This is really 2 questions. One is an exact answer. If you know the area under a curve for any value of x, then (if you know calculus) you can take the derivative of the area function to get the curve. That is, if you have the curve f(x) you know the area under it from 0 to x, A(x), is the integral from 0 to x of f(x), so A'(x) = f(x).
For approximate values: of course, if you know n points you can always find an n - 1 degree polynomial that goes through those points (by setting up a standard polynomial and substituting the points in to get n equations in n unknowns. If the curve is not a polynomial this is not exact, but it is often good enough for data analysis.
Otherwise, if you know something about the curve, then the same technique holds: set up an equation with variables for the coefficients, plug in your points, and solve for the coefficients.
There are also some special cases, such as fourier analysis for periodic functions, but that is probably more than you care about now.
2007-02-24 00:49:59
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answer #1
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answered by sofarsogood 5
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Uhm...
Well, if you have the function A(x) for the area under the curve from a (some constant) to x (note that A(a) = 0, since the area under the curve from a to a is just 0). One just has to differentiate A(x) with respect to x, to obtain the original function. (FTC)
For example: The area under some curve from 0 to x can be expressed by the function g(x) = x ^ 2 + x, so the original function is f(x) = g'(x) = 2x + 1
However, if you only know the VALUE of the area under the curve from a to b of some function, you do not have enough information to solve the original function. e.g: The area under some curve from 0 to 1 is 2(units), you can have infinite possibilities:
1. f(x) = 2
2. f(x) = 4x
3. Or even piecewise function:
f(x) = 1 for 0<= x <= 0.5
f(x) = 8x - 3 for x > 0.5
4. ...
Can you get it? :)
2007-02-23 23:54:13
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answer #2
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answered by ? Woc Viet ? 2
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The number of points needed varies with the complexity of the curve. Assume an equal spacing of points, you would need to know as many points as so that you have at least two points for the shortest period corresponding to the fastest Fourier frequency component in the curve--otherwise aliasing would occur and the reconstruction would have errors.
The area only gives you one specific segment's mean which is the constant to add to the Fourier series to move it from the zero mean to the actual curve's vertical level.
2007-02-23 23:42:07
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answer #3
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answered by sciquest 4
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I'm not entirely sure but I would imagine you would have to have some idea of what sort of curve you were dealing with... eg exponential, polynomial (and of what order?), trigonometric etc, otherwise there would just be too many possibilities. I suppose this is a modelling problem so the type of curve would be one of your modelling asumptions.
Eg
A parabola can be drawn through any 3 (non-co-linear) points, so I would think that if you knew the area under the curve and you knew it was a parabola, two points would be sufficient...
I don't know for sure, but I'm adding your question to my watch list as it's got me thinking!
2007-02-23 23:34:12
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answer #4
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answered by _Jess_ 4
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needed for a type of purposes. Take the conventional x-y axes and take any curve y=f(x) Say you commence at x=a as a lot as x=b Divide the area a to b into very small "products" (delta x) because delta x is particularly-very small the genuine (the curve) is virtually a immediately line. so that you've many rectangles, the first one having a community of delta x circumstances y (starting up at x=a) and ending at x=b. you would have a limiteless type of those delta x products(!) until eventually you've determined the curve. The reduce of the sum of those sorts of (as delta x has a tendency to 0) is the integral of y* dx (from a to b) I absolutely have made it extremely straight forward for you to comprehend! i wish I absolutely have succeeded!
2016-12-04 21:21:05
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answer #5
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answered by Anonymous
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As you put you have just a number.So you can´t say anything
If you had the area underneath as a function of x its derivative would give you the function
A = Integral from a to x of f(x)dx
2007-02-24 06:38:20
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answer #6
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answered by santmann2002 7
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