If we let r = rms of f(t), then
4r^2 = Int(15 - 15e^-1/2t)^2 dt[0,4]
= 15Int(1 - e^-t/2)^2 dt[0,4]
= 15Int(1 - 2e-t/2 + e^-t)[0,4]
= 15(t + 4e^-t/2 - e^-t)[0,4]
= 15[(4 + 4/e^2 - 1/e^4) - (0+4-1)]
= 15(1 + 4/e^2 - 1/e^4)
so A.
see wikipedia on root mean square
My apologies to Alan who posted this question last week. I forgot to square f(t) in the integral.
Find the RMS value of the function
i=15[1-e^(-1/2t)] from t=0 to t=4
A. 7.5[sqrt(1+4e^-2 -e^-4)]
B. 7.5[sqrt(4-2e^-2 +e^-4)]
C. 7.5[sqrt(4+4e^-2 -e^-4)]
D. 7.5[sqrt(5-2e^-2 +e^-4)]
1 answer