Question

Approximate the integral \(\int_{R} \int f(x, y) d A\) by dividing the rectangle \(R\) with vertices \((0,0),\) \((4,0),(4,2),\) and (0,2) into eight equal squares and finding the sum \(\sum_{i=1}^{8} f\left(x_{i}, y_{i}\right) \Delta A_{i}\) where \(\left(x_{i}, y_{i}\right)\) is the center of the \(i\) th square. Evaluate the iterated integral and compare it with the approximation. $$ \int_{0}^{4} \int_{0}^{2}(x+y) d y d x $$

Short answer

The approximate value of the integral using the given method is 24 while the exact value calculated from the double integral is 32.

Step-by-step solution

Approximation

First, divide the rectangle into eight equal squares. Each square will have dimensions of 1x1, hence, \(\Delta A_{i} = 1\). The centers of these squares (where \(x_i, y_i\) will be evaluated) are \((0.5,0.5), (1.5,0.5), (2.5,0.5), (3.5,0.5), (0.5,1.5), (1.5,1.5), (2.5,1.5), (3.5,1.5)\). Then, find the sum \(\sum_{i=1}^{8} f(x_{i}, y_{i}) \Delta A_{i}\), where \(f(x, y) = x+y\).

Evaluate the Sum

The sum from the first step simplifies to \(0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5 = 24\). This is the approximation result.

Evaluate the Double Integral

Proceed to evaluate the double integral \(\int_{0}^{4} \int_{0}^{2} (x+y) dy dx\). The inner integral \(\int_{0}^{2} (x+y) dy\) simplifies to \(2x + y^2 \Bigl|_0^2 = 2x + 4\). The outer integral is then \(\int _{0}^{4} (2x + 4) dx = x^2 + 4x \Bigl|_0^4 = 16 + 16 = 32\). This is the exact value.

Comparison

Comparing the results, though the approximation method produces a close result, it provides a result that is smaller than the exact value.

Key concepts

Riemann Sums

Understanding Riemann sums is crucial in grasping the foundational concept of integral calculus. These sums allow us to approximate the area under a curve within a certain interval, thereby providing an estimation of the integral.

Riemann sums work by dividing a region into smaller parts, typically rectangles or trapezoids, and then calculating the area of these parts to get a total sum. This approach comes in handy when we're dealing with a non-uniform or complex region. The rectangle heights can be determined by the function's value at specific points such as the left endpoint, right endpoint, midpoints, or any other point within the subintervals.

In the provided exercise, eight squares with equal areas are created within the rectangular region, and the midpoints are used to calculate the height of the rectangles - a common way to implement Riemann sums. As the number of rectangles increases (or equivalently, as their width decreases), the sum becomes a better approximation of the integral, leading ideally to the exact area as the limit approaches infinity.

Approximation of Integrals

The process of approximating integrals is a practical application that surfaces in many fields including physics, engineering, and economics where exact values may not always be required or feasible to determine.

Approximation methods, including Riemann sums, the trapezoidal rule, and Simpson's rule, are used when we cannot find an antiderivative or when we deal with empirical data. Each method has its own level of accuracy and ease of calculation. For example, mid-point Riemann sums, which were utilized in the given exercise, provide a balance between simplicity and accuracy, making them a popular choice in many scenarios.

The choice of an approximation method depends on the function's behavior, the desired level of accuracy, and computational resources. In educational settings, approximation exercises foster a deeper understanding of how integrals describe 'accumulated change' and prepare students for practical situations where numerical methods are frequently necessary.

Iterated Integrals

Iterated integrals extend the concept of integration into higher dimensions. While a single integral represents the area under a curve, double integrals allow us to find the volume under a surface over a region in two-dimensional space.

The fundamental principle at work is breaking down the complex problem of finding the volume under a two-dimensional surface into a series of simpler one-dimensional integrals. This is done by first fixing one variable and integrating with respect to the other, and then integrating the result with respect to the first variable.

In the example solution, the double integral is computed iteratively. First, the integral is taken with respect to y, treating x as a constant, and then with respect to x. The result gives the exact volume under the surface created by the function f(x, y) = (x + y) over the specified region. This specific calculation exemplifies how iterated integrals can yield precise results, contrasting with the approximation which, albeit close, does not give the exact volume.