Problem 58 - Applied Calculus

Question

BUSINESS: Warranties When a company sells a product with a lifetime guarantee, the number of items returned for repair under the guarantee usually decreases with time. A company estimates that the annual rate of returns after \(t\) years will be \(800 e^{-0.2 t}\) Find the total number of returns by summing (integrating) this rate from 0 to \(\infty\)

Step-by-Step Solution

Answer

The total number of returns is 4000.

Step 1: Identify the Function

The function representing the rate of returns is given as \( R(t) = 800 e^{-0.2 t}\). Our goal is to find the total number of returns, which requires integrating this function over time from 0 to \(+\infty\).

Step 2: Set Up the Integral

To find the total number of returns, set up the integral of \( R(t) \) from 0 to \(+\infty\). This is written as: \[\int_{0}^{\infty} 800 e^{-0.2 t} \, dt\]

Step 3: Solve the Integral

To integrate the function \( 800 e^{-0.2 t} \), we use the substitution method. Let \( u = -0.2t \), then \( du = -0.2 \, dt \), or equivalently, \( dt = \frac{du}{-0.2} \). Substitute these into the integral:\[800 \int e^{u} \cdot \left(\frac{du}{-0.2}\right)\]Simplify to:\[-4000 \int e^{u} \, du\]The integral of \( e^{u} \) is \( e^{u} \), thus:\[-4000 e^{u} + C\]Returning to the variable \( t \), substitute back to get \(-4000 e^{-0.2t} + C\).

Step 4: Evaluate the Definite Integral

To find the total number of returns, evaluate the definite integral:\[-4000 e^{-0.2t} \Bigg|_{0}^{\infty}\]Calculate at the bounds:- At \( t = 0 \), \(-4000 e^{0} = -4000\)- At \( t = \infty \), \(-4000 e^{- ext{very large number}} \approx 0\)Thus, the evaluation is:\(0 - (-4000) = 4000\).

Step 5: Interpret the Result

The total number of items returned over an infinite time period is \( 4000 \). This means that over time, 4000 items will be returned, as expected by the company's guarantee.

Key Concepts

Integration Technique

Integration techniques are mathematical methods used to find the antiderivatives of functions. When we integrate a function, we're essentially summing up an infinite number of tiny pieces to find the total change. In this exercise, we're using integration to determine the total number of product returns over an infinite time period.

For the function given by the company, we start with the annual return rate described by an exponential function: \( R(t) = 800 e^{-0.2 t} \). This is a perfect candidate for integration because it describes a continuous rate over time. The primary technique used here is the **substitution method**, which simplifies integration of compositions of functions by turning them into easier forms.

• Begin by identifying parts of the function that resemble basic integral forms, such as \( e^{cx} \), commonly integrated via substitution.
• Set a substitution variable, say \( u \) for the exponent in \( e \) to simplify the integral into a standard form.
• Work through the substitution to rewrite the integral in terms of \( u \) and solve it.

This process transforms the original complex problem into easier subtasks involving simpler functions. Using substitution correctly is key to solving integrals that involve exponential decay or growth.

Definite Integrals

Definite integrals allow us to calculate the total accumulation of a quantity, such as product returns, over a specific interval—here from \( t = 0 \) to \( t = \infty \). When performing a definite integral, not only do you integrate the function, but you also evaluate it at the upper and lower bounds of the interval.

The notation \( \int_{a}^{b} f(x) \, dx \) signifies integration from \( a \) to \( b \). In our problem, \( a = 0 \) and \( b = \infty \). This type of evaluation is crucial in understanding real-world scenarios like calculating infinite accumulations.

Steps include:

• Setting up the integral with the given bounds, \([0, \infty)\).
• Solving the indefinite integral using the chosen technique, in this case, substitution.
• Evaluating the function at the upper bound (often simplifying as \( x \rightarrow \infty \) to find limiting behavior) and at the lower bound, then subtracting these values.

In this exercise, because \( \lim_{t \rightarrow \infty} e^{-0.2t} = 0 \), the evaluation simplifies to a finite result despite the infinite upper limit.

Exponential Decay

Exponential decay describes a process where quantities decrease at a rate proportional to their current value. This is often used to model real-world phenomena such as radioactive decay or depreciation in economics.

In the context of this exercise, the formula \( 800 e^{-0.2 t} \) reflects an **exponential decay**, where the number of returns decreases over time but never completely disappears. Understanding this function involves recognizing:

• **Decay Rate**: Represented by the constant \(-0.2\), demonstrating that each year the return count falls by about \( 20\% \) of its rate at the start.
• **Initial Value**: The coefficient \( 800 \), which is the estimated return rate immediately after purchase.

This decay captures how warranties lead to fewer returns as products age, aligning with the pattern businesses typically observe. The exponential nature ensures that while returns decrease over time, they never quite reach zero, which is consistent with reality, given that some products will always fail, albeit at diminishing frequency.