Calculus Question HEEELP!!

Question

Let,

x^2

f(x) = ∫ e^(-t^2) dt

x

find the critical points of f(x) WITH DETAILS .

note: integral from (x) to (x^2).

Answer 1

(x) =∫_x^{x^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.