We find expected revenue in first bid auction by following method.
let us say $V_1$ and $V_2$ denotes maximum amount that player 1 and player 2 willing to pay.
$V_1,V_2 \in [0,1]$
In case when we have only 2 players.
Nash equilibrium is $(\frac{V_1}{2},\frac{V_2}{2})$.
Expected revenue = $P(V_1>V_2)\frac{V_1}{2} + P(V_2>V_1) \frac{V_2}{2}$
we replace $V_1$ by $\frac{1+V_2}{2}$ (mean in range $V_2$ to 1).
Now Expected revenue = $\frac{V_2^{2}}{2} +(1-V_2)\frac{1+V_2}{4}$ = $\frac{1+V_2^{2}}{4}$
Now by integrating from 0 to 1 we get $\frac{1}{3}$ answer.
I did not Understand why we take mean.
Can anyone please explain.
Your formula is not quite correct. The expected revenue for the auctioneer is (using the fact that valuations are independently drawn)
$$ \int_{v_1 \geq v_2} \frac{v_1}{2} + \int_{v_2 > v_1} \frac{v_2}{2} = \int_0^1 \int_0^{v_1}\frac{v_1}{2} dv_2 dv_1 + \int_0^1 \int_0^{v_2}\frac{v_2}{2} dv_1 dv_2 = \frac{1}{3}. $$