Let f : ℝ → ℝ be a differentiable function such that f'(x) + f(x) = ∫₀² f(t) dt. If f(0) = e⁻², then

Q: Wait, why can I just replace the integral with a constant k? Isn't the integral dependent on the function f(t)?

That is a brilliant question! You are absolutely right that the integral depends on f(t), but look closely at the limits of integration: 0 and 2. Because these are fixed numbers, the result of the definite integral is a specific, fixed value—a constant. Even though we don't know what that value is yet, we know it doesn't change with x. By calling it k, we turn a scary-looking integral equation into a simple algebraic one. It is a classic JEE strategy: simplify the unknown into a manageable parameter.

Q: I got stuck while integrating the term with e^{-t}. How do I handle the negative sign?

It is very common to trip over signs in integration! Remember that the integral of e^{ax} is (1/a)e^{ax}. Here, a = -1. So, the integral of e^{-t} is (1/-1)e^{-t}, which is -e^{-t}. When you evaluate this at the limits 0 and 2, be extremely careful with your brackets. The negative sign from the integration and the negative signs from the limits often cancel out or combine in ways that are easy to miss. Take it one step at a time!

Q: Why did we need to solve for k? Couldn't we just find f(x) directly?

The problem is that f(x) is defined in terms of the integral, and the integral is defined in terms of f(x). This is a self-referential loop! We cannot find the explicit form of f(x) without knowing the value of that integral. By solving for k, we break the loop. Once k is known, the function f(x) becomes fully determined, and we can calculate any value we need, like f(2).

Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x) + f(x) = \int_0^2 f(t) dt$. If $f(0) = e^{-2}$, then $2f(0) - f(2)$ is equal to ______.

Sigma Percentile · Accepted Answer

Welcome, future engineer! Today, we are going to dismantle a problem that looks like a monster but is actually a masterpiece of symmetry.

When you first see the equation $f'(x) + f(x) = \int_0^2 f(t) dt$, it is natural to feel a bit of hesitation. It looks like a differential equation, but that integral on the right side seems to be holding the function hostage.

But here is the secret: in the world of JEE Advanced, whenever you see an integral with constant limits, you are looking at a hidden constant. Let us embark on this journey together.

### Phase 1: The Great Simplification

We look at the term $\int_0^2 f(t) dt$. Since the limits are $0$ and $2$, this integral evaluates to a single, unchanging number.

Let us call this number $k$. Suddenly, the equation transforms from a terrifying integral-differential hybrid into a clean, elegant first-order linear differential equation:

$$f'(x) + f(x) = k$$

This is the moment the problem shifts from impossible to solvable.

### Phase 2: The Machinery of Calculus

Now that we have $f'(x) + f(x) = k$, we recognize the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Here, our $P(x) = 1$ and our $Q(x) = k$.

To solve this, we need the Integrating Factor (I.F.). The formula is:

$$I.F. = e^{\int P(x) dx} = e^{\int 1 dx} = e^x$$

We multiply both sides of our equation by $e^x$, which gives us $e^x f'(x) + e^x f(x) = k e^x$. Notice the left side? It is the derivative of the product $f(x)e^x$.

So, we have:

$$\frac{d}{dx} [f(x)e^x] = k e^x$$

Integrating both sides, we get $f(x)e^x = k e^x + C$. Dividing by $e^x$, we find our general solution:

$$f(x) = k + C e^{-x}$$

### Phase 3: The Loop of Logic

We have a general solution, but we have two unknowns: the constant $k$ and the integration constant $C$. We use the initial condition $f(0) = e^{-2}$ to find $C$.

Substituting $x = 0$, we get $f(0) = k + C e^0$, which means $e^{-2} = k + C$. Therefore, $C = e^{-2} - k$.

Now, our function is fully defined in terms of $k$:

$$f(x) = k + (e^{-2} - k)e^{-x}$$

But what is $k$? We go back to our original definition: $k = \int_0^2 f(t) dt$. We substitute our expression for $f(t)$ into this integral:

$$k = \int_0^2 [k + (e^{-2} - k)e^{-t}] dt$$

We integrate term by term: the integral of $k$ is $kt$, and the integral of $(e^{-2} - k)e^{-t}$ is $-(e^{-2} - k)e^{-t}$. Evaluating this from $0$ to $2$:

$$k = [kt - (e^{-2} - k)e^{-t}]_0^2$$

$$k = (2k - (e^{-2} - k)e^{-2}) - (0 - (e^{-2} - k))$$

After careful algebraic expansion, we find that the $k$ terms on both sides cancel out, leaving us with a clean equation for $k$:

$$k = e^{-2} - 1$$

### Phase 4: The Victory Lap

With $k$ in hand, we can write the final, explicit form of our function:

$$f(x) = e^{-2} - 1 + e^{-x}$$

The question asks for $2f(0) - f(2)$. We know $f(0) = e^{-2}$. We calculate $f(2)$:

$$f(2) = e^{-2} - 1 + e^{-2} = 2e^{-2} - 1$$

Finally, we compute the expression:

$$2f(0) - f(2) = 2(e^{-2}) - (2e^{-2} - 1) = 1$$

The exponential terms vanish, leaving us with the final answer of **1**. It is a beautiful, clean finish. You see, the problem wasn't about brute force; it was about identifying the structure, respecting the constants, and trusting the process.

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Identify the Constant Integral

Formulate the Differential Equation

Calculate the Integrating Factor

Solve the Differential Equation

Apply Initial Condition

Define the General Function

Set up the Integral Equation for $k$

Integrate to Solve for $k$

Evaluate Limits for $k$

Simplify and Find $k$

Determine the Final Function

Calculate the Final Result

The Sigma Insight: Linear Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

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