Question

Reverse the order of integration in the following integrals. $$\int_{0}^{3} \int_{0}^{6-2 x} f(x, y) d y d x$$

Short answer

Question: Reverse the order of integration in the given integral: $$\int_{0}^{3} \int_{0}^{6-2 x} f(x, y) d y d x$$ Answer: The integral with the reversed order of integration is: $$\int_{0}^{6} \int_{0}^{3-\frac{1}{2}y} f(x, y) dxdy$$

Step-by-step solution

1. Determine the region of integration

Given the limits of integration \(0 \le x \le 3\) and \(0 \le y \le 6-2x\), we can rewrite these inequalities as: \(x \ge 0\) \(x \le 3\) \(y \le 6-2x\) \(y \ge 0\) By solving these inequalities simultaneously, we find that the region of integration is a triangle with vertices at (0, 0), (3, 0), and (3, 6).

2. Reverse the order of integration

Now, we will determine the new limits of integration when we integrate with respect to x first and then y. The range of y should be from 0 to 6 because the triangle ranges from the origin (0, 0) to the vertex (3, 6). For the given vertex coordinates, we have the equation of the hypotenuse of the triangle, which is \(y = 6 - 2x\). To reverse the order of integration, we will solve for x: \(x = \frac{6-y}{2}\) or \(x = 3 - \frac{1}{2}y\). So, the new limits of integration with respect to x will be from 0 to \(3 - \frac{1}{2}y\).

3. Write the integral with the new order of integration

The reversed integral will be: $$\int_{0}^{6} \int_{0}^{3-\frac{1}{2}y} f(x, y)dxdy$$ The order of integration is now reversed, and so is the integral.

Key concepts

Integration Limits

Understanding integration limits is crucial for evaluating double integrals. In this exercise, the original integration limits are set up as:

• For the inner integral, which is with respect to \(y\): \(0 \leq y \leq 6 - 2x\).
• For the outer integral, which is with respect to \(x\): \(0 \leq x \leq 3\).

These limits define how the function \(f(x, y)\) is being integrated within the given region. They show that as \(x\) moves from 0 to 3, the value of \(y\) is constrained between 0 and \(6 - 2x\).
Reversing the order of integration involves switching the roles of \(x\) and \(y\) in these limits. This swap requires insights into the shape and boundaries of the integration region to correctly describe the range of \(x\) when \(y\) spans across a particular interval. This process ensures the same region is covered, maintaining the integrity of the integral.

Region of Integration

The region of integration is the area bounded by the limits given in the integral. In this problem, analyzing these limits helps to sketch out the region.

• The boundaries are formulated by the equations \(y = 6 - 2x\), \(x = 0\), \(x = 3\), and \(y = 0\).
• This forms a triangular region when plotted in the xy-plane.
• The vertices of this triangle are at points: (0, 0), (3, 0), and (3, 6).

Understanding this spatial formation is essential. This triangular shape indicates where the function \(f(x, y)\) exists and provides insight into reversing the order of integration.
The hypotenuse is defined by \(y = 6 - 2x\), acting as a crucial diagonal that helps adjust which variable is integrated first when the order is reversed.

Double Integrals

Double integrals allow you to compute the accumulation of quantities across two dimensions.
In this case, we are evaluating \(\int_0^3 \int_0^{6-2x} f(x, y) \ dy \ dx\), where the integration takes place with respect to \(y\) first, followed by \(x\). This means we first sum up the values of \(f(x, y)\) along the line starling from \(y=0\) to \(y=6-2x\), for a fixed \(x\), and then integrate over the range of \(x\).
Reversing the order to integrate with respect to \(x\) first involves changing the limits to \(\int_0^6 \int_0^{3-\frac{1}{2}y} f(x, y) \ dx \ dy\). This involves evaluating the function along \(x=0\) to \(x = 3 - \frac{1}{2}y\) for a given \(y\), and then integrating with respect to \(y\).
Changing integration order is not just a mathematical exercise but a powerful tool that simplifies complex calculations and adapts integrals to make them more manageable while retaining the same region.