Find the average value of the function N(t) = 200e^(kt) on the interval [2,7]. Include an integral and appropriate units.
2007-01-07 15:40:59 · 4 answers · asked by Taryn 2 in Science & Mathematics ➔ Mathematics
Integrate it over the interval [2,7] to get the area under the curve. Then divide by the width of the domain.
∫{200e^(kt)}dt = (200/k)e^(kt) | on [2,7]
= (200/k)[e^(7k) - e^(2k)]
Average = (200/k)[e^(7k) - e^(2k)]/(7 - 2)
= (200/k)[e^(7k) - e^(2k)]/5 = (40/k)[e^(7k) - e^(2k)]
2007-01-07 15:51:34 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Graphically, average value means average height. It equals to the total area enclosed by the curve divided by the total length of the domain of the curve.
[1/(7-2)]â«200e^(kt) dt [t:2...7]
= 40[e^(7k) - e^(2k)]/k
2007-01-07 23:49:17 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
Average value of N(t) over interval [a,b] = (integral (from a to b) N(t) dt)/(b-a).
Here, N(t) = 200e^(kt), whose antiderivative is (200/k)e^(kt) + C ==> average value = (200/k)(e^7k - e^2k)/5 = (40/k)(e^7k - e^2k)
2007-01-07 23:48:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Huh... I dunno how to do this one... I get to learn something ^_^
2007-01-07 23:44:03 · answer #4 · answered by Anonymous · 0⤊ 1⤋