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Chapter 5: Problem 67

GENERAL: Average Value The population of a city is expected to be \(P(x)=x\left(x^{2}+36\right)^{-1 / 2}\) million people after \(x\) years. Find the average population between year \(x=0\) and year \(x=8\)

Short Answer

Expert verified
The average population is 0.5 million people.

Step by step solution

01

Understand the Formula for Average Value

The average value of a continuous function \(f(x)\) over the interval \([a, b]\) is given by \( \frac{1}{b-a} \int_{a}^{b} f(x)\, dx \). For our problem, the function \(P(x)\) represents the population in millions.
02

Set Up the Integral

To find the average population, set up the integral of \(P(x) = x(x^2 + 36)^{-1/2}\) from \(x = 0\) to \(x = 8\). We need to compute: \( \int_{0}^{8} x (x^{2}+36)^{-1 / 2}\, dx \).
03

Simplify the Integral

This integral involves a substitution. Let \(u = x^2 + 36\). Then, \(du = 2x\, dx\), or \(x\, dx = \frac{1}{2}\, du\). Change the limits of integration from \(x=0\) to \(u=36\), and \(x=8\) to \(u=100\).
04

Apply Substitution and Evaluate the Integral

Replace the integral with new limits: \( \int_{36}^{100} \frac{1}{2} u^{-1/2}\, du \). This becomes \( \frac{1}{2} \cdot 2 \left[u^{1/2}\right]_{36}^{100} \), which simplifies to evaluating \([\sqrt{100} - \sqrt{36}]\).
05

Solve the Integral

Calculate \([\sqrt{100} - \sqrt{36}]\), which is \([10 - 6] = 4\).
06

Calculate the Average Value

Compute the average value by dividing the integral result by the interval length: \( \frac{1}{8-0} \times 4 = \frac{4}{8} = 0.5\).
07

Conclude

The average population between year \(x=0\) and year \(x=8\) is 0.5 million people.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculation
Integral calculation is a fundamental concept in calculus, often used to find the total accumulation of a quantity over an interval. In our problem, we are looking at finding the average value of the population over a period of years.
  • The formula used is: \( \frac{1}{b-a} \int_{a}^{b} f(x)\, dx \), where \( f(x) \) is our function.
  • This involves evaluating the integral of the function \( P(x) = x(x^2 + 36)^{-1/2} \) from \( x = 0 \) to \( x = 8 \).
The integral represents the total population added up over the interval \([0, 8]\), and the division by the length of the interval gives us the average.
Substitution Method
The substitution method is a technique used to simplify complex integrals by substituting a part of the integral with a new variable.In this problem, since the integral involves a complex expression, substitution helps in simplifying the calculation:
  • Set \( u = x^2 + 36 \). This changes the expression inside the integral to a simpler form.
  • Calculate the derivative: \( du = 2x\, dx \), thus \( x\, dx = \frac{1}{2}\, du \).
This substitution reduces the original integral to one that is easier to solve: \( \int \frac{1}{2} u^{-1/2}\, du \). Adjusting the limits as \( x \) changes helps align the bounds with our new variable \( u \).
Continuous Functions
Continuous functions are an essential concept in calculus, particularly when dealing with integrals. A function is continuous at a point if it does not have breaks, jumps, or holes at that point. Over an interval, a function is continuous if it is continuous at every point in the interval.For the average value calculation:
  • The function \( P(x) = x(x^2 + 36)^{-1/2} \) is continuous over the interval \( x = 0 \) to \( x = 8 \).
  • Continuity ensures the integral accurately represents the population change over this time span, without any discontinuities disrupting the calculation.
This property is key because only continuous functions over an interval are subject to such average value calculations.
Interval Analysis
Interval analysis involves evaluating a function over a specified range, which is critical for tasks like finding the average value of a function.In our exercise:
  • The interval is \([0, 8]\), representing the first eight years.
  • By integrating over this interval, we can find total population changes, subsequently calculating the average.
The limits of integration change when using substitution, from \( x \) values to \( u \) values:
  • When \( x = 0 \), \( u = 36 \).
  • When \( x = 8 \), \( u = 100 \).
This transformation maintains accuracy in the calculation and ensures we correctly evaluate the integral over the intended period.

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Most popular questions from this chapter

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found in this way (as well as by the methods of Section 6.1). $$ \int x(x+4)^{7} d x $$

Find the area bounded by the given curves. $$ y=3 x^{2}-x-1 \text { and } y=5 x+8 $$

Suppose that for a demand function \(d(x)\) we have \(d(0)=1000 .\) Describe in everyday language what this means about the number 1000 .

ECONOMICS: Balance of Trade A country's annual imports are \(I(t)=30 e^{0.2 t}\) and its exports are \(E(t)=25 e^{0.1 t}\), both in billions of dollars, where \(t\) is measured in years and \(t=0\) corresponds to the beginning of 2000 . Find the country's accumulated trade deficit (imports minus exports) for the 10 years beginning with 2000 .

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found in this way (as well as by the methods of Section 6.1). $$ \int \frac{x-1}{x-2} d x $$
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