Let's calculate the value of the following integral with an accuracy of 0.1 units:

$$
\int_{-1}^{1} \frac{x^{2}}{x^{2}+2 x+2} d x
$$
For brevity, let $f(x)=\frac{x^{2}}{x^{2}+2 x+2}, I=\int_{-1}^{1} f(x) d x$. By writing the denominator as $(x+1)^{2}+1$, it can be seen that $f(x)$ is a continuous function for all $x$ and thus the above integral exists. To approximately calculate the value of $I$, we seek upper and lower sums $S$ and $s$ such that $\frac{S-s}{2} \leq 0.1$. Since $s \leq I \leq S$, the estimate $\left|I-\frac{S+s}{2}\right|<\frac{S-s}{2}$ holds, and thus $\frac{S+s}{2}$ will be the desired approximation. Estimating the difference $S-s$ is convenient if the function in question is monotonic. Indeed, if some function $g(x)$ is monotonically decreasing on the interval $[a, b]$, then by dividing the interval into $n$ equal parts with points $a=x_{0}<x_{1}<\ldots<x_{n}=b$,

$$
\begin{gathered}
S=\sum_{k=1}^{n} g\left(x_{k-1}\right)\left(x_{k}-x_{k-1}\right), \quad s=\sum_{k=1}^{n} g\left(x_{k}\right)\left(x_{k}-x_{k-1}\right), \\
S-s=\frac{b-a}{n} \sum_{k=1}^{n}\left[g\left(x_{k-1}\right)-g\left(x_{k}\right)\right]=\frac{b-a}{n}(g(a)-g(b)), \quad \text { and }
\end{gathered}
$$

a similar statement holds for increasing functions. Let's therefore examine whether our function $f(x)$ is monotonic on certain segments. Since

$$
f^{\prime}(x)=\frac{2 x(x+2)}{\left(x^{2}+2 x+2\right)^{2}}, \quad \text { if } \quad-1 \leq x \leq 0, \quad \text { then } \quad f^{\prime}(x) \leq 0
$$

$f(x)$ is decreasing, while for $0 \leq x \leq 1$, $f^{\prime}(x) \geq 0$, meaning $f(x)$ is monotonically increasing. Therefore, we apply the decomposition

$$
I=\int_{-1}^{0} f(x) d x+\int_{0}^{1} f(x) d x=I_{1}+I_{2}
$$

Divide the interval $[-1,0]$ into 10 equal parts, the approximation of $I_{1}$ is

$$
\begin{aligned}
\frac{S_{1}+s_{1}}{2} & =\sum_{k=1}^{10} \frac{1}{2}\left[f\left(\frac{k-10}{10}\right)+f\left(\frac{k-1-10}{10}\right)\right] \cdot \frac{1}{10}= \\
& =\frac{1}{10}\left[\frac{1}{2}(f(0)+f(-1))+\sum_{k=1}^{9} f\left(\frac{-k}{10}\right)\right]
\end{aligned}
$$

and the error of this is at most

$$
\frac{S_{1}-s_{1}}{2}=\frac{1 \cdot 1}{2 \cdot 10}=\frac{1}{20}
$$

Divide the interval $[0,1]$ into 3 equal parts, the approximation of $I_{2}$ is

$$
\frac{S_{2}+s_{2}}{2}=\sum_{k=1}^{3} \frac{1}{2}\left[f\left(\frac{k}{3}\right)+f\left(\frac{k-1}{3}\right)\right] \cdot \frac{1}{3}=\frac{1}{3}\left[\frac{1}{2}(f(0)+f(1))+f\left(\frac{1}{3}\right)+f\left(\frac{2}{3}\right)\right]
$$

and the error of this is at most

$$
\frac{S_{2}-s_{2}}{2}=\frac{1}{2 \cdot 3} \cdot \frac{1}{5}=\frac{1}{30}
$$

(since now $f(1)-f(0)=\frac{1}{5}$).

Thus, the value of $I$ is approximated by

$$
\frac{S_{1}+s_{1}+S_{2}+s_{2}}{2}
$$

with an error of at most

$$
\frac{1}{20}+\frac{1}{30}=\frac{1}{10} \cdot \frac{5}{6}=\frac{1}{10}-\frac{1}{60}
$$

and in the calculation, an additional error of $\frac{1}{60}$ can be made. Therefore, the value of $f\left(\frac{-k}{10}\right)$ for $(k=1, \ldots, 9)$, as well as the numbers $f\left(\frac{1}{3}\right)$ and $f\left(\frac{2}{3}\right)$, need only be calculated to 3 decimal places, because then the error committed is at most $\frac{11}{1000}<\frac{1}{60}$.

Performing the calculation, we get the value $I=0.4$.

Remark. Those who know that the derivative of the $\lg x$ function is $\frac{\lg e}{x}$ (where $e=2.71828 \ldots$) can easily calculate that $I=2-\frac{\lg 5}{\lg e}=0.391 \ldots$. Thus, our error is significantly less than 0.1, which is not surprising, since we were quite generous in all our estimates.