Problem 25 - Thomas Calculus

Question

In Exercises 25 and \(26,\) integrate \(f\) over the given region. $$\begin{array}{l}{\text { Square } f(x, y)=1 /(x y) \quad \text { over } \quad \text { the square } \quad 1 \leq x \leq 2} \\ {1 \leq y \leq 2}\end{array}$$

Step-by-Step Solution

Answer

The integral evaluates to \( (\ln 2)^2 \).

Step 1: Understand the Problem

The problem asks us to integrate the function \( f(x, y) = \frac{1}{xy} \) over a square region defined by \( 1 \leq x \leq 2 \) and \( 1 \leq y \leq 2 \). We need to compute the double integral of this function over the given region.

Step 2: Set Up the Double Integral

Write the expression for the double integral as follows:\[\int_{1}^{2} \int_{1}^{2} \frac{1}{xy} \, dy \, dx\]Here, we first integrate with respect to \( y \) and then with respect to \( x \).

Step 3: Integrate with Respect to y

Integrate the inner integral with respect to \( y \):\[\int_{1}^{2} \frac{1}{xy} \, dy = \frac{1}{x} \int_{1}^{2} \frac{1}{y} \, dy\]The antiderivative of \( \frac{1}{y} \) is \( \ln |y| \). Evaluating from 1 to 2 gives:\[\int_{1}^{2} \frac{1}{y} \, dy = \ln 2 - \ln 1 = \ln 2\]

Step 4: Integrate with Respect to x

Now, substitute the result of the inner integral back into the outer integral:\[\int_{1}^{2} \frac{1}{x} \ln 2 \, dx\]Factor out the constant \( \ln 2 \), and integrate \( \frac{1}{x} \, dx \), whose antiderivative is \( \ln |x| \):\[\ln 2 \left[ \ln |x| \right]_{1}^{2} = \ln 2 \cdot (\ln 2 - \ln 1) = \ln 2 \cdot \ln 2\]

Step 5: Simplify the Result

The final expression for the double integral evaluates to:\[(\ln 2)^2\]This is the result of our integration over the given region.

Key Concepts

Integration Techniques

When we talk about integration techniques, especially for a double integral, we are dealing with a more complex process that extends the concept of single integrals to functions of two variables. The double integral allows us to compute the volume under a surface over a particular region. The basic idea is to integrate one variable at a time. In our problem, the function of two variables is \( f(x, y) = \frac{1}{xy} \), which we need to integrate over a square region between \( x=1 \) to \( x=2 \) and \( y=1 \) to \( y=2 \).

The integration process for such scenarios can be broken down into smaller steps, starting with setting up the iterated integrals and then carefully integrating with respect to each variable. Remember, for any double integral, you must correctly set the bounds and choose an order of integration. The order can sometimes simplify the integration work significantly. Stick to your setup like a roadmap, first integrating with respect to the inner variable and then the outer.

Iterated Integrals

In the case of iterated integrals, which is an approach used for calculating double integrals, you perform two separate integrations, one nested inside the other. This method involves integrating the function first with respect to one variable while keeping the other constant. Once that's done, you integrate the resulting expression with respect to the second variable.
For instance, in the given problem, we first computed the integral of \( \frac{1}{xy} \) with respect to \( y \), treating \( x \) as a constant. This results in the integral \( \frac{1}{x} \int \frac{1}{y} \, dy \). After calculating the inner integration, the result is then integrated with respect to \( x \). It's crucial to evaluate limits of integration carefully at each stage to ensure proper results. Iterated integrals allow flexibility in calculating complex areas and volumes, providing insight into the multidimensional space a surface occupies.

Logarithmic Function

The logarithmic function often comes into play during integration, especially when dealing with functions of the form \( \frac{1}{y} \) or \( \frac{1}{x} \). The antiderivative of these types of expressions is the natural logarithm function, \( \ln |y| \) or \( \ln |x| \). The logarithmic function is defined for positive real numbers, which suits our scenario because the region limits \(1 \leq x \leq 2\) and \(1 \leq y \leq 2\) maintain the function's domain.
In our integration problem, when calculating \( \int_{1}^{2} \frac{1}{y} \, dy \), it results in \( \ln 2 - \ln 1 \). It's important to note that \( \ln 1 \) is \(0\), simplifying this to \( \ln 2 \). This simplicity in the logarithmic properties allows us to handle such integrations efficiently, converting complicated rational expressions to easily interpretable results like \( \ln(\text{something}) \).

Antiderivatives

The concept of antiderivatives is fundamental when performing any integration, as it refers to the function you would differentiate to obtain the integrand. For instance, the antiderivative of \( \frac{1}{y} \) is \( \ln |y| \), a crucial part of solving the given integration problem. Understanding antiderivatives helps bridge the gap between differentiation and integration, making it easy to reverse the differentiation process.

In our exercise, finding and evaluating antiderivatives correctly allows us to compute the integral of \( \frac{1}{x} \cdot \ln 2 \) by noting that \( \ln |x| \) is the antiderivative of \( \frac{1}{x} \). Evaluating this from 1 to 2 gives us \( \ln 2 - \ln 1 \), simplified to \( \ln 2 \), as \( \ln 1 \) equals 0. Determining accurate antiderivatives ensures all calculations in an integral are seamless, leading to precise results like the final \( (\ln 2)^2 \) in our problem.