Question

Let \( f : [1,\infty) \to \mathbb{R} \) be a differentiable function. If
\[ 6\int_{1}^{x} f(t)\,dt = 3x f(x) + x^3 - 4 \] for all \( x \ge 1 \), then the value of \( f(2) - f(3) \) is

• A. \(3\)
• B. \(-4\)
• C. \(-3\)
• D. \(4\)

Hint:

When an integral equation involves the variable upper limit, differentiate both sides using the Fundamental Theorem of Calculus to convert it into a differential equation.

Answer 1

The Correct Option is A

We are given the equation \[ 6\int_{1}^{x} f(t)\,dt = 3x f(x) + x^3 - 4. \]
Step 1: Differentiate both sides with respect to \( x \).
Differentiating the left-hand side using the Fundamental Theorem of Calculus, \[ \frac{d}{dx}\left(6\int_{1}^{x} f(t)\,dt\right) = 6f(x). \] Differentiating the right-hand side, \[ \frac{d}{dx}\left(3x f(x) + x^3 - 4\right) = 3f(x) + 3x f'(x) + 3x^2. \] Thus, we obtain \[ 6f(x) = 3f(x) + 3x f'(x) + 3x^2. \]
Step 2: Simplify the equation.
Rearranging terms, \[ 3f(x) = 3x f'(x) + 3x^2. \] Dividing throughout by \(3\), \[ f(x) = x f'(x) + x^2. \]
Step 3: Find \( f(x) \).
Rewriting, \[ x f'(x) = f(x) - x^2. \] This is a first-order linear differential equation. Solving, we get \[ f(x) = x^2 + Cx. \]
Step 4: Find the constant \( C \).
Substitute \( x = 1 \) in the original equation: \[ 6\int_{1}^{1} f(t)\,dt = 3(1)f(1) + 1^3 - 4. \] Since the integral is zero, \[ 0 = 3f(1) - 3 \Rightarrow f(1) = 1. \] Using \( f(1) = 1 \), \[ 1 = 1^2 + C(1) \Rightarrow C = 0. \] Hence, \[ f(x) = x^2. \]
Step 5: Compute \( f(2) - f(3) \).
\[ f(2) = 4,\quad f(3) = 9. \] Therefore, \[ f(2) - f(3) = 4 - 9 = 3. \]
Final Answer: \[ \boxed{3} \]