Question

If \(\mathrm{F}\) is a continuous function and \(\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{F}(\mathrm{t}) \mathrm{dt}\), show that \(\mathrm{F}(\mathrm{x})=0\) for every \(\mathrm{x} .\)

Short answer

Question: Prove that if F(x) is a continuous function and F(x) = ∫(from 0 to x) F(t) dt, then F(x) = 0 for every x. Answer: By using the Fundamental Theorem of Calculus and properties of integrals, we derived the differential equation F'(x) = F(x) with the initial condition F(0) = 0. Using an integrating factor and solving the initial value problem, we found that F(x) = 0 for every x.

Step-by-step solution

Rewrite the given equation using the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we can rewrite the given equation as: $$F(x) = \int_{0}^{x} F(t) dt = \frac{d}{dx}\left[\int_{0}^{x} F(u)du\right]$$ where we changed the integration variable from t to u to avoid confusion with the differentiation variable x.

Differentiate both sides of the equation with respect to x

By differentiating both sides of the equation, we have: $$\frac{d}{dx} F(x) = \frac{d}{dx} \left[\frac{d}{dx}\int_{0}^{x} F(u) du\right]\Rightarrow \Rightarrow F'(x) = F(x)$$

Convert the differential equation to the initial value problem

The differential equation \(F'(x) = F(x)\) is the same as the initial value problem: $$F'(x) - F(x) = 0 \Rightarrow F(0) = \int_0^0 F(t) dt = 0$$

Solve the initial value problem

To solve this initial value problem, we can use an integrating factor. The integrating factor is given by \(e^{-\int -1 dx}\), which is simply \(e^x\). We can now multiply both sides of the equation by the integrating factor: $$e^x (F'(x)-F(x))=0$$ Now, we observe that the left side of the equation is the derivative of \(F(x)e^x\). Thus, we integrate both sides of the equation with respect to x: $$\int \frac{d}{dx} \left[F(x)e^x\right] dx = \int 0 dx$$ The integral of the left side is simply \(F(x)e^x + c_1\) and the integral of the right side is \(c_2\). Therefore, we have: $$F(x)e^x + c_1 = c_2$$

Apply the initial condition

Using the initial condition \(F(0) = 0\), we substitute x = 0 into the equation to get: $$F(0)e^0 + c_1 = c_2 \Rightarrow c_2 = c_1$$ So, our expression now becomes: $$F(x)e^x = c_1 - c_1 = 0$$

Divide by \(e^x\) and conclude

Finally, we divide both sides of the equation by \(e^x\) (which is nonzero for all x) to get our result: $$F(x) = 0$$ Thus, we have shown that F(x) = 0 for every x.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a pivotal concept in integral calculus, connecting the seemingly disparate ideas of differentiation and integration. It serves as the backbone for solving many calculus problems. Essentially, this theorem asserts that the process of accumulation (integration) and the process of rate of change (differentiation) are opposite operations.

It is composed of two parts, with the first part providing an explicit link between a definite integral and an antiderivative of a function. This is what allows us to evaluate the integral \[\int_{a}^{b} f(x) dx = F(b) - F(a)\] using antiderivatives, where F is an antiderivative of f. The second part states that when we have a continuous function on an interval, then its integral over that interval can be computed using any of its infinitely many antiderivatives, thus simplifying the process of finding the area under a curve markedly.

The application of this theorem is not just theoretical; it is instrumental in solving practical problems as well, including those found in physics, engineering, and beyond. For students preparing for competitive exams like the IIT JEE, mastering this fundamental theorem is indispensable.

Initial Value Problem

In differential equations, an 'Initial Value Problem' (IVP) is a specific type of problem where you not only have a differential equation to solve, but you're also given an initial condition to satisfy. This condition typically specifies the value of the function or its derivatives at a particular point, which allows for a unique solution to be found.

For example, the equation \(dy/dt = f(y, t)\), with the initial condition \(y(t_0) = y_0\), constitutes an Initial Value Problem. Solutions to such problems are essential in physics and engineering, where they model a wide range of dynamic systems, from the motion of particles to the behavior of electrical circuits.

When tackling initial value problems, it's crucial to integrate the given information wisely to pinpoint that one solution which aligns perfectly with the provided initial conditions. In many IIT JEE problems, students will find themselves utilizing this concept to tailor a specific solution to a system's behavior at a given time.

Integrating Factor

An essential technique in solving linear first-order differential equations is the use of an 'Integrating Factor'. This magic multiplier is a function which, when multiplied by both sides of the differential equation, transforms it into a perfect derivative, thus making it integrable.

An integrating factor generally has the form \(e^{\int Pdx}\), where P is a function of x. Intuitively, you can think of the integrating factor as a balancing weight that stabilizes the equation to a form that's easily solvable.

The integrating factor does not just help in integrating the equations; it also ensures that the solution abides by the initial conditions provided. A proper understanding of this technique is a huge boon for IIT JEE aspirants, as many differential equations in the syllabus can be neatly solved by this method. It bridges the gap between complex differentials and solvable equations, emphasizing the interplay between various calculus elements.

Differential Equations

A 'Differential Equation' is an equation that relates a function with its derivatives. In the realm of calculus, these equations are the cornerstone, as they allow us to describe various physical phenomena where change is inherent -- growth rates, motion, forces, and other rate-related concepts.

There are many types of differential equations, ranging from simple separable equations to more complex linear and nonlinear equations. They can be ordinary (ODEs), involving functions of a single variable and their derivatives, or partial (PDEs), involving functions of multiple variables and their partial derivatives.

For IIT JEE students, differential equations form a significant part of the curriculum. Solving these involves identifying the type of equation, applying the appropriate strategy, and using initial or boundary conditions for finding particular solutions. Mastering differential equations requires practice and an in-depth understanding of integral calculus, making it a fascinating and challenging topic for those preparing for competitive exams. It is the key that unlocks the door to modeling and solving real-world scenarios mathematically.