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Chapter 16: Problem 3

Which order of integration is preferable to integrate \(f(x, y)=x y\) over \(R=\\{(x, y): y-1 \leq x \leq 1-y, 0 \leq y \leq 1\\} ?\)

Short Answer

Expert verified
Answer: The preferable order of integration for this exercise is to integrate with respect to y first and then x (dy dx).

Step by step solution

01

1. Visualize the region R

First, draw the boundaries of the region R by sketching the lines: \(x = y-1, x = 1-y, y = 0, \text{and}\, y = 1\). This will help visualize the ordered pairs (x, y) and inequalities that make up the region R.
02

2. Rewrite the inequalities

We currently have the inequalities as a function of x: \(y-1 \leq x \leq 1-y\) However, we need to write them as functions of y. Rearrange the inequalities to obtain these functions: \(y \geq x+1 \text{ and } y \leq 1-x\)
03

3. Determine the order of integration for dx dy and dy dx

We have two main possibilities for integrating in this exercise: either integrate first with respect to x and then with respect to y (dx dy), or first with respect to y and then with respect to x (dy dx). Here, we will find the limits of integration for each case: a) For dx dy: \(x\) ranges from \(y-1\) to \(1-y\), and \(y\) ranges from \(0\) to \(1\). b) For dy dx: \(y\) ranges from \(x+1\) to \(1-x\), and \(x\) ranges from \(-1\) to \(0\). Given these range expressions, it appears that both orders of integration contain simple bounds, but the dy dx integration seems simpler due to lower bound being \(-1\) and upper bound being \(0\) for x.
04

4. Choose the preferable order of integration

Comparing the two possibilities discussed above, the dy dx order has simpler boundaries. Therefore, the preferable order of integration for this exercise is to integrate with respect to y first and then x (dy dx).

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