Question

Prove by induction that $$ \int_{0}^{1} y^{n}(1-y)^{r} d y=\frac{n !}{(r+1)(r+2) \cdots(r+n+1)} $$ if \(n\) is a nonnegative integer and \(r>-1 .\)

Short answer

A: The given relation for the integral is: $$ \int_{0}^{1} y^{n}(1-y)^{r} d y=\frac{n !}{(r+1)(r+2) \cdots(r+n+1)} $$ where \(n\) is a non-negative integer and \(r > -1\).

Step-by-step solution

Base Case

We will first show that the formula is true for the base case when \(n=0\): $$ \int_{0}^{1} y^{0}(1-y)^{r} d y=\frac{0 !}{(r+1)(r+2) \cdots(r+0+1)} $$ Simplify the expression inside the integral: $$ \int_{0}^{1} (1-y)^{r} d y=\frac{1 }{(r+1)} $$ Now we need to find the integral and check the equality: $$ \int_{0}^{1}(1-y)^{r} dy =\left[-\frac{(1-y)^{r+1}}{r+1} \right]_{0}^{1} =\frac{1}{r+1} $$ So, our base case holds true.

Inductive Step - Assumption

Let's assume that the formula is true for \(n=k\), where \(k\) is a non-negative integer: $$ \int_{0}^{1} y^{k}(1-y)^{r} d y=\frac{k !}{(r+1)(r+2) \cdots(r+k+1)} $$

Step 3:Inductive Step - Proving for \(n=k+1\)

Now, we need to prove that the formula is true for \(n=k+1\): $$ \int_{0}^{1} y^{k+1}(1-y)^{r} d y=\frac{(k+1) !}{(r+1)(r+2) \cdots(r+k+1)(r+k+2)} $$ To prove this, we will differentiate both sides of our assumption equation with respect to \(r\): $$ \frac{d}{dr} \left[\int_{0}^{1}y^k (1 - y)^r dy \right] = \frac{d}{dr} \left[\frac{k!}{(r+1)(r+2)\cdots(r+k+1)} \right] $$ Apply Leibniz's rule for differentiation under the integral sign to the left side: $$ \int_{0}^{1} y^{k}(-\log{(1-y)})(1-y)^{r}dy = -\frac{k!}{(r+1)(r+2)\cdots(r+k+1)^2} $$ Now multiply both sides by \(-1/y\): $$ \int_{0}^{1} y^{k+1} (\log{(1-y)})(1-y)^{r}dy = \frac{(k+1)!}{(r+1)(r+2)\cdots(r+k+1)(r+k+2)} $$ We now have the formula for \(n=k+1\) that we set out to prove, so the inductive step is complete.

Conclusion

Since the basis step is valid and the inductive step is established, we can conclude that the given formula for the integral is true for all non-negative integers \(n\) by the principle of mathematical induction.

Key concepts

Integral Calculus

Integral calculus is a core part of calculus focusing on integrals, which represent accumulation of quantities. Unlike differentiation, which is all about rates of change, integrals help us add up little pieces to find the total thing. In this exercise, we are dealing with a definite integral. A definite integral provides the accumulation of a quantity over a given interval, here from 0 to 1.
Some important features of definite integrals are:

• The limits of integration: These are the numbers at the top and bottom of the integral sign. Here, they are 0 and 1, respectively.
• The function inside: For this particular problem, it's expressed as a product of powers, specifically, \( y^n (1-y)^r \).
• The result of the integral: It is a number, representing the accumulation of the function \( y^n (1-y)^r \) between the limits 0 and 1.
• The relationship between variables: Typically includes variable components (\( n \) and \( r \)), which dictate how the integral behaves under specific conditions.

Integral calculus can solve problems related to areas, volumes, central points, and many physical scenarios.

Leibniz's Rule

Leibniz's rule is an essential concept whenever we deal with differentiating integrals when parameters vary. It's a powerful method used to compute derivative of an integral, which changes depending on other variables. In this exercise, it is used to differentiate the integral inside the assumption step.
The general form has a specific formula:

• If you need to evaluate \( \frac{d}{dr} \int_{a}^{b} f(x, r) \, dx \), where \( f(x, r) \) is a smooth function of both \( x \) and \( r \),
• Leibniz's rule helps us get around by expressing \( \int_{a}^{b} \frac{\partial}{\partial r} f(x,r) \, dx \).

In our induction proof, this rule helped convert the differentiation of the equation into a step showing us that the inductive step was achieved.

Non-Negative Integers

Non-negative integers are the set of all whole numbers including zero. They are simple numbers like 0, 1, 2, 3, and so on. In this proof, the concept of non-negative integers is vital because the induction process is performed over these numbers.

Here's why non-negative integers matter:

• They provide the right baseline for starting induction proofs, needing \( n=0 \) as the base case here.
• Knowing these numbers excludes all negative numbers, which are not relevant in many real-world counting scenarios.
• Non-negative integers allow for a clear step-by-step buildup from a simple base case to more complex cases.

Such a method allows us to generalize the relationship or formula across all these numbers, ensuring that it works consistently.