Mixed strategies: Nash equilibrium I'm working on a game theory problem.I'm having trouble understanding what the mixed strategy nash equilibrium is exactly in this game.

The game is :Two players have to choose how distribute a piece of land ($size = 1$). Each player (1, 2) send his decision $x_i \in{[0, 1]}$ to an external agent.

If $x_1 +x_2 \leq 1$, each player gain a land portion equal to $x_i +\frac{1-x_1-x_2}{2}$

If $x_1+x_2 &gt; 1$, they end up with nothing.

The first part of the problem is to solve the game in pure strategies. I figured out that the answer of this part is:

Player 1 should pick $1- x_2$, any higher and he gets 0. The game is symmetric so the same hold from player2's perspective. The optimal strategy is the intersection of these two functions.

The second part is to solve the game in mixed strategies (if exist). But I don't know how to aproach this part.
Hint 1: Suppose a mixed strategy NE $(\sigma_1,\sigma_2)$ exists, where $\sigma_i$ is a distribution over $[0,1]$. By the property of MSNE, 
$$u_i(x_i,\sigma_j)=u_i(x_i',\sigma_j)$$ 
for all pure strategies $x_i,x_i'$ in the support of $\sigma_i$. 

Hint 2: Note that a segment of $u_i$ is linear in strategies, and thus 
$$
x_i+\frac12(1-x_i-\mathbb E[x_j])=x_i'+\frac12(1-x_i'-\mathbb E[x_j])
$$