bribing in second price auctions In this paper, they analyse a second-price auction game with an option for the first player to bribe the second player to stay out of he game.
In proposition 2, they show that for any $b\in(0, \Bbb{E}[\theta_i])$ there exists a sequential equilibrium in which bribing occurs, and there exist $A^b$ and $B^b$ such that the sets of bribers and acceptors are $[B^b, 1]$ and $[0, A^b]$ 
I am interested in finding those $A^b$ and $B^b$, more specifically I am interested in how Figure 1 was generated (the equilibrium values of A and B as a function of b under the uniform distribution). I wish to calculate the area between A and B, and the difference between them also as function of b.
I also wish to better understand how did they achieve this results as I am trying to develop similar results for a more general case (with 3 players)
 A: Figure $1$ has two parts – one where $A$ and $B$ are both in equilibrium, and one where $A$ takes on its maximal value $1$ and only $B$ is in equilibrium.
For $B$ to be in equilibrium, the bribing player must be indifferent at $B$ whether to bribe or not. If she bribes, the integral of her payoff over the accepting types of the bribed player is just $A(B-b)$, whereas if she doesn't, it's
$$
\int_0^B\mathrm d\theta_i(B-\theta_i)=\frac12B^2\;,
$$
so the equilibrium condition for $B$ is
$$
A(B-b)=\frac12B^2\;.
$$
(The integral over the rejecting types of the bribed player isn't affected by whether she bribes.)
For $A$ to be in equilibrium, the bribed player must be indifferent at $A$ whether to accept the bribe or not. If he accepts, the integral of his payoff over the bribing types of the bribing player is just $(1-B)b$, whereas if he rejects, it's
$$
\int_B^A\mathrm d\theta_j(A-\theta_j)=\frac12(A-B)^2\;,
$$
so the equilibrium condition for $A$ is
$$
(1-B)b=\frac12(A-B)^2\;.
$$
(Again, the integral over the non-bribing types isn't affected by whether he'd hypothetically accept.)
Now we can use these to treat the two cases in Figure $1$. In the right-hand part, $A=1$ is fixed, and only the equilibrium condition for $B$ applies; substituting $A=1$ into it and solving the resulting quadratic equation for $B$ yields
$$
B=1\pm\sqrt{1-2b}\;.
$$
In the left-hand part, both $A$ and $B$ are in equilibrium, so we can use both equilibrium conditions to find $A$ and $B$. Solving the equilibrium condition for $B$ for $A$,
$$
A=\frac{B^2}{2(B-b)}\;,
$$
and substituting into the equilibrium condition for $A$ yields a quartic equation for $B$:
$$
(B-b)^2(1-B)b=\frac18B^2(B-2b)^2\;.
$$
Here's a plot. I'll leave it to you to figure out which of the branches is a stable equilibrium in each case. In view of the quartic equation for $B$ (or cubic for $b$), calculating the area between $A$ and $B$ will probably best be done numerically.
