Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 812: 57

$0.0059$

$arctan(x)=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)}$ $\int_{0}^{1/2}arctan(x)dx=\int_{0}^{1/2}x^{3}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)}dx$ $\int_{0}^{1/2}arctan(x)dx=\int_{0}^{1/2}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+4}}{(2n+1)}dx$ Then $=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+5}}{(2n+5)(2n+1)}$ $=\frac{0.5^{2}}{5}-\frac{0.5^{7}}{21}+\frac{0.5^{9}}{45}$ $\approx 0.0059$