Answer
$\approx 0.0996676643$ with an error of about $4.6 \times 10^{-12}$
Work Step by Step
We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $0.1$.
$\int_0^{0.1} e^{-x^2} dx=\int_0^{0.1} [1-x^2+\dfrac{x^4}{2!}-\dfrac{x^{6}}{3!}+...] dx $
or, $=0.1-\dfrac{(0.1)^3}{3 }+\dfrac{(0.1)^5}{5 \cdot 2!}-...$
But $\dfrac{(0.1)^9}{9 \cdot 4!}\approx 4.6 \times 10^{-12}$ and the proceeding term is greater than $10^{-8}$.
so, the first three terms of the series would give the accuracy.
$\int_0^{0.1}e^{-x^2} dx \approx 0.0996676643$ with an error of about $4.6 \times 10^{-12}$
