Let f: R → R be a twice differentiable function such that f(2)=1. If F(x)=x f(x) for all x ∈ R, ∫₀² x F′(x) d x=6 and ∫₀² x² F^(′ ′)(x) d x=40, then F′(2)+∫₀² F(x) d x is equal to :

Solution

Maths sir · 4 steps

Integrate by parts twice: the first integral gives ∫ F, the second gives F'(2).

1. Value at 2. F(2) = 2f(2) = 2 × 1 = 2. At x = 0, every boundary term below has a factor of x, so it vanishes.
2. First integral. By parts, ∫₀² xF'(x) dx = [xF(x)]₀² - ∫₀² F(x) dx = 4 - ∫₀² F(x) dx. This equals 6, so ∫₀² F(x) dx = -2.

   ∫₀² x²F''(x) dx = [x²F'(x)]₀² - 2∫₀² xF'(x) dx = 4F'(2) - 12

3. Second integral. This equals 40, so 4F'(2) = 52 and F'(2) = 13.
4. So. F'(2) + ∫₀² F(x) dx = 13 + (-2) = 11, which is option (b).

The idea to keep. When an integral has a power of x times a derivative, integrate by parts to bring the derivative down one level. Each round trades one power of x for a boundary term.

Answer from the NTA final answer key. Solution by PrepWiser.