Starting this problem is similar to question 3 in homework 1.
$ Y = min(x_1, x_2) $
$ f(x_1) = c_1\cdot e^{-c_1x1} $
$ f(x_2) = c_2\cdot e^{-c_2x2} $
From here we can use properties of integration to expand our min function into integrals.
$ \int_0^1 \int_0^1 min(x_1, x_2) $
I believe we can set the bounds from 0 to 1 since probabilities can not exceed this range.
$ \int_0^1 \int_0^{f(x_2)} f(x_1) dx_1 dx_2 + \int_0^1 \int_{f(x_2)}^1 f(x_2) dx_1 dx_2 $
This equation is derived through the property that if $ x_1 $ is smaller than $ x_2 $ then $ x_1 $ will be equal to Y and vise-versa
