Integrals · Question 49

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) dx = 6 and 02 x2 F''(x) dx = 40, then F'(2) + 02 F(x) dx is equal to:

1. 13
2. 11
3. 9
4. 15

Answer

(2) 11

Solution

F(2) = 2f(2) = 2

Integrating by parts:

02 xF'(x) dx = [xF(x)]02 - 02 F(x) dx

6 = 2F(2) - 02 F(x) dx = 4 - 02 F(x) dx

02 F(x) dx = -2

Again by parts:

02 x2F''(x) dx = [x2F'(x)]02 - 2 02 xF'(x) dx

40 = 4F'(2) - 2(6)

F'(2) = 13

F'(2) + 02 F(x) dx = 13 - 2 = 11

Answer: (2)