Let f : [0, ) R be a differentiable function such that f(x) = 1 - 2x + 0x ex-t f(t) dt for all x [0, ). Then the area of the region bounded by y = f(x) and the coordinate axes is
- 5
- 2
- 2
- 12
Answer
(4) 12
Solution
f(x) = 1 - 2x + ex 0x e-t f(t) dt ⇒ f(0) = 1
Differentiating (Leibnitz rule):
f'(x) = -2 + ex 0x e-t f(t) dt + f(x)
= -2 + (f(x) - 1 + 2x ) + f(x)
⇒ f'(x) - 2f(x) = 2x - 3 (linear, I.F. = e-2x)
⇒ ddx (f e-2x ) = (2x - 3)e-2x
⇒ f e-2x = (2x - 3)e-2x dx = (1 - x)e-2x + C
⇒ f(x) = 1 - x + Ce2x; f(0) = 1 ⇒ C = 0
∴ f(x) = 1 - x, a line cutting the axes at (1, 0) and (0, 1).
Area = 12 1 1 = 12
Answer: (4)