# Thread: Area under a Curve

1. ## Area under a Curve

Hi

Is it possible to find the area under a curve if all I have is the x and y co-ordinate? I.e. I dont have the equation if the curve, for example, I know that the y value for the beginning of my curve is 3 and the x-value for the end of my curve is 5.

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1..
1...
1....
1.....
1......
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1..................................... and so on (it's sposed to be smooth)

-I hope you get the idea of what I'm looking at

2. I might just guess the curve based on what it looks like...it sort of looks like a $\frac{1}{x}$ curve

3. I'm going to think on this one...may get back to you. Thanks!

4. Yes, you can use your points to run a regression and then integrate the resulting expression/function you derive.

Or, you can use data points and use Simpson's rule or another approximating method.

Let's do an example:

Suppose we want to build a road through a hill and have to move dirt. That's

volume. The road is 75 feet wide and 600 feet long. We get the following horizontal distances(x) and heights(y) through the hill.

x=0, 100, 200, 300, 400, 500, 600

y=0,7,16,24,25,26,0

We can run a quartic regression through a good calculator and derive the following equation

$f(x)=(9.09X10^{-10})x^{4}-(1.59X10^{-6})x^{3}+(5.93X10^{-4})x^{2}+.02x+.11$

$75\int_{0}^{600}f(x)=673,707.60 \;\ ft^{2}$

Or, we can use simpson's rule without deriving an equation.

$V\approx{75}\cdot\frac{600}{18}\left[0+4(7)+2(16)+4(24)+2(25)+4(16)+0\right]=675,000 \;\ ft^{2}$

As you can see, they're both pretty close. The first would be the more accurate though, I would think, because R^2 = .998515

Post some points and we'll derive the equation