Question

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{2} \int_{x / 2}^{1} d y d x $$

Short answer

The sketch of region \(R\) defined by the limits of the double integral contains a triangular area in the xy-plane. The order of integration can indeed be switched to form a new double integral as \(\int_{0}^{1} \int_{0}^{2y} d x d y\). Calculting both integrals produces the same final result confirming they represent the same region \(R\).

Step-by-step solution

Interpretation of the original order of integration

Firstly, interpret the given integral \(\int_{0}^{2} \int_{x / 2}^{1} d y d x\). The limits for \(x\) are from 0 to 2. The limits for \(y\) are dependent on \(x\) and range between \(x / 2\) and 1. The region \(R\) is defined by these limits.

Define the region R

Define the region \(R\) as \({(x, y) | 0 ≤ x ≤ 2 and x / 2 ≤ y ≤ 1}\). The lower limit \(y = x / 2\) indicates a line which passes through the origin with a slope of half. The upper limit \(y = 1\) is a horizontal line.

Sketch the region

Draw these lines on a two-dimensional coordinate system. The area between these lines for \(0 ≤ x ≤ 2\) gives the region \(R\).

Change the order of integration

Now, change the order of integration such that we’re integrating first with respect to \(x\) and then with respect to \(y\). To proceed with effectively, observe the restrictions on the values of \(x\) and \(y\). The transformed double integral becomes \(\int_{0}^{1} \int_{0}^{2y} d x d y\).

Analyze the new limits

Here, \(y\) ranges from 0 to 1. For each fixed \(y\), \(x\) ranges from 0 to \(2y\). Note that both integrals represent the same region \(R\) and should yield the same area.