Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 410: 64

1800 liters of water flows from the tank during the first 10 minutes.

We can find the total amount of water that flows from the tank during the first 10 minutes: $\int_{0}^{10}r(t)~dt$ $=\int_{0}^{10}(200-4t)~dt$ $=(200t-2t^2)\vert_{0}^{10}$ $=[200(10)-2(10)^2]-[200(0)-2(0)^2]$ $=(2000-200)-(0)$ $= 1800~liters$ 1800 liters of water flows from the tank during the first 10 minutes.