Problem 11 - Calculus Early Transcendentals

Question

Sketch each region and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d y d x\). $$R=\\{(x, y): 1 \leq x \leq 2, x+1 \leq y \leq 2 x+4\\}$$

Step-by-Step Solution

Answer

Answer: The iterated integral representing the integration of a continuous function over the region R in the order dy dx is given by: $$\int_{1}^{2}\int_{x+1}^{2x+4} f(x, y) \, dy \, dx$$

Step 1: Identify the boundaries of the region R

The region R is defined by four inequalities: 1. \(x \geq 1\) 2. \(x \leq 2\) 3. \(y \geq x + 1\) 4. \(y \leq 2x + 4\) These inequalities can be used to obtain the equations of the boundary lines. The boundary lines are: 1. \(x = 1\) 2. \(x = 2\) 3. \(y = x + 1\) 4. \(y = 2x + 4\)

Step 2: Sketch the region R

To sketch the region R, we first plot the boundary lines on the coordinate plane: 1. Vertical line at \(x = 1\) 2. Vertical line at \(x = 2\) 3. Line with slope 1 and y-intercept 1: Passes through points A(0,1),B(1,2) and continues with slope 1 4. Line with slope 2 and y-intercept 4: Passes through points C(0,4),D(1,6) and continues with slope 2 The region R is enclosed by these boundary lines. To find the vertices of R, we check for intersections between these lines: - Intersection of lines 1 and 3: \((1,2)\) - Intersection of lines 1 and 4: \((1,6)\) - Intersection of lines 2 and 3: \((2,3)\) - Intersection of lines 2 and 4: \((2,8)\) The region R is a trapezoid with vertices (1,2), (1,6), (2,3), and (2,8).

Step 3: Set up the iterated integral

Now that we have identified the boundaries of the region R, we can set up the iterated integral with the order dy dx. The limits of integration for x are given by the inequalities \(1 \leq x \leq 2\). The limits of integration for y depend on x and are given by the inequalities \(x+1 \leq y \leq 2x+4\). The iterated integral is: $$\int_{1}^{2}\int_{x+1}^{2x+4} f(x, y) \, dy \, dx$$ This is the iterated integral representing the integration of a continuous function over the region R in the order dy dx.

Key Concepts

Region Sketching

Understanding and sketching the region for iterated integrals is crucial for visualizing the area over which you're integrating. The exercise gives us a region defined by specific inequalities involving variables \(x\) and \(y\). These inequalities correspond to equations of lines that create boundaries on the coordinate plane.

- **Vertical Lines**: In this case, they are at \(x=1\) and \(x=2\), creating the lateral boundaries of the region.- **Sloped Lines**: Two lines define the top and bottom boundaries. These are \(y = x + 1\), which is a line with slope 1, and \(y = 2x + 4\), a line with a steeper slope of 2.
The intersection points of these lines, such as \((1,2)\), \((1,6)\), \((2,3)\), and \((2,8)\), become the vertices of the trapezoid region. Plotting these accurately on a graph allows us to see the exact shape and position of our region of interest. This visual understanding sets the stage for correctly setting up the integral.

Integral Boundaries

Setting the correct boundaries for an iterated integral is essential for proper integration across a region. In our exercise, the boundaries for the region \(R\) are determined by the given inequalities:

• The \(x\) boundaries are straightforward: from 1 to 2, as defined by \(1 \leq x \leq 2\).
• For \(y\), the boundaries alter based on \(x\). They are defined by \(y \geq x + 1\) and \(y \leq 2x + 4\). As \(x\) moves within its range, \(y\) is dynamically limited by these equations.

The process of identifying these boundaries involves:- Analyzing the inequalities to understand both the minimum and maximum y-values for each x-value.- Ensuring the y-values are expressed in terms of x so they can fit into an iterated integral efficiently.Correctly determining these limits guarantees the integration accurately covers the entire region \(R\), without excess or missing parts.

Integration Order

When setting up an iterated integral, deciding the order of integration impacts both the formulation and complexity of the problem. For this exercise, we use the order \(dy \, dx\). This means:

• First, you integrate with respect to \(y\), keeping \(x\) constant.
• Next, you integrate the resulting expression with respect to \(x\).

Using this order clarifies the region's shape dictates how we express \(y\)'s boundaries as functions of \(x\). Thus, the integral becomes \(\int_{1}^{2}\int_{x+1}^{2x+4} f(x, y) \, dy \, dx\), where:- The outer integral \(\int_{1}^{2}\) represents \(x\)'s limits from 1 to 2.- The inner integral \(\int_{x+1}^{2x+4}\) adjusts \(y\)'s limits.

This approach allows for a structured way to handle the integration step by step, ensuring that the function \(f(x, y)\) is evaluated correctly over the entire region \(R\) as defined.