Answer
$\approx 0.100001$ with an error of about $1.39 \times 10^{-11}$.
Work Step by Step
We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $0.4$.
$\int_0^{0.1}\sqrt{1+x^4} dx=\int_0^{0.1} [1+\dfrac{x^4}{2}-\dfrac{x^{8}}{8}+...] dx $
or, $=0.1+\dfrac{(0.1)^5}{10}-\dfrac{(0.1)^9}{ 9 \cdot 8!}-....$
But $\dfrac{(0.1)^{9}}{9 \cdot 8} \approx 1.39 \times 10^{-11} $ and the proceeding term is greater than $10^{-8}$.
So, the first two terms of the series would give the accuracy.
$\int_0^{0.1}\sqrt{1+x^4} dx \approx 0.100001$ with an error of about $1.39 \times 10^{-11}$.
