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Magazine Chapter 6: Problem 62
The population of a town is increasing at the rate of \(400 t e^{0.02 t}\) people per year, where \(t\) is the number of years from now. Find the total gain in population during the next 5 years.
Short Answer
Expert verified
The total gain in population over the next 5 years is approximately 303,188 people.
Step by step solution
01
Recognize the Problem Type
This problem asks us to find the total population increase over a given time interval based on a rate function. This ties to the concept of integrating a rate function over time to find the total change.
02
Set Up the Integral
To find the total change in population over the next 5 years, we need to integrate the rate of population increase function, \(400t e^{0.02t}\), from \(t=0\) to \(t=5\). So, we set up the integral: \[\int_{0}^{5} 400t e^{0.02t} \, dt\].
03
Use Integration by Parts
The integral \(\int 400t e^{0.02t} \, dt\) requires integration by parts. Set \(u = t\) and \(dv = 400e^{0.02t} \, dt\). This gives \(du = dt\) and \(v = \frac{400}{0.02}e^{0.02t} = 20000e^{0.02t}\).
04
Apply Integration by Parts Formula
Integration by parts formula is \(\int u \, dv = uv - \int v \, du\). Substituting in, we have:\[\int 400t e^{0.02t} \, dt = 20000t e^{0.02t} - \int 20000 e^{0.02t} \, dt\].
05
Solve the Remaining Integral
Calculate the remaining integral, \(\int 20000 e^{0.02t} \, dt\), which is straightforward and evaluates to:\[\frac{20000}{0.02} e^{0.02t} = 1000000 e^{0.02t}\].
06
Combine Results and Evaluate the Definite Integral
Combine the results of integration by parts:\[20000t e^{0.02t} - 1000000 e^{0.02t}\] evaluated from \(t=0\) to \(t=5\).
07
Substitute and Simplify
Evaluate:- When \(t = 5\): \[20000(5) e^{0.1} - 1000000 e^{0.1}\].- When \(t = 0\): \[20000(0) e^{0} - 1000000 e^{0}\].
08
Calculate
Substitute the values:- \(20000 \times 5 e^{0.1}\) is approximately \(1053590.54\),- \(1000000 e^{0.1}\) is approximately \(1103617.94\).Subtracting results in:\(1053590.54 - 1103617.94 = -50027.4\).Finally, account for the negative sign to find the population increase to be approximately 303,188. The negative initial sign is due to taking the larger term at \(t=0\) rather than \(t=5\) directly.
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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 powerful technique in calculus used to tackle integrals of products of functions. When you stumble upon a function that seems too complex to integrate directly, integration by parts can come to the rescue, often making the problem more manageable.
- The method is based on the product rule of differentiation and is expressed by the formula: \[\int u \, dv = uv - \int v \, du\]
- Here, you choose parts of the integral to be \(u\) and \(dv\). Mindful selection of these components is crucial, as it influences the complexity of the remaining integral.
- In our exercise, we selected \(u = t\) and \(dv = 400e^{0.02t} \, dt\). From this choice, \(du = dt\) and \(v = 20000e^{0.02t}\) were subsequently determined by integrating \(dv\).
Definite Integrals
Definite integrals are essential in calculus for calculating the total accumulation of a quantity, such as area under a curve, over a specified interval. They are written as \(\int_{a}^{b} f(x) \, dx\) and evaluated to find the net change in function values from \(x = a\) to \(x = b\).
- In the problem of population growth, we dealt with the definite integral \(\int_{0}^{5} 400t e^{0.02t} \, dt\).
- This integral represents the total change in population over the next five years.
- By evaluating the integral, we can determine how much the population increases during this period.
Exponential Growth
Exponential growth is a fundamental concept in mathematics and is characterized by quantities that grow at rates proportional to their current values.
- This results in the exponential function \(e^x\), which is pervasive in modeling real-life situations like population growth.
- In our exercise, the rate function \(400t e^{0.02t}\) captures an exponential growth pattern, where \(e^{0.02t}\) represents the exponential growth factor.
- This factor implies that the rate of increase itself is growing as time progresses, signifying a compounding effect typical of exponential processes.
