Let f, g:(0, ) R be two functions defined by f(x)= -xx (|t|-t2 )e-t2dt and g(x)= 0x2t1/2e-tdt. Then, the value of 9 (f ( e9 )+g ( e9 ) ) is equal to :
- 10
- 9
- 8
- 6
Answer
(3) 8
Solution
By the Newton–Leibniz rule (for x > 0, |x| = |-x| = x):
f'(x) = (x - x2 )e-x2 1 - (x - x2 )e-x2 (-1) = 2 (x - x2 )e-x2
g'(x) = (x2 )1/2e-x2 2x = 2x2e-x2
Adding, the x2e-x2 terms cancel: f'(x) + g'(x) = 2xe-x2
Integrating from 0 (where both integrals are zero): f(x) + g(x) = 0x 2t e-t2dt = 1 - e-x2
At x = e9, e-x2 = 19, so 9 (1 - 19 ) = 8
Answer: (3)