INT((I-1)·((H-1)/(B-1))+1)
reduces to:
((H-1)/(B-1))·(I-1) + 1.
Transforming the formula so that the result includes expressions for length in inches requires substituting the corresponding number of points expressed in inches times the number of points per inch. To obtain an area result which is in terms of square inches, the point area formula must be divided by the number of points in a square inch, namely K2. The new area formula becomes:
[((K·H-1)/(K·B-1))·(I-1) + (1)]/K2
Performing the summation and some algebra yields:
[(K·H-1)/(K·B-1))·K·B·(K·B-1)/2 + K·B]/K2.
[(K·H-1))·K·B/2 + K·B]/K2.
(H-1/K))·B/2 + B/K.
Taking the limit of this as K approaches yields: ½B·H. text