Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(\int_{a}^{b}[f(x)-g(x)] d x=A,\) then \(\int_{a}^{b}[g(x)-f(x)] d x=-A .\)

Short answer

The statement is true. If \(\int_{a}^{b}[f(x)-g(x)] d x=A,\) then indeed \(\int_{a}^{b}[g(x)-f(x)] d x=-A .\) This is due to the key property of definite integrals that reversing the order of the functions in the integral reverses the sign of the integral's value.

Step-by-step solution

Understanding the statement

The statement is suggesting that if the integral of [f(x)-g(x)] from 'a' to 'b' is 'A', then the integral of [g(x)-f(x)] from 'a' to 'b' should be '-A'. This implies that changing the order of functions in the integral should reverse the sign of the integral's value.

Recalling the properties of integrals

One important property of definite integrals is that reversing the order of the functions in the integral reverses the sign of the integral's value. It means if we have an integral of the form \(\int_{a}^{b}[f(x)-g(x)] d x\), and we switch the functions to get \(\int_{a}^{b}[g(x)-f(x)] d x\), the value of the integral will be the negative of the original.

Verifying the statement

By applying the stated property of integrals, one can see that the statement is true. If \(\int_{a}^{b}[f(x)-g(x)] d x=A\), then indeed \(\int_{a}^{b}[g(x)-f(x)] d x=-A .\)

Key concepts

Integral Properties

Understanding the properties of definite integrals is crucial for mastering the concepts of integral calculus. As you delve into the world of integrals, you will encounter various rules that simplify the process of integration. One such essential property is related to the linearity of integrals. Essentially, this property states that the integral of a difference of functions \(f(x) - g(x)\) over a specific interval can be decomposed into separate integrals, expressed as \(\int_{a}^{b} [f(x) - g(x)] dx = \int_{a}^{b} f(x) dx - \int_{a}^{b} g(x) dx\).

Additionally, if the positions of the functions within the integral are reversed, the result is the negation of the original value. This is represented mathematically as \(\int_{a}^{b} [g(x) - f(x)] dx = -(\int_{a}^{b} [f(x) - g(x)] dx)\).

These integral properties are invaluable for solving complex integral calculus problems and facilitate the application of more advanced techniques, such as integration by parts or substitution. By internalizing these basic properties, students can construct a solid foundation for their understanding of integral calculus.

Antiderivatives

In integral calculus, antiderivatives play a significant role as they represent the reversal of differentiation. An antiderivative of a function \(f(x)\) is another function \(F(x)\) whose derivative is \(f(x)\). In other words, \(F'(x) = f(x)\). When it comes to definite integrals, antiderivatives are utilized to calculate the area under the curve of a function between two points.

This process is formalized by the Fundamental Theorem of Calculus, which connects antiderivatives with definite integrals. The theorem states that if \(F(x)\) is an antiderivative of \(f(x)\), then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). The ability to find antiderivatives is essential for evaluating definite integrals and is a core skill in calculus.

The concept of antiderivatives extends beyond merely finding the area beneath curves. It includes understanding the behavior of accumulative quantities and solving first-order differential equations, underscoring its significance in both theoretical and applied mathematics.

Integral Calculus

Integral calculus, alongside differential calculus, forms the fundamental building blocks of calculus. The key purpose of integral calculus is to find the accumulation of quantities and the area beneath curves. These problems are often represented by definite integrals, calculations that provide the total accumulation within a specified range, from point \(a\) to point \(b\), denoted as \(\int_{a}^{b} f(x) dx\).

Integral calculus is not only about computation but also about understanding the concept of accumulation and the graphical interpretation of areas and volumes. It plays a pivotal role in various fields, including physics, engineering, economics, and beyond. Mastery of integral calculus demands a firm grasp of its foundational properties and the ability to apply them effectively. This includes recognizing the function of antiderivatives, applying the Fundamental Theorem of Calculus, and utilizing specific techniques such as substitution or integration by parts to solve more challenging integrals. By piecing together these critical concepts, students and practitioners alike can unlock the full potential of integral calculus for analytical and practical problem-solving.