Area under trendline
Two ways:
1 Divide into relatively small intervals, & in each interval muliply the
width by the average height.
2 As you've got a polynomial in x, you can use the integration formula:
integral(x^n)dx = (x^(n+1))/(n+1)
... so the integral from 0 to A2 of your formula (3E+09x6 - 7E+08x5 +
8E+07x4 - 4E+06x3 + 42077x2 + 1197.x) becomes
=3E+09*A2^7/7 - 7E+08*A2^6/6 + 8E+07*A2^5/5 - 4E+06*A2^4/4 +
42077*A2^3/3 + 1197*A2^2/2
--
David Biddulph
"Josh" wrote in message
...
Does anybody know how to find the area under the curve of a trendline (6th
degree polynomial in my case).
Thanks in advance for any help.
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