Asked by Kailey
The rate at which water flows into a tank, in gallons per hour, is given by a differentiable, increasing function R of time t. The table below gives the rate as measured at various times in an 8-hour time period.
t (hours) 0 2 3 7 8
R(t) (gallons per hour) 1.95 2.5 2.8 4 4.26
The rate of water flow R(t) can be estimated by W(t) = ln(t2 + 7). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.
I solved this and got 0.2986, but my teacher told me this was wrong. I've tried redoing it but keep getting the same answer. Any help on how to do this would be very much appreciated!!
t (hours) 0 2 3 7 8
R(t) (gallons per hour) 1.95 2.5 2.8 4 4.26
The rate of water flow R(t) can be estimated by W(t) = ln(t2 + 7). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.
I solved this and got 0.2986, but my teacher told me this was wrong. I've tried redoing it but keep getting the same answer. Any help on how to do this would be very much appreciated!!
Answers
Answered by
Damon
integral from t = 0 to 7 = 8 (8 hours), then divide by 8
integral ln (t^2 + 7) dt
note:
http://www.wolframalpha.com/input/?i=integrate+ln(x%5E2%2B7)
t [ln(t^2+7)-2] + 2 sqrt 7 tan^-1(t/sqrt 7)
at t = 8 - at t = 0
divide by 8
integral ln (t^2 + 7) dt
note:
http://www.wolframalpha.com/input/?i=integrate+ln(x%5E2%2B7)
t [ln(t^2+7)-2] + 2 sqrt 7 tan^-1(t/sqrt 7)
at t = 8 - at t = 0
divide by 8
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