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November 15th, 2016, 11:57 PM   #1
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Calculating future revenue through integration

Hi,
My goal is to calculate the future revenue for a marketing campaign for the time period between 140 and 115 days.
The equation is f(x) = -30ln(x) + 200, where x represents the days and f(x) is the revenue.
The integral must be (as far as I know):
F(x) = -30*(xln(x) - x) + 200x + C = -30x*ln(x) + 230x + C.

My next step would be to calculate the constant C. As far as I know, I need a known coordinate in F(x) to do so. I calculate the sales for day 1, as that number should be the same as the area for day 1, f(1) = F(1). f(1) = 200.
F(1) = 230. So using my logic, the constant C would be -30.
The final equation would be: F(x) = -30x*ln(x) + 230x -30.
To verify above, I am calculating the revenue using f(x) for day 1-3, so f(1)+f(2)+f(3) = 546.25
However F(3) = 561.12, which means something is wrong.
Likewise I tested f(3)+f(4) = 167.04+158.41 = 325.45 (day 3 + 4). This should be equivalent to F(4)-F(2) = 723.64-388.41 = 335.23. So again the numbers do not match.

Does anyone know what I am doing wrong here?
Any help is greatly appreciated.

Kind regards
Thomas
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November 16th, 2016, 05:40 AM   #2
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If f(x) is a rate of revenue in units of currency per day, then revenue earned between day 115 and day 140 would be represented by the definite integral ...

$\displaystyle \int_{115}^{140} f(x) \, dx$
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November 16th, 2016, 05:47 AM   #3
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Hi Skeeter,
Thanks - very true, and this is why I did the integration.
However upon validating the calculations, I cannot make the numbers match. I think something is wrong in my (definite) integral equation?
I hope anyone can help out.
Thomas
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November 16th, 2016, 06:16 AM   #4
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$\displaystyle \int_{115}^{140} 200-30\ln{x} \, dx = \bigg[10x(23-3\ln{x})\bigg]_{115}^{140}$

$\bigg[1400(23-3\ln{140})\bigg]-\bigg[1150(23-3\ln{115})\bigg]$

$23(250)+3450\ln{115}-4200\ln{140} \approx 1365.12$
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November 17th, 2016, 07:26 AM   #5
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Thank you again Skeeter - however if I validate the result of i.e. 1365.12 through the accumulated sum of (f(116)+f(117)..f(140)) I still do not get the same number (1362.171). Likewise if I apply your math on the 6 vs 2 days, I get 639 using the integral and 623 using (f(3)+f(4)+f(5)+f(6)), which unfortunately is quite a significant deviation.

This once more leads me to believe that something is wrong with my integral equation.
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November 17th, 2016, 07:40 AM   #6
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if f(x) is a discrete rather than a continuous function, then a definite integral will not work. You would need to calculate

$\displaystyle 10 \sum_{n=115}^{140} 20-3\ln(n) \approx 1419.82$
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