91Ó°ÊÓ

Calculus is a branch of mathematics that studies continuous change. It's divided into two main parts: differential calculus and integral calculus. Differential calculus focuses on the rates at which quantities change, while integral calculus is concerned with the accumulation of quantities.
For example, in the context of our exercise, we use integral calculus to find the total area under a curve over a specific interval. This is shown in the double integral, where we're not just finding the area under a single curve, but under a surface over a certain region in the xy-plane.
Calculus is essential for understanding and describing many natural phenomena and for solving various problems in fields like physics, engineering, and economics.

Integration is the process of finding the integral of a function. It's the reverse operation of differentiation. When we integrate a function, we're essentially summing up an infinite number of infinitesimally small quantities.
In the problem we've worked on, we perform a double integration. This involves two steps:

• First, we integrate with respect to one variable (in this case, x).
• Then, we integrate the resulting expression with respect to another variable (here, y).

Both integrations help in calculating the total 'volume' under a surface defined by the function over a given region.
Practicing integration techniques, such as u-substitution and integration by parts, is crucial for mastering more complex integrals.

An antiderivative, also called an indefinite integral, is a function whose derivative is the original function. Finding the antiderivative is the first step in computing definite integrals.
For the inner integral in our exercise, we first needed to determine the antiderivative of the function 2x. Since the derivative of \(x^2 \) is 2x, we know that the antiderivative of 2x is \(x^2\).
When we find antiderivatives, we typically add a constant of integration (denoted as C). However, this constant cancels out when evaluating definite integrals, which focus on a specific interval.
Understanding how to find antiderivatives is fundamental to solving both definite and indefinite integrals.

Definite integrals give us the exact area under a curve between two points. Unlike antiderivatives, definite integrals compute the net accumulation over an interval.
In our problem, the definite integral \( \int_{0}^{1} \int_{0}^{1} 2 x \, d x \, d y \) was evaluated over the region from 0 to 1 for both x and y.
This process involved:

• Integrating 2x with respect to x over \[0,1\], giving us \(1 \).
• Then, integrating the result (which is 1) with respect to y over \[0,1\]. This also gave us \(1 \).

So, the final result of our double integral is 1.
Definite integrals are widely used in physics and engineering to calculate quantities like areas, volumes, and total accumulations.