Let f(x) = (2 - x2 ) ex ( 1 + x )(1 - x)3/2 dx. If f(0) = 0, then f ( 12 ) is equal to:
- 2e - 1
- 2e + 1
- 3e - 1
- 3e + 1
Answer
(3) 3e - 1
Solution
Try the form ex [g(x) + g'(x) ]dx = ex g(x) + C with g(x) = 1 + x1 - x.
g(x) = 1 - x2 1 + x (1 - x)3/2
g'(x) = 12 1 - x1 + x (1 - x) + (1 + x)(1 - x)2 = 1 1 + x (1 - x)3/2
g(x) + g'(x) = 2 - x2 1 + x (1 - x)3/2 (the given integrand without ex)
f(x) = ex 1 + x1 - x + C
f(0) = 0 1 + C = 0 C = -1
f ( 12 ) = e1/2 3/21/2 - 1 = 3e - 1
Answer: (3)