Problem 44
Question
Let \(m, n \in \mathbb{Z}\) with \(m, n \geq 0 .\) Show that $$ \int_{0}^{1} x^{m}(1-x)^{n} d x=\frac{m ! n !}{(m+n+1) !} $$ (Hint: If \(n \in \mathbb{N}\) and \(I_{m, n}\) denotes the given integral, then using Integration by Parts, \(I_{m, n}=[n /(m+1)] I_{m+1, n-1}\), and \(\left.I_{m+n, 0}=1 /(m+n+1) .\right)\)
Step-by-Step Solution
Verified Answer
Using the hint provided, we can apply the recursive formula and integration by parts to find the closed-form solution for the integral. The basis case for the given integral is \(I_{m,0} = \frac{1}{m+1}\). Applying the recursive formula \(n\) times and substituting the basis case, we obtain:
\[
I_{m, n}=\frac{m ! n !}{(m+n+1) !}.
\]
Hence, we have shown that
\[
\int_{0}^{1} x^{m}(1-x)^{n} d x=\frac{m ! n !}{(m+n+1) !}.
\]
1Step 1: Basis Case - Setting up Integration by Parts
Let \(I_{m, n}\) denote the given integral. The first step is to evaluate the integral with \(n = 0\):
\[
I_{m, 0} = \int_{0}^{1} x^{m}(1-x)^{0} d x.
\]
We need to find an antiderivative of \(x^{m}\). Here, we can use integration by parts with \(u = x^m\) and \(dv = dx\). So, we get:
\[
du = m x^{m - 1} dx, \quad v = x.
\]
Now, we can apply integration by parts to the integral:
\[
I_{m,0} = \left[ u v \right]_0^1 - \int_{0}^{1} v du,
\]
which simplifies to
\[
I_{m,0} = \left[ x^m x \right]_0^1 - \int_{0}^{1} m x^{m} dx.
\]
2Step 2: Evaluating the Basis Case
Now we can evaluate the simplified basis case to get:
\[
I_{m,0} = \left[ x^{m+1} \right]_0^1 - m \int_{0}^{1} x^{m} dx.
\]
As \(x^{m+1}\) is \(0\) when \(x = 0\), and \(1\) when \(x = 1\), we get
\[
I_{m,0} = 1 - m I_{m,0},
\]
which implies that
\[
I_{m,0} = \frac{1}{m+1}.
\]
3Step 3: Applying the Recursive Formula
Now that we have a formula for the basis case, let's use the given recursive formula:
\[
I_{m, n} = \frac{n}{m+1} I_{m+1, n-1} .
\]
We will use this formula repeatedly to get to the \(I_{m, 0}\) case, which we already have a formula for. After each step, we will rewrite the formula until we get to our desired closed-form solution.
4Step 4: Iterating on the Recursive Formula
Applying the recursive formula \(n\) times, we get:
\[
I_{m,n} = \frac{n(n-1)\cdots 1}{(m+1)(m+2)\cdots(m+n)} I_{m+n,0}.
\]
Now, we plug in the basis case formula for \(I_{m+n,0}\), which is \(\frac{1}{m+n+1}\):
\[
I_{m,n} = \frac{n!(m!)}{(m+n+1)(m)!n!}.
\]
5Step 5: Simplifying and Concluding
Finally, we simplify the expression to obtain the closed-form solution for the integral:
\[
I_{m, n}=\frac{m ! n !}{(m+n+1) !}.
\]
So, we have shown that
\[
\int_{0}^{1} x^{m}(1-x)^{n} d x=\frac{m ! n !}{(m+n+1) !}.
\]
Key Concepts
Integral CalculusRecursive FormulaFactorial NotationBinomial Theorem Application
Integral Calculus
Integral calculus is a branch of mathematical analysis that deals with the theory and applications of integrals. In particular, it allows us to find the accumulation of quantities and is intimately related to the concept of an antiderivative.
Understanding integral calculus involves learning about different methods of integration. Integration by parts is one key technique, which is based on the product rule for differentiation and is used to integrate products of two functions. In cases where one of the functions reduces to a simpler form after differentiation (as in the exercise), this method becomes particularly handy. It serves as an essential tool for handling integrals that are not immediately solvable using standard antiderivatives.
When approaching an integral like the one given in the exercise, careful application of integration by parts allows us to break down a seemingly complex integral into more manageable pieces, eventually leading us to a solution.
Understanding integral calculus involves learning about different methods of integration. Integration by parts is one key technique, which is based on the product rule for differentiation and is used to integrate products of two functions. In cases where one of the functions reduces to a simpler form after differentiation (as in the exercise), this method becomes particularly handy. It serves as an essential tool for handling integrals that are not immediately solvable using standard antiderivatives.
When approaching an integral like the one given in the exercise, careful application of integration by parts allows us to break down a seemingly complex integral into more manageable pieces, eventually leading us to a solution.
