Chapter 81: Problem 5
\(\int_{0}^{1} x^{p}(1-x)^{q} \mathrm{~d} x=\frac{p ! q !}{(p+q+1) !}\) für \(p, q \in \mathrm{N}_{0}\) Hinweis: Wiederholte Produktintegration.
Short Answer
Expert verified
The integral evaluates to \(\frac{p! q!}{(p+q+1)!}\).
Step by step solution
01
Set up the integral
We need to evaluate the integral \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x \). Here, \(p\) and \(q\) are non-negative integers.
02
Use substitution
We can simplify this integral using the substitution \( t = 1-x \), which gives \( \mathrm{d}t = - \mathrm{d}x \). Then the integral becomes \( \int_{1}^{0} (1-t)^p t^q (-\mathrm{d}t) \), which simplifies to \( \int_{0}^{1} t^q (1-t)^p \, \mathrm{d}t \).
03
Apply the Beta function
The integral now resembles the Beta function \( B(q+1, p+1) = \int_{0}^{1} t^q (1-t)^p \, \mathrm{d}t \). We use the fact that \( B(q+1, p+1) = \frac{q!p!}{(p+q+1)!} \).
04
Conclude with the result
Since the integral evaluates perfectly to the expression for the Beta function, we confirm that \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x = \frac{p!q!}{(p+q+1)!} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus. It is the process of finding the integral of a function and is used to determine areas under curves and solutions to differential equations. In simple terms, integration is like finding the total sum or the total accumulation of a function's output over an interval.
For the given integral, \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x \), where \( p \) and \( q \) are non-negative integers, we are aiming to calculate the total accumulation of the function \( x^p (1-x)^q \) from \( x = 0 \) to \( x = 1 \).
This specific integral is solved through repeated product integration, which is a technique often used to handle products of functions within an integral.
For the given integral, \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x \), where \( p \) and \( q \) are non-negative integers, we are aiming to calculate the total accumulation of the function \( x^p (1-x)^q \) from \( x = 0 \) to \( x = 1 \).
This specific integral is solved through repeated product integration, which is a technique often used to handle products of functions within an integral.
Beta Function
The Beta function is a special function in mathematics, closely related to gamma functions and factorials. It is defined for positive real numbers \( x \) and \( y \) by the integral:
\[ B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \mathrm{d}t \]
This function can evaluate integrals involving products of powers similar to our problem. In this exercise, the Beta function \( B(q+1, p+1) \) is used as a solution to the integral \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x \).
The result \( \frac{q!p!}{(p+q+1)!} \) is obtained directly using the properties of the Beta function, showing how powerful this tool can be in simplifying complex integrations.
\[ B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \mathrm{d}t \]
This function can evaluate integrals involving products of powers similar to our problem. In this exercise, the Beta function \( B(q+1, p+1) \) is used as a solution to the integral \( \int_{0}^{1} x^p (1-x)^q \, \mathrm{d}x \).
The result \( \frac{q!p!}{(p+q+1)!} \) is obtained directly using the properties of the Beta function, showing how powerful this tool can be in simplifying complex integrations.
Substitution Method
The substitution method is a technique used to simplify integrals. It involves changing the variable of integration to make the integration process easier.
In this exercise, we used the substitution \( t = 1-x \). This means the new variable \( t \) takes the value \( 1 \) when \( x=0 \) and \( 0 \) when \( x=1 \). The differential \( \mathrm{d}t = -\mathrm{d}x \) adjusts the limits of integration, turning the integral into \( \int_{0}^{1} t^q (1-t)^p \, \mathrm{d}t \).
By applying this substitution, we transform the original integral into a form that matches the Beta function, thus making it easier to evaluate.
In this exercise, we used the substitution \( t = 1-x \). This means the new variable \( t \) takes the value \( 1 \) when \( x=0 \) and \( 0 \) when \( x=1 \). The differential \( \mathrm{d}t = -\mathrm{d}x \) adjusts the limits of integration, turning the integral into \( \int_{0}^{1} t^q (1-t)^p \, \mathrm{d}t \).
By applying this substitution, we transform the original integral into a form that matches the Beta function, thus making it easier to evaluate.
Factorial Function
The factorial function is a mathematical function denoted by \( n! \), where \( n \) is a non-negative integer. The factorial of \( n \) is the product of all positive integers less than or equal to \( n \).
For example:
Using factorials allows us to express the result of the Beta function and delivers the final solution in a compact form, highlighting the factorial's significance in mathematical expressions and solutions.
For example:
- \( 0! = 1 \)
- \( 1! = 1 \)
- \( 2! = 2 \times 1 = 2 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
Using factorials allows us to express the result of the Beta function and delivers the final solution in a compact form, highlighting the factorial's significance in mathematical expressions and solutions.