The right "weigh" to do integrals Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, painstakingly cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.
One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.
I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the "below" group from the "above" group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.
Which one of us was correct?
 A: I guess we speak of the area "under a curve" when the piece of curve is above the $x$ axis. Sign does matter in the definition of the proper integral, so I would say you were right, and your boss was wrong. Doing integrals by "weighing" things should allow for negative weights.
A: In my opinion, this is not really a math question.  Which procedure is correct depends what you're going to do with your calculation.
As a matter of definition, the integral indeed measures the signed area (positive area minus negative area), as you suggest.  So your approach is computing an approximation to the definite integral $\int_a^b f$.
But maybe you want the total (unsigned) area.  E.g. if you're going to lay concrete along (some real-world space corresponding to) the region bounded by the curve and the x-axis then surely you want the total area -- there's no such thing as negative concrete.
Without knowing what you're using the calculation for, it's impossible to say.  (Essentially, you're asking us: "Which of these is mathematically correct: $A-B+C$ or $A+B+C$?"  Of course it depends upon what you're trying to do.) I would like to think that your boss knew what the point of it all was, so without further information I guess I would trust him.
In fact, your story arouses my curiosity.  I suppose you're not putting us on, but weighing paper cutouts is just about the last method I would ever think of for computing area (aren't you going to need a very sensitive scale or an awfully big piece of paper to get anywhere with this?).  How long ago are we talking?  What was the job?  You don't have to answer these questions, but it would be interesting to know...
A: you are correct. Actually, the integral is equal to: "(total above weight) minus (total below weight)", which I'm sure is what you meant. What your boss is calculating is actually
$$
\int_a^b |f(x)| dx
$$
