)\cdot dr$) and set $cos\theta$ to $cos(180)$
4) Setup the energy integral:
$$
\Delta KE =  \int_{-\infty}^{r} (KQ/r^2) dr * cos(180)
$$
4 Multiply by $-1$ to find delta potential energy. My answer is $-KQ/r$ and not the actual $KQ/r$
Why?
$$
\Delta V = -1 \int_{-\infty}^{r} (KQ/r^2) dr * cos(180)
$$
The math:
$$
\Delta V = -1 * cos(180)\int_{-\infty}^{r} (KQ/r^2) dr 
$$
$$
\Delta V = \int_{-\infty}^{r} (KQ/r^2) dr 
$$
$-\infty$ might as well be $\infty$
$$
\Delta V = \frac{-KQ}{r} \Big|_\infty^r
$$
$$
\frac{-KQ}{r} - \frac{-KQ}{\infty}
$$
 A: Thanks laying out your work so neatly in the question.
I think the solution is the following
$$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$
where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge).
so, for example, if the point charge goes from $r$ to $2r$ we have two positive charges and the change in kinetic energy will be positive as the point charge is pushed away from the source chare. - lets try that in the equation above....
$$\Delta KE= \int_{r}^{2r}{ KQq \over r^2} dr$$
$$ = -\left[{KQq \over r}\right]_r^{2r} = -\left({KQq \over 2r}-{KQq \over r}\right)$$
$$ = -{KQq \over r}\left({1 \over 2} - 1\right) = -{KQq \over r}\left({-1 \over 2}\right) = + {KQq \over 2r}$$
so we expected a positive $\Delta KE$ and we got one.
Now this equation for kinetic energy works from any $r_a$ to any other $r_b$. If we go in the positive direction where $r_b > r_a$ it works fine as we saw for $r$ to $2r$. If we go in the other direction towards the origin (so $r_b < r_a$) we do not need to put in the $cos 180$ term because that is looked after in the integral because
$$\int_{r_a}^{r_b}{ KQq \over r^2} dr = -\int_{r_b}^{r_a}{ KQq \over r^2} dr$$
[more generally of course $\int_a^b f(x)dx=-\int_b^a f(x)dx$ ]
Thus the solution to your question is that you do not need the $cos(180)$ term which is $-1$ - take this out and your solution is perfect.
apologies that my first answer missed this - thanks for showing more working.
so to put it all together
$$\Delta PE = - \Delta KE = -\int_{\infty}^{r}{ KQq \over r^2} dr$$
$$ = +\left[{KQq \over r}\right]_\infty^r = {KQq \over r}-0$$
to get the potential we divide by $q$ the charge of the point charge.
