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.

 A: 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.  
A: 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.
A: 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.
