Question

If \(f(x)\) is a function such that \(x \int_{0}^{x}(1-t) f(t) d t\) \(=\int_{0}^{x} t-f(t) d t\) and \(f(1)=1\) then find \(f(x)\).

Short answer

Question: Given the equation \(x \int_{0}^{x}(1-t) f(t) d t = \int_{0}^{x} t-f(t) d t\) and \(f(1) = 1\), find an expression for the function \(f(x)\). Answer: \(f(x) = \frac{\int_{0}^{x}(1-t) f(t) d t - x}{1-x^2}\)

Step-by-step solution

Simplify the given equation

We are given the equation: \(x \int_{0}^{x}(1-t) f(t) d t = \int_{0}^{x} t-f(t) d t\) We can simplify this by noticing that if we differentiate both sides of the equation with respect to \(x\), the integral on the right side would integrate only the term \(t\). Now we will move to the next step to differentiate both sides of the given equation.

Differentiate both sides of the equation

Differentiate both sides of the equation with respect to \(x\): \(\frac{d}{dx} \left(x \int_{0}^{x}(1-t) f(t) d t\right) = \frac{d}{dx} \left(\int_{0}^{x} t-f(t) d t\right)\) Using the Leibniz rule, we can differentiate the integral on the left side: \(\int_{0}^{x}(1-t) f(t) d t + x \cdot (1-x) f(x) = x - f(x)\) Now, reorganize the terms on both sides of the equation to get everything involving \(f(x)\) on one side and everything else on the other side:

Solve the resulting differential equation

After reorganizing the terms, we obtain: \(x \cdot (1-x) f(x) + f(x) = \int_{0}^{x}(1-t) f(t) d t - x\) Notice that the left side has the form \((1 - x)(f(x) + xf(x))\). Now we can solve for \(f(x)\): \((1-x)(1+x)f(x) = \int_{0}^{x}(1-t) f(t) d t - x\) \((1-x^2) f(x) = \int_{0}^{x}(1-t) f(t) d t - x\) To find the function \(f(x)\), we need to consider the given value of the function at \(x=1\), which is \(f(1)=1\). Substitute this value into the above equation: \((1-1^2) f(1) = \int_{0}^{1}(1-t) f(t) d t - 1\) \(f(1) = \int_{0}^{1}(1-t) f(t) d t - 1\) Since \(f(1)=1\), the left side is equal to 1: \(1 = \int_{0}^{1}(1-t) f(t) d t - 1\) Therefore, for any \(x\): \(f(x) = \frac{\int_{0}^{x}(1-t) f(t) d t - x}{1-x^2}\) Hence, the function \(f(x)\) is given by: \(f(x) = \frac{\int_{0}^{x}(1-t) f(t) d t - x}{1-x^2}\)

Key concepts

Leibniz Rule

Leibniz Rule is a powerful tool in integral calculus, particularly when dealing with the differentiation of integrals. It provides a method for finding the derivative of an integral, whose limits of integration are themselves functions.

For an integral in the form of \(\int_{a(x)}^{b(x)} f(t, x) dt\), where both \(a(x)\) and \(b(x)\) are functions of \(x\), the Leibniz Rule states that the derivative of this integral with respect to \(x\) is given by:
\[\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t, x) dt\right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \partial_x f(t, x) dt\]
It's crucial to understand this rule for solving problems that involve the derivative of integrals, such as our textbook problem where the differentiation of both sides was required to solve for \(f(x)\). Leibniz Rule allows us to deftly separate the variable outside the integral from the integration process, simplifying complex expressions into solvable equations.

Differential Equations

Differential equations are equations that relate a function with its derivatives. They are a cornerstone of mathematics because they can be used to model a vast range of phenomena in engineering, physics, economics, biology, and beyond. In the context of our textbook exercise, a differential equation was formed after applying the differentiation to both sides of the given equation using Leibniz Rule.

There are two main types of differential equations: ordinary differential equations (ODEs), which contain one independent variable and its derivatives, and partial differential equations (PDEs), which involve multiple independent variables and their partial derivatives.
To solve a differential equation, one typically isolates the function and integrates, sometimes requiring additional techniques like separation of variables, integrating factors, or even more advanced methods like Laplace transforms or series solutions. The goal is to find an expression for the unknown function that satisfies the equation.

Integration Techniques

Integration is the process of finding the integral of a function, and it has several techniques that can be applied depending on the form of the function. When faced with complex integrals, techniques such as substitution, integration by parts, partial fractions, and trigonometric substitution are often used.

For example, the substitution method involves changing the variable in an integral to simplify it, which could be useful if we suspect that parts of our function could form a simpler expression. Integration by parts, on the other hand, is typically used when the product of two functions is being integrated.

• Substitution: \(u = g(x)\), then \(du = g'(x)dx\)
• Integration by parts: \(\int u dv = uv - \int v du\)
• Partial Fractions: used when integrating rational functions, breaking them down into simpler fractions
• Trigonometric Substitution: useful when integrating functions containing \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\)

Our exercise solution would require one to be familiar with these techniques to maneuver through the steps required to find \(f(x)\). Understanding the integration process and techniques allows students to solve even the most challenging integral calculus problems by breaking them down into manageable parts.