Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{0}^{1} e^{x^{3}}(x-\alpha) d x=0\), then (A) \(1<\alpha<2\) (B) \(\alpha<0\) (C) \(0<\alpha<1\) (D) \(\alpha=0\)

Step-by-Step Solution

Answer

The correct answer is (C) \(0 < \alpha < 1\).

Step 1: Understanding the Problem

We are given a definite integral \(\int_{0}^{1} e^{x^{3}}(x-\alpha) dx = 0\), and we need to find the correct range for \(\alpha\). The integral of a function over an interval equals zero when the function has equal positive and negative areas over that interval.

Step 2: Analyzing the Integrand

The function \(f(x) = e^{x^3}(x - \alpha)\) changes sign at \(x = \alpha\). This is the candidate point where \(x - \alpha = 0\). To solve this, we need to ensure that the contributions from the area where \(x > \alpha\) equal the contributions where \(x < \alpha\), with opposite signs.

Step 3: Choosing Possible Values for \(\alpha\)

Since we are integrating from 0 to 1, we consider the behavior of \(x - \alpha\). If \(\alpha\) is larger than 1, \(x - \alpha\) is negative for all \(x\), which cannot result in zero integral. If \(\alpha\) is smaller than 0, \(x - \alpha\) is always positive, which also does not yield a zero integral.

Step 4: Finding the Correct Range of \(\alpha\)

For \(\int_{0}^{1} e^{x^3}(x-\alpha) dx = 0\), \(\alpha\) must be such that the areas of the positive and negative parts cancel out. With \(x\) ranging from 0 to 1, \(x\) must be greater than \(\alpha\) initially and less than \(\alpha\) in the remaining part of the interval for it to switch signs. This solution agrees when \(\alpha\) is between 0 and 1.

Key Concepts

Definite Integration

In calculus, definite integration denotes the process of calculating the area under the curve of a given function, between two specified limits of integration. For the expression \( \int_{0}^{1} e^{x^3}(x - \alpha) \, dx = 0 \), this integral assesses the total accumulation of values from \( x = 0 \) to \( x = 1 \) for the function \( e^{x^3}(x - \alpha) \). The requirement for this value to be zero implies a balance between positive and negative areas under the curve.

There are key characteristics of definite integrals to remember:

• The limits of integration (lower limit to upper limit) specify the interval over which integration is performed.
• The result is a numerical value representing the net area under the curve from the lower to upper limit.
• If the integrand has parts where it crosses the horizontal axis, portions of the area will be positive or negative, depending on whether the curve is above or below the axis.

Integrand Analysis

Analyzing the integrand is crucial to solving integrals. In the context of this problem, we have the integrand \( f(x) = e^{x^3}(x - \alpha) \). The term \( e^{x^3} \) is always positive because the exponential function never zeroes or turns negative.

The function changes its behavior based on the term \( x - \alpha \), which determines when the expression becomes positive or negative. Notably, \( x - \alpha = 0 \) at \( x = \alpha \), making \( x = \alpha \) a critical point where the sign changes:

• For \( x > \alpha \), the term \( x - \alpha \) is positive, resulting in a positive contribution to the integral.
• For \( x < \alpha \), the term \( x - \alpha \) is negative, contributing negatively to the integral.

This contrasting behavior highlights how the function balances its integral over the interval \([0, 1]\) to achieve a net result of zero.

Zero Integrals

Zero integrals occur when the total value from positive contributions exactly cancels out the total from negative contributions in a definite integral. For this problem’s definite integral to equal zero, the positive and negative regions in \( e^{x^3}(x - \alpha) \) need to perfectly offset each other over the interval 0 to 1.

The condition \( \int_{0}^{1} f(x) \, dx = 0 \) implies an interesting equilibrium where:

• The sub-area of the curve where \( x > \alpha \) produces positive values.
• The sub-area of the curve where \( x < \alpha \) registers negative values.

Understanding zero integrals is vital in solving problems involving symmetry and balance in calculus. It provides insights into how specific parameters, like \( \alpha \), need to be set to maintain this balance. In this exercise, \( \alpha \) must lie between 0 and 1 to ensure this delicate equilibrium is met, leading to option (C): \( 0 < \alpha < 1 \).