Problem 19 - University Calculus: Early Transcendentals

Question

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 double integral is \( \frac{1}{2} \).

Step 1: Understand the Region of Integration

The region \( R \) is defined by the rectangular bounds \( 0 \leq x \leq \ln 2 \) and \( 0 \leq y \leq \ln 2 \). This indicates that \( R \) is a square in the \( xy \)-plane with vertices at the points \( (0,0), (\ln 2, 0), (\ln 2, \ln 2), (0, \ln 2) \).

Step 2: Set Up the Double Integral

The double integral to evaluate is given by:\[\int_{0}^{\ln 2} \int_{0}^{\ln 2} e^{x-y} \, dx \, dy\]The limits of integration for \( x \) are from \( 0 \) to \( \ln 2 \), and for \( y \), they are also from \( 0 \) to \( \ln 2 \).

Step 3: Integrate with Respect to x

First, integrate \( e^{x-y} \) with respect to \( x \), treating \( y \) as a constant:\[\int_{0}^{\ln 2} e^{x-y} \, dx = e^{-y} \int_{0}^{\ln 2} e^{x} \, dx\]Calculate the integral:\[e^{-y} \left[ e^{x} \right]_{0}^{\ln 2} = e^{-y} \left( e^{\ln 2} - e^{0} \right) = e^{-y} (2 - 1) = e^{-y} \cdot 1 = e^{-y}\]

Step 4: Integrate with Respect to y

Now integrate the resulting expression \( e^{-y} \) with respect to \( y \):\[\int_{0}^{\ln 2} e^{-y} \, dy\]Calculate the integral:\[- \left[ e^{-y} \right]_{0}^{\ln 2} = - \left( e^{-ln 2} - e^{0} \right) = - \left( \frac{1}{2} - 1 \right) = - \left( -\frac{1}{2} \right) = \frac{1}{2}\]

Step 5: Conclusion

The value of the double integral \( \iint_{R} e^{x-y} \, dA \) over the region \( R \) is \( \frac{1}{2} \).

Key Concepts

Region of Integration

When we speak about the "Region of Integration" in a double integral, we refer to the specific area over which we integrate. In our exercise, this region is defined as a square on the xy-plane. To be precise, for this problem, the region \( R \) is confined within the boundaries \( 0 \leq x \leq \ln 2 \) and \( 0 \leq y \leq \ln 2 \).

These bounds create a square whose corners or vertices are \( (0,0), (\ln 2, 0), (\ln 2, \ln 2), \) and \( (0, \ln 2) \). This conceptualizes a two-dimensional area for integration. Identifying this region correctly is essential because it determines the limits for our double integral, ensuring we compute the integral over the correct area.

• Region is a square with equal side lengths of \( \ln 2 \).
• Boundaries are drawn parallel to the x-axis and y-axis.
• Integration is only computed within this specific area.

Understanding the region of integration helps in setting up the integral correctly, ensuring that we are capturing the right measurements for evaluating the problem at hand.

Exponential Function Integration

Exponential functions are pervasive in calculus, and integrating them requires some distinct approaches. In our problem, we deal with the integrand \( e^{x-y} \). The integration process involves handling exponential functions, which can initially seem challenging.

We first integrate \( e^{x-y} \) with respect to \( x \), treating \( y \) as a constant. This simplification leads us to the task of integrating \( e^{x} \), which is relatively straightforward:
\[\int e^{x} \, dx = e^{x} + C\]

After integrating with respect to \( x \), we obtain an expression that still contains \( y \). We continue by integrating this result with respect to \( y \). This second integration provides:
\[\int e^{-y} \, dy = -e^{-y} + C\]

Key points to consider for exponential function integration include:

• Understand that \( y \) becomes a constant during the first integration.
• Solve the integral systematically, tackling one variable at a time.
• Apply standard integration rules for exponential functions.

By breaking it down step by step, exponential function integration becomes manageable and straightforward.

Definite Integrals

The concept of definite integrals plays a pivotal role in calculus, particularly in evaluating areas and accumulations. Unlike indefinite integrals, a definite integral provides a numeric value that represents the area under a curve within specified bounds.

In this exercise, after determining the region of integration and performing the integration steps, we proceed to evaluate the definite integral over a specific interval. For the inner integral with respect to \( x \), we compute:
\[\left[ e^{x-y} \right]_{0}^{\ln 2} = e^{-y} \left( e^{\ln 2} - e^{0} \right) = e^{-y} \cdot 1 = e^{-y}\]

Next, we move to the outer integral with respect to \( y \):
\[- \left[ e^{-y} \right]_{0}^{\ln 2} = - \left( \frac{1}{2} - 1 \right)= \frac{1}{2}\]

Essential highlights for definite integrals involve:

• Definite integrals yield the total accumulation over the region.
• Limits of integration are applied only after performing the integration.
• Results express a summation over the defined interval or area.

By focusing on the definite integral definitions and properties, it becomes easier to grasp their application within double integrals and achieve precise results.