Problem 63 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int_{0}^{\infty} e^{-a x} d x=\frac{1}{a}\), then \(\int_{0}^{\infty} x^{n} e^{-a x} d x\) is (A) \(\frac{(-1)^{n} n !}{a^{n+1}}\) (B) \(\frac{(-1)^{n}(n-1) !}{a^{n}}\) (C) \(\frac{n !}{a^{n+1}}\) (D) None of these

Step-by-Step Solution

Answer

Option C: \(\frac{n!}{a^{n+1}}\).

Step 1: Understanding the Problem

We are given a definite integral from 0 to \(\infty\) that represents the repeated integration involving an exponential function. We need to find the expression for \(\int_{0}^{\infty} x^{n} e^{-a x} d x\) using the given information that \(\int_{0}^{\infty} e^{-a x} d x=\frac{1}{a}\).

Step 2: Using Integration by Parts

To solve \(\int_{0}^{\infty} x^{n} e^{-a x} d x\), we apply the technique of integration by parts: \(\int u \, dv = uv - \int v \, du\). Let \(u = x^n\) and \(dv = e^{-ax} \, dx\). Then \(du = n x^{n-1} \, dx\) and \(v = -\frac{1}{a}e^{-ax}\).

Step 3: Setting up Integration by Parts Formula

Substitute into the integration by parts formula: \(\int_{0}^{\infty} u \, dv = [x^n (-\frac{1}{a} e^{-a x})]_{0}^{\infty} - \int_{0}^{\infty} (-\frac{1}{a} e^{-a x}) n x^{n-1} \, dx\). The term \([x^n (-\frac{1}{a} e^{-a x})]_{0}^{\infty}\) evaluates to 0 - 0 = 0.

Step 4: Reducing the Problem

This simplifies the integral to \(\frac{n}{a} \int_{0}^{\infty} x^{n-1} e^{-a x} \, dx\). Notice this is similar to our original integral but one power lower.

Step 5: Recursive Relationship & Simplification

By recursively applying integration by parts \(n\) times, we continue reducing the power of \(x\) until it reaches zero. This results in the relationship \(\int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}}\). This accounts for all powers from \(x^n\) down to \(x^0\) each contributing a factor of \(-\frac{1}{a}\) and counting through \(n!\).

Step 6: Final Answer

Thus the expression is \(\frac{n!}{a^{n+1}}\). This corresponds to option C.

Key Concepts

Integration by Parts

Integration by parts is a powerful tool in calculus used to solve integrals involving the product of functions. It's especially helpful when dealing with integrals that combine polynomials and exponential functions, like the exercise at hand. The formula for integration by parts is given by:
\[ \int u \, dv = uv - \int v \, du \]
Choosing the right functions for \( u \) and \( dv \) is crucial for simplifying the integral. Typically, we select \( u \) as the algebraic part (polynomial), because its derivative gets simpler, and \( dv \) as the exponential part, because its integral remains straightforward.

• In this exercise, \( u = x^n \) and \( dv = e^{-ax} \, dx \) were chosen. Consequently, the derivative \( du = n x^{n-1} \, dx \) simplifies the problem when combined with the integral of \( dv \).
• This choice transforms the original integral into a recursive form, paving the way for solving problems with higher powers of \( x \) by repetition.

Recursive Integration

Recursive integration leverages the concept of applying a method repeatedly to break down a complex integral into simpler parts. Using integration by parts multiple times reveals a pattern or sequence that can simplify the problem. Here, the integral \( \int_{0}^{\infty} x^n e^{-ax} \, dx \) becomes easier to solve as we observe a repetitive structure.

• Each application of integration by parts reduces the power of \( x \) from \( n \) to \( n-1 \), forming a sequence of integrals.
• This recursive strategy culminates in reaching the simplest form when the power of \( x \) reaches zero, namely \( \int_{0}^{\infty} e^{-ax} \, dx \).

In this exercise, each step in the recursion builds on the previous one until it reaches a solution that encompasses all conditions given, arriving at the final answer efficiently and systematically.

Exponential Function Integration

Integrating exponential functions is a common component in calculus and is usually manageable due to the nature of the exponential term. The exponential decay or growth, represented as \( e^{-ax} \), plays a critical role in diminishing the effect of the polynomial term \( x^n \) over an infinite range. The formula used here:
\[ \int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a} \]
provides an essential condition that simplifies the evaluation of more complex integrals involving exponential functions.

• In these problems, the exponential function's integral over \([0, \infty)\) leads to a convergence that is invaluable for calculations.
• The exponential's simple derivative and anti-derivative properties make these functions compatible with techniques like integration by parts and recursive approaches.

In essence, understanding exponential function integration allows us to effectively solve integrals that may initially appear complex by employing natural logarithmic properties and convergence behavior.