Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

We can use a computer to approximate \(\int_{1}^{\infty} f(x) d x\) by taking \(b\) very large in \(\int_{1}^{b} f(x) d x\) provided we know that the first integral converges. Calculate \(\int_{1}^{100}\left(1 / x^{p}\right) d x\) for \(p=2,1.1,1.01\), 1 , and \(0.99 .\) Note that this gives no hint that the integral \(\int_{1}^{\infty}\left(1 / x^{p}\right) d x\) converges for \(p>1\) and diverges for \(p \leq 1\)

Step-by-Step Solution

Answer

The integrals are 0.99, 9, 90, 4.605, and 151.1 for p=2, 1.1, 1.01, 1, and 0.99.

Step 1: Understanding the Integral

We need to evaluate the integral \( \int_{1}^{100} \frac{1}{x^p} \, dx \) for different values of \( p \). This integral is part of the formula \( \int_{1}^{b} \frac{1}{x^p} \, dx \) where \( b \) is a large number used to approximate the integral from 1 to infinity. We are examining how the value of \( b = 100 \) affects the outcome for different values of \( p \).

Step 2: Evaluate the Integral for p=2

For \( p = 2 \), we need to calculate the integral \( \int_{1}^{100} \frac{1}{x^2} \, dx \). The antiderivative of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \). Therefore, the definite integral becomes:\[ \int_{1}^{100} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{100} = -\frac{1}{100} - (-1) = 1 - \frac{1}{100} = 0.99 \]

Step 3: Evaluate the Integral for p=1.1

For \( p = 1.1 \), calculate the integral \( \int_{1}^{100} \frac{1}{x^{1.1}} \, dx \). The antiderivative is \( \frac{x^{-0.1}}{-0.1} = -10 x^{-0.1} \). Thus:\[ \int_{1}^{100} \frac{1}{x^{1.1}} \, dx = \left[ -10 x^{-0.1} \right]_{1}^{100} = -10 (100^{-0.1}) + 10 (1^{-0.1}) = -10(0.1) + 10(1) = 10 - 1 = 9 \]

Step 4: Evaluate the Integral for p=1.01

For \( p = 1.01 \), compute \( \int_{1}^{100} \frac{1}{x^{1.01}} \, dx \). The antiderivative is \( \frac{x^{-0.01}}{-0.01} = -100 x^{-0.01} \). Therefore:\[ \int_{1}^{100} \frac{1}{x^{1.01}} \, dx = \left[ -100 x^{-0.01} \right]_{1}^{100} = -100 (100^{-0.01}) + 100(1^{-0.01}) = -100(0.1) + 100(1) = 100 - 10 = 90 \]

Step 5: Evaluate the Integral for p=1

For \( p = 1 \), calculate the integral \( \int_{1}^{100} \frac{1}{x} \, dx \), which is a logarithm function. So:\[ \int_{1}^{100} \frac{1}{x} \, dx = \ln(100) - \ln(1) = \ln(100) = 2 \ln(10) \approx 4.605 \]

Step 6: Evaluate the Integral for p=0.99

For \( p = 0.99 \), compute \( \int_{1}^{100} \frac{1}{x^{0.99}} \, dx \). The antiderivative is \( \frac{x^{0.01}}{0.01} = 100 x^{0.01} \). Thus:\[ \int_{1}^{100} \frac{1}{x^{0.99}} \, dx = \left[ 100 x^{0.01} \right]_{1}^{100} = 100 (100^{0.01}) - 100(1^{0.01}) = 100(2.511) - 100 = 151.1 \]

Key Concepts

Antiderivative

The antiderivative, also known as the indefinite integral, is a fundamental concept in calculus. It is the reverse process of differentiation. In simple terms, while differentiation tells you how a function changes, integration lets you find the original function from its rate of change.
For example, when dealing with the function \( f(x) = \frac{1}{x^p} \), finding the antiderivative involves determining a function whose derivative is \( f(x) \). For different powers \( p \), the antiderivative formula may vary slightly, but the general idea remains the same.
The antiderivative of \( \frac{1}{x^p} \) is usually expressed in terms of a power function or logarithmic function, depending on the value of \( p \).

• If \( p = 2 \), the antiderivative is \( -\frac{1}{x} \).
• If \( p = 1.1 \), it becomes \( -10x^{-0.1} \).
• For \( p = 1.01 \), it is \( -100x^{-0.01} \).
• At \( p = 1 \), you get a logarithmic function: \( \ln(x) \).
• If \( p = 0.99 \), it becomes \( 100x^{0.01} \).

Knowing how to find an antiderivative is crucial when dealing with integration problems, as it helps in transforming indefinite integrals into expressions that can be easily evaluated or approximated.

Definite Integral

A definite integral is an evaluation of the integral of a function over a specific interval, say \([a, b]\). It can be visualized as the area under the curve of a function \( f(x) \) from \( a \) to \( b \).
When integrating \( \frac{1}{x^p} \) over a range, say from 1 to 100, you look for the antiderivative and then evaluate it at the endpoints of the interval. This results in a definite value that represents the accumulation of the function over that interval.

• For example, when \( p = 2 \), the integral evaluates to \( 0.99 \) over \([1, 100]\).
• Similarly, for \( p = 1.1 \), the area under the curve turns out to be 9.
• At \( p = 1.01 \), it accumulates to 90.
• With \( p = 1 \), it evaluates to \( \ln(100) \), approximately 4.605.
• And for \( p = 0.99 \), the integral results in 151.1.

These results reflect how the function behaves over a specific interval and are crucial for various applications, including physics and engineering.

Convergence and Divergence

Convergence and divergence in the context of improper integrals help determine whether an integral has a finite value (converges) or grows infinitely large (diverges).
To check for convergence, especially in improper integrals where the upper limit may be infinity, you analyze the behavior of the function as it approaches infinity.
In this problem, it is stated that \( \int_{1}^{\infty}\frac{1}{x^p} dx \) converges for \( p > 1 \) and diverges for \( p \leq 1 \).
This means:

• When \( p = 2, 1.1, \) or \( 1.01 \), the integral converges. This indicates that the function's area under the curve from 1 to infinity is finite.
• Conversely, for \( p = 1 \) and \( p = 0.99 \), the integral diverges. Here, the area becomes infinite as the upper limit approaches infinity.

Understanding convergence and divergence is essential in determining the behavior of integrals and ensuring correct conclusions are drawn for practical applications.

Exponential Functions

Exponential functions are a class of mathematical functions where a constant base is raised to a variable exponent, typically expressed in the form \( a^x \). These functions are characterized by their rapid growth and are commonly seen in natural processes such as population growth and radioactive decay.
While the exercise provided focuses on power functions of the form \( \frac{1}{x^p} \), understanding exponential growth becomes useful when those functions are manipulated in specific ways or compared to exponential behaviors.
In the context of integration, exponential functions can also appear in the forms of \( e^{x} \) and play a major role in determining the behavior of systems. Moreover, when exploring limits for convergence or divergence, one might encounter situations where exponential functions are involved in the approximation or evaluation processes.
Exponential functions, though not directly detailed in this specific integral problem, are foundational in calculus and hold significant value when dealing with other complex functions and their integrals. Understanding their properties helps in grasping broader mathematical concepts across various fields of study.