Answer
$\int_0^{0.1}\sqrt{1+x^4} dx \approx 0.100001$ (Error of about $1.39 \times 10^{-11}) $
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{0.1}\sqrt{1+x^4} dx=\int_0^{0.1} [1+\dfrac{x^4}{2}-\dfrac{x^{8}}{8}+...] \space dx \\=0.1+\dfrac{(0.1)^5}{10}-\dfrac{(0.1)^9}{ 9 \cdot 8!}-....$
But $\dfrac{(0.1)^{9}}{9 \cdot 8} \approx 1.39 \times 10^{-11} $ and this means that the proceeding term is greater than $10^{-5}$ and the first five terms of the series yield the accuracy.
Thus, $\int_0^{0.1}\sqrt{1+x^4} dx \approx 0.100001$ (Error of about $1.39 \times 10^{-11}) $
