Find the average value of the function over the given interval and all values s in the interval for which function equals its average value.
f(s)=(s^2+3)/(s^2), 2<=s<=7
use graphing utility to verify your result!
Thanks in advance!
f(s)=(s^2+3)/(s^2), 2<=s<=7
use graphing utility to verify your result!
Thanks in advance!
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Sorry, read the question wrong
Average Value = 1/(b-a)∫f(s)ds where the integral is between 2 and 7
= 1/(7-2)∫(s^2+3)/(s^2)ds
1/5∫(s^2+3)/(s^2)ds
1/5[∫(1)ds + ∫3/(s^2)]ds
1/5[s + -3/s] between 2 and 7
1/5[7 -3/7 -2 + 3/2]
= 1.21428571
1.21428571 = (s^2+3)/(s^2)
1.21428571 = 1 + 3/(s^2)
0.21428571 = 3/(s^2)
s^2 = 3/0.21428571
s^2 = 14 <- ain't that convenient
s = +-SQRT(14)
s = +-3.74165742
so s = 3.74165742 since it has to be in between 2 and 7
Average Value = 1/(b-a)∫f(s)ds where the integral is between 2 and 7
= 1/(7-2)∫(s^2+3)/(s^2)ds
1/5∫(s^2+3)/(s^2)ds
1/5[∫(1)ds + ∫3/(s^2)]ds
1/5[s + -3/s] between 2 and 7
1/5[7 -3/7 -2 + 3/2]
= 1.21428571
1.21428571 = (s^2+3)/(s^2)
1.21428571 = 1 + 3/(s^2)
0.21428571 = 3/(s^2)
s^2 = 3/0.21428571
s^2 = 14 <- ain't that convenient
s = +-SQRT(14)
s = +-3.74165742
so s = 3.74165742 since it has to be in between 2 and 7