If value of LP relaxation of s-t minimum cuts is P ,then wen can find a s-t cut at most P edges?
It's important to first understand how $\delta(S_l)$ changes. When $l = 0$, $S_l$ contains all vertices, and when $l = 1$, $S_l$ contains no vertices. In either case, there is no edge going between $S_l$ and $\overline{S_l}$, so $\delta(S_l) = 0$.

For values of $l$ between $0$ and $1$, $\delta(S_l)$ contains exactly the edges with one endpoint in $S_l$ and one endpoint in $\overline{S_l}$. In particular, if we define

$$f_{ij}(l) =\left\{ 
 \begin{array}{ll}
  1  &amp; \mbox{if } l \mbox{ is between } x_i \mbox{ and } x_j \\
  0 &amp; \mbox{otherwise.}
 \end{array}
\right.$$

then we have $$|\delta(S_l)| = \sum_{(i, j) \in \delta(S_l)} 1 = \sum_{\substack{(i,j) \in E \\ i \in S_l \\ j \not\in S_l}} 1 = \sum_{(i, j) \in E} f_{ij}(l).$$ 

Now let's consider one particular edge, from $u$ to $v$. Because $f_{uv}(\cdot)$ is just $0$ everywhere except for on one interval where it takes on the value $1$, the value of $\int_0^1 f_{uv}(l)\,\mathrm{d}l$ is just the length of the interval on which $f_{uv}(l) = 1$, that is, $|x_u - x_v|$. So 

$$\sum_{(u, v) \in E} |x_u - x_v| = \sum_{(u, v) \in E} \int_0^1 f_{uv}(l)\,\mathrm{d}l = \int_0^1 \sum_{(u, v) \in E} f_{uv}(l)\,\mathrm{d}l = \int_0^1 |\delta(S_l)|\,\mathrm{d}l.$$