Area under a curve

My Calculus may be rusty

Well, If you have an equations (I'm assuming based on a trendline) you're almost there.

I believe you just need the integral of that equation, then plug whatever x you need.

Example I. Simple example (line equation y=4)

x y
0 4
1 4
2 4
3 4

the Integral of y=4 with respect to x is: Area = 4x/1 (or just =4x)

so the area under the line at x=4 = 4*(4) =16

Example II: straight diagnol line (y=x)

x y
0 0
1 1
2 2
3 3

Integral of y=x: Area = (x^2)/2

so at x=4, [(4)^2]/2 = (16)/2 = 8 ..sounds good so far

Example 3; a Curve equation: y=(x^2) + 3x + 1

x y
0 1
1 5
2 11
3 19

Integral: Area = [(x^3)/3] + [3(x^2)/2] + x

so at x=4, the area under the curve is equal to:

[(4^3)/3] + [3*(4^2)/2] + 4 simplified to
[(64/3) + (3*16/2) + 4] simplified to
21 + 24 + 4 =49

Feel free to tell me if I'm off a bit somewhere..it's been awhile.

Hope that helps out somewhat,
Adam S.

Have a look at http://ucs.orst.edu/~haggertr/487/integrate.htm (NT)