Answer:
15.47 gallons
Explanation:
The pumps remove the oil at the rate modelled by
R(t) = 2 + cos(πt / 11) in gallon per hour
The oil removes in 6hours period is given by integral of R(t)
from t = 0 to t = 6
[tex]Q = \int\limits^6_0 {R(t)} \, dt = \int\limits^6_0 {(2+ cos(\frac{\pi t}{11} })) \, dt[/tex]
[tex]Q = [2t + \frac{sin(\frac{\pi t}{11}) }{\frac{\pi }{11} } ]_0^6[/tex]
[tex]Q = [2(6) + \frac{11}{\pi } sin(\frac{6\pi }{11}) ] - [0 + \frac{11sin (0)}{\pi } ][/tex]
[tex]Q = [12 + \frac{11}{\pi } (0.9898)]\\\\Q = 12 + 3.465694\\\\Q = 15.465694\\\\Q = 15.47 gallons[/tex]
