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.

Question

t (hours)
0 2 3 7 8
R(t) (gallons per hour)
1.95 2.5 2.8 4 4.26

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

2. Is there some time t, 0 < t < 8, such that R′(t) = 0? Justify your answer.

3. 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.

Answer 1

#1. There are many handy online calculators you can use to verify your results.
#2. I doubt it, since the values are increasing
#3. The average rate of flow over the interval is

(∫[0,8] W(t) dt)/(8-0)
= 1/8 ∫[0,8] ln(t^2+7) dt = 3.09 gal/hr

Doing the integral takes some maneuvering.

Using integration by parts,
u = ln(t^2+7)
du = 2t/(t^2+7) dt

dv = dt
v = t

∫ln(t^2+7) dt = t ln(t^2+7) - 2∫t^2/(t^2+7) dt
Now, t^2/(t^2+7) = 1 - 1/(t^2+7) so we have
∫ln(t^2+7) dt = t ln(t^2+7) - 2(∫1 - 1/(t^2+7)) dt
= t ln(t^2+7) - 2t + 2√7 arctan(t/√7)