Problem 163 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f\) defined on \([a, b]\) is bounded above if there exists a real number \(K\) such that \(|f(x)| \leq K, \forall x \in[a, b]\). The real number \(K\) is called an upper bound of \(f . f\) is called bounded below if there exists a real number \(k\) such that \(|f(x)| \geq k, \forall x \in[a, b]\). The real number \(k\) is called a lower bound of \(f\). It may be observed that if \(K\) is an upper bound of \(f\), then every real number greater than or equal to \(K\) is also an upper bound of \(f\). The smallest of all the upper bounds of \(f\) is called the least upper bound or the supremum (sup.) of \(f\). Similarly, the greatest of all the lower bounds of \(f\) is called the greatest lower bound or the infimum (Inf.) of \(f\). If the functions \(f\) and \(g\) are bounded and continuous on \([a, b]\) and \(g\) keeps the same sign on \([a, b]\), then there exists a number \(\mu\) between the infimum and supremum of \(f\) on \([a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=\mu \int_{a}^{b} g(x) d x $$. \(\int_{0}^{1} \frac{x^{2}}{\sqrt{1+x^{2}}}\) lies between (A) \(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) (B) \(\frac{1}{\sqrt{2}}\) and 1 (C) \(\frac{1}{3 \sqrt{2}}\) and \(\frac{1}{3}\) (D) None of these

Step-by-Step Solution

Answer

The answer is option (A). However, there is discrepancy since 0-\(\frac{1}{\sqrt{2}}\) does not exactly fit the range \(\frac{1}{\sqrt{2}}\) to \(\frac{1}{\sqrt{3}}\).

Step 1: Define the Function

To solve the problem, consider the function to be integrated, \( f(x) = \frac{x^2}{\sqrt{1+x^2}} \), defined on the interval \([0, 1]\). We need to evaluate the integral \( \int_{0}^{1} \frac{x^{2}}{\sqrt{1+x^{2}}} \, dx \).

Step 2: Analyze Endpoint Behavior

Evaluate \( f(x) \) at the endpoints of the interval. At \( x = 0 \), \( f(0) = 0 \). At \( x = 1 \), \( f(1) = \frac{1}{\sqrt{2}} \). Thus, \( f(x) \) ranges from 0 to \( \frac{1}{\sqrt{2}} \) over the interval \([0, 1]\).

Step 3: Estimate the Integral Bounds

Since \( f(x) = \frac{x^2}{\sqrt{1+x^2}} \) is increasing on \([0, 1]\), estimate bounds for the integral using \( g(x) = 1 \). We have \( \int_{0}^{1} 0 \cdot g(x) \, dx \leq \int_{0}^{1} f(x) \, dx \leq \int_{0}^{1} \frac{1}{\sqrt{2}} \cdot g(x) \, dx \).

Step 4: Calculate the Integral Bounds

Calculate the bounds. For the lower bound: \( \int_{0}^{1} 0 \, dx = 0 \). For the upper bound, since \( g(x) = 1 \): \( \int_{0}^{1} \frac{1}{\sqrt{2}} \, dx = \frac{1}{\sqrt{2}} \). Thus, the value lies between 0 and \( \frac{1}{\sqrt{2}} \).

Step 5: Compare to Options

Compare the calculated range with the given options. Since \( 0 \leq \int_{0}^{1} \frac{x^{2}}{\sqrt{1+x^{2}}} \, dx \leq \frac{1}{\sqrt{2}} \) and checking the given options, the only viable choice from those available is (A) \( \frac{1}{\sqrt{2}} \) and \( \frac{1}{\sqrt{3}} \), however, the calculated bounds suggest the range might not fully encapsulate the option.

Key Concepts

Bounded Functions

In calculus, understanding bounded functions is essential for analyzing function behavior. A function is termed "bounded" if it is confined within certain limits across its domain. More specifically:

• A function \( f(x) \) is bounded above on an interval \([a, b]\) if there is a real number \( K \) such that \( |f(x)| \leq K \) for all \( x \) in \([a, b]\).
• A function is bounded below if there exists a real number \( k \) satisfying \( |f(x)| \geq k \) for all \( x \) in the interval.

The real numbers \( K \) and \( k \), respectively, are called the upper and lower bounds. The importance lies in controlling the function's output within a specific range, which is crucial in further computations such as integration. This ensures that functions don't have unpredictable increases or decreases, making them manageable and predictable for mathematical operations within a given interval.
In the context of integration, boundedness helps in determining feasible limits for integrals, especially when calculations are conducted over a closed and finite interval.

Definite Integral

The concept of a definite integral is central to calculus and provides a powerful way to calculate the accumulation of quantities. Essentially, a definite integral over an interval \([a, b]\) represents the total accumulation of a function’s values from \(a\) to \(b\). This is different from an indefinite integral, which is primarily concerned with anti-differentiation and doesn’t have specific bounds. To elaborate:

• The definite integral is denoted by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• It reflects the area under the curve \( f(x) \) between \( a \) and \( b \), when \( f(x) \) is plotted on a graph.
• The Fundamental Theorem of Calculus links the concept of differentiation with integration, supporting the evaluation of definite integrals using antiderivatives.

By applying these principles, we can estimate integrals even without finding their exact values, as shown in the example with bounds estimation. This leads to integral approximation, which is widely utilized when exact solutions are challenging to obtain.

Supremum and Infimum

Supremum and infimum are pivotal concepts in real analysis that deal with the generalization of maximum and minimum values. While particular functions or sets might not have maxima or minima in traditional terms, supremum and infimum offer a broader approach:

• The supremum (or least upper bound) of a set is the smallest real number that is greater than or equal to every number in that set. It may not be a part of the set itself.
• Conversely, the infimum (or greatest lower bound) is the largest real number that is less than or equal to every number in the set.

In the instance of integration over a function like \( f(x) = \frac{x^2}{\sqrt{1+x^2}} \) on the interval \([0, 1]\), using supremum and infimum helps understand the range within which the integral lies. This provides a comprehensive manner to label function behavior and supports bounding integrals based on the function's calculated range. While this assures us of the integral's limits, it guides the choice of appropriate approximations and solutions as seen in integration exercises and ensuring calculative confidence.