Approaching to Integral from Limit of Sum ->Finding voltage drop due to uniformly distributed load current on a distributor
I can't say that I feel I completely understood your question. But if I gather it correctly, the value of \$i\$ doesn't change as you move from a finite number of taps to an infinite number of them. This, as you will see, creates a problem that won't be resolvable with a finite answer.

I'm going to label each tap, left to right, as \$k\$ starting with \$k=1\$ on the left and \$k=n\$ at the rightmost tap. You have \$n\$ taps. The current removed at each tap is a constant \$i\$. I'm going to call the entire length, \$L\$ (and not \$l\$.) But the length of each segment of \$L\$ I'll call \$l_k=\frac{L}{n}\$. Length \$L\$ exhibits a certain resistance per unit length \$r\$. The resistance of any segment length is \$r_k=r\cdot l_k=r\frac{L}{n}\$.

The current entering from the left must be the sum of all the branch currents, or \$I=n\cdot i\$. This current must pass through the first segment \$l_1\$ of length \$L\$ and drop a voltage \$v_1=n\cdot i\cdot r\cdot\frac{L}{n}\$. The next segment will drop a voltage \$v_2=\left(n-1\right)\cdot i\cdot r\cdot\frac{L}{n}\$. Etc.

So the question you are asking, "what is the voltage drop across length \$L\$ as \$n\to\infty\$," can be written out this way:

$$\begin{align*}
\Delta V=V-V^{'}&amp;= \lim_{n\to\infty} \sum_{k=1}^n \left(n-k+1\right)\cdot i\cdot r\cdot \frac{L}{n}\\\\
&amp;=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \left(n-k+1\right)\\\\
&amp;=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\left[\left(n+1\right)\sum_{k=1}^n 1-\sum_{k=1}^n k\right]\\\\
&amp;=i\cdot r\cdot L\:\lim_{n\to\infty} \frac{1}{n}\left[\left(n+1\right)n-\frac{\left(n+1\right)n}{2}\right]\\\\
&amp;=i\cdot r\cdot L\:\lim_{n\to\infty} \left[\frac{1}{n}\frac{\left(n+1\right)n}{2}\right]\\\\
&amp;=i\cdot r\cdot L\:\lim_{n\to\infty} \left[\frac{n+1}{2}\right]\\\\
&amp;=\frac{i\cdot r\cdot L}{2}\:\left[1+\lim_{n\to\infty} n\right]
\end{align*}$$

And I don't think I need to tell you how to figure out that limit.



Of course, if you also diminished \$i\$ as a function of \$n\$ then perhaps there might be a different answer. But I didn't see that behavior in your question.