2007-05-16 18:21:28 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The average population is the integral of p(t) between those limits divided by 15 years.
p(t)= 281e^(0.012t)
Since starts from 2000, we are integrating from 0 to 15
int p(t) = 281 / 0.012 * e^(0.012t)
= 23416.7 * (e^0.18 - e^0)
= 4618.2
Dividing this by 15 we get
307.8
2007-05-16 18:26:50 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋
If it were a discrete problem, you'd add up the population in each year and divide by 15 to get the average. Since this is continuous, you find the integral of p(t), which is 23416.7e^(0.012t), call it P(t), evaluate at 15 and 0, subtract (should get 4618.17), and divide by 15. End up with 307.878.
Check it simply: p(0) = 281 and p(15) = 336, so 307 is a reasonable average.
2007-05-16 18:36:55 · answer #2 · answered by Philo 7 · 0⤊ 0⤋
Just integrate the equation with limits of 2000 and 2015 and divide the result by 15. The integration does not look all that difficult (this looks like a census extrapolation with 281 representing the population of the US in 2000.)
2007-05-16 18:33:01 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
Take the definite integral between the limits specified, and divide by 15. That's all there is to it. The result will be slightly larger than 281 -- perhaps about 300.
2007-05-16 18:26:19 · answer #4 · answered by Anonymous · 0⤊ 0⤋