Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 812: 59

$0.401024$

$\sqrt {1+x^{4}}=(1+x^{4})^{1/2}=\Sigma_{n=0}^{\infty}\dfrac{(\frac{1}{2})(-\frac{1}{2}).....(\frac{1}{2})-n+1)x^{4n}}{(n)!}$ $\int_{0}^{0.4}\sqrt {1+x^{4}}dx=\int_{0}^{0.4}\Sigma_{n=0}^{\infty}\dfrac{(\frac{1}{2})(-\frac{1}{2}).....(\frac{1}{2})-n+1)x^{4n}}{(n)!}$ Then $=[\Sigma_{n=0}^{\infty}\dfrac{(\frac{1}{2})(-\frac{1}{2}).....(\frac{1}{2})-n+1)x^{4n+1}}{(4n+1)(n)!}]_{0}^{0.4}$ $\approx 0.401024$