Problem 74 - Calculus Early Transcendentals

Question

Consider the family of functions \(f(x)=1 / x^{p},\) where \(p\) is a real number. For what values of \(p\) does the integral \(\int_{0}^{1} f(x) d x\) exist? What is its value?

Step-by-Step Solution

Answer

Short answer: The integral \(\int_{0}^{1} \frac{1}{x^p} dx\) exists for \(0 \leq p \leq 1\). When \(0 < p < 1\), the integral converges and its value is \(\frac{1}{1-p}\). When \(p = 0\), the integral converges and its value is \(1\). If \(p > 1\), the integral diverges.

Step 1: Identify the integral to solve

The integral to solve is given by $$ \int_{0}^{1} \frac{1}{x^p} dx $$

Step 2: Rewrite the expression for clarity

To make it easier to work with, rewrite the expression as follows: $$ \int_{0}^{1} x^{-p} dx $$

Step 3: Analyze the behavior of the function within the given interval

The function \(x^{-p}\) has a singularity (i.e., it is undefined) at \(x = 0\). To find the values of \(p\) for which the integral converge, we need to analyze the behavior of the integrand as \(x\) approaches \(0\). When \(x \rightarrow 0^+\), the expression \(x^{-p}\) behaves as follow: 1. If \(p > 1\), then \(x^{-p} \rightarrow \infty\). 2. If \(p = 1\), then \(x^{-p} = x^{-1}\), which is still infinite as \(x \rightarrow 0^+\), but less than the previous case. 3. If \(0 < p < 1\), then \(x^{-p} \rightarrow 1\) as \(x \rightarrow 0^+\), which looks better. 4. If \(p \leq 0\), then \(x^{-p}\) stays finite, since the exponent is non-positive. From the analysis above, we can conclude that the integral converges if \(p \leq 1\) and diverges if \(p > 1\).

Step 4: Calculate the integral for p <= 1

Now we need to find the integral for the values of \(p\) for which it converges. So, let \(0 < p \leq 1\), and calculate the integral: $$ \int_{0}^{1} x^{-p} dx $$ Using the power rule for integration, we get: $$ \bigg[ \frac{x^{(1-p)}}{1-p} \bigg]_{0}^{1} $$ Now, we need to evaluate the integral at the bounds: $$ \frac{1^{(1-p)}}{1-p} - \frac{0^{(1-p)}}{1-p} $$ Since \(1^{(1-p)} = 1\) and \(0^{(1-p)} = 0\), we have: $$ \frac{1}{1-p} - 0 = \frac{1}{1-p} $$ For \(p = 0\), the integral becomes: $$ \int_{0}^{1} x^{-0} dx = \int_{0}^{1} 1 dx $$ Evaluating the integral gives: $$ x \bigg|_{0}^{1} = 1 - 0 = 1 $$ Finally, the integral of \(\int_{0}^{1} f(x) dx\) exists for \(0 \leq p \leq 1\), and its value is given by \(\frac{1}{1-p}\) for \(0 < p < 1\) and \(1\) for \(p = 0\).

Key Concepts

Convergence

In the realm of improper integrals, understanding when they converge or not is crucial. Convergence determines whether the integral has a finite value. For this integral, we look at the function \[ \int_{0}^{1} \frac{1}{x^p} \, dx \]where convergence is influenced by the value of \( p \). Specifically, the function \[ x^{-p} \]behaves differently based on \( p \) as \( x \) nears zero:

• For \( p > 1 \), the function approaches infinity, leading to divergence.
• When \( p = 1 \), the function becomes \( x^{-1} \), which also diverges as \( x \rightarrow 0^+ \), although the rate of divergence is slower than in the case of \( p > 1 \).
• When \( 0 < p < 1 \), \( x^{-p} \) converges smoothly to 1, suggesting convergence.
• For \( p \leq 0 \), the function remains finite, as the exponent being non-positive makes it less impactful under the bounds.

So, convergence occurs when \( 0 \leq p \leq 1 \). This means the integral can be evaluated to a number rather than tending to infinity under these conditions.

Singularity

Singularities in functions are points where they become undefined or reach infinity. These can often pose challenges when evaluating integrals. In \[ \int_{0}^{1} x^{-p} \, dx \],the singularity occurs at \( x = 0 \) when \( p \geq 1 \) as the function becomes undefined or tends to infinite values. Understanding this behavior is crucial for assessing convergence. At precisely \( p = 1 \), we encounter a specific type of singularity where the integral resembles the improper integral \( \int dx/x \),which is known to be divergent as it tends to infinity approaching zero. This kind of singularity requires careful treatment usually via limiting processes or transformations to handle it correctly in calculations.

Power Rule for Integration

The Power Rule for Integration is a fundamental tool necessary for solving integrals involving power functions. It states that for any real number \( n eq -1 \),\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \( C \) is a constant of integration. In our integral \[ \int_{0}^{1} x^{-p} \, dx \],we apply this rule to find solutions for different \( p \) values. Let's break it down: - For \( 0 < p < 1 \), using \( n = -p \), the integral becomes \[ \frac{x^{1-p}}{1-p} \] evaluated from 0 to 1.- Substituting the limits, we compute:\[ \frac{1}{1-p} - \frac{0}{1-p} = \frac{1}{1-p} \]This result highlights the elegance and utility of the Power Rule, simplifying what might initially seem a complex calculation. When \( p = 0 \), the formula confirms that \[ \int_{0}^{1} 1 \, dx = 1 \],reinforcing the Power Rule's reliability in handling integrals.