Problem 63 - Applied Calculus

Question

BUSINESS: Total Sales A company's sales rate is \(x^{2} e^{-x}\) million sales per month after \(x\) months. Find a formula for the total sales in the first \(x\) months. [Hint: Integrate the sales rate to find the total sales and determine the constant \(C\) so that total sales are zero at time \(x=0 .]\)

Step-by-Step Solution

Answer

The formula for total sales in the first \(x\) months is \(T(x) = -x^2 e^{-x} + 2x e^{-x} - 2e^{-x} + 2\).

Step 1: Understand the Problem

We need to find a formula for the total sales in terms of months, which means we need to integrate the given sales rate function to determine the total sales over the first \(x\) months.

Step 2: Set Up the Integral

The sales rate function is given as \(f(x) = x^2 e^{-x}\). To find the total sales over the first \(x\) months, we set up the integral of this function from 0 to \(x\): \[ T(x) = \int_{0}^{x} t^2 e^{-t} \, dt \]

Step 3: Integrate the Sales Rate

To solve the integral \(\int t^2 e^{-t} \, dt\), we'll use integration by parts method twice. Let \(u = t^2\) and \(dv = e^{-t} \, dt\). Then \(du = 2t \, dt\) and \(v = -e^{-t}\). Apply integration by parts: \[ \int t^2 e^{-t} \, dt = -t^2 e^{-t} - \int -2t e^{-t} \, dt.\] Let \(u = 2t\) and \(dv = e^{-t} \, dt\) again, then \(du = 2 \, dt\) and \(v = -e^{-t}\) for the second integration by parts:\[ - \int -2t e^{-t} \, dt = 2 \int t e^{-t} \, dt = 2(-t e^{-t} - \int -e^{-t} \, dt),\]\[ = 2(-t e^{-t} + e^{-t}) = -2t e^{-t} + 2e^{-t}.\]Thus, the integration becomes:\[ \int t^2 e^{-t} \, dt = -t^2 e^{-t} + 2t e^{-t} - 2e^{-t}. \]

Step 4: Evaluate the Definite Integral

Evaluate the definite integral from 0 to \(x\):\[ \left[-t^2 e^{-t} + 2t e^{-t} - 2e^{-t}\right]_{0}^{x} = \left[-x^2 e^{-x} + 2x e^{-x} - 2e^{-x}\right] - \left[0 - 0 + 2\right] = -x^2 e^{-x} + 2x e^{-x} - 2e^{-x} + 2.\]

Step 5: Determine the Constant C

Since the problem states that total sales are zero at time \(x=0\), set \(T(0)\) to zero and find the constant \(C\):\[ T(0) = -0^2 e^{-0} + 2(0)e^{-0} - 2e^{-0} + 2 = 0 \]This simplifies to the correct equation and thus begins at zero as expected.

Key Concepts

Integral Calculus

Integral Calculus is all about finding the total amount or accumulation of something. When given a rate of change, integrating that rate gives us the total quantity. In our problem, the rate of sales is defined by the function \(x^2 e^{-x}\), which represents the sales rate per month.

By integrating the function of sales rate over a certain period, we can calculate the cumulative sales over time. This is key in businesses to understand overall performance, as it gives the entire picture of sales up to a specific time.

The fundamental idea here is that the integral sums up all of the small pieces (in this case, monthly sales) across a given time to give the total sales figure.

Integration by Parts

Integration by Parts is a powerful technique used to solve integrals, especially when the integrand is a product of two different kinds of functions. This method is based on the formula: \[ \int u \, dv = uv - \int v \, du \]For our exercise, the function \(t^2 e^{-t}\) was integrated using this technique. Here's how it works in simpler terms:

• First, choose parts of the integrand to be \(u\) and \(dv\).
• Then, calculate \(du\) by differentiating \(u\) and \(v\) by integrating \(dv\).

In our case, this process was applied twice, since it led to another integration challenge that could again be addressed by integration by parts. Choosing \( t^2 \) as \( u \), and \( e^{-t} \, dt \) as \( dv \), we found \( du = 2t \, dt \) and \( v = -e^{-t} \). This allowed us to successively break down and solve our integral.

Definite Integrals

Definite Integrals provide the accumulated value of a function across a particular range or interval. In our business sales exercise, we set up a definite integral with limits from 0 to \(x\). This determines total sales from the beginning to any point \(x\) months forward.

The notation of a definite integral, \(\int_{a}^{b} f(x) \, dx\), signifies summing the function \(f(x)\) from \(a\) to \(b\). After integrating, we plug in the bounds into the integrated function, and subtract to find the accumulated quantity between those bounds.

Evaluating the given sales function, we applied these principles and found that the result respected the condition set by the problem: total sales should be zero at \(x=0\). This is crucial for setting the constant \(C\) correctly when calculating real-world problems.