2007-02-26 22:49:16 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Break up the curve into sections, say 1 section for each unit along the curve (in the x direction). Then, figure out the y value of the curve in the MIDDLE of that section.
Your rectangle will be 1 unit wide and the height will be from 0 to the value at the middle of that section.
The area of a rectangle is of course width*height, so if we use a section width of 1 unit, and for example the height at the middle you measured was 5, then the area of that section would be 1*5 = 5.
Keep doing this for each section of the curve and add up all the answers to get the total area under the curve.
The larger the sections you use (1 unit, 2 unit, 5 units, etc) the easier it will be to calculate it, but the less accurate the number will be. The smaller the sections you use (0.5 unit, 0.1 unit, 0.0001 unit, etc), it will take a long time to add up everything, but it will be more accurate.
When your unit size is infinitely small, then you will have exact precision. Performing this with an infinitely small unit size is called Integrating, it's one of the basics of Calculus.
2007-02-26 23:02:09 · answer #1 · answered by vanchuck 2 · 0⤊ 0⤋
That's essentially what Calculus I is all about.
For a quick example... Draw a curve above a line.
Trace that onto graph paper. You can now count the boxes between the curve and line. and you can estimate the portions of boxes...
Divide all the boxes along the curve into quarters... you are now more accurate.
divide the resulting boxes in quarters... more accurate again.
Calculus deals with finding the "limit" of how small you can make those boxes... thus being more accurate in determining the area.
2007-02-26 22:58:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Its a bit hard to explain at the moment for me as I am at work but you do need to know basic integration for that. And I hope the following links help! The second one has an applet that shows you how they do it.
2007-02-26 22:58:44 · answer #3 · answered by | e v e | 2 · 0⤊ 0⤋
dividing it into n trapezia with the width infinitesimally small so the
trapezia becomes rectangle and sum the areas of the n rectangles in the interval
area=Lt. h[f(a)+f(a+h)+f(a+2h)+....
h->0 .... f[a+(n-1)h]
n->infinity
2007-02-26 22:59:43 · answer #4 · answered by raj 7 · 0⤊ 1⤋