1466. Calculate $\int_{0}^{1} e^{-x^{2}} d x$ with an accuracy of 0.001.
Solution. As mentioned above, this integral cannot be expressed in a closed form through elementary functions. Nevertheless, it can be calculated with any degree of accuracy using power series.

Let's expand the integrand into a power series. We will use the known series

$$
e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{n}}{n!}+\ldots
$$

By replacing $x$ with $-x^{2}$ in this series, we get

$$
e^{-x^{2}}=1-\frac{x^{2}}{1!}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\ldots+(-1)^{n} \frac{x^{2 n}}{n!}+\ldots
$$

from which

$$
\begin{gathered}
\int_{0}^{1} e^{-x^{2}} d x=\int_{0}^{1}\left(1-\frac{x^{2}}{1!}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\frac{x^{8}}{4!}-\frac{x^{10}}{5!}+\ldots\right) d x= \\
=\left.\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{2!5}-\frac{x^{7}}{3!7}+\frac{x^{9}}{4!9}-\frac{x^{11}}{5!11}+\ldots\right)\right|_{0} ^{1}= \\
=1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\frac{1}{4!9}-\frac{1}{5!11}+\ldots
\end{gathered}
$$

Calculating the terms of this series to an accuracy of 0.001, we notice that the sixth term of the series in absolute value is less than 0.001. Therefore, for the solution of this problem, according to the Leibniz criterion, we need to take the sum of the first five terms, which will ensure the required accuracy:

$$
\int_{0}^{1} e^{-x^{2}} d x \approx 0.747
$$