91Ó°ÊÓ

Double integral calculus is a powerful tool used to compute the volume under a surface in a two-dimensional space. It differs from a single integral by considering not just a curve but a surface over a given area. To evaluate a double integral, we work with an integral within an integral, hence the term 'double'.

For example, the double integral exercise provided involves the integral \[\int_{0}^{4} \int_{y}^{2 y} x y d x d y\] which signifies that we want to find the volume under the surface described by the function \(x y\) over the region in the xy-plane bounded by \(y = 0\) and \(y = 4\), as well as along the x-direction from \(x = y\) to \(x = 2y\). The process involves integrating with respect to one variable while keeping the other constant and then integrating the result with respect to the second variable.

The concept of integration bounds involves determining the limits between which the integration should be performed. In double integrals, these bounds can be values of either x or y and they define the region over which the function is to be integrated.

From the exercise's double integral \[\int_{0}^{4} \int_{y}^{2 y} x y d x d y\], we can observe that the bounds for \(y\) are from 0 to 4 and the bounds for \(x\) are dependent on the value of \(y\), specifically from \(y\) to \(2y\). It's vital to identify these bounds correctly, as they not only dictate the limits of integration but also ensure that we are integrating over the correct region for the problem.

An antiderivative of a function is another function that, when differentiated, yields the original function. In the context of both single and double integrals, finding the antiderivative is a critical step. It is often referred to as the 'indefinite integral' when no bounds are specified. The main goal in integrating a function is to calculate the antiderivative before applying the limits of integration.

The exercise asks for the antiderivative of \(xy\) with respect to \(x\), which is \[\frac{1}{2}x^2 y\], and subsequently, the antiderivative of the resultant expression with respect to \(y\), which is \[\frac{3}{8}y^4\]. Both steps are crucial to get to the final answer, and each involves applying the fundamental theorem of calculus, which links differentiation with integration.