Answer
$\approx 0.48491743$
Work Step by Step
Here, $e^{(-x^3)}=1+(-x^3)+\dfrac{(-x^3)^2}{2!}+\dfrac{(-x^3)^3}{3!}+\dfrac{(-x^3)^4}{4!}+...$
$\implies e^{(-x^3)}=1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+... $
Now, $\int_0^{1/2} e^{(-x^3)}=\int_0^{1/2} [1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+..]$
and $[x-\dfrac{x^4}{4}+\dfrac{x^7}{(7)2!}-\dfrac{x^{10}}{(10)3!}+\dfrac{x^{13}}{(13)4!}+..]_0^{1/2}=(\dfrac{1}{2})-(\dfrac{1}{64})+(\dfrac{1}{1792})+....\approx 0.48491743$
