# How do I calculate the area under this curve from the graph?

so I've researched and found one method for this type of graph which is simply counting the squares but then aren't we supposed to apply the area of the squares to get the full answer? here is the part of the question I'm trying to solve for

A 3.0-kg object moving along the x axis has a velocity of 2.4 m/s as it passes through the origin. It is acted on by a single force, Fx, that varies with x, as shown in Figure 6-31.

(a) Find the work done by the force from x = 0.0 m to x = 2.0 m.

• I imagine yo don't know the function (if you know, it's just a problem of integration). Without knowing it, you can integrate with numeric methods, such as "counting squares" method or some Montecarlo method. – MatMorPau22 Nov 6 '17 at 17:36
• Based upon this I assuming the function is the Integration of work correct? from the specified displacement intervals? – AceNinja1101 Nov 6 '17 at 17:38
• What sort of answer is expected? Multiple choice? I wouldn't expect them to require you to calculate it exactly. "aren't we supposed to apply the area of the squares to get the full answer?" I don't understand what this means. – Acccumulation Nov 6 '17 at 17:39
• Well no but what I mean is, since the curve is not an exact line we can't simply find the area under the curve easily, and I found that counting the squares is another form of approximation – AceNinja1101 Nov 6 '17 at 17:42
• Yes... if you don't have the F(x) function, then most you can do is approximate. You can't calculate the work done by a force if you don't know how much force you are applying. Unless, its multiple choice question, the exact answer would be a result of few assumptions. – LostCause Nov 6 '17 at 17:50

this method works well if you have access to a sensitive weighing device. first print the graph out on a sheet of paper. measure the sheet to determine its area. weigh the sheet of paper to get its weight; you now know its weight per square centimeter. take scissors and cut out the portion whose area you require, and weigh it. divide its weight by its weight per square centimeter. the answer is square centimeters of area.

• Typical printer papers weighs approximately $4.5 \ g$ per sheet. A sheet is $8.5" \times 11"$, $93.5 \ in^2$, for $0.0481 \frac g{in^2}$, since $1 \ in^2 = 6.4516 \ cm^2$, the paper will weigh $0.00746 \frac g{cm^2}$. You're going to have a hard time finding a scale that measures that accurately; let alone controlling your measuring procedure enough to get good numbers, I think. Novel way to solve it, but not reliable IMO. – JMac Nov 6 '17 at 20:04
• I worked with Mettler microbalances every day in a previous lifetime, these were more than adequate for the task. In fact, they were sufficiently sensitive that if you wrote on a piece of paper with liquid ink and weighed it, the evaporation rate of the ink solvent was clearly discernable. – niels nielsen Nov 6 '17 at 22:14
• Seems like it's a pretty expensive and complicated solution. – JMac Nov 6 '17 at 23:11
• Careful, this won't differentiate between negative work and positive work-- you'll have to weigh those separately and subtract! – Jahan Claes Nov 7 '17 at 0:37
• jahan is right! – niels nielsen Nov 7 '17 at 0:57

I see two options here:
1) Count/estimate the squares, also count partial squares, and note that 8 squares is a Joule (Nm).
2) Find a program on the internet that can read graphs from a photo and let it do the math for you. I have used such programs before to read values from a graph, maybe you can find one that can integrate numerically too. If you find a nice one, please share it here.

There are two problems to solve.

First, you need the area under the curve in "arbitrary units" - like the squares on the graph paper. You could count them - but a more accurate approach would be to use Simpson's Rule to make a fairly accurate integration, based on using weights of 1-4-2-4-2-4-2-4-1 on the 9 data points from 0 to 2.0.

Eyeballing the height of the curve for each intersection, you would get get following table:

point value weight product
0    0.0    1      0.0
1    1.2    4      4.8
2    2.0    2      4.0
3    2.6    4     10.4
4    3.0    2      6.0
5    3.3    4     13.2
6    3.5    2      7.0
7    3.8    4     15.2
8    3.9    1      3.9
______
Sum 64.5


The weighted sum is 64.5 (namely 0.0 * 1 + 1.2 * 4 + 2.0 * 2 + ... + 3.9 * 1); the integral in squares is 1/3rd of that (per Simpson's rule) - or 21.5 squares. You can convince yourself that counting squares gives you roughly the same number.

The second step:
You then need to figure out the scaling. Since the x divisions are 0.25 and the Y divisions are 0.5, one square is 1/8th of a Nm - or Joule.

Now you know the number of squares, and the work done per square. Multiply.

Be careful not to quote more significant figures than is justified. I am eyeballing the division of each graph square; over 8 points that is going to give me an error on the order of a fraction of a square, so 21.5 probably has an error of about +- 0.5 on it.

• Now that makes sense! and yeah I roughly counted about 22 squares between those intervals. The scaling are the values in between the intervals such as 0 to 1 right for both X and Y direction? and my professor doesn't really check for that from what I know of especially since he didn't teach any of this in lecture so I highly doubt others will follow the same. So what would I do if I wanted to calculate the entire area from 0 to 4? – AceNinja1101 Nov 6 '17 at 20:47
• To get the result from 0 to 4 you could keep doing the same thing - note that when the force opposes the direction of motion you will be doing negative work so the method of weighing the graph would have to be adjusted. You would subtract the weight of the "negative" area from the weight of the "positive" area. The advantage of the Simpson method is that you just need to write the height of the curve with the sign and math takes care of the rest . – Floris Nov 6 '17 at 20:52
• What do you mean by weight? As in the number of partial squares that occur more than once? – AceNinja1101 Nov 6 '17 at 21:21
• Sorry - I was referring to the accepted answer that suggests determining the weight of the piece of paper as a proxy for the area. But in my answer I use weight to determine how much a data point counts. I see that might have confused you. Did you read about the Simpson rule on the link? Do you understand how it works? It's a good way to estimate the area under a curve - well worth knowing. – Floris Nov 6 '17 at 21:28
• Okay that's what I assumed actually and I shall read it after my last class but I did skim over it and it does makes things easier especially for this type of problem – AceNinja1101 Nov 6 '17 at 21:52