Let f: [1, ∞) → [2, ∞) be a differentiable function such that f(1) = 2. If 6 ∫₁ˣ f(t) dt = 3xf(x) -

Q: Why can't we just integrate the right side directly to solve for $f(x)$?

That is a classic trap! You cannot integrate the right side because $f(x)$ is the unknown function itself. You are not integrating a known function; you are solving for an unknown one. Differentiating both sides is the only way to transform this integral equation into a differential equation, which we can then solve using standard techniques.

Q: Why does the derivative of $6 \int_1^x f(t) dt$ become $6f(x)$?

This is the beauty of the Newton-Leibniz Theorem! It states that $\frac{d}{dx} \int_a^x f(t) dt = f(x)$. Think of the integral as an 'accumulator' of the function's values, and the derivative as the 'rate of change' of that accumulation. They are inverse operations, so they effectively cancel each other out, leaving you with the integrand evaluated at the upper limit $x$.

Q: How do I handle the $e^{-\ln x}$ term in the Integrating Factor? I always get stuck there.

Remember your logarithmic identities! $e^{-\ln x}$ is the same as $e^{\ln(x^{-1})}$. Since $e^{\ln(u)} = u$, this simplifies beautifully to $x^{-1}$, or $\frac{1}{x}$. It is a classic JEE trick to make the integration step trivial. Never let the exponential-logarithm combination intimidate you; it is usually designed to simplify into something very clean.

Question

Let $f: [1, \infty) 	o [2, \infty)$ be a differentiable function such that $f(1) = 2$. If $6 \int_1^x f(t) dt = 3xf(x) - x^3$ for all $x \geq 1$, then the value of $f(2)$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the equation: 
$$6 \int_1^x f(t) dt = 3xf(x) - x^3$$

Imagine this equation as a locked room. The function $f(x)$ is inside, hidden behind the integral sign. To extract it, we use the **Newton-Leibniz Theorem**, which acts as the bridge between integration and differentiation.

### Phase 1: The Newton-Leibniz Liberation

We differentiate both sides with respect to $x$. On the left side, the constant $6$ remains, and the derivative of the integral $\int_1^x f(t) dt$ is simply $f(x)$.

Thus, the left side becomes $6f(x)$.

On the right side, we apply the **Product Rule**: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. The derivative of $3xf(x)$ is $3f(x) + 3xf'(x)$, and the derivative of $-x^3$ is $-3x^2$.

Equating the two sides, we obtain:
$$6f(x) = 3f(x) + 3xf'(x) - 3x^2$$

### Phase 2: Transformation into a Linear Differential Equation

We have successfully moved from an integral equation to a differential equation. Subtracting $3f(x)$ from both sides yields:
$$3f(x) = 3xf'(x) - 3x^2$$

Dividing the entire equation by $3$ simplifies it to:
$$f(x) = xf'(x) - x^2$$

Rearranging to standard form, we get:
$$xf'(x) - f(x) = x^2$$

To reach the standard **Linear Differential Equation (LDE)** form, $\frac{dy}{dx} + P(x)y = Q(x)$, we divide by $x$:
$$f'(x) - \frac{1}{x}f(x) = x$$

Here, our $P(x) = -\frac{1}{x}$ and our $Q(x) = x$.

### Phase 3: The Integrating Factor (I.F.)

Every LDE is solved using an **Integrating Factor**, defined as $I.F. = e^{\int P(x) dx}$.

Calculating this, we find:
$$I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = e^{\ln(x^{-1})} = \frac{1}{x}$$

Multiplying our differential equation by this factor, we get:
$$\frac{1}{x} f'(x) - \frac{1}{x^2} f(x) = 1$$

The left side is now the derivative of the product $\frac{f(x)}{x}$. This is the magic of the Integrating Factor—it collapses the expression into a perfect derivative.

### Phase 4: The Final Reveal

Integrating both sides with respect to $x$:
$$\int \frac{d}{dx} \left( \frac{f(x)}{x} \right) dx = \int 1 dx$$
$$\frac{f(x)}{x} = x + C$$

Multiplying by $x$, we find the general solution:
$$f(x) = x^2 + Cx$$

We use the initial condition $f(1) = 2$ to find $C$:
$$2 = (1)^2 + C(1) \Rightarrow C = 1$$

Our specific function is **$f(x) = x^2 + x$**.

Finally, calculating $f(2)$:
$$f(2) = (2)^2 + 2 = 4 + 2 = 6$$

The final answer is **$6$**.

Animated Solution for Mathematics - Differential Equations: Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Enter Numerical Value:

Visualized Solution

Analyze the Integral Equation

Apply Newton-Leibniz Theorem

Differentiate the Right-Hand Side

Simplify the Differential Equation

Identify the Linear Form

Calculate the Integrating Factor (I.F.)

Solve the Differential Equation

Apply Initial Condition

Find the Final Value $f (2)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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