Problem 198 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(l_{1}=\int_{0}^{1} 2^{x^{2}} d x, l_{2}=\int_{0}^{1} 2^{x^{\prime}} d x, l_{3}=\int_{1}^{2} 2^{x^{2}} d x\), and \(l_{4}=\int_{1}^{2} 2^{x^{3}} d x\) then (A) \(l_{2}>l_{1}\) (B) \(l_{1}>l_{2}\) (C) \(l_{3}=l_{4}\) (D) \(l_{3}>l_{4}\)

Step-by-Step Solution

Answer

(A) \( l_{2} > l_{1} \) is correct.

Step 1: Analyze the Given Integrals

We are provided with four definite integrals: - \( l_{1} = \int_{0}^{1} 2^{x^2} \, dx \)- \( l_{2} = \int_{0}^{1} 2^{x'} \, dx \) (assuming \( x' = x \))- \( l_{3} = \int_{1}^{2} 2^{x^2} \, dx \)- \( l_{4} = \int_{1}^{2} 2^{x^3} \, dx \)These are to be compared to determine which inequality or equality holds true.

Step 2: Compare Integrals Between Same Limits

First, compare \( l_{1} \) and \( l_{2} \). Since \( 2^{x^2} \) grows slower than \( 2^{x} \) for \( x \in [0, 1] \), \( l_{1} < l_{2} \). Thus, \( l_{2} > l_{1} \).

Step 3: Compare Integrals Between Different Functions

Next, compare \( l_{3} \) and \( l_{4} \). Notice that for \( x \geq 1 \), \( 2^{x^3} \) grows faster than \( 2^{x^2} \). Thus, \( l_{3} < l_{4} \), in contradiction with \( l_{3} = l_{4} \) or \( l_{3} > l_{4} \). Therefore, none of the options involving \( l_{3} \) are suitable.

Key Concepts

Definite Integrals Comparison

When we evaluate definite integrals, our goal is to find the area under a curve on a specific interval. In this particular exercise, comparing definite integrals means looking at the area under the curves of different functions between the same or different limits.

For instance, consider the comparison between \( l_1 = \int_{0}^{1} 2^{x^2} \, dx \) and \( l_2 = \int_{0}^{1} 2^{x} \, dx \). Here, you need to analyze the behavior of the functions \( 2^{x^2} \) and \( 2^{x} \) in the interval from 0 to 1.

• The function \( 2^{x} \) increases at a faster rate than \( 2^{x^2} \) as \( x \) moves from 0 to 1. This implies that the area under the curve \( 2^{x} \) will be larger than the area under \( 2^{x^2} \).
• Therefore, the integral \( l_2 \) will be greater than \( l_1 \), expressing the relation \( l_2 > l_1 \).

Understanding how these functions behave over the given interval is crucial in determining the relationships between the integrals.

Exponential Functions

Exponential functions like \( 2^{x} \) and \( 2^{x^2} \) exhibit rapid growth. Exponential functions are widely studied because their rates of change increase proportionally relative to their current value.

The exercises use two forms of exponential functions:

• \( 2^{x} \) represents a simple exponential growth, where the exponent is directly proportional to \( x \).
• \( 2^{x^2} \) involves a polynomial term \( x^2 \), causing changes in the growth rate. It grows slower on certain intervals compared to its linear counterpart \( 2^{x} \).

Grasping how the exponent affects the rate of growth is essential to understanding the behavior of definite integrals involving such exponential functions. This insight aids in tasks like predicting the changes in the integral values as seen when comparing \( l_1 \) and \( l_2 \).

Integral Limits Analysis

In calculus, analyzing integral limits helps us understand where to evaluate functions for finding the area under curves. The limits of integration provide boundaries over which we calculate the integral.

For this exercise, consider these intervals for different integrals:

• \( l_1 \) and \( l_2 \) are defined on the interval from 0 to 1, so comparison depends solely on the functions within this boundary, making the relative behavior of \( 2^{x^2} \) and \( 2^{x} \) crucial.
• For \( l_3 = \int_{1}^{2} 2^{x^2} \, dx \) and \( l_4 = \int_{1}^{2} 2^{x^3} \, dx \), the interval is from 1 to 2. Here, you compare functions within these limits to determine which grows faster.

Always observe that even small changes in the limits can deeply influence the outcome of an integral. Therefore, understanding the function's rate of change and how it interacts with its limits is key to solving and analyzing integrals effectively.