AP Calculus BC: 6.11 Integrating Using Integration by Parts bc only – Exam Style questions with Answer- FRQ

Question

The rate at which raw sewage enters a treatment tank is given by E  (t ) gallons per hour for \(0\leq t\leq 4\), hours. Treated sewage is removed from the tank at the constant rate of 645 gallons per hour. The treatment tank is empty at time t =0.
(a) How many gallons of sewage enter the treatment tank during the time interval \(0\leq t\leq 4\) Round your answer to the nearest gallon.
(b) For\( 0\leq t\leq 4,\) at what time t is the amount of sewage in the treatment tank greatest? To the nearest gallon, what is the maximum amount of sewage in the tank? Justify your answers.
(c) For \(0\leq t\leq 4?\) the cost of treating the raw sewage that enters the tank at time t is ( 0.15 -0.02t )  dollars per gallon. To the nearest dollar, what is the total cost of treating all the sewage that enters the tank during the time interval\( 0\leq t\leq 4?\)

Answer/Explanation

(a) \(\int_{0}^{4}E(t) dt=3981 \)gallons

(b) Let S (t ) be the amount of sewage in the treatment tank at time t. Then \(S'(t)=E(t)-645 and S'(t)=0\) when
E ( t) = 645. On the interval \(0\leq t\leq 4\)= 645
when  t = 2.309  and t= 3.559.  

The amount of sewage in the treatment tank is greatest at
t = 2.309 hours. At that time, the amount of sewage in the tank, rounded to the nearest gallon, is 1637 gallons.

(c) Total cost= \(\int_{0}^{4}(0.15-0.02t)E(t)dt=474.320\) The total cost of treating the sewage that enters the tank
during the time interval \(0\leq t\leq 4\) to the nearest dollar, is $474.

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