Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 812: 60

$0.035$

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