1666. Let the random variable $X$ have the probability density function

$$
p(x)=|x| e^{-x^{2}}
$$

What is the probability that this random variable will take a value in the interval $(0,1)$?
Solution. The desired probability $P(0<X<1)$ can be found using property 2 of the probability density function

$$
P\left(x_{1}<X<x_{2}\right)=\int_{x_{1}}^{x_{2}} p(x) d x
$$

In our case

$$
\begin{gathered}
P(0<X<1)=\int_{0}^{1}|x| e^{-x^{2}} d x=\int_{0}^{1} x e^{-x^{2}} d x=-\frac{1}{2} \int_{0}^{-1} e^{t} d t= \\
=\frac{1}{2}\left(1-e^{-1}\right)=\frac{1}{2}(1-0.3679)=\frac{1}{2} \cdot 0.6321=0.3160
\end{gathered}
$$

(here the substitution $t=-x^{2}$ is made).