1. **State the problem:** We need to find the total amount of water that flows out of the faucet during the first two minutes. The flow rate is given by the function $$v(t) = t^3 - \frac{1}{2}t^2 + 4$$ gallons per minute, where $t$ is the time in minutes. 2. **Set up the integral:** The total volume of water that flows out from $t=0$ to $t=2$ is the definite integral $$\int_0^2 v(t) \, dt = \int_0^2 \left(t^3 - \frac{1}{2}t^2 + 4\right) dt.$$ 3. **Integrate the function:** $$\int \left(t^3 - \frac{1}{2}t^2 + 4\right) dt = \frac{t^4}{4} - \frac{1}{2} \cdot \frac{t^3}{3} + 4t + C = \frac{t^4}{4} - \frac{t^3}{6} + 4t + C.$$ 4. **Evaluate the definite integral:** Calculate at the upper limit $t=2$: $$\frac{2^4}{4} - \frac{2^3}{6} + 4 \cdot 2 = \frac{16}{4} - \frac{8}{6} + 8 = 4 - \frac{4}{3} + 8 = 12 - \frac{4}{3} = \frac{36}{3} - \frac{4}{3} = \frac{32}{3} \approx 10.67.$$ Calculate at the lower limit $t=0$: $$\frac{0^4}{4} - \frac{0^3}{6} + 4 \cdot 0 = 0.$$ 5. **Find the total volume:** $$\int_0^2 v(t) dt = \frac{32}{3} - 0 = \frac{32}{3} \approx 10.67$$ gallons. 6. **Round to the nearest gallon:** $$10.67 \approx 11$$ gallons. 7. **Compare with the options:** The closest option is d. 10 2/3 gallons, which is exactly $\frac{32}{3}$ gallons. **Final answer:** d. 10 2/3 gallons.