Problem 33 - Thomas Calculus

Question

In Exercises \(33-46,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{1} \int_{2}^{4-2 x} d y d x \end{equation}

Step-by-Step Solution

Answer

Reversed integral: \(\int_{2}^{4} \int_{0}^{\frac{4-y}{2}} dx \, dy\).

Step 1: Understand the Given Integral

The given integral is \(\int_{0}^{1} \int_{2}^{4-2x} dy \, dx\). It describes a region in the \(xy\)-plane. The outer integral limits \(x\) from \(0\) to \(1\), and for each \(x\), \(y\) is integrated from \(2\) to \(4 - 2x\). Our task is to sketch this region and reverse the order of integration.

Step 2: Sketch the Region of Integration

First, consider the inequalities from \(2 \leq y \leq 4 - 2x\) and \(0 \leq x \leq 1\). When \(x = 0\), \(y\) ranges from \(2\) to \(4\). When \(x = 1\), \(y = 4 - 2 \times 1 = 2\). Thus, the line segment starts at \(y = 4\) for \(x=0\) and ends at \(y=2\) for \(x=1\) on the line \(y = 4 - 2x\). The region is bounded by the lines \(y = 2\), \(y = 4 - 2x\), \(x = 0\), and \(x = 1\).

Step 3: Determine New Integration Limits

To reverse the order, express \(x\) in terms of \(y\). From \(y = 4 - 2x\), rearranging gives \(x = \frac{4 - y}{2}\). The region's vertical strip integrates \(x\) from 0 to \(\frac{4-y}{2}\), and \(y\) ranges from 2 to 4 based on the sketch.

Step 4: Write the Reversed Double Integral

With the new limits, the reversed order integral becomes \(\int_{2}^{4} \int_{0}^{\frac{4-y}{2}} dx \, dy\). This integral has the same region as the original but with \(x\) and \(y\) integration orders swapped.

Key Concepts

Order of Integration

The order of integration refers to the sequence in which the integration is performed in a double integral. Typically, we perform integration with respect to one variable first, followed by the other. In the original exercise, the order of integration is given as \(\int_{0}^{1} \int_{2}^{4-2x} dy \, dx\). Here, we integrate with respect to \(y\) first and then \(x\).
Reversing the order involves swapping these variables and their limits. It's important to clearly understand what each order means:

• Integrate \(dy\) first: This means for each \(x\), you sum over an interval of \(y\) values before considering the next \(x\).
• Integrate \(dx\) first in the reversed order: Now, you sum over \(x\) for each fixed \(y\) value first.

Changing the order isn't arbitrary, it's typically done to simplify calculations or match the constraints of the problem. Understanding the order of integration helps in visualizing the region of integration and setting up your integrals correctly, especially when changing orders.

Region of Integration

The region of integration is the area over which we perform our integration in the \(xy\)-plane. It's defined by the bounds in our double integral. In this exercise,

• The outer integral bounds \(x\) from \(0\) to \(1\).
• The inner bounds \(y\) from \(2\) to \(4-2x\) within each slice of \(x\).

To visualize this, let's consider the defined line bounds: * \(y = 4 - 2x\) gives a sloped line starting at \((0,4)\) and ending at \((1,2)\). * Horizontal bounds: \(y = 2\) runs constant, while \(x\) limits from \(0\) to \(1\).Sketching the region results in a triangular shape with vertices at \((0,2)\), \((0,4)\), and \((1,2)\).This triangle is where all possible solutions for \(y\) and \(x\) exist within the given limits. This visualization is essential to successfully swapping the integration order, confirming that our new limits capture the same region.

Limits of Integration

Limits of integration dictate the precise range over which you are summing the area under your function. They are crucial in defining the region and the order of integration. In the original integral:

• The outer limits suggest \(x\) varies from \(0\) to \(1\).
• The inner limits describe \(y\) ranging from \(2\) to \(4 - 2x\) for each \(x\).

Reversing integration order demands adjusting these limits to suit the new order. We need to solve the condition \(y = 4 - 2x\) for \(x\), resulting in \(x = (4-y)/2\). Hence, the revised limits for the reversed integral:

• \(y\) now ranges from \(2\) to \(4\), the outermost dimension
• \(x\) varies from \(0\) to \(\frac{4-y}{2}\), inner limits in the changed order.

Correctly setting these limits ensures the integral covers the correct area in the \((x, y)\) plane and is equivalent to the initial one. Always verify new bounds by checking against original constraints.