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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?

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@J. M.: I guess you, in the end, did it like according to your boss? –  AD. Sep 3 '10 at 3:48
Well, I was young and stupid back then, so I let the boss do it his way (I couldn't help but feel I was right, but hey, he's the boss). Now I am no longer so young, but as to whether the other condition still applies is apparently still a matter of contention amongst my peers. –  Guess who it is. Sep 3 '10 at 5:44
*sigh* If there were only a way to accept two answers... –  Guess who it is. Sep 3 '10 at 9:58
I had to upvote just because of the title :) –  anon Sep 3 '10 at 17:18
This is really cool! –  bobobobo Nov 8 '10 at 1:14

3 Answers 3

up vote 8 down vote accepted

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 $$ \int_a^b |f(x)| dx $$

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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...

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My curiosity is also roused. As soon as I read this question, I started wondering about the points posed in Pete's last paragraph. –  Derek Jennings Sep 3 '10 at 9:25
I wouldn't even dream of putting anyone on on a mathematics Q&A site, Pete. :) As I recall, it had something to do with the integral of the set of empirically-defined measurements, since they're correlated with a certain quantity we were trying to determine from our data. Without a plausible functional form, and none of us knowing about things like splines back then (hey, I only barely learned calculus in school!), it was what was written in the "manual". We had, however, an exquisitely sensitive Sartorius analytical balance, which was meant for weighing very light objects. –  Guess who it is. Sep 3 '10 at 9:43
Apparently it's been done for a long time, too: mathworld.wolfram.com/Cycloid.html says Galileo found areas under the curve by a similar technique, only that he used metal instead of paper. If you look at various engineering handbooks, you'll see this estimation method mentioned, too. I will be the first to admit that it's very crude, and very much dependent on the skill of the operator in drawing and cutting. –  Guess who it is. Sep 3 '10 at 9:48
@J.M.: very interesting. As a lifetime academic my ideas about how things might be done in the real world are rather...theoretical. –  Pete L. Clark Dec 3 '10 at 5:55

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.

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