A factory produces bicycles at a rate of 100+0.4t2−0.7t bicycles per week (t in weeks). How many bicycles were produced from day 15 to 28?
Hint: the lower limit of integration should be the integer number of weeks preceding day 15, i.e. a=2, and the final answer should be an integer number of bicycles.
To find the number of bicycles produced from day 15 to day 28, you need to calculate the definite integral of the production rate function over that time period.
First, let's find the number of full weeks (integer number of weeks) that are included in the time period from day 15 to day 28. Since each week has 7 days, we have:
Number of full weeks = (28 - 15) / 7 = 13 / 7 = 1 week
So, there is at least 1 full week from day 15 to day 28.
Next, we need to determine the starting and ending points for the integral in terms of weeks. We know that day 15 is in the second week, so we'll start the integral at t = 2. Since we want to include day 28, we'll end the integral at t = 4, which represents the end of the fourth week.
Now, let's set up the definite integral to find the total number of bicycles produced from day 15 to day 28:
∫[2 to 4] (100 + 0.4t^2 - 0.7t) dt
To find the integral, you need to evaluate the antiderivative of the integrand and then subtract the value at the lower limit from the value at the upper limit.
Let's evaluate the integral step by step:
∫[2 to 4] (100 + 0.4t^2 - 0.7t) dt
= [100t + (0.4/3)t^3 - (0.7/2)t^2] from 2 to 4
Now, let's substitute the values of t to get:
= [100(4) + (0.4/3)(4^3) - (0.7/2)(4^2)] - [100(2) + (0.4/3)(2^3) - (0.7/2)(2^2)]
= [400 + (0.4/3)(64) - (0.7/2)(16)] - [200 + (0.4/3)(8) - (0.7/2)(4)]
Simplifying further, we get:
= 400 + (8/3)(4) - (7/2)(4) - 200 - (8/3)(2) + (7/2)(2)
= 400 + 32/3 - 28 - 200 - 16/3 + 7
= 400 - 196 + 32/3 - 16/3 + 7
= 11 + 16/3 - 16/3
= 11
Therefore, the number of bicycles produced from day 15 to day 28 is 11.