Problem 5 - Thomas Calculus

Question

The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{-1}^{1} \frac{d x}{x^{2 / 3}}$$

Step-by-Step Solution

Answer

The integral diverges because it is improper and unbounded near zero.

Step 1 - Analyze the Integral for Convergence

The given integral is \( \int_{-1}^{1} \frac{dx}{x^{2/3}} \). Since \( x^{2/3} \) results in an undefined expression at \( x = 0 \), the integral is improper at this point. We need to verify the convergence by splitting the integral from \(-1\) to \(0\) and from \(0\) to \(1\).

Step 2 - Evaluate the Integral from -1 to 0

Consider the integral \( \int_{-1}^{0} \frac{dx}{x^{2/3}} \). Use the substitution \( x = u^3 \), implying \( dx = 3u^2 du \). Adjust bounds accordingly: when \( x = -1 \), \( u = -1 \) and when \( x = 0 \), \( u = 0 \). The integral becomes \( \int_{-1}^{0} 3 \frac{du}{u^2} \) which simplifies to \( -3u^{-1} \). Evaluating gives \( \left[-3(-1)^{-1} \right]_{-1}^{0} = -3 \cdot (-1 - 0) = 3 \).

Step 3 - Evaluate the Integral from 0 to 1

Consider the integral \( \int_{0}^{1} \frac{dx}{x^{2/3}} \). Use the substitution \( x = u^3 \), implying \( dx = 3u^2 du \). Adjust bounds accordingly: when \( x = 0 \), \( u = 0 \) and when \( x = 1 \), \( u = 1 \). The integral becomes \( 3 \int_{0}^{1} \frac{du}{u^{2}} \), which evaluates to \([-3u^{-1}]_{0}^{1} = -3(1 - \infty) = \infty \), implying the integral diverges.

Step 4 - Conclusion on Divergence

The integral from \( -1 \) to \( 0 \) is convergent, however, the integral from \( 0 \) to \( 1 \) is divergent. Therefore, the original integral is divergent because any part of the integral being divergent indicates the whole integral does not converge.

Key Concepts

Integral Convergence

When dealing with integrals, it is essential to determine whether they converge or diverge. In simple terms, an integral converges if it results in a finite value. In contrast, it diverges if the value tends to infinity or becomes undefined. In our exercise, the function \[\int_{-1}^{1} \frac{dx}{x^{2/3}}\] presents a challenge because at \(x = 0\), the function is undefined. Here, we are dealing with an improper integral.

To evaluate convergence, we split the integral into two segments: one from \(-1\) to \(0\), and the other from \(0\) to \(1\). It's important to analyze each segment separately:

• From \(-1\) to \(0\): This part was found to converge, as it evaluates to a finite number.
• From \(0\) to \(1\): This segment diverges, as it results in an infinite value.

For the whole integral to converge, both parts need to result in finite values. Since one part diverges, the entire integral is considered divergent.

Substitution Method

Substitution method, often referred to as \(u\)-substitution, is a powerful tool for evaluating integrals, especially when dealing with complex functions or improper integrals. This method involves transforming the original integral into an easier form by substituting variables.

For the problem at hand:

• Identify the substitution: We replaced \(x\) with \(u^3\). This helps simplify the expression \(x^{2/3}\), making integrals easier to evaluate.
• Calculate the derivative: With \(x = u^3\), we find \(dx = 3u^2 \, du\). The integral bounds change as well, requiring adjustment to reflect this new variable.

By using substitution, the integral \( \int \frac{dx}{x^{2/3}} \) transformed into something more approachable such as \( 3 \int \frac{du}{u^2} \). Solving these newer expressions is often straightforward, utilizing well-known antiderivatives.

Definite Integrals

Definite integrals are those defined over a finite interval \([a, b]\), calculating the net area between the function and the \(x\)-axis. Evaluating definite integrals often involves substituting the boundary values into the antiderivative, then finding the difference.

In this exercise:

• Evaluating from \(-1\) to \(0\): After substitution and simplification, the integration bounds shift accordingly. This computes as a finite number indicating convergence.
• Evaluating from \(0\) to \(1\): Despite substitution, this yields an infinite result indicating divergence.

Even if portions appear simple, remember that defining an integral's behavior over its entire range is crucial. As demonstrated, any part of an improper integral sequence that is divergent will cause the whole expression to diverge despite other convergent segments.