Answer
$0.035$
Work Step by Step
$x^{2}e^{-x^{2}}=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+2}}{(n)!}$
$\int_{0}^{0.5}x^{2}e^{-x^{2}}dx=\int_{0}^{0.5}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+2}}{(n)!}$
Then
$\approx \Sigma_{n=0}^{1}\dfrac{(-1)^{n}(0.5)^{2n+3}}{(2n+3)(n)!}$
$\approx 0.035$
