| 
  ||
|  
       | 
  ||
 
      
  | 
  
For a rectangle: By the ordinary fundamental theorem of calculus, we have

For a right triangle: for convenience we 
  choose a triangle bounded by line x = 0, y = 0 and 
.
  We similarly get:

Rearrangement of the right hand side gives the theorem for rectangles and right 
  triangles.
  It means that for R a rectangle or right triangle in the x-y plane, (for which 
  dS = dSk) we have
![]()
Both sides of this equation are finitely additive: that is, if we take two 
  disjoint regions, and evaluate either one over both, you get the sum of their 
  values on the two regions separately. This is true even if the regions share 
  a common boundary, because the line integrals will cancel out over the common 
  boundary which ceases to be a boundary.
  The result follows from additivity for any region that can be broken up into 
  rectangles and triangles, which accounts for most regions we will encounter.