A new college textbook edition typically generates most of its sales in the year of its publication. Sales drop off in subsequent years as a result of competition from the used book market. Suppose that the annual sales of a particular textbook may be modeled by
S(t) = 30,000 te−1.5t, textbooks, where t is the number of years since the edition was published.
Use integration by parts to determine how many textbooks will be sold in the first three years of the edition?
To determine the number of textbooks sold in the first three years, we need to find the definite integral of the sales function S(t) over the interval [0, 3]. This can be done using the integration by parts technique.
The formula for integration by parts is:
∫u dv = uv - ∫v du
In this case, we can consider S(t) = 30,000 te^(-1.5t) as the product of two functions:
u = t
dv = 30,000e^(-1.5t) dt
Now, let's find du and v.
Taking the derivative of u with respect to t gives us:
du = dt
To find v, we need to integrate dv. The integral of dv can be evaluated by applying the two steps of integration by substitution. Let's define u = -1.5t, then du = -1.5dt, and we can rewrite the integral as:
v = ∫30,000e^(-1.5t) dt = ∫-1.5e^u du = -20,000e^u
Now, we can use the integration by parts formula to find the integral of S(t):
∫S(t) dt = uv - ∫v du
Plugging in the values:
∫(30,000te^(-1.5t)) dt = t(-20,000e^(-1.5t)) - ∫(-20,000e^(-1.5t)) dt
Simplifying, we have:
∫(30,000te^(-1.5t)) dt = -20,000te^(-1.5t) + 20,000∫e^(-1.5t) dt
Now, let's evaluate the remaining integral:
∫e^(-1.5t) dt
To solve this, we can apply integration by substitution. Let u = -1.5t, then du = -1.5dt. The integral becomes:
∫e^u (-1.5 du) = -1.5 ∫e^u du = -1.5e^u = -1.5e^(-1.5t)
Now, we substitute this back into the main expression:
∫(30,000te^(-1.5t)) dt = -20,000te^(-1.5t) + 20,000(-1.5e^(-1.5t)) + C
Where C is the constant of integration.
To find the number of textbooks sold in the first three years, we evaluate the definite integral from 0 to 3:
∫[0,3] (30,000te^(-1.5t)) dt = [-20,000te^(-1.5t) + 20,000(-1.5e^(-1.5t))] from 0 to 3
Plugging in the values:
= [-20,000(3)e^(-1.5(3)) + 20,000(-1.5e^(-1.5(3)))] - [-20,000(0)e^(-1.5(0)) + 20,000(-1.5e^(-1.5(0)))]
Simplifying, we have:
= [-60,000e^(-4.5) + 30,000e^(-4.5)] - [0 + 30,000]
= -30,000e^(-4.5) + 30,000
The total number of textbooks sold in the first three years is approximately -30,000e^(-4.5) + 30,000.