The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. The table below gives the rate as measured at various times in an 8-hour time period.

t---------0-----2------3-------7----8
(hours)

R(t)--1.95---2.5---2.8----4.00--4.26
(gallons per
hour)

Use a trapezoidal sum with the four sub-intervals indicated by the data in the table to estimate definite integral 0 to 8 of R(t) dt. Using correct units, explain the meaning of your answer in terms of water flow.

(2) (2.5+1.95)/2)+(1)(2.8+2.5)/2)+(4)(4+2.8)...
Simplified gives a water flow of 24.83 gallons over eight hours.

Is there some time t, 0 < t < 8, for which we are guaranteed that R' (t) = 0? Justify your answer.
No and this is shown by a graph of the function.

The rate of water flow R(t) can be estimated by W(t) = ln( t^2 + 7 ). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.
[W(8)-W(0)]/(8-0) =(4.26268-1.94591)/8 = 0.2986 gallons/hour

f is a continuous function with a domain [−3, 9] such that
f(x) =
3 for -3 <= x < 0
-x+3 for 0 <= x <= 6
-3 for 6 < x <= 9

when f(x)>0 due to the fundamental theorem

For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
6 + (-x^2/2 + 3x)+3

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