>>320 >>322 >>331

 D = {(x,y) | a<x<b, c<y<d}
より
 ∬_D (x-y)dxdy = (bb-aa)(d-c)/2 - (b-a)(dd-cc)/2
   = (b-a)(d-c)(a+b-c-d)/2,

たとえば
 a_n = A/nn,
 b_n = n + A/nn,
 c_n = C/nn,
 d_n = n + C/nn,
ならば
 ∬_D (x-y)dxdy = A - C,