1. **State the problem:** Calculate the definite integral $$F(10) - F(0) = \int_0^{10} 20te^{-0.8t} \, dt$$ using integration by parts. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = 20t \implies du = 20 \, dt$$ $$dv = e^{-0.8t} dt \implies v = \int e^{-0.8t} dt = \frac{e^{-0.8t}}{-0.8} = -\frac{5}{4} e^{-0.8t}$$ 4. **Apply integration by parts:** $$\int_0^{10} 20te^{-0.8t} dt = \left. u v \right|_0^{10} - \int_0^{10} v \, du = \left. 20t \left(-\frac{5}{4} e^{-0.8t} \right) \right|_0^{10} - \int_0^{10} \left(-\frac{5}{4} e^{-0.8t} \right) 20 dt$$ 5. **Simplify the integral:** $$= \left. -25 t e^{-0.8t} \right|_0^{10} + 100 \int_0^{10} e^{-0.8t} dt$$ 6. **Evaluate the remaining integral:** $$\int_0^{10} e^{-0.8t} dt = \left. \frac{e^{-0.8t}}{-0.8} \right|_0^{10} = -\frac{5}{4} \left(e^{-8} - 1 \right) = \frac{5}{4} (1 - e^{-8})$$ 7. **Substitute back:** $$F(10) - F(0) = -25 \cdot 10 \cdot e^{-8} + 25 \cdot 0 \cdot e^{0} + 100 \cdot \frac{5}{4} (1 - e^{-8}) = -250 e^{-8} + 0 + 125 (1 - e^{-8})$$ 8. **Simplify:** $$= 125 - 125 e^{-8} - 250 e^{-8} = 125 - 375 e^{-8}$$ 9. **Calculate numerical value:** Since $e^{-8} \approx 0.00033546$, $$F(10) - F(0) \approx 125 - 375 \times 0.00033546 = 125 - 0.1258 = 124.8742$$ 10. **Interpret units:** Given the integral of flow rate over time, the units are volume (e.g., liters). **Final answer:** $$\boxed{F(10) - F(0) \approx 124.87 \text{ units of volume}}$$