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

We are given that: \[ F(x) = x f(x) \] Now, we first calculate \( \int_0^2 x F'(x) \, dx \): \[ \int_0^2 x F'(x) \, dx = \int_0^2 x \left( f(x) + x f'(x) \right) \, dx = 6 \] We split the integral into two parts: \[ \int_0^2 x f(x) \, dx + \int_0^2 x^2 f'(x) \, dx = 6 \] Step 1: Using the given information. \[ F(2) = 2 \times f(2) = 2 \quad {(since \( f(2) = 1 \))} \] Substituting back: \[ \int_0^2 x F(x) \, dx = -2 \quad {(using the result from integration step)} \] Step 2: Compute the sum. We can now calculate the sum of \( F'(2) + \int_0^2 F(x) \, dx \) by adding the results from the two equations: \[ F'(2) + \int_0^2 F(x) \, dx = 15 \]