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 integrate the given function from 0 to 3 years.

The formula for integration by parts is:
∫u dv = uv - ∫v du

Let's denote:
u = t
dv = 30,000e^(-1.5t) dt

To find du and v, we take the derivatives and antiderivatives, respectively:
du = dt
v = ∫30,000e^(-1.5t) dt

To find v, we can use the substitution method:
Let u = -1.5t, then dv = 30,000e^u du

∫30,000e^(-1.5t) dt = v = ∫e^u du

Integrating e^u, we have:
v = -20,000e^(-1.5t)

Substituting the values back into the integration by parts formula, we have:
∫(t)(30,000e^(-1.5t)) dt = (t)(-20,000e^(-1.5t)) - ∫-20,000e^(-1.5t) dt

Now, we can integrate the second term on the right-hand side:
∫-20,000e^(-1.5t) dt = -20,000 ∫e^(-1.5t) dt

To find this integral, we can once again use substitution:
Let u = -1.5t, then du = -1.5 dt

So, the integral becomes:
-20,000 ∫e^u(-1.5) du = 30,000e^(-1.5t)

Now, we substitute the values back into the integration by parts formula:
(t)(-20,000e^(-1.5t)) - ∫-20,000e^(-1.5t) dt
= -20,000te^(-1.5t) + 30,000e^(-1.5t)

Evaluating this expression from t = 0 to t = 3 gives us the total number of textbooks sold in the first three years:
(-20,000(3)e^(-1.5(3)) + 30,000e^(-1.5(3))) - (-20,000(0)e^(-1.5(0)) + 30,000e^(-1.5(0)))

Simplifying this expression gives us the final answer.