Problem 21 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R} e^{x-y} d A, \quad R : \quad 0 \leq x \leq \ln 2, \quad 0 \leq y \leq \ln 2$$

Step-by-Step Solution

Answer

The value of the integral is \(-\frac{1}{4}\).

Step 1: Define the Double Integral

The problem requires us to evaluate the double integral \(\iint_{R} e^{x-y} \, dA\) over the region \(R\) defined by \(0 \leq x \leq \ln 2\) and \(0 \leq y \leq \ln 2\). This means we integrate with respect to \(x\) and \(y\) over the rectangular region bounded by these inequalities.

Step 2: Set Up the Integral

Since the region \(R\) is a rectangle defined in terms of \(x\) and \(y\), the double integral can be written as an iterated integral: \[ \int_{0}^{\ln 2} \int_{0}^{\ln 2} e^{x-y} \, dx \, dy. \] We'll integrate first with respect to \(x\).

Step 3: Integrate with Respect to x

To evaluate the inner integral \(\int_{0}^{\ln 2} e^{x-y} \, dx\), treat \(y\) as a constant.The antiderivative of \(e^{x-y}\) with respect to \(x\) is \(e^{x-y}\). Evaluate this from \(x = 0\) to \(x = \ln 2\):\[ \left[ e^{x-y} \right]_{0}^{\ln 2} = e^{\ln 2 - y} - e^{0 - y} = \frac{1}{2}e^{-y} - e^{-y}. \]

Step 4: Simplify the Expression

Simplify the result of the inner integral:\[ \frac{1}{2}e^{-y} - e^{-y} = -\frac{1}{2}e^{-y}. \]This expression will be used in the outer integral.

Step 5: Integrate with Respect to y

Now integrate the simplified expression with respect to \(y\):\[ \int_{0}^{\ln 2} -\frac{1}{2}e^{-y} \, dy. \] The antiderivative of \(-\frac{1}{2}e^{-y}\) is \(\frac{1}{2}e^{-y}\). Evaluate this from \(y = 0\) to \(y = \ln 2\):\[ \left[ \frac{1}{2}e^{-y} \right]_{0}^{\ln 2} = \frac{1}{2}e^{-\ln 2} - \frac{1}{2}e^{0} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2}. \]

Step 6: Calculate the Final Result

Calculating the differences yields:\[ \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}. \] Thus, the value of the double integral is \(-\frac{1}{4}\).

Key Concepts

Iterated Integral

A double integral is often expressed as an iterated integral. This simply means performing two successive integration steps, one for each variable.

• When dealing with a double integral, it is crucial to understand the order in which we integrate the two variables.
• We can perform the integration either with respect to one variable first, then the other, or vice versa.

In our exercise, the integration is executed over the region defined for variables \( x \) and \( y \), where each variable spans from 0 to \( \ln 2 \). The iterated integral form becomes:\[ \int_{0}^{\ln 2} \int_{0}^{\ln 2} e^{x-y} \, dx \, dy. \]We begin by integrating with respect to \( x \). Once completed, we then integrate the resulting expression with respect to \( y \). Each function and its relative domain play a significant role in determining the boundaries of integration.

Region of Integration

The region of integration refers to the specific area over which the double integral is calculated. For our problem, this region is defined by constraints on both \( x \) and \( y \).

• In this case, the region \( R \) is a rectangle bounded by the values \( 0 \leq x \leq \ln 2 \) and \( 0 \leq y \leq \ln 2 \).
• This is a specific, finite domain which simplifies the calculation of the integral as we work with fixed bounds.

Understanding the region of integration allows us to correctly set up our iterated integral. For finite regions like this, the limits of integration are constants, which simplifies the integral process significantly.

Exponential Function

An exponential function, especially in calculus and integrals, involves expressions where variables appear in the exponent. In our exercise, we deal with the exponential function \( e^{x-y} \).

• Exponential functions have unique properties such as rapid growth and decay, and a simple rule for differentiation and integration: the function itself is its own derivative.
• For integrals involving exponential functions, knowing its antiderivative is essential. Here, for \( e^{x-y} \), the antiderivative with respect to \( x \) is the same expression: \( e^{x-y} \).

The integration process for exponential functions is often simplified by their predictable patterns.

Antiderivative

Finding the antiderivative of a function is a necessary step in evaluating integrals. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).

• For our problem, with \( e^{x-y} \), treating \( y \) as a constant, its antiderivative with respect to \( x \) is \( e^{x-y} \) itself.
• The antiderivative is used to evaluate definite integrals by finding the difference \( F(b) - F(a) \) over given bounds \( a \) and \( b \).

Using the antiderivative to solve our iterated integral is crucial because it allows us to leverage known formulas for computing the area under the curve, especially for complex functions like exponentials.