345. Find the value of the integral $\int_{0}^{1} x d x$, considering it as the limit of integral sums.
Solution. Divide the interval $[0,1]$ (Fig. 194) into $n$ equal parts; in this case, the length of each part is $1 / n$. Draw lines through the division points parallel to the ordinate axis. Then each strip, for example, $A B C D$, will represent a trapezoid. For each trapezoid, construct a rectangle $A B E D$. If $A$ is the $k$-th division point of the segment $O Q$, then the length of the segment $O A$ is equal to the length of the segment $A B$ and is equal to $k / n$. Therefore, the area of the rectangle $A B E D$ is $k / n^{2}$.

Then, according to the method described above for calculating the area of the figure $O P Q$, we have

$$
\begin{gathered}
\int_{0}^{1} x d x=\lim _{n \rightarrow \infty} S_{O P Q}=\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\cdots+\frac{n}{n^{2}}\right)= \\
=\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}(1+2+3+\cdots n)\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}} \cdot \frac{1+n}{2} \cdot n\right)= \\
=\lim _{n \rightarrow \infty}\left(\frac{1+n}{2 n}\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{2 n}+\frac{n}{2 n}\right)=\frac{1}{2}
\end{gathered}
$$