Tom K. answered 07/06/19
Knowledgeable and Friendly Math and Statistics Tutor
The area such that Z <= z includes the rectangle [0, z] x [0,1] and the region with boundary points [z, 0], [z, 1], [1, z]. and y = z/x for x from 0 to 1.
Using I[a,b] for the integral from a to b and E[a,b] for evaluation from a to b,
I[0,z]I[0,1] 4x/3 + 2y/3 dy dx = I[0,z] 4xy/3 + y2/3 E[0,1] dx =
I[0,z] 4x/3 + 1/3 dx = 2x2/3 + 1/3x E[0,z] = 2z2/3 + 1/3z
I[z,1]I[0,z/x]4x/3+2y/3 dy dx =
I[z,1] 4xy/3 + y2/3 E[0, z/x] dx =
I[z,1] 4z/3 + z2/3x2 dx = 4zx/3 - z2/3x E[z,1] =
4z/3 - z2/3 -4z2/3 + z/3
Adding these together, we get 2z2/3 + 1/3z + 4z/3 - z2/3 -4z2/3 + z/3 =
2z-z2
As a check, note that this equals 0 at z=0 and 1 at z = 1
f(z) = F'(z) = 2 - 2z.