assume the rate at which the world's oil is being consumed is
r = 27 e^0.06 t.
Here, just as in your book, r is in billions of barrels per year, t is in years, and t = 0 is January 1, 1990.
(a) Fill in the blanks below to express the definite integral which measures the total quantity of oil used between the start of 1990 and the start of 1996:
∫ab dt,
where a = and b = .
(b) Use the Fundamental Theorem of Calculus to evaluate the integral.
Answer: billion barrels.
(a) To express the definite integral which measures the total quantity of oil used between the start of 1990 and the start of 1996, we need to determine the correct values for a and b.
First, let's establish the time values for the start of 1990 and the start of 1996:
- The start of 1990 is t = 0 years (given in the problem statement).
- The start of 1996 can be calculated by finding the value of t when the year is 1996. Since t = 0 corresponds to January 1, 1990, we need to calculate how many years have passed from January 1, 1990, to January 1, 1996. There are 6 years between these two dates.
Therefore, a = 0 and b = 6.
Now, we can set up the definite integral:
∫ab 27e^0.06t dt.
(b) To evaluate the integral using the Fundamental Theorem of Calculus, we can take the antiderivative of the integrand and evaluate it at the upper and lower limits of integration.
Let's start by finding the antiderivative of 27e^(0.06t):
∫ 27e^0.06t dt = (27/0.06) e^0.06t + C,
where C is the constant of integration.
Next, we evaluate this antiderivative at the upper and lower limits:
[(27/0.06) e^(0.06t)]|a to b.
Plugging in the values for a and b, we get:
[(27/0.06) e^(0.06(6))] - [(27/0.06) e^(0.06(0))].
Calculating further:
[(27/0.06) e^(0.36)] - [(27/0.06) e^(0)].
Simplifying:
[(27/0.06) e^(0.36)] - (27/0.06).
Now, we can calculate this expression to find the answer in billion barrels.