The electric field in a region is given by $$\vec{E} = (Ax + B)\hat{i}$$, where E is in NC$$^{-1}$$ and $$x$$ is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at $$x = 1$$ is $$V_1$$ and that at $$x = -5$$ is $$V_2$$, then $$V_1 - V_2$$ is:

We have the electric field in the region given as $$\vec E = (Ax + B)\,\hat i$$, where the constants are $$A = 20\text{ (SI units)}$$ and $$B = 10\text{ (SI units)}$$.

For one-dimensional situations, the relation between electric field and electric potential is stated first:

$$E_x = -\frac{dV}{dx}.$$

This can be rewritten as

$$dV = -E_x\,dx.$$

The potential difference between two points $$x = x_1$$ and $$x = x_2$$ is therefore obtained by integrating:

$$V(x_2) - V(x_1) = -\int_{x_1}^{x_2} E_x \, dx.$$

In our problem the two positions are $$x_1 = -5\text{ m}$$ and $$x_2 = 1\text{ m}$$, with corresponding potentials $$V_2$$ and $$V_1$$. Substituting these limits,

$$V_1 - V_2 = -\int_{-5}^{1} (Ax + B)\,dx.$$

Now we substitute the numerical values of $$A$$ and $$B$$:

$$V_1 - V_2 = -\int_{-5}^{1} \bigl(20x + 10\bigr)\,dx.$$

We carry out the algebraic integration term by term. The integral of $$20x$$ is $$20 \cdot \dfrac{x^2}{2} = 10x^2$$, and the integral of $$10$$ is $$10x$$. So,

$$\int (20x + 10)\,dx \;=\; 10x^2 + 10x.$$

Placing the limits gives

$$V_1 - V_2 = -\Bigl[\,10x^2 + 10x\,\Bigr]_{x=-5}^{x=1}.$$

First we evaluate the bracket at the upper limit $$x = 1$$:

$$10(1)^2 + 10(1) = 10 + 10 = 20.$$

Next we evaluate at the lower limit $$x = -5$$:

$$10(-5)^2 + 10(-5) = 10(25) - 50 = 250 - 50 = 200.$$

Now we find the difference of the two bracketed values:

$$\Bigl(10x^2 + 10x\Bigr)_{1} - \Bigl(10x^2 + 10x\Bigr)_{-5} = 20 - 200 = -180.$$

Finally, we remember the overall minus sign outside the bracket:

$$V_1 - V_2 = -(-180) = 180\ \text{V}.$$

Hence, the correct answer is Option C.