Problem 170 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: \(I_{n}=\int_{0}^{\infty} x^{n} e^{-x} d x(n\) is a positive integer) \(\begin{aligned} &=n ! \\ \text { Reason: } I_{n}=& n I_{n-1} \end{aligned}\)

Step-by-Step Solution

Answer

The assertion is true: \( I_n = n! \).

Step 1: Define the Integral

The integral in question is \( I_{n} = \int_{0}^{\infty} x^{n} e^{-x} \, dx \). This integral is used to find properties related to gamma functions and factorials.

Step 2: Use Integration by Parts

Apply integration by parts to the integral \( I_n \). Let \( u = x^n \) and \( dv = e^{-x} \, dx \), then \( du = n x^{n-1} \, dx \) and \( v = -e^{-x} \). The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).

Step 3: Calculate Using Integration by Parts

Substituting into the integration by parts formula, we have: \(\begin{align*}I_n = \left[ -x^n e^{-x} \right]_0^{\infty} + n \int_{0}^{\infty} x^{n-1} e^{-x} \, dx.\end{align*} \)The first term at infinity is zero because \( x^n e^{-x} \to 0 \). At zero, the term is also zero, so:\( I_n = n \int_{0}^{\infty} x^{n-1} e^{-x} \, dx = n I_{n-1} \).

Step 4: Verify the Reason

The reason states \( I_n = n I_{n-1} \). We derived this using integration by parts, confirming that the reason is correct.

Step 5: Establish the Base Case

For the factorials, consider \( I_0 = \int_{0}^{\infty} e^{-x} \, dx \). This results in \( I_0 = 1 \), which is consistent with the fact that \( 0! = 1 \).

Step 6: Deduce the General Formula

We have \( I_n = n \cdot (n-1) \cdot I_{n-2} \) and continuing in this pattern yields \( I_n = n! \). Thus, the assertion \( I_n = n! \) holds true.

Key Concepts

Integration by Parts

The technique of integration by parts is a strong tool in calculus that is used to integrate products of functions more comfortably. It is derived from the product rule for differentiation. Integration by parts is particularly useful when dealing with integrals that involve a polynomial multiplied by an exponential, logarithmic, or trigonometric function.
The formula is given by:

• \(\int u \, dv = uv - \int v \, du\)

Here's a breakdown of how it works:

• Select \(u\) and \(dv\) from the original integral. Typically, \(u\) is chosen to be a function that becomes simpler when differentiated, like a polynomial.
• Compute \(du\) by differentiating \(u\), and find \(v\) by integrating \(dv\).
• Substitute into the formula to simplify the integral.

In our exercise, \(u\) was chosen as \(x^n\), simplifying to \(nx^{n-1}\) upon differentiation. The choice of \(dv = e^{-x} \) simplifies to \(-e^{-x}\) on integration, leading to the simplification and evaluation of the integral.

Factorial

Factorials are a basic yet crucial part of mathematics, notably used in permutations, combinations, and series expansion. A factorial of a non-negative integer \(n\), denoted by \(n!\), is defined as the product of all positive integers less than or equal to \(n\).
For instance:

• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

Factorials often appear in calculus, especially when working with series and the gamma function. The gamma function itself generalizes the concept of factorial for non-integer values. In our exercise, the factorial helps express a recursive formula for integrals. We deduce that \(I_n = n!\) because each step in the recursion multiplies \(I_{n-1}\) by \(n\), forming a sequence akin to the factorial progression.
Factorials also have a simple base case: \(0!\) is defined to be \(1\). This base case is foundational when modeling recurrence relationships within different mathematical contexts.

Improper Integral

Improper integrals are used to extend the concept of integration to functions that are unbounded or over an infinite interval. These occur when the limits of integration are infinite or the function approaches infinity within the interval.
Let's consider:

• \(\int_{0}^{\infty} e^{-x} \, dx\)

This integral evaluates a function over an infinite domain, requiring us to take limits. The function \(e^{-x}\) decreases sharply, allowing the integral to converge even though the upper limit is infinity. For convergence:

• Calculate the limit as \(b \rightarrow \infty\): \(\lim_{b \to \infty} [-e^{-x}]_0^b = 1\).

Our exercise involves an improper integral \( I_n = \int_{0}^{\infty} x^n e^{-x} \, dx \). The integration by parts technique applied ensures that even as \(x\) approaches infinity, we achieve finite results. This process is crucial in validating solutions and deriving properties like those relating to the gamma function.