I Computing the expectation of the minimum difference between the 0th i.i.d.r.v. and ith i.i.d.r.v.s where 1 ≤ i ≤ n

[WMDhamnekar]

Problem :Let $X_0,X_1,\dots,X_n$ be independent random variables, each distributed uniformly on [0,1].Find $E\left[ \min_{1\leq i\leq n}\vert X_0 -X_i\vert \right]$.

Would any member of Physics Forum take efforts to explain with all details the following author's solution to this question?

Author's solution:
Let L be the expression in question. Then $$L=\displaystyle\int_0^1 E \left[ \min_{1 \leq i \leq n}\vert x- X_i \vert dx\right] =\displaystyle\int_0^1\displaystyle\int_0^1\left[P(\vert X_0 - x\vert\geq u )\right]^ndu dx $$
Since $P(\vert X_0 -x \vert \geq u ) = \max(1-u-x,0) + \max(x-u ,0), x,u \in [0,1]$ we have $$ P(\vert X_0 -x \vert \geq u )=\begin{cases} 1- 2u & 0 \leq u < x\\ 1-u -x & x \leq u < 1-x
& x \in[0,\frac12 ]\\ 0, & 1-x \leq u \leq 1\end{cases}$$
So,
$$L = 2\displaystyle\int_0^\frac12\left[\displaystyle\int_0^x (1-2u)^n du + \displaystyle\int_x^{1-x}(1-u-x)^n du\right]dx = \frac{n+3}{2(n+1)(n+2)}$$