Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{0}^{1} \int_{0}^{1} x e^{x y} d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is \( e - 2 \).

Step 1: Identify the Order of Integration

The given iterated integral is \( \int_{0}^{1} \int_{0}^{1} x e^{xy} \ d y \, d x \). This means we first integrate with respect to \( y \) while keeping \( x \) constant, and then integrate the resulting function with respect to \( x \).

Step 2: Integrate with Respect to y

First, consider the inner integral \( \int_{0}^{1} x e^{xy} \ d y \). Since \( x \) is treated as a constant, substitute \( u = xy \), giving \( du = x \, dy \), or \( dy = \frac{du}{x} \). Convert the limits for \( u \) from \( y = 0 \) to \( y = 1 \), which implies \( u = 0 \) to \( u = x \). Thus, the integral becomes \( \int_{0}^{x} e^{u} \frac{du}{x} \). \(\int e^{u} \, du = e^{u} \), so \( \int_{0}^{x} e^{u} \, du = e^{x} - e^{0} = e^{x} - 1 \). Therefore, the inner integral becomes \( \frac{1}{x}(e^{x} - 1) \).

Step 3: Simplify the Inner Integral Result

Apply the computed result of the inner integral: \( x \left( \frac{1}{x}(e^{x} - 1) \right) = e^{x} - 1 \). This gets us the simplified result \( \int_{0}^{1} (e^{x} - 1) \ d x \) for the outer integral.

Step 4: Evaluate the Outer Integral

Now evaluate the integral \( \int_{0}^{1} (e^{x} - 1) \ d x \). This becomes \( \int_{0}^{1} e^{x} \ d x - \int_{0}^{1} 1 \ d x \). The integral \( \int e^{x} \, dx = e^{x} + C \), so from 0 to 1, this is \( [e^{1} - e^{0}] = e - 1 \). The integral \( \int 1 \, dx = x + C \), so from 0 to 1, this becomes \( [1 - 0] = 1 \). Therefore, the outer integral is \( e - 1 - 1 = e - 2 \).

Step 5: Provide the Final Result

The final result of evaluating the iterated integrals is \( e - 2 \).

Key Concepts

Double Integration

Double integration is an important technique in calculus, particularly when dealing with areas and volumes under curves or surfaces.

In double integration, the process involves integrating a function with respect to two variables, commonly denoted as \( x \) and \( y \). This is done in succession, which is why it's called iterated integration: you perform one integral (the inner integral), and then another one (the outer integral) on the result.

• First, choose the order of integration. This is crucial, as it affects the limits you need to apply to each integral.
• In most practical exercises, you'll encounter integrals like \( \int \int f(x, y) \, dy \, dx \), indicating \( y \) is integrated first.

For the current exercise, the approach begins with integrating with respect to \( y \), treating \( x \) as a constant. Once this is done, the resultant expression is integrated with respect to \( x \).

Each level of integration simplifies the problem step by step, which is helpful when dealing with more complex equations. Understanding the sequence and progression of these steps is essential to mastering calculus problem-solving.

Exponential Functions

Exponential functions are powerful functions where a constant base is raised to a variable exponent, commonly written as \( e^x \), where \( e \) is approximately 2.71828, known as Euler's number.

These functions come into play in many areas, notably in calculating growth and decay processes.

• The derivative of any function in the form \( f(x) = e^{x} \) is always \( e^{x} \), highlighting their unique property.
• When integrating exponential functions, the integral \( \int e^{u} \, du = e^{u} + C \) is pivotal, where \( C \) is the constant of integration.

In the given problem, the focus is on integrating an expression with the form \( x e^{xy} \). By performing a substitution, \( u = xy \), it transforms the expression into a form where it's easier to integrate, using the facts mentioned above.

The continuous and smooth nature of exponential functions makes them easy to work with, often resulting in elegant solutions, especially when combined with other calculus operations like differentiation and integration.

Calculus Problem-Solving

Problem-solving in calculus typically involves breaking down complex equations into more manageable parts through a series of methodical steps.

Solving calculus problems, like iterated integrals, requires a solid foundation in mathematics and an understanding of how different mathematical operations interact.

• Begin by analyzing the given problem. Identify the function and determine the variables and range of integration involved.
• Formulate a plan of action: Choose the correct order of integration, make necessary substitutions, and simplify expressions where possible.
• Evaluate each integral step by step, checking for common mistakes, like incorrect limits or missed constants of integration.

In this exercise, the step-by-step approach is demonstrated starting with how to handle nested integrals, progressively simplifying the equations until the desired result is achieved.

By the end of the process, the problem is solved, resulting in \( e - 2 \). Practicing these steps will hone your skills in efficiently managing related calculus problems.