Problem 126 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of a positive integer \(n \leq 5\) such that \(\int_{0}^{1} e^{x}(x-1)^{n} d x=16-6 e\) is (A) 1 (B) 2 (C) 3 (D) 4

Step-by-Step Solution

Answer

n = 3

Step 1: Understand the Problem

We need to find the positive integer \( n \) such that \( n \leq 5 \) and \( \int_{0}^{1} e^{x}(x-1)^{n} \, dx = 16 - 6e \). We'll evaluate the integral for each value of \( n \) from 0 to 5 and determine which matches the value 16 - 6e.

Step 2: Set Up the Integral

The given integral is \( \int_{0}^{1} e^{x}(x-1)^{n} \, dx \). The idea is to integrate by parts or use a known formula for specific cases when n is small. Let's start evaluating for small values of \( n \).

Step 3: Evaluate Integral for Small n

For \( n=0 \), the integral is \( \int_{0}^{1} e^x \, dx \). This evaluates to \( [e^x]_0^1 = e - 1 \). It doesn't match 16 - 6e, so we try next values of \( n \).

Step 4: Calculate for n=1

For \( n=1 \), \( \int_{0}^{1} e^x (x-1) \, dx \) is chosen to be solved, and results using integration by parts. Solving gives us \( 1 - e \). This doesn't match, so continue to \( n=2 \).

Step 5: Test for n=2

Repeat for \( n=2 \). Use integration by parts again: \( \int_{0}^{1} e^x (x-1)^2 \, dx = 2 - e \), still not matching.

Step 6: Check for n=3

For \( n=3 \), evaluate \( \int_{0}^{1} e^x (x-1)^3 \, dx \). Solving this integral gives \( 16 - 6e \), which matches the result required.

Step 7: Confirm and Conclude

We needed \( n \) where the integral equals \( 16 - 6e \). We found this occurs for \( n=3 \). Double-check the integral solution if time allows but our value for \( n \) matches given conditions.

Key Concepts

Definite Integrals

A definite integral is an integral with upper and lower limits that calculates the net area under the curve of a function from one point to another. In our exercise, the definite integral is evaluated from 0 to 1. This means we're looking at the area under the curve of the function

• The exponential function
• Multiplied by a polynomial function raised to the power of a given integer n

This is expressed as \( \int_0^1 e^x(x-1)^n \, dx \).
In our exercise, we needed to solve this integral to find a specific value of \( n \) that matches the result 16 - 6e. The definite integral helps us find exact values, giving us a specific numeric outcome rather than a general function. It's particularly useful for finding precise areas or other cumulative quantities over a set interval.

Exponential Functions

Exponential functions are a fundamental concept in calculus and mathematics. They have the form \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. This type of function grows very quickly and is unique due to its property: the rate of growth is proportional to its size.In the context of our exercise, the exponential function \( e^x \) is a part of the integrand. When multiplied by another function, such as \((x-1)^n\), it influences the shape and the area under the curve we are trying to find.
Exponential functions are continuous and differentiable everywhere, making them relatively straightforward to integrate or differentiate in calculus. In our task, it was necessary to manipulate the function using integration techniques to achieve the desired result. These functions are critical in both theoretical and practical applications, modeling real-world phenomena like population growth and radioactive decay.

Polynomial Functions

Polynomial functions consist of variables raised to whole number exponents, summed together. A simple example is \( x^n \), but it can also be a combination like \( x^3 - x^2 + x - 1 \). In the exercise, we used \((x-1)^n\), a polynomial form where n is an integer. This expression shifts the basic polynomial \( x^n \) to the left by one unit.Understanding polynomial functions is essential for many mathematical operations, especially as they often serve as building blocks for more complex expressions. They are very flexible and can approximate a wide range of other functions.
In the problem, changing n altered the function we integrated. Since polynomial functions have predictable behavior, we could systematically test each potential value of n to see which resulted in the target integral value. This is a great exercise in exploring how polynomial transformations affect results in definite integrals.