Answer
$\int_0^{0.1}e^{-x^2} dx \approx 0.0996676643$ (Error of about $4.6 \times 10^{-12})$
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{0.1} e^{-x^2} dx=\int_0^{0.1} [1-x^2+\dfrac{x^4}{2!}-\dfrac{x^{6}}{3!}+...] \space dx \\=0.1-\dfrac{(0.1)^3}{3 }+\dfrac{(0.1)^5}{5 \cdot 2!}-...$
But $\dfrac{(0.1)^9}{9 \cdot 4!}\approx 4.6 \times 10^{-12}$ and this means that the proceeding term is greater than $10^{-8}$ and the first 3 terms of the series yield the accuracy.
Thus, $\int_0^{0.1}e^{-x^2} dx \approx 0.0996676643$ (Error of about $4.6 \times 10^{-12})$
