Two disks with seek times x1 and x2
Each x1 and x2 is uniformly distributed from 0 to 12
p(x1,2) = 1/12

Read = Minimum seek time is INT(0,12) p(x1) dx1 INT(0,12) p(x2) dx2 min(x1,x2)

Write = Maximum seek time is INT(0,12) p(x1) dx1 INT(0,12) p(x2) dx2 max(x1,x2)

Mean read = 1/144 INT(0,12) dx2 [INT(0,x2) x1 dx1 + INT(x2,12) x2 dx1]

= 1/144 INT(0,12) dx2 [ x2**2/2 +x2*(12-x2) ]
= 1/144 INT(0,12) dx2 [ 12 x2 - x2*x2/2 ]
= 6 - 2 = 4

Mean write = 1/144 INT(0,12) dx2 [INT(0,x2) x2 dx1 + INT(x2,12) x1 dx1]
           = 1/144 INT(0,12) dx2 [ x2*x2 +.5*(144-x2*x2) ]
           =  1/144 INT(0,12) dx2 [ .5*(144 + x2*x2) ]
           = 6 + 2 = 8