Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice-differentiable function such that \( f(2) = 1 \). If \( F(x) = x f(x) \) for all \( x \in \mathbb{R} \), and the integrals \( \int_0^2 x F'(x) \, dx = 6 \) and \( \int_0^2 x^2 F''(x) \, dx = 40 \), then \( F'(2) + \int_0^2 F(x) \, dx \) is equal to:

• A. 11
• B. 15
• C. 9
• D. 13

Hint:

For complex integrals, splitting the problem into smaller parts can help simplify the computation and provide clarity in solving.

Answer 1

The Correct Option is B

Step 1: Calculate \( \int_0^2 x F'(x) \, dx \)

We are given: \[ F(x) = x f(x) \] Differentiating both sides: \[ F'(x) = f(x) + x f'(x) \] Now consider the given integral: \[ \int_0^2 x F'(x) \, dx = \int_0^2 x \left( f(x) + x f'(x) \right) \, dx \] Splitting the integral into two parts: \[ \int_0^2 x f(x) \, dx + \int_0^2 x^2 f'(x) \, dx = 6 \]

Step 2: Using the Given Information

From the given condition: \[ F(2) = 2 \times f(2) = 2 \quad \text{(since \(f(2) = 1\))} \] Substituting back into the integration results: \[ \int_0^2 x F(x) \, dx = -2 \]

Step 3: Compute the Final Sum

Using the given condition: \[ F'(2) + \int_0^2 F(x) \, dx = 15 \]

Final Answer: 15