91Ó°ÊÓ

Understanding the integration of functions is foundational to solving problems involving areas under curves and total accumulated values. In calculus, an integral of a function represents the area under the curve of that function within certain limits.

The given problem presents us with two separate integrals, each with distinct limits. The first, \(\int_{1}^{3} f(x) dx = -2\), signifies the net area under the curve \(f(x)\) from \(x = 1\) to \(x = 3\). Similarly, \(\int_{-5}^{-4} g(y) dy = 4\) represents the net area under the curve \(g(y)\) between \(y = -5\) and \(y = -4\).

To enhance comprehension, visualize the graph of \(f(x)\) along the x-axis, and \(g(y)\) along the y-axis, each occupying their respective intervals. The concept becomes crucial when these individual integrals form a component of a larger composite function, particularly within a double integral over a rectangular region.

In the given exercise, the region of integration, denoted as \(D\), is described as a rectangle. A rectangular region in the coordinate plane is defined by a set of inequalities representing the range of \(x\) and \(y\) values within the rectangle's boundaries. For our exercise, \(1 \leq x \leq 3\) and \( -5 \leq y \leq -4\) delineate the corners of this rectangle.

When evaluating a double integral over such a region, you should visualize a rectangle on the coordinate plane with sides parallel to the axes. This imagery simplifies the setup of the double integral, as it allows us to neatly separate the integration for \(x\) and for \(y\) into distinct intervals corresponding to the sides of the rectangle. The simplicity of this region's geometry often enables us to decouple the double integral into the product of two single-variable integrals, facilitating easier computation.

In specific scenarios, such as the one in our exercise, the double integral of a product can be split into the product of two separate integrals when the functions involved depend on different variables. The principle that allows us to do so stems from the independence of the variable of integration in each of the functions \(f(x)\) and \(g(y)\).

Illustratively, since \(f(x)\) is a function of \(x\) alone and \(g(y)\) is a function of \(y\) alone, their integration over a rectangular region can be disentangled and carried out independently of one another. After computing these integrals separately, as shown in the step-by-step solution, their product gives us the final result of the original double integral.

Thus, the value of \(\iint_{D} f(x) g(y) dA\) translates to \(\int_{1}^{3} f(x) dx\) multiplied by \(\int_{-5}^{-4} g(y) dy\), which simplifies the computation to a mere multiplication of the two given integral values.