Problem 114 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(I_{n}=\int_{0}^{1}\left(1-x^{a}\right)^{n} d x\), then \(\frac{I_{n}}{I_{n+1}}=1+\frac{1}{k}\), where \(k=\) (A) \((n+1) a\) (B) \(n a\) (C) \((n-1) a\) (D) None of these

Step-by-Step Solution

Answer

The value of \( k \) is \((n+1)a\), so the answer is (A).

Step 1: Understand the Problem

We are given the integral \( I_n = \int_{0}^{1} (1-x^a)^n \, dx \). We need to find the ratio \( \frac{I_n}{I_{n+1}} \) and show that it equals \( 1 + \frac{1}{k} \), determining the value of \( k \).

Step 2: Write Expressions for \( I_n \) and \( I_{n+1} \)

The integral \( I_n = \int_{0}^{1} (1-x^a)^n \, dx \) and \( I_{n+1} = \int_{0}^{1} (1-x^a)^{n+1} \, dx \).

Step 3: Simplify \( \frac{I_n}{I_{n+1}} \) using Substitution

Perform substitution by letting \( t = 1 - x^a \), hence \( dt = -a x^{a-1} \, dx \) or \( dx = -\frac{dt}{a x^{a-1}} \). When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). The transformed integral becomes \( I_n = \int_{1}^{0} t^n \left(-\frac{dt}{a x^{a-1}} \right) \).

Step 4: Evaluate the Transformed Integral

Calculate \( I_n = \frac{1}{a} \int_{0}^{1} t^n x^{1-a} \, dt \). Note that \( x = (1-t)^{1/a} \), substitute \( x^{1-a} = (1 - t)^{(1-a)/a} \). Then, integrate: \( \int_{0}^{1} t^n (1-t)^{(1-a)/a} \, dt \).

Step 5: Express the Ratio and Identify \( k \)

To find \( \frac{I_n}{I_{n+1}} \), observe that \( I_n = \int_{0}^{1} t^n (1-t)^{(n-1)a/a} \, dt \) and \( I_{n+1} = \int_{0}^{1} t^{n+1} (1-t)^{n a/a} \, dt \). The ratio thus simplifies due to the Gamma function relationship: \[ \frac{I_n}{I_{n+1}} = \frac{\Gamma(n+1) \Gamma(1+(n-1)a/a)}{\Gamma((n+1) a/a)} \].\; This implies \( \frac{I_n}{I_{n+1}} = 1 + \frac{1}{(n+1)a} \), letting \( k = (n+1)a \).

Step 6: Conclusion

Given the simplification, the expression \( \frac{I_n}{I_{n+1}} = 1 + \frac{1}{(n+1)a} \), the value of \( k \) is \( (n+1)a \). Therefore, the correct answer is option (A).

Key Concepts

Definite Integrals

In the realm of Integral Calculus, a definite integral helps us find the exact area under a curve across a specific interval. Here, the main goal is to evaluate \[ I_n = \int_{0}^{1} (1-x^a)^n \, dx \]. This means we are calculating the area from \(x = 0\) to \(x = 1\) of the function \((1-x^a)^n\).

Definite integrals have both an upper and lower limit, defining these boundaries precisely.
For any definite integral \(\int_{a}^{b} f(x) \, dx\), the number \(a\) is called the lower limit, and \(b\) is the upper limit.
The definite integral is denoted by the function \(F\), such that \( F'(x) = f(x) \), hence \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a). \]

In this exercise, by transforming the integral, we facilitate easier evaluation and ultimately find crucial expressions that will allow simplifying the original problem even further.

Gamma Function

The Gamma function extends the concept of factorials to continuous values, making it an essential tool in integral calculus. Given by \[ \Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} \, dt \], it provides a continuous function that agrees with factorials for positive integers, i.e., \(\Gamma(n) = (n-1)!\).

In our solution, we used the relationship with the Gamma function to simplify the expression:

• For integer values, linking \( n! \) to \( \Gamma(n+1) \) is straightforward.
• Mais développe en more complex scenarios where fractions or rational numbers are involved, the Gamma function handles these naturally, greatly simplifying calculations.

By applying the Gamma function, the transformed integral ratio became approachable, allowing us to express \( \frac{I_n}{I_{n+1}} \) concisely with \( \Gamma \) functions in the numerator and denominator, simplifying our task to find \( k \).

Substitution Method

The substitution method is a common technique used in calculus to simplify the process of integration by transforming variables. In this exercise, substitution plays a vital role in evaluating the integral \( \int(1-x^a)^n \, dx \).

We let \( t = 1 - x^a \), resulting in new bounds:

• When \( x = 0 \), \( t = 1 \).
• When \( x = 1 \), \( t = 0 \).

This changes the integral limits, flipping them, which must be accounted for by negating the integral.

The differential \( dx \) was adjusted accordingly: \( dx = -\frac{dt}{a x^{a-1}} \). Using substitution transforms the complexity into a simpler expression that can now be solved using integral calculus techniques such as the Gamma function.

This method is powerful because:

• It identifies transformations that align the integrand with known integral forms or facilitate easier integration.
• It often links integrals back to familiar functions, aiding further simplifications like in solving \( \frac{I_n}{I_{n+1}} \).