Recursive Formula
A recursive formula is a way to define the terms of a sequence with respect to the previous terms. Its beauty lies in its ability to process each piece of a sequence step by step while building up to a more comprehensive solution.
In the context of our integral calculus exercise, the recursive formula provided, namely \(I_{m, n} = \frac{n}{m+1} I_{m+1, n-1}\), is vital. It creates a bridge from one integral to another, simplifying the problem by reducing the power of the second factor by one and increasing the power of the first, iteratively, until an easily calculable integral remains.
Use of recursive formulas can significantly simplify computation, especially when paired with a known base case, as seen with \(I_{m+n, 0}\). By applying this technique, the original problem is translated into a format where it can be solved through successive substitutions.
In the context of our integral calculus exercise, the recursive formula provided, namely \(I_{m, n} = \frac{n}{m+1} I_{m+1, n-1}\), is vital. It creates a bridge from one integral to another, simplifying the problem by reducing the power of the second factor by one and increasing the power of the first, iteratively, until an easily calculable integral remains.
Use of recursive formulas can significantly simplify computation, especially when paired with a known base case, as seen with \(I_{m+n, 0}\). By applying this technique, the original problem is translated into a format where it can be solved through successive substitutions.
Factorial Notation
Factorial notation is succinctly symbolized by the exclamation mark \( ! \), which represents the product of a series of descending positive integers. For example, \( 5! \), or '5 factorial,' is equal to \( 5 \times 4 \times 3 \times 2 \times 1 \).
This notation is not just for counting permutations; it plays a significant role in many areas of mathematics, including integral calculus, as exemplified by the formula presented in our exercise \( \frac{m! \times n!}{(m+n+1)!} \). Factorials are essential when working with series expansions, probabilities, and in this case, providing the final result of the integral of a product of polynomial factors.
Factorial expressions also arise naturally when dealing with the coefficients in binomial expansions, and they are central to many combinatorial identities, which are often used to solve integrals involving polynomial terms.
This notation is not just for counting permutations; it plays a significant role in many areas of mathematics, including integral calculus, as exemplified by the formula presented in our exercise \( \frac{m! \times n!}{(m+n+1)!} \). Factorials are essential when working with series expansions, probabilities, and in this case, providing the final result of the integral of a product of polynomial factors.
Factorial expressions also arise naturally when dealing with the coefficients in binomial expansions, and they are central to many combinatorial identities, which are often used to solve integrals involving polynomial terms.
Binomial Theorem Application
The binomial theorem is a powerful tool that allows us to expand expressions of the form \( (x + y)^n \) into a sum involving terms of the form \( \binom{n}{k} x^{n-k} y^k \), where \( \binom{n}{k} \) signifies the binomial coefficients, which are symmetrical and are found in Pascal's triangle.
Although not explicitly invoked in the exercise, the binomial theorem application is implicitly present. The integrand \( x^m(1-x)^n \) resembles the binomial expansion, which often leads to integrals in the same form as our exercise when dealing with probability density functions or moments of a distribution. Understanding the binomial theorem enriches our comprehension of the underlying structure of the integrand and provides additional strategies for manipulating and solving complex integrals involving binomial expressions.
Moreover, recognizing the parallels between binomial expansions and factorials permits a deeper insight into the nature of the solution, as the final form of the integral involves the factorial notation. The combined knowledge of factorial notation and the binomial theorem application thus proves immensely beneficial for students delving into the realm of integral calculus.
Although not explicitly invoked in the exercise, the binomial theorem application is implicitly present. The integrand \( x^m(1-x)^n \) resembles the binomial expansion, which often leads to integrals in the same form as our exercise when dealing with probability density functions or moments of a distribution. Understanding the binomial theorem enriches our comprehension of the underlying structure of the integrand and provides additional strategies for manipulating and solving complex integrals involving binomial expressions.
Moreover, recognizing the parallels between binomial expansions and factorials permits a deeper insight into the nature of the solution, as the final form of the integral involves the factorial notation. The combined knowledge of factorial notation and the binomial theorem application thus proves immensely beneficial for students delving into the realm of integral calculus.
Other exercises in this chapter
Problem 42
Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and \(\phi:[m(f), M(f)] \rightarrow \mathbb{R}\) be continuous. Show that \(\phi \circ f:[a, b] \rightarro
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Let \(f_{1}, \ldots, f_{m}:[a, b] \rightarrow \mathbb{R}\) be integrable functions and let \(r_{j}:=\int_{a}^{b} f_{j}(x) d x\) for \(j=1, \ldots, m .\) Show th
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Let \(a \in \mathbb{R}\) and \(n \in \mathbb{Z}\) with \(n \geq 0 .\) Show that $$ \int_{0}^{a}\left(a^{2}-x^{2}\right)^{n} d x=\frac{\left(2^{n} n !\right)^{2}
View solution Problem 46
(Taylor's Theorem with Integral Remainder) Let \(n\) be a nonnegative integer and let \(f:[a, b] \rightarrow \mathbb{R}\) be such that \(f^{\prime}, f^{\prime \
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