745. Output the formula $I_{2}^{*}=\frac{1}{n} \sum_{i=1}^{n} \frac{\varphi\left(x_{i}\right)}{f\left(x_{i}\right)}$ for estimating the definite integral $I=\int_{a}^{b} \varphi(x) \mathrm{d} x$, where $f(x)$ is the density function of the "auxiliary" random variable $X$ in the integration interval $(a, b) ; x_{i}$ are possible values of $X$, sampled according to the known density $f(x) ; n$ is the number of trials.
Solution. Let $f(x)$ be the probability density function of some random variable $X$ over the interval of integration $(a, b)$, i.e.,

$$
\int_{a}^{b} f(x) d x=1.
$$

We can represent the integral $I$ as follows:

$$
I=\int_{a}^{b} \frac{\varphi(x)}{f(x)} \cdot f(x) d x
$$

Thus, the integral $I$ is represented as the expected value of the function $F(X)=\varphi(X) / f(X)$. As an estimate of this expected value, and therefore of the integral $I$, we take the sample mean

$$
l_{i}^{*}=\frac{1}{n} \sum_{i=1}^{n} \frac{\varphi\left(x_{i}\right)}{f\left(x_{i}\right)}
$$

where $x_{i}$ are the possible values of the random variable $X$, which are generated according to the known density $f(x)$; $n$ is the number of trials. The desired estimate is obtained.

Note that it is desirable to choose $f(x)$ such that the ratio $f(x) /|\varphi(x)|=$ const, if possible.