A minimum value of integral from 0 to x te-t2 dt is

1) 1

2) 2

3) 3

4) 0

Solution: (4) 0

f (x) = ∫0x te-t^2 dt

f’ (x) = xe-x^2 = 0

x = 0, f’’ (x) = e-x^2 (1 – 2x2)

f’’ (0) = 1 > 0

The minimum value is f (0) = 0.