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 \(\iint_{R_{2}} x^{-n} d A ?\)

Short answer

A) $$\iint_{R_1} x^{-n} dA, \: R_1 =\{(x, y) : x\ge 1, 1\le y \le 2\}$$ B) $$\iint_{R_2} x^{-n} dA, \: R_2 =\{(x, y) : 1\le x\le 2, y\ge 1\}$$ Answer: Only the integral A, $$\iint_{R_1} x^{-n} dA$$, has a finite value for n>1.

Step-by-step solution

Setup the integral for \(R_1\)#

For the region \(R_1=\{(x, y) : x\ge 1, 1\le y \le 2\},\) the bounds of integration are \(x\in[1,\infty)\) and \(y\in[1,2].\) Now we can setup the integral for this region: $$I_1 = \iint_{R_1} x^{-n} dA = \int_1^\infty\int_1^2 x^{-n} dy \, dx.$$

Evaluate \(I_1\)#

We first evaluate the inner integral: $$\int_1^2 x^{-n} dy = x^{-n}\int_1^2 dy = x^{-n}(2-1) = x^{-n}.$$ Now, we can evaluate the outer integral: $$I_1 = \int_1^\infty x^{-n} dx$$ Since \(n > 1,\) this integral converges to a finite value. We can prove this by computing its antiderivative: $$\int x^{-n} dx = \frac{x^{-n+1}}{-n+1}|_1^\infty = \frac{1}{1-n}\left(\frac{1}{\infty^{n-1}} - 1\right) = \frac{1}{n-1},$$ which is finite.

Setup the integral for \(R_2\)#

For the region \(R_2=\{(x, y) : 1\le x\le 2, y\ge 1\},\) the bounds of integration are \(x\in[1,2]\) and \(y\in[1,\infty).\) Now we can setup the integral for this region: $$I_2 = \iint_{R_2} x^{-n} dA = \int_1^2\int_1^\infty x^{-n} dy \, dx.$$

Evaluate \(I_2\)#

We first evaluate the inner integral: $$\int_1^\infty x^{-n} dy = x^{-n}\int_1^\infty dy = x^{-n}\cdot\infty = \infty \quad (\text{since } x^{-n} > 0 \text{ for } x \in [1,2])$$ Now, we can evaluate the outer integral: $$I_2 = \int_1^2 \infty \, dx$$ The integral is infinite, so its value is not finite.

Conclusion#

In conclusion, the integral $$\iint_{R_1} x^{-n} dA$$ has a finite value for \(n>1,\) while the integral $$\iint_{R_2} x^{-n} dA$$ does not.

Key concepts

Convergence of Improper Integrals

Improper integrals arise when we deal with infinite limits or unbounded integrands. In this exercise, analyzing the double integral over the region \( R_1 \) requires us to examine infinite limits for \( x \). When we set up the integral \( \int_1^\infty x^{-n} \, dx \), improper integrals come into play since the upper limit is \( \infty \).

To determine if an improper integral converges, we often complete a process called "evaluation by limits." This means calculating antiderivatives, and applying limits as they approach infinity.

For the integral to converge, the value must be finite after evaluation. When \( n > 1 \), the integral \( \int_1^\infty x^{-n} \, dx \) has a finite result, \( \frac{1}{n-1} \), demonstrating convergence. This behavior confines any possible indefinite growth, making it possible to evaluate it to a unique value.

Integration Bounds

Integration bounds define the region over which we integrate. Establishing bounds is crucial to proper integral evaluation. For \( R_1 \), the bounds are \( x \in [1, \infty) \) and \( y \in [1, 2] \). Here, \( x \) extends towards infinity while \( y \) remains bounded between 1 and 2.

These bounds direct us to set up the integral expression \( \int_1^\infty \int_1^2 x^{-n} \, dy \, dx \). The outer integral captures the entirety of the function along the \( x \)-axis starting from 1 to infinity, while the inner integral addresses the specified \( y \)-range.

For \( R_2 \), the restriction shifts: \( x \) is integrated over \([1, 2]\), while \( y \) trends towards infinity. So the integration setup becomes: \( \int_1^2 \int_1^\infty x^{-n} \, dy \, dx \), exhibiting variations in how the region spans over the coordinate axes.

Evaluation of Iterated Integrals

Evaluating iterated integrals involves performing integration across multiple variables consecutively. Initially, we focus on the inner integral, holding the other variable as constant. For \( R_1 \), the inner integral \( \int_1^2 x^{-n} \, dy \) computes first, leaving us with \( x^{-n} \times 1 \).

Next, the outer integral \( \int_1^\infty x^{-n} \, dx \) finalizes the evaluation. This step-by-step approach importantly transitions from integrating with respect to \( y \) to \( x \), treating each iteration as a separate simple integral process.

It's crucial to adhere to the bounds provided for each variable, ensuring accurate calculations. The order of integration often significantly impacts the solvability and outcome of a double or iterated integral, manifesting when different regions or integrands are analyzed.

Antiderivatives

Finding antiderivatives is a key step in evaluating integrals. It involves reversing differentiation to obtain the original function. In the solution, to evaluate \( \int_1^\infty x^{-n} \, dx \), we determine the antiderivative of \( x^{-n} \).

The antiderivative, \( \frac{x^{-n+1}}{-n+1} \), gives a formula that can be evaluated at particular limits. By calculating \( \left[ \frac{1}{1-n} \left( \lim_{x \to \infty} x^{-n+1} - 1 \right) \right] \), we end up with \( \frac{1}{n-1} \) for the limit result, crucial to determine convergence as discussed.

Understanding how to derive these antiderivatives is fundamental, as it enables us to solve the integral and assess the convergence, especially within the context of improper integrals where direct computation isn’t always feasible.