Mark M. answered • 05/11/17

Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

Let A(t) = amount of water in the reservoir at time t hours

Then A'(t) = rate of change in the amount of water (in gal/hour) at time t

= (rate in) - (rate out)

= (1500 + 7t

^{2}) - (4t + t^{2}) = 6t^{2 }- 4t + 1500(net change in amount of water from t=1 to t=5) = A(5) - A(1)

= ∫

_{from 1 to 5}A'(t)dt (using the Fundamental Theorem of Calculus) = ∫

_{from 1 to 5)}(6t^{2}- 4t + 1500) dt = (2t

^{3}- 2t^{2}+ 1500t)_{from 1 to 5} = 7700 - 1500 = 6200 gal