Question about doing the integral to find the electric potential I am trying to find the electric potential which is: $- \int_{a}^{b}  E \cdot dR$. 
Assume I am in a constant electric field and I go directly parallel through it across from point A to point B (B being farther away), I get $ E*a - E*b$ and if I traverse the opposite direction ($ - \int_{b}^{a} E \cdot dR$), I'll get the same result because when I go in the opposite direction, I get $ - 
\int_{b}^{a} E \cdot dR$ = $- \int_{b}^{a} -E*dR$. That then reverses the limits and the result turns out to be the same.
 A: Hint:
You are wrong in the second integral: the sigh of $\mathbf{E}$ does not change sign, it is always the same field!
A: In general, we can express the potential $\Phi(\vec r_2)-\Phi(\vec r_1)$ between $\vec r_2$ and $\vec r_1$ as 
$$\Phi(\vec r_2)-\Phi(\vec r_1)=-\int_{\vec r_1}^{\vec r_2}\vec E\cdot d\vec \ell \tag 1$$
Now, suppose that the path from $\vec r_1$ to $\vec r_2$ is given parametrically by
$$\vec r=\vec r(t)$$
where $\vec r_1=\vec r(t=0)$ and $\vec r_2=\vec r(t=1)$.  Then, we can write $(1)$ as 
$$\Phi(\vec r(1))-\Phi(\vec r(0))=-\int_0^1 \vec E(\vec r(t))\cdot \frac{d\vec r(t)}{dt}\,dt \tag 2$$
If the path from $\vec r_1$ to $\vec r_2$ is traversed in the opposite direction, then the only change we need to make to $(2)$ is to interchange the limits of integration.  Then, we have
$$\begin{align}
-\int_1^0 \vec E(\vec r(t))\cdot \frac{d\vec r(t)}{dt}\,dt&=-\left(-\int_0^1 \vec E(\vec r(t))\cdot \frac{d\vec r(t)}{dt}\,dt\right)\\\\
&=-\left(\Phi(\vec r_2)-\Phi(\vec r_1)\right)
\end{align}$$
as expected.

Let's now apply $(2)$ to the problem of interest.
Assume without loss of generality that the electric field is given by 
$$\vec E=\hat xE_x$$
where $\hat x$ is a unit vector parallel to the $x$ axis and $E_x$ is uniform (i.e., a constant).  
CASE $1$:
Assume that the direction of the path is aligned in the direction of the electric field.  
Let's parameterize the path as $\vec r(t) = \vec r_1+(\vec r_2-\vec r_1)t$, starting at $t=0$ where $\vec r(0)=\vec r_1$ and ending at $t=1$ where $\vec r(1)=\vec r_2$.  Then 
$$\vec E(\vec r(t))\cdot \frac{d\vec r(t)}{dt}\,dt=E_x(x_2-x_1)\,dt$$
and $(2)$ becomes
$$\Phi(\vec r_2)-\Phi(\vec r_1)=-E_x(x_2-x_1)$$  
CASE $2$:
Assume that the direction of the path is aligned in the opposite direction of the electric field.  
Using the same parameterized description of the path as in Case $1$, we have $\vec r(t) = \vec r_1+(\vec r_2-\vec r_1)t$.
However, now we start at $t=1$ where $\vec r(1)=\vec r_2$ and end at $t=0$ where $\vec r(1)=\vec r_2$.  We still have 
$$\vec E(\vec r(t))\cdot \frac{d\vec r(t)}{dt}\,dt=E_x(x_2-x_1)\,dt$$
but we reverse the limits in $(2)$ and obtain
$$\Phi(\vec r_1)-\Phi(\vec r_2)=E_x(x_2-x_1)$$  
