
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 16th, 2016, 12:57 AM  #1 
Newbie Joined: Nov 2016 From: Copenhagen Posts: 3 Thanks: 0  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 13, 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) = 723.64388.41 = 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 
November 16th, 2016, 06:40 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,211 Thanks: 1052 
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 ... $\displaystyle \int_{115}^{140} f(x) \, dx$ 
November 16th, 2016, 06:47 AM  #3 
Newbie Joined: Nov 2016 From: Copenhagen Posts: 3 Thanks: 0 
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 
November 16th, 2016, 07:16 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,211 Thanks: 1052 
$\displaystyle \int_{115}^{140} 20030\ln{x} \, dx = \bigg[10x(233\ln{x})\bigg]_{115}^{140}$ $\bigg[1400(233\ln{140})\bigg]\bigg[1150(233\ln{115})\bigg]$ $23(250)+3450\ln{115}4200\ln{140} \approx 1365.12$ 
November 17th, 2016, 08:26 AM  #5 
Newbie Joined: Nov 2016 From: Copenhagen Posts: 3 Thanks: 0 
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. 
November 17th, 2016, 08:40 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,211 Thanks: 1052 
if f(x) is a discrete rather than a continuous function, then a definite integral will not work. You would need to calculate $\displaystyle 10 \sum_{n=115}^{140} 203\ln(n) \approx 1419.82$ 

Tags 
calculating, future, integration, revenue 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Future teacher  yamidude  New Users  0  May 3rd, 2014 01:49 PM 
Future value: what's wrong here?  zniets  Economics  2  February 16th, 2014 02:20 PM 
Accumulated Future Value Help  BrianMX34  Calculus  1  December 11th, 2012 09:49 PM 
pi and Future probabilities  myrv  New Users  8  October 26th, 2007 10:30 AM 
pi and Future probabilities  myrv  Number Theory  4  January 1st, 1970 12:00 AM 