Problem 95 - Applied Calculus

Question

A dealer predicts that new cars will sell at the rate of \(8 x e^{-0.1 x}\) sales per week in week \(x\). Find the total sales in the first half year (week 0 to week 26 ).

Step-by-Step Solution

Answer

Total sales from week 0 to 26 are approximately -787 cars.

Step 1: Understand the problem

We need to find the total sales of cars from week 0 to week 26, given the sales function \(8x e^{-0.1x}\). This means we need to find the definite integral of this function within the given interval to get the total sales.

Step 2: Set up the integral

The problem asks us to find the total sales over a period, which requires integration of the sales function \(8x e^{-0.1x}\). We set up the definite integral as \(\int_0^{26} 8x e^{-0.1x} \, dx\).

Step 3: Integrate the function

We use integration by parts, a technique where we choose parts of the integrand to apply the formula:\[\int u \, dv = uv - \int v \, du\]Let \(u = 8x\) (then \(du = 8 \, dx\)) and \(dv = e^{-0.1x} \, dx\) (then \(v = -10 e^{-0.1x}\).Substitute in:\[8x (-10)e^{-0.1x} - \int -10 e^{-0.1x} (8) \, dx\]

Step 4: Solve the integral

Simplify and solve the expression:\[-80x e^{-0.1x} - \int -80 e^{-0.1x} \, dx\]Solve the remaining integral:\(-80(-10)e^{-0.1x} = 800 e^{-0.1x}\)Thus, the full integral is \(-80x e^{-0.1x} + 800 e^{-0.1x}\).

Step 5: Evaluate the definite integral

Substitute the limits of integration (0 to 26) into the result:\([-80(26)e^{-0.1 \cdot 26} + 800 e^{-0.1 \cdot 26}] - [-80(0)e^{-0.1 \cdot 0} + 800 e^{-0.1 \cdot 0}]\)Calculate:\(-2080 e^{-2.6} + 800 e^{-2.6} - (0 + 800)\) to find the total sales.

Step 6: Calculate the final answer

Compute the values using the exponential function's values (approximations if necessary). Calculate\[(-2080 + 800) e^{-2.6} - 800 \]Use a calculator for precise measures:\(F(26) - F(0) = (-2080 + 800) (0.0742736) - 800\)Conclude with the total number of sales.

Key Concepts

Definite Integral

A definite integral helps us calculate the total accumulation of a quantity, for instance, sales over a certain period. In this exercise, we need to find the total sales from week 0 to week 26. Hence, we use the definite integral to sum up the sales rates of the car function provided:

• The function given is the sales rate over time: \(8x e^{-0.1x}\).
• To find the total sales, integrate this function from 0 to 26.

This process involves setting up the integral \( \int_0^{26} 8x e^{-0.1x} \, dx \). Calculating a definite integral gives us a single number, representing the total sales over the specified period. This is different from an indefinite integral, which provides a family of functions without such specific limits.

Integration by Parts

Integration by parts is a method used to solve integrals where the standard procedures don't easily apply. In this exercise, since the integrand \((8x e^{-0.1x})\) involves both a polynomial \((8x)\) and an exponential function \((e^{-0.1x})\), we choose integration by parts:

• First, we pick functions for \(u\) and \(dv\):
  • Let \(u = 8x\), making \(du = 8 \, dx\).
  • Let \(dv = e^{-0.1x} \, dx\), hence \(v = -10 e^{-0.1x}\).
• Apply the formula: \(\int u \, dv = uv - \int v \, du\).

By substituting our choices, we get \(-80x e^{-0.1x} + \int 80 e^{-0.1x} \, dx\), which leads us to a simpler integral that we can solve.

Exponential Functions

Exponential functions are crucial in this problem since they determine how sales decrease over time. The exponential function \(e^{-0.1x}\) introduces the concept of a decay or reduction in sales rate as weeks pass:

• The base, \(e\), is a constant approximately equal to 2.718.
• The exponent \(-0.1x\) gives us the rate of decay; higher \(x\) values reduce the function's value faster.

These functions are prevalent in modeling real-life situations where quantities increase or decrease quickly – like radioactive decay or, in this case, decreasing sales rate as demand diminishes over time. Exponential functions simplify capturing growth or decay trends, helping businesses and industries predict future behavior effectively. Calculations involving these functions often require a calculator or computational tool, particularly when dealing with irrational numbers like \(e\).