Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 58

$0.1876$

$sin(x^{4})=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{8n+4}}{(2n+1)!}$ $\int_{0}^{1}sin(x^{4})dx=\int_{0}^{1}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{8n+4}}{(2n+1)!}$ Then $=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{8n+5}}{(8n+5)(2n+1)!}]_{0}^{1}$ $=[\frac{x^{5}}{5}-\frac{x^{13}}{13(3!)}+\frac{x^{21}}{21(5!)}]_{0}^{1}$ $\approx 0.1876$