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Need urgent math help???
I couldnt find an answer for this question
Find avrage value of the function over the given interval
F(t)= e^.09t on [0,10]
3 Answers
- TomVLv 75 years agoFavorite Answer
The average value of any function over a given interval is the integral of the function over that interval, (1/0.09)e^(0.09t) + C, divided by the length of the interval, 10.
Avg = (1/10)∫e^(0.09t)dt = (e^0.9 - 1)/0.9 ≈ 1.62178
- PolyhymnioLv 75 years ago
Integral from 0 to 10 of F(x) over (10 - 0) = 10
(1/0.09)(e^0.9 - 1)/10 = 1.11111(e^0.9 - 1)
