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

Given that:\[F(x) = x f(x)\]First, we evaluate \( \int_0^2 x F'(x) \, dx \):\[\int_0^2 x F'(x) \, dx = \int_0^2 x \left( f(x) + x f'(x) \right) \, dx = 6\]This integral is separated into two parts:\[\int_0^2 x f(x) \, dx + \int_0^2 x^2 f'(x) \, dx = 6\]Step 1: Apply the given information. \[F(2) = 2 \times f(2) = 2 \quad {(given \( f(2) = 1 \))}\]Substitution yields:\[\int_0^2 x F(x) \, dx = -2 \quad {(from integration result)}\]Step 2: Compute the total. The sum \( F'(2) + \int_0^2 F(x) \, dx \) is found by combining the results from the two derived equations:\[F'(2) + \int_0^2 F(x) \, dx = 15\]