Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(\iint_{R}(1+x) d A\), where \(R=\\{(x, y): 0 \leq x \leq 2\), \(0 \leq y \leq 1\\} .\) See the hint in Problem \(21 .\)

Step-by-Step Solution

Answer

The value of the integral is 4.

Step 1: Understand the Region R

The region \( R \) is defined by two sets of inequalities: \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \). This describes a rectangle in the xy-plane, where \( x \) ranges from 0 to 2 and \( y \) ranges from 0 to 1.

Step 2: Set Up the Double Integral

We need to evaluate the double integral \( \iint_{R} (1+x) \, dA \). This can be expressed as an iterated integral, since the region \( R \) is rectangular: \[\int_{0}^{2} \int_{0}^{1} (1 + x) \, dy \, dx\]

Step 3: Evaluate the Inner Integral

First, integrate with respect to \( y \). The inner integral is:\[\int_{0}^{1} (1 + x) \, dy = (1 + x) \int_{0}^{1} \, dy = (1 + x) [y]_{0}^{1} = (1 + x) (1 - 0) = 1 + x.\]

Step 4: Evaluate the Outer Integral

Now, integrate the result from the inner integral with respect to \( x \):\[\int_{0}^{2} (1 + x) \, dx = \left[ x + \frac{x^2}{2} \right]_{0}^{2} = \left(2 + \frac{4}{2}\right) - \left(0 + \frac{0}{2}\right) = 2 + 2 = 4.\]

Step 5: Conclude the Calculation

The double integral \( \iint_{R} (1 + x) \, dA \) over the region \( R \) equals 4. Therefore, the integral evaluates to 4.

Key Concepts

Iterated Integral

An iterated integral is a way to evaluate a double integral by breaking it down into two successive single integrals. When you perform an iterated integral, you first integrate one variable while keeping the other constant. For instance, the region of integration might be a rectangle in the xy-plane, allowing us to convert the double integral into two single integrals.To perform an iterated integral, you usually follow these steps:

• Identify the limits of integration for each variable, typically from a geometric perspective like a rectangle or other shape in the xy-plane.
• Set up the integral by choosing an order of integration; that is, decide whether to integrate with respect to x or y first.
• Evaluate the inner integral first, treating all other variables as constants.
• Then, evaluate the outer integral using the result from the inner integral.

In our specific exercise, the iterated integral was set up as \(\int_{0}^{2} \int_{0}^{1} (1 + x) \, dy \, dx\), where integrating with respect to \( y \) came first, followed by \( x \). This method effectively handles the complexity of double integrals.

Region of Integration

Understanding the region of integration is crucial for setting up a double integral correctly. The region of integration is the area over which the function is being integrated. In Cartesian coordinates, it is typically defined by inequalities that describe the limits for x and y.In this specific problem, the region R is described by the inequalities \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\). This means that R is a rectangle located in the first quadrant of the xy-plane:

• The line \(0 \leq x \leq 2\) defines the horizontal boundaries.
• The line \(0 \leq y \leq 1\) defines the vertical boundaries.

Visualizing this region is often helpful. Imagine a rectangle with corners at the coordinates (0,0), (0,1), (2,0), and (2,1). Identifying and understanding these boundaries ensure that the integration covers the correct area, thus leading to the correct result.

Evaluation of Integrals

The evaluation of integrals, especially in the context of the problem statement, involves computing integrals systematically. For an iterated integral, this process includes breaking down the integral into simpler parts: the inner and outer integrals.Consider the steps:

• First, evaluate the inner integral by integrating with respect to the variable y. This integral is done while keeping x constant.
• After solving the inner integral, the expression is integrated with respect to x. This result of the outer integral provides the final answer.

In our exercise:

1. The inner integral: \(\int_{0}^{1} (1 + x) \, dy = 1 + x \) (since \( y \) ranges from 0 to 1).
2. The outer integral: Next, integrate \(1 + x \) with \( x \) from 0 to 2 to get \( x + \frac{x^2}{2} \).

The final evaluation yielded the result of 4, illustrating how breaking down the integrals step-by-step aids in obtaining the solution.

Integral Calculus

Integral Calculus is an essential branch of mathematics focusing on integration, which helps calculate areas, volumes, and other quantities across different fields. The idea is to generalize the process of summing up small quantities. Integral calculus uses the concept of the integral to:

• Accumulate quantities continuously over regions.
• Determine areas under curves, which are crucial for finding lengths, volumes, and other spatial properties.
• Solve differential equations by finding anti-derivatives, which directly relate to the concept of integration.

In the case of a double integral, like our problem, you are essentially computing a volume under a surface described by a function of two variables over a specified region. This problem showed how integral calculus is applied by breaking it down into simple, calculable parts using iterated integrals over a defined region. The systematic process simplifies solving what might first appear to be complex calculations into manageable steps.