Problem 45 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{0}^{x} f(t) d t=x+\int_{x}^{1} t f(t) d t\), then the value of \(f(1)\) is (A) \(\frac{1}{2}\) (B) 0 (C) 1 (D) \(\frac{-1}{2}\)

Step-by-Step Solution

Answer

The value of \( f(1) \) is (A) \( \frac{1}{2} \).

Step 1: Understand the Problem

We are given that \( \int_{0}^{x} f(t) \, dt = x + \int_{x}^{1} tf(t) \, dt \). We need to find the value of \( f(1) \).

Step 2: Find a Derivative Expression

Differentiate both sides of the equation with respect to \( x \). The left side will be \( f(x) \). The right side, by using the Leibniz rule for differentiation of an integral, is \( 1 - xf(x) \). Thus we have the equation \( f(x) = 1 - x f(x) \).

Step 3: Simplify the Equation

From \( f(x) = 1 - x f(x) \), solve for \( f(x) \):\[ f(x) + x f(x) = 1 \]Which simplifies to:\[ f(x)(1 + x) = 1 \]Thus, \[ f(x) = \frac{1}{1 + x} \].

Step 4: Evaluate \( f(1) \)

Substitute \( x = 1 \) into the expression we found for \( f(x) \):\[ f(1) = \frac{1}{1 + 1} = \frac{1}{2} \].

Step 5: Verify Solution

Recheck the differentiation and integration steps to ensure correctness. Since everything checks out, \( f(1) \) is confirmed to be \( \frac{1}{2} \).

Key Concepts

Leibniz Rule

The Leibniz Rule is a powerful tool in calculus, especially when dealing with differentiation under the integral sign. Imagine you have a function that depends on another variable and is inside an integral. The Leibniz Rule helps you differentiate this function with respect to a variable:

• Consider an integral of the form: \( \int_{a(x)}^{b(x)} g(t, x) \, dt \)
• The Leibniz Rule tells us how to differentiate this integral with respect to \( x \):

The rule states that the derivative of this integral is given by:\[\frac{d}{dx} \int_{a(x)}^{b(x)} g(t, x) \, dt = g(b(x), x) \cdot \frac{d}{dx} b(x) - g(a(x), x) \cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} g(t, x) \, dt\]Using it, you can find derivatives even when the limits of the integral depend on \( x \). This rule was cleverly applied in the exercise to differentiate both sides of the equation, stepping closer to finding \( f(x) \).
It is particularly useful and simplifies the analysis significantly when compared with attempting to manually handle the differentiation of a nested integral.

Integral Calculus

Integral Calculus is all about understanding accumulation—whether it's area, volume, or some other quantity that adds up continuously. At its core:

• It helps find the total accumulation of a quantity given a rate of accumulation over an interval.
• Definite integrals, like the ones in the exercise \( \int_{0}^{x} f(t) \, dt \) and \( \int_{x}^{1} t f(t) \, dt \), calculate the net accumulation from one point to another.
• The Fundamental Theorem of Calculus states that differentiation and integration are inverse processes.

This exercise showcases the use of both definite integrals and derivatives to solve for a function. To solve the problem, you not only integrate over a fixed interval but also differentiate the resulting expression to analyze how it changes. Integral calculus offers tools to understand both the shape and changing nature of functions within boundaries, making it invaluable in creating models or solving real-world problems.

Differentiation

Differentiation is a fundamental concept that deals with how functions change at any given point. Here are the basic ideas:

• By differentiating a function, you determine the rate and direction of change.
• In calculus, one of the primary uses of differentiation is to find the slope of the tangent to a curve at a point, giving insight into the behavior of the function around that point.
• In the given exercise, differentiation is pivotal in forming the equation needed to solve for \( f(x) \).

The problem involves differentiating both sides of an equation featuring integrals, showcasing the dynamic interplay between these operations. Differentiation was used to transition from the general integral form to a specific expression of \( f(x) \). Once the relationship was deduced, solutions could be reached by evaluating the function at specific values.
In practical terms, differentiation helps us predict how something grows or shrinks based on certain conditions or inputs.