Answer
$\approx 0.48491743$
Work Step by Step
We know that the Taylor Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Now, $e^{(-x^3)}=1+(-x^3)+\dfrac{(-x^3)^2}{2!}+\dfrac{(-x^3)^3}{3!}+\dfrac{(-x^3)^4}{4!}+...=1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+... ...(1)$
Take the integral of equation (1).
$\int_0^{1/2} e^{(-x^3)}=\int_0^{1/2} [1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+..]$
or, $=[x-\dfrac{x^4}{4}+\dfrac{x^7}{(7)2!}-\dfrac{x^{10}}{(10)3!}+\dfrac{x^{13}}{(13)4!}+..]_0^{1/2}$
or, $\dfrac{1}{2}-\dfrac{1}{64}+\dfrac{1}{1792}+....\approx 0.48491743$
