Problem 103 - Calculus Early Transcendentals

Question

Let \(R_{1}=\\{(x, y): x \geq 1,1 \leq y \leq 2\\}\) and \(R_{2}=\\{(x, y): 1 \leq x \leq 2, y \geq 1\\} .\) For \(n>1,\) which integral(s) have finite values: \(\iint_{R_{1}} x^{-n} d A\) or \(\iint_{R_{2}} x^{-n} d A ?\)

Step-by-Step Solution

Answer

Answer: Both regions, \(R_1\) and \(R_2\), have finite values for the double integral when n>1.

Step 1: Understand the given regions

The regions \(R_1\) and \(R_2\) are defined as follows: \(R_1 = \{(x, y): x \geq 1, 1 \leq y \leq 2\}\) \(R_2 = \{(x, y): 1 \leq x \leq 2, y \geq 1\}\)

Step 2: Set up the double integral for region \(R_1\)

For region \(R_1\), we have: \(\iint_{R_{1}} x^{-n} d A = \int_{1}^{2} \int_{1}^{\infty} x^{-n} dx \ dy\)

Step 3: Evaluate the double integral for \(R_1\)

First we evaluate the inner integral with respect to \(x\): \(\int_{1}^{\infty} x^{-n} dx = [\frac{x^{-n+1}}{-n+1}]_{1}^{\infty}\) Since \(n>1\), this evaluates to a finite value: \(\frac{1}{n-1}\) Now evaluate the outer integral with respect to \(y\): \(\int_{1}^{2} \frac{1}{n-1} dy = [\frac{y}{n-1}]_{1}^{2}\) This evaluates to a finite value of: \(\frac{1}{n-1}\)

Step 4: Set up the double integral for region \(R_2\)

For region \(R_2\), we have: \(\iint_{R_{2}} x^{-n} d A = \int_{1}^{\infty} \int_{1}^{2} x^{-n} dy \ dx\)

Step 5: Evaluate the double integral for \(R_2\)

First we evaluate the inner integral with respect to \(y\): \(\int_{1}^{2} x^{-n} dy = [x^{-n}y]_{1}^{2} = x^{-n}(2-1) = x^{-n}\) Now evaluate the outer integral with respect to \(x\): \(\int_{1}^{\infty} x^{-n} dx = [\frac{x^{-n+1}}{-n+1}]_{1}^{\infty}\) Since \(n>1\), this evaluates to a finite value: \(\frac{1}{n-1}\)

Step 6: Conclusion

For \(n>1\), both integrals: \(\iint_{R_{1}} x^{-n} d A\) and \(\iint_{R_{2}} x^{-n} d A\) have finite values.

Key Concepts

Region of Integration

In calculus, the region of integration refers to the specific area over which a double integral is evaluated. It determines the boundaries and limits within which the variables of integration operate. For example, in the problem, we have two regions, \(R_1\) and \(R_2\). Each region defines where the variables \(x\) and \(y\) lie.

• Region \(R_1\): It's defined by \(x \geq 1\) and \(1 \leq y \leq 2\). This means \(x\) can take any value from 1 to infinity, while \(y\) can only vary between 1 and 2.
• Region \(R_2\): Here, \(1 \leq x \leq 2\) and \(y \geq 1\) are the restrictions. Consequently, \(x\) is bounded between 1 and 2, while \(y\) can extend infinitely upwards from 1.

In each case, these regions influence the setup of the double integral and the subsequent steps in solving it. Understanding these constraints is critical for proper evaluation.

Improper Integrals

Improper integrals occur when one or both of the limits of an integral extend to infinity. This presents a challenge since infinity is not a tangible number. We use limits to handle these situations, allowing us to evaluate integrals that involve unbounded regions.

In our exercise, region \(R_1\) uses infinity as an upper limit for \(x\):

• \( \int_{1}^{\infty} x^{-n} dx \)

This exemplifies an improper integral, as \(x\) stretches towards infinity. Evaluating the integral requires finding the antiderivative and then appraising its behavior as \(x\) approaches infinity.

The technique typically involves:

• Evaluating the definite integral from a finite bound to a symbol representing infinity.
• Taking the limit as this symbol approaches infinity.

Assessing improper integrals is essential to ensure any potential issues with convergence are addressed.

Finite Values

Finite values refer to a scenario where, after calculating an integral, the result is a definite number. This concept contrasts with integrals that might diverge, leading to infinite values. In our context, once the improper integrals for both regions \(R_1\) and \(R_2\) are evaluated,

• \( \iint_{R_{1}} x^{-n} d A = \frac{1}{n-1} \)
• \( \iint_{R_{2}} x^{-n} d A = \frac{1}{n-1} \)

Both return finite results, provided \(n > 1\). Evaluating whether an integral yields a finite value is key to understanding

• the behavior of a function at extremes,
• and how it varies over different domains of integration.

In practical terms, knowing that an integral produces a finite value means it's possible to comprehensively understand and interpret the quantity represented by the integral within its specified domain.