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Magazine Chapter 6: Problem 61
A company begins advertising a new product and finds that after \(t\) weeks the product is gaining customer recognition at the rate of \(t^{2} \ln t\) thousand customers per week (for \(t \geq 1\) ). Find the total gain in recognition from the end of week 1 to the end of week 6 .
Short Answer
Expert verified
The total gain in recognition is approximately 105.92 thousand customers.
Step by step solution
01
Define the Problem
We are given a rate of customer recognition, \(f(t) = t^2 \ln t\), in thousands of customers per week. We need to find the total gain in recognition from the end of week 1 to the end of week 6.
02
Identify the Integral Limits
The problem asks for the total gain in recognition from the end of week 1 to the end of week 6. Thus, the limits for the integral will be from \(t = 1\) to \(t = 6\).
03
Set Up the Integral
The total gain in recognition is the integral of the rate function from \(t = 1\) to \(t = 6\): \[\int_{1}^{6} t^2 \ln t \, dt\]
04
Perform Integration by Parts
To solve the integral \(\int t^2 \ln t \, dt\), use integration by parts. Let \(u = \ln t\) and \(dv = t^2 \, dt\). Then \(du = \frac{1}{t} \, dt\) and \(v = \frac{t^3}{3}\). Apply the formula \(\int u \, dv = uv - \int v \, du\).
05
Calculate Each Part
First calculate \(uv\):\[uv = \ln t \cdot \frac{t^3}{3} = \frac{t^3 \ln t}{3}\]Next, calculate \(\int v \, du\):\[\int \frac{t^3}{3} \cdot \frac{1}{t} \, dt = \int \frac{t^2}{3} \, dt = \frac{1}{3} \cdot \frac{t^3}{3} = \frac{t^3}{9}\]
06
Substitute Back into Integration by Parts Formula
Substitute back into the integration by parts formula:\[\int t^2 \ln t \, dt = \frac{t^3 \ln t}{3} - \frac{t^3}{9} = \frac{t^3}{3}(\ln t - \frac{1}{3})\]
07
Evaluate the Definite Integral
Evaluate the expression \(\frac{t^3}{3}(\ln t - \frac{1}{3})\) from \(t = 1\) to \(t = 6\):\[\left[ \frac{t^3}{3}(\ln t - \frac{1}{3}) \right]_1^6 = \frac{6^3}{3}(\ln 6 - \frac{1}{3}) - \frac{1^3}{3}(\ln 1 - \frac{1}{3})\]
08
Simplify the Result
Calculating each part, we have:1. When \(t = 6\): \[\frac{6^3}{3}(\ln 6 - \frac{1}{3}) = 72(\ln 6 - \frac{1}{3})\]2. When \(t = 1\): \[\frac{1}{3}(0 - \frac{1}{3}) = -\frac{1}{9}\]The final result is:\[72(\ln 6 - \frac{1}{3}) + \frac{1}{9}\]
09
Final Calculation
To find the numerical value, approximate: \[72 \left( \ln 6 - \frac{1}{3} \right) + \frac{1}{9} \approx 72 (1.79176 - 0.3333) + 0.1111 = 72 \times 1.45846 + 0.1111 = 105.80832 + 0.1111 = 105.91942\]
10
Interpret the Result
Therefore, the total gain in recognition from the end of week 1 to the end of week 6 is approximately 105.92 thousand customers.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by Parts is a method used to solve more complex integrals by breaking them down into simpler parts. This technique is particularly useful when the integral is a product of two functions. The formula for integration by parts is given as:\[\int u \, dv = uv - \int v \, du,\]where \( u \) is usually chosen as a function that simplifies upon differentiation, and \( dv \) is chosen as a function that is easy to integrate.
For instance, when integrating a function like \( t^2 \ln t \), selecting \( u = \ln t \) because its derivative simplifies to \( \frac{1}{t} \), and \( dv = t^2 \ dt \) because it can be easily integrated to \( \frac{t^3}{3} \), makes the problem manageable.
For instance, when integrating a function like \( t^2 \ln t \), selecting \( u = \ln t \) because its derivative simplifies to \( \frac{1}{t} \), and \( dv = t^2 \ dt \) because it can be easily integrated to \( \frac{t^3}{3} \), makes the problem manageable.
- Choose \( u \) and \( dv \) effectively to make \( du \) and \( v \) as straightforward as possible.
- Compute \( uv \) and \( \int v \, du \), substituting back into the formula for a solution.
Definite Integral
A Definite Integral provides the total accumulation of a quantity over a specific interval. For the function \( f(t) = t^2 \ln t \), integrating from \( t = 1 \) to \( t = 6 \) gives the total recognition gained over that period.
The definite integral \[\int_{1}^{6} t^2 \ln t \, dt\]calculates the area under the curve of the rate function from the start to the end of this time frame.
The definite integral \[\int_{1}^{6} t^2 \ln t \, dt\]calculates the area under the curve of the rate function from the start to the end of this time frame.
- The limits of integration are crucial鈥攖hey specify the exact region of interest.
- It helps in determining the net effect of a rate over a period, such as customer gain over weeks.
Customer Recognition Rate
Customer Recognition Rate is a measure of how well a product is gaining visibility and acceptance in the market, often changing over time. In this context, the rate is expressed as \( t^2 \ln t \) thousand customers per week, indicating it grows as time progresses.
Understanding this growth is crucial for businesses:
Understanding this growth is crucial for businesses:
- It assists in assessing the effectiveness of marketing campaigns over time.
- Allows businesses to strategically plan further initiatives to maintain or enhance customer growth.
- Gives insight into customer behavior and how quickly they adapt to new products.
