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