Let f : [0, ∞) → R be a continuous function such that f(x)=1-2x+∫(from 0 to x) e^(x-t)f(t)dt for all

Q: Wait, why can't I just treat the integral $\int_0^x e^{x-t}f(t)dt$ as a constant $C$ and solve it?

That is a very common trap! You can only treat an integral as a constant if the limits of integration are constants (like $\int_0^1$). Here, the upper limit is $x$. This means the integral is a function of $x$, not a constant. It changes as $x$ changes, so you must treat it as a variable term.

Q: I'm confused about the Leibniz Rule. When I differentiate $\int_0^x e^{-t}f(t)dt$, do I just get $e^{-x}f(x)$?

Exactly! The Leibniz Rule states that $\frac{d}{dx} \int_{a}^{x} g(t) dt = g(x)$. Since your integrand is $e^{-t}f(t)$, substituting the upper limit $x$ for $t$ gives you $e^{-x}f(x)$. You've got the concept perfectly—just be careful to combine it with the Product Rule when you have that $e^x$ sitting outside the integral!

Q: Why did we assume $f(0) = 1$ to find the constant $C$?

We didn't assume it; we derived it from the original equation! If you look at the original equation $f(x) = 1 - 2x + \int_0^x e^{x-t}f(t)dt$ and plug in $x=0$, the integral becomes $\int_0^0 \dots dt$, which is zero. Thus, $f(0) = 1 - 2(0) + 0$, which simplifies to $f(0) = 1$. Always check the original equation for initial conditions before you start solving the differential equation.

Question

Let $f : [0, \infty) 	o R$ be a continuous function such that $f(x)=1-2x+\int_0^x e^{x-t}f(t)dt$ for all $x \in [0, \infty)$. Then, which of the following statement(s) is (are) TRUE ?

Options:

(A) The curve $y = f(x)$ passes through the point (1, 2)

(B) The curve $y = f(x)$ passes through the point (2, -1)

(C) The area of the region $\{(x, y) \in [0, 1] 	imes R : f(x) \le y \le \sqrt{1-x^2} \}$ is $\frac{\pi-2}{4}$

(D) The area of the region $\{(x, y) \in [0, 1] 	imes R : f(x) \le y \le \sqrt{1-x^2} \}$ is $\frac{\pi-1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a problem that often strikes fear into the hearts of students: the integral equation.

You see an expression like $f(x) = 1 - 2x + \int_0^x e^{x-t}f(t)dt$, and your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, these problems are not monsters; they are puzzles waiting for the right key.

### Phase 1: The Algebraic Surgery

The first step is to look at the kernel of the integral: $e^{x-t}$. Notice that the variable of integration is $t$. This means $e^x$ is effectively a constant with respect to $t$.

We can pull it out of the integral sign. This transforms our equation into:

$$f(x) = 1 - 2x + e^x \int_0^x e^{-t}f(t)dt$$

Why did we do this? Because now, the $x$ dependency is clearly separated. We have an $e^x$ multiplying an integral that depends on $x$.

This is the perfect setup for the Newton-Leibniz Rule. We are about to perform calculus surgery to turn this integral equation into a differential one.

### Phase 2: The Calculus Transformation

Now, we differentiate both sides with respect to $x$. On the left, we have $f'(x)$. On the right, we have the derivative of a sum.

The derivative of $1$ is $0$, and the derivative of $-2x$ is $-2$. The tricky part is the product $e^x \int_0^x e^{-t}f(t)dt$. We must use the Product Rule:

$$f'(x) = -2 + \left[ \frac{d}{dx}(e^x) \cdot \int_0^x e^{-t}f(t)dt + e^x \cdot \frac{d}{dx} \left( \int_0^x e^{-t}f(t)dt ight) ight]$$

Applying the Leibniz Rule to the integral part, the derivative of $\int_0^x e^{-t}f(t)dt$ is simply $e^{-x}f(x)$. Our equation becomes:

$$f'(x) = -2 + e^x \int_0^x e^{-t}f(t)dt + e^x(e^{-x}f(x))$$

Notice the beauty of the math here: $e^x \cdot e^{-x} = 1$. The term simplifies to just $f(x)$.

And look at the integral term that remains—it is exactly the same integral we had in our original equation! We can substitute it back: $e^x \int_0^x e^{-t}f(t)dt = f(x) - 1 + 2x$.

### Phase 3: The Linear Differential Equation

Substituting that back, we get:

$$f'(x) = -2 + (f(x) - 1 + 2x) + f(x)$$

Rearranging this, we arrive at a standard Linear Differential Equation (LDE):

$$f'(x) - 2f(x) = 2x - 3$$

This is the heart of the problem. To solve this, we need an Integrating Factor (I.F.). The coefficient of $f(x)$ is $-2$, so our I.F. is $e^{\int -2 dx} = e^{-2x}$.

Multiplying the entire equation by $e^{-2x}$ allows us to write the left side as the derivative of a product:

$$\frac{d}{dx} (f(x)e^{-2x}) = (2x - 3)e^{-2x}$$

Integrating both sides, we use Integration by Parts on the right. After careful calculation, we find $f(x) = 1 - x + Ce^{2x}$.

Using our initial condition $f(0) = 1$, we find $C=0$. The function is simply $f(x) = 1 - x$.

### Phase 4: The Geometric Revelation

We have found the function, but the problem asks for an area. We are looking for the area between $f(x) = 1 - x$ and $y = \sqrt{1-x^2}$ for $x \in [0, 1]$.

Visualize the coordinate plane. $y = \sqrt{1-x^2}$ is the upper half of a unit circle. For $x \in [0, 1]$, this is a quarter circle in the first quadrant with radius $1$.

The line $y = 1 - x$ connects $(0, 1)$ and $(1, 0)$. The region bounded between them is the area of the quarter circle minus the area of the right-angled triangle formed by the line and the axes.

$$	ext{Area} = 	ext{Area(Quarter Circle)} - 	ext{Area(Triangle)} = \frac{\pi(1)^2}{4} - \frac{1}{2}(1)(1) = \frac{\pi - 2}{4}$$

And there you have it. From a terrifying integral equation to a simple geometric shape. This is the elegance of JEE Advanced mathematics.

You didn't just solve an equation; you visualized a space. Keep this clarity, and no problem will ever be too big for you. The final answer is **$\frac{\pi - 2}{4}$**.

Animated Solution for Mathematics - Differential Equations: Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TRUE ?

Select Answer:

Visualized Solution

Analyzing the Integral Equation

Simplifying the Integral

Differentiating Both Sides

Applying Product & Newton-Leibniz Rule

Simplifying the Derivative

Isolating the Integral Term

Forming the Differential Equation

Finding the Integrating Factor

Solving the Differential Equation

Applying Integration by Parts

The General Solution

Using the Initial Condition

The Final Function $f (x)$

Visualizing the Area Region

Calculating the Final Area

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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