9514 1404 393
Answer:
6200 gallons
Step-by-step explanation:
The rate of change of volume is the difference between the inflow rate and the outflow rate.
v'(t) = (1500 +7t^2) -(4t +t^2) = 6t^2 -4t +1500
The change in volume over the time interval of interest is the integral of this expression over that interval:
[tex]\displaystyle\Delta V=\int_1^5{(6t^2-4t+1500)}\,dt=\left.(2t^3-2t^2+1500t)\right|^5_1\\\\=2(5^3-1^3)-2(5^2-1^2)+1500(5-1) = 2\cdot124-2\cdot24+1500\cdot4\\\\\boxed{\Delta V=6200\qquad\text{gallons}}[/tex]
The amount of water in the reservoir increases by 6200 gallons over the time interval.
