Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{0}^{1} \int_{0}^{1} \frac{y}{(x y+1)^{2}} d x d y $$

Step-by-Step Solution

Answer

The integral evaluates to \(1 - \ln(2)\).

Step 1: Understand the Inner Integral

We start by evaluating the inner integral with respect to variable \(x\). The inner integral is: \[ \int_{0}^{1} \frac{y}{(xy + 1)^{2}} \, dx.\] In this integral, treat \(y\) as a constant since we are integrating with respect to \(x\).

Step 2: Substitute and Simplify Inner Integral

We aim to solve \[\int \frac{y}{(xy + 1)^{2}} \, dx.\]Let \( u = xy + 1 \), then \( du = y \, dx \). Hence, the integral becomes:\[\int \frac{1}{u^2} \, du\] over the corresponding limits. When \(x = 0, u = 1\) and when \(x = 1, u = y + 1\).

Step 3: Evaluate the Substituted Integral

Evaluate \[\int_{1}^{y+1} u^{-2} \, du = - u^{-1} \bigg|_{1}^{y+1}.\] Calculating this, we have:\(-\frac{1}{y+1} + 1 = \frac{y}{y+1}.\)

Step 4: Evaluate the Outer Integral

Next, integrate the result from Step 3 with respect to \(y\):\[\int_{0}^{1} \frac{y}{y+1} \, dy.\]Use substitution \(v = y + 1\), then \(dv = dy\), so the limits become \(v = 1\) to \(v = 2\). This transforms the integral into:\[\int_{1}^{2} \left( 1 - \frac{1}{v} \right) \, dv.\]

Step 5: Integrate the Transformed Function

Evaluate the integral from Step 4:\[\int_{1}^{2} 1 \, dv - \int_{1}^{2} \frac{1}{v} \, dv.\]Calculating individually, we have:\[(v)\bigg|_{1}^{2} = 1 \, \text{and} \, \ln(v)\bigg|_{1}^{2} = \ln(2).\]Thus, the result is:\[ 1 - \ln(2). \]

Step 6: Compile the Final Answer

Summarize your calculations to get the final value:The evaluated iterated integral is \[ 1 - \ln(2). \]

Key Concepts

Double Integral

A double integral is a powerful tool used in calculus to compute the volume under a surface over a given region. It involves integrating a function of two variables, such as \(f(x, y)\), over a specified area.The double integral\(\int\int f(x, y) \, dx \, dy\), or equivalently\(\int\int f(x, y) \, dy \, dx\), computes the cumulative effect of\(f(x, y)\) throughout the region. In practice, double integrals are approached using an iterated integral method, where you integrate one variable first, treating the other as constant, and then perform the second integration with respect to the remaining variable. Iterated integrals have extensive applications, including finding volumes, mass, and other quantities based on density or area functions. When working with double integrals, it's essential to visualize the physical or geometrical interpretation to understand the limits and the function involved.

Substitution Method

The substitution method is an essential integration technique used to simplify complex integrals. It involves changing variables to make an integral easier to solve. In the context of our exercise, when we need to evaluate\(\int \frac{y}{(xy+1)^{2}} \, dx\), we use a substitution to transform the integrand into a more familiar form.The primary steps involve:

• Identifying a suitable substitution, such as\(u = xy + 1\), which simplifies the integrand significantly.
• Recalculating the differential in terms of the new variable, giving\(du = y \, dx\).
• Replacing all original variables in the integrand and adjusting the limits of integration accordingly.

Using substitution helps in managing integrals that are otherwise challenging to integrate directly, turning them into more recognizable forms for which standard integration techniques apply easily.

Integral Calculus

Integral calculus is the branch of calculus that deals with the concepts of integration and its applications. Its core idea is to find functions given their rate of change, known as their derivative, or to calculate the accumulation of quantities, such as areas under curves or volumes under surfaces. In this exercise, integral calculus helps us solve the problem by:

• Evaluating integrals both by direct integration and using substitutions.
• Breaking down complex expressions step by step to find specific solutions.
• Applying fundamental theorems of calculus that connect differentiation and integration.

The focus of integral calculus on anti-differentiation allows us to tackle a wide range of mathematical, physical, and engineering problems, providing a systematic way to analyze changes over continuous domains.

Integration Techniques

Integration techniques are strategies used to solve integrals that are not straightforward to solve directly. They range from simple formulas to more advanced methods like substitution, integration by parts, and partial fraction decomposition.For this problem, the primary technique employed is:

• Substitution:This technique simplifies an integrand by changing variables, as seen in transforming\(\frac{y}{(xy+1)^{2}}\) into a form that's easier to integrate.The key to a successful substitution is choosing the right substitution that makes the integral manageable.
• Simplifying the integral to break them down into manageable pieces.In our solution's final steps, we split\(\int_{1}^{2} \left( 1 - \frac{1}{v} \right) \, dv\) into two simpler integrals: constant integration and logarithmic integration.

Mastering these techniques is about practice and developing a sense of which method best suits a particular integral. Each strategy opens the door to transforming and resolving complex integrals efficiently.