(x) =∫xx^2 e^(- t^2) dt. By second fundamental theorem of Calculus f'(x) = e^(- x4)·2x - e^(- x2) = 0.
We set it to 0 to get critical points of f(x). We get equation: 2x/e^(x4) - 1/e^(x2) = 0.
We can solve this equation approximatly using TI-84: x = 1.2581 or x = 0.4300.
This is 2 critical numbers.