Problem 3 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The preferable order of integration is to integrate with respect to x first and then y, as follows: 1. Integrate with respect to x: \(\int_{y-1}^{1-y}\) 2. Integrate with respect to y: \(\int_{0}^{1}\) This order simplifies the process and minimizes the need for substitutions or trigonometric substitutions while integrating.

Step 1: Visualize the region R

Start by sketching the region R, which is defined by the given inequalities. Draw the lines \(x = y-1\), \(x = 1-y\), \(0 \leq y \leq 1\) on the x-y plane, and identify the region enclosed by the boundaries.

Step 2: Determine the limits of integration for each order

We need to determine the limits of integration for both orders. Let's start with integrating with respect to x first and then y. From the given inequalities, we have \(y-1 \leq x \leq 1-y\), and \(0 \leq y \leq 1\). Therefore, we get the limits of integration as follows: For x: \(\int_{y-1}^{1-y}\) For y: \(\int_{0}^{1}\) Now, let's determine the limits of integration for the reverse order, i.e., integrating with respect to y first and then x. Observe the region and the given inequalities. We can rewrite the inequalities as \(x-1 \leq y \leq 1-x\) and \(0 \leq x \leq 1\). Therefore, we get the limits of integration as follows: For y: \(\int_{x-1}^{1-x}\) For x: \(\int_{0}^{1}\)

Step 3: Compare the orders of integration

Now that we have the limits of integration for both orders, we can see that the preferable order of integration is the one which simplifies the process and doesn't require much manipulation. In this case, the preferable order is: 1. Integrate with respect to x: \(\int_{y-1}^{1-y}\) 2. Integrate with respect to y: \(\int_{0}^{1}\) The reason is that the limits of integration for x are already provided in a simple form, which allows for a more straightforward integration process. The second order (integrating with respect to y first) involves limits that may require some substitutions or trigonometric substitution to simplify, making the first order more preferable.