Question

The value of \(\beta(2,1)+\beta(1,2)\) is \ldots.....

Short answer

The value is 1.

Step-by-step solution

Understanding the Beta Function

The beta function, denoted as \( \beta(x, y) \), is a special function defined by the integral \( \int_0^1 t^{x-1} (1-t)^{y-1} \, dt \). We will use this definition to calculate \( \beta(2,1) \) and \( \beta(1,2) \).

Calculating \( \beta(2,1) \)

In this case, plug \( x = 2 \) and \( y = 1 \) into the definition of the beta function. This gives us \( \beta(2,1) = \int_0^1 t^{2-1} (1-t)^{1-1} \, dt = \int_0^1 t \, dt \), which evaluates to \( \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2} \).

Calculating \( \beta(1,2) \)

Now, to calculate \( \beta(1,2) \), substitute \( x = 1 \) and \( y = 2 \) into the beta function definition. This results in \( \beta(1,2) = \int_0^1 t^{1-1} (1-t)^{2-1} \, dt = \int_0^1 (1-t) \, dt \), which evaluates to \( \left[ t - \frac{t^2}{2} \right]_0^1 = \frac{1}{2} \).

Adding the Results

Now add the two evaluated functions: \( \beta(2,1) = \frac{1}{2} \) and \( \beta(1,2) = \frac{1}{2} \). Thus, \( \beta(2,1) + \beta(1,2) = \frac{1}{2} + \frac{1}{2} = 1 \).

Key concepts

Integral Calculus

Integral calculus is one of the major branches of calculus, focused on understanding and applying the process of integration. Unlike differentiation, which deals with finding the rate of change of a function, integration is concerned with finding the total accumulation of quantities.
It involves calculating the area under curves or accumulating quantities, such as finding the work done by a force over a distance.
When working with integral calculus, you often encounter definite integrals, which have limits, like 0 and 1 in our exercise. These limits define the interval over which you are integrating.

• The definite integral gives the net area between the curve and the x-axis within those limits.
• It can be thought of as a sum of infinitesimally small products of function values and interval lengths, resulting in a total accumulation.

In this exercise, calculating \( \beta(2,1) \) and \( \beta(1,2) \) involved evaluating such definite integrals. Mastering integral calculus is crucial for understanding many mathematical concepts, including special functions like the Beta Function.

Special Functions

Special functions are a significant part of mathematical analysis, often emerging in various problems in physics and engineering. They are termed 'special' because they aren't necessarily elementary like polynomials but possess unique and often very useful properties in mathematical models.

• The Beta Function, denoted by \( \beta(x, y) \), is one of these special functions and is defined by the integral \( \int_0^1 t^{x-1} (1-t)^{y-1} \, dt \).
• Interestingly, the Beta Function is symmetric, meaning \( \beta(x, y) = \beta(y, x) \).
• It also relates to the Gamma Function, another special function, through the equality: \( \beta(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \).

In solving our exercise, knowing the properties and definition of the Beta Function was crucial in calculating \( \beta(2,1) \) and \( \beta(1,2) \). Understanding these functions enhances your capability to tackle complex mathematical problems.

Mathematical Analysis

Mathematical analysis is a branch of mathematics dealing with limits and related theories, such as differentiation, integration, measure, infinite series, and analytic functions.
In this exercise, mathematical analysis comes into play through the detailed study of integrals in the Beta Function.

• This branch seeks to rigorously establish the properties of real numbers, sequences, and functions.
• It serves as the foundation for understanding more complex topics such as real and complex analysis.
• In our context, employing mathematical analysis helped formalize the steps to evaluate the definite integrals for \( \beta(2,1) \) and \( \beta(1,2) \).

By engaging deeply with these concepts, students can enhance their critical thinking skills and ability to solve intricate mathematical problems. It's a doorway to higher mathematics, preparing you for advanced studies in calculus and beyond.