When a factory operates from 6:00 a.m.6:00 a.m. to 6:00 p.m.6:00 p.m., its total fuel consumption varies according to the formula f(t)=0.9t2−0.3t√+10𝑓(𝑡)=0.9𝑡2−0.3𝑡+10, where t is the time in hours after 6:00 a.m.6:00 a.m. and f(t)𝑓(𝑡) is the number of barrels of fuel oil. What is the average rate of consumption from 6:00 a.m.6:00 a.m. to 4:00 p.m.4:00 p.m.? Round your answer to two decimal places and, if necessary, round any intermediate calculations to six decimal places.

To find the average rate of fuel consumption from 6:00 a.m. to 4:00 p.m., we need to follow these steps:

1. Identify the time interval:

   • The factory operates from 6:00 a.m. to 4:00 p.m.
   • This is a total of 10 hours (from 6:00 a.m. to 4:00 p.m.).

2. Set up the integral for total fuel consumption:

   • The formula for fuel consumption is $f(t) = 0.9t^2 - 0.3t + 10$.
   • We need to integrate this function from $t = 0$ to $t = 10$.

3. Calculate the definite integral: $\int_{0}^{10} (0.9t^2 - 0.3t + 10) \, dt$

4. Evaluate the integral:

   • Integrate each term separately: $\int 0.9t^2 \, dt = 0.9 \cdot \frac{t^3}{3} = 0.3t^3$ $\int -0.3t \, dt = -0.3 \cdot \frac{t^2}{2} = -0.15t^2$ $\int 10 \, dt = 10t$
   • Combine the results: $\int_{0}^{10} (0.9t^2 - 0.3t + 10) \, dt = \left[ 0.3t^3 - 0.15t^2 + 10t \right]_{0}^{10}$

5. Evaluate the expression at the bounds:

   • At $t = 10$: $0.3(10)^3 - 0.15(10)^2 + 10(10) = 0.3(1000) - 0.15(100) + 100 = 300 - 15 + 100 = 385$
   • At $t = 0$: $0.3(0)^3 - 0.15(0)^2 + 10(0) = 0$
   • The total fuel consumption over the interval is: $385 - 0 = 385 \text{ barrels}$

6. Calculate the average rate of consumption:

   • The average rate of consumption is the total fuel consumption divided by the time interval: $\text{Average rate} = \frac{385 \text{ barrels}}{10 \text{ hours}} = 38.5 \text{ barrels per hour}$

Therefore, the average rate of fuel consumption from 6:00 a.m. to 4:00 p.m. is $38.50$ barrels per hour.