Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{-1}^{1} \int_{0}^{1} x e^{x^{2}} d x d y $$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( e - 1 \).

Step 1: Understand the Iterated Integral

The given integral \( \int_{-1}^{1} \int_{0}^{1} x e^{x^2} \, dx \, dy \) represents an iterated integral where the inner integral with respect to \( x \) is computed first, followed by the integral with respect to \( y \). Our task is to evaluate this iterated integral.

Step 2: Evaluate the Inner Integral

Firstly, evaluate the inner integral \( \int_{0}^{1} x e^{x^2} \, dx \). Use a substitution method: let \( u = x^2 \), then \( du = 2x \, dx \), implying \( x \, dx = \frac{1}{2} \, du \). Change the limits of integration accordingly: when \( x = 0, u = 0 \), and when \( x = 1, u = 1 \). The inner integral becomes: \[ \int_{0}^{1} x e^{x^2} \, dx = \frac{1}{2} \int_{0}^{1} e^u \, du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e^1 - e^0) = \frac{1}{2} (e - 1). \]

Step 3: Evaluate the Outer Integral

With the result from the inner integral \( \frac{1}{2} (e - 1) \), substitute it into the outer integral: \( \int_{-1}^{1} \frac{1}{2} (e - 1) \, dy \). Since the integrand is a constant with respect to \( y \), the integral simplifies to: \[ \frac{1}{2} (e - 1) \times \int_{-1}^{1} 1 \, dy = \frac{1}{2} (e - 1) \times [y]_{-1}^{1} = \frac{1}{2} (e - 1) \times (1 - (-1)) = (e - 1). \]

Step 4: Final Answer

After completing both integrals, we find that the value of the iterated integral is \( e - 1 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique for evaluating integrals, particularly useful when dealing with composite functions. In our example, the inner integral \( \int_{0}^{1} x e^{x^2} \, dx \) involves a function inside of an exponential, making direct integration challenging. By using substitution, we simplify this problem.Here's how it works:

• Identify a substitution: In this case, let \( u = x^2 \). This choice transforms the exponent inside \( e^{x^2} \) into a single variable, \( u \).
• Differentiate to substitute properly: Since \( u = x^2 \), we have \( du = 2x \, dx \). Hence, \( x \, dx = \frac{1}{2} \, du \).
• Change the limits of integration: When \( x = 0 \), \( u = 0 \), and when \( x = 1 \), \( u = 1 \).

With these substitutions, the integral transforms into a simpler form: \( \frac{1}{2} \int_{0}^{1} e^u \, du \), which is straightforward to compute. By simplifying the integrand in terms of \( u \), integration becomes easier, yielding the result \( \frac{1}{2} (e - 1) \). This illustrates how substitution can transform complex integrals into manageable tasks.

Limits of Integration

Understanding and correctly applying limits of integration is critical in solving definite integrals. These limits define the interval over which the integration is performed, and their careful adjustment is especially crucial when substitution changes the variable of integration.In the context of our problem:

• The original inner integral \( \int_{0}^{1} x e^{x^2} \, dx \) spans from \( x = 0 \) to \( x = 1 \).
• After substituting \( u = x^2 \), the limits are recalculated: \( u \) changes as \( x \) changes from 0 to 1, giving limits \( u = 0 \) and \( u = 1 \).
• For the outer integral \( \int_{-1}^{1} \, dy \), the limits are constant \( y = -1 \) to \( y = 1 \), as they dictate the integration range for \( y \).

Adjusting the limits ensures that they remain consistent with the transformed variable, maintaining the integral's value and allowing for accurate solutions.

Outer Integral

The outer integral is the integration performed after the inner integral has been evaluated. It provides the final step, which, in our case, involves the variable \( y \). In the iterated integral process:

• Once the inner integral \( \int_{0}^{1} x e^{x^2} \, dx \) is evaluated to \( \frac{1}{2} (e - 1) \), we substitute this result into the outer integral, \( \int_{-1}^{1} \frac{1}{2} (e - 1) \, dy \).
• The outer integral often appears simpler. Here, it becomes the product of the constant \( \frac{1}{2} (e - 1) \) and the integral of \( 1 \, dy \), evaluated over the interval \([-1, 1]\).
• This is straightforward because \( \int_{-1}^{1} 1 \, dy = [y]_{-1}^1 = 1 - (-1) = 2 \).

Multiplying gives the final solution: \( (e - 1) \). Thus, the outer integral consolidates the results of the inner integral across its defined limits, providing the overall value for the iterated integral.

Inner Integral

The inner integral is the first step in evaluating an iterated integral. It focuses on a specific variable while treating others as constants, allowing for a step-wise simplification.For this exercise:

• The inner integral \( \int_{0}^{1} x e^{x^2} \, dx \) is executed by treating \( y \) as a constant, simplifying our integrand.
• After applying the substitution \( u = x^2 \), the complex expression is simplified to a single variable integral \( \frac{1}{2} \int_{0}^{1} e^u \, du \).
• Evaluate this integral: Since \( \int e^u \, du = e^u \), it yields \( \frac{1}{2} [e^u]_{0}^{1} = \frac{1}{2} (e - 1) \).

Completing the inner integral calculation simplifies the iterative process, setting the stage for solving the outer integral. Understanding and accurately solving the inner integral is crucial, as it directly impacts the result of the entire problem.