Q: Calculating electric potential from electric field 
The Electric field in  a region is given as $\vec{E}=-2x\hat{i}+3y^2\hat{j}-4z^3\hat{k}$. Calculate the electric potential at point $(1,2,3)m$

Now we know that electric potential at point $A$ is defined as $$V_A=-\frac{W_{elc}}{q_0}\bigg]_{\infty\to A}$$
which evaluates to $$V_A=-\int_{\infty}^{(1,2,3)}\vec{E}.d\vec{r}$$
Now this integral evaluates to an inderteminate form $(\infty-\infty)$
Where have I gone wrong?

A: The lower limit on the integral for the potential is not always $\infty$. It's the position where the electric field is zero, that is where one "starts pushing against it" so as so to do work, which then becomes energy stored in the potential.
Physically, charges and currents are localised, which give you (physical) boundary conditions $|\mathbf{E}| \rightarrow 0$ as $r \rightarrow \infty$, hence why $\infty$ is usually taken as the "starting" point (e.g. for a point charge).
In this case, the electric field is $0$ at $r = (0,0,0)$, so you should start the integration there.

A: The electric potential at position $\vec{r}_A$ is defined to be
$$V(\vec{r}_A)=V(\vec{r}_0)
 -\int_{\vec{r}_0}^{\vec{r}_A}\vec{E}(\vec{r})\cdot d\vec{r}$$
The definition allows you to choose any $\vec{r}_0$ you like.
You have already noticed that choosing $\vec{r}_0=(\infty,\infty,\infty)$
doesn't work, because then your integral diverges.
So choose another one. I suggest to use $\vec{r}_0=\vec{0}=(0,0,0)$,
because then the integration becomes most easy.
Notice that your final result will still contain $V(\vec{0})$
as an unknown constant. But this is unavoidable.
