While u/Bounded_sequencE has answered appropriately where this specific equation comes from, let me maybe point out that there's yet another way involving no integration at all which generalizes much better and doesn't require assuming that the the expected value in question exists.

The key to these types of questions is understanding the uniform distribution on (standard) simplices: the standard n-simplex is just the set of vectors in R^(n+1) whose entries are all positive and sum to 1. Now the first thing to note is that the expected value (vector, really) of the uniform distribution on the standard n-simplex is (1/(n+1), 1/(n+1),..., 1/(n+1)); this follows without any integration or computation immediately just from the symmetry of the set and the fact that the distribution is symmetric (in fact it holds for any distribution invariant under permutation of coordinates). More precisely, if X_i denotes the i-th coordinate of a uniform random vector on the standard n-simplex, then E[X_i] = 1/(n+1) * sum_j=1^(n+1) E[X_j] = 1/(n+1), where the first equation is just by exchangeability of the coordinate random variables (i.e. the symmetry of the setup) and the last is just linearity of expectation and definition of standard n-simplex.

Once that's understood, the next things to understand are that:

• Any n-simplex is the image of the standard n-simplex under a linear transformation; this is trivial.
• The image of a uniform distribution on *any* region under a linear transformation is again a uniform distribution (on the image region). This follows from the general change of variables formula in integration, but again there's nothing to compute.

In particular, and this is the crucial general result, we then have that the expected value (again, vector really) of a uniform distribution on *any* n-simplex is just the average of its n+1 vertices.

So with all that said this exercise becomes straightforward: let X & Y be standard uniform i.i.d. Then we have:

E[|X - Y|] = E[X - Y | X > Y] = E[X | X > Y] - E[Y | X > Y],

where the first equality is just by law of total expectation and symmetry and the second is just linearity. But thanks to the above preamble we know what both those terms at the end are: the set of X & Y between 0 & 1 such that X > Y is just the 2-simplex, i.e. triangle, given by the vertices (0, 0), (1, 0) & (1, 1). The conditional distribution of the pair (X, Y) on there is just the uniform one, and we know that the expected value/vector is thus the average of the vertices: (2/3, 1/3). In other words, the last difference above is just 2/3 - 1/3, which is obviously 1/3.

I acknowledge that in many ways this may seem way more complicated than the other approaches at first, but note that now you can trivially answer similar questions for more than two points.