第一部分：纯粹数学 [40 分]

1. Provide a sketch showing the curve and the rectangles.

[Sketch a continuous curve $y = \mathrm{f}(x)$ in the first quadrant from $x=1$ to $x=2$. Divide the interval $[1, 2]$ on the $x$-axis into $n$ equal sub-intervals, each of width $\Delta x = \frac{2-1}{n} = \frac{1}{n}$. Draw right-endpoint rectangles for each sub-interval, such that their heights are $\mathrm{f}\left(1+\frac{1}{n}\right)$, $\mathrm{f}\left(1+\frac{2}{n}\right)$, ..., $\mathrm{f}(2)$.]

2. Relate the algebraic sum to the total area of the rectangles in the sketch.

The width of each rectangle is $\frac{1}{n}$. The right endpoint of the $r$-th rectangle is $x_r = 1 + \frac{r}{n} = \frac{n+r}{n}$, and its height is $\mathrm{f}\left(\frac{n+r}{n}\right)$.

The area of the $r$-th rectangle is $\frac{1}{n} \mathrm{f}\left(\frac{n+r}{n}\right)$.

The given sum $\frac{1}{n} \left\{ \mathrm{f}\left(\frac{n+1}{n}\right) + \mathrm{f}\left(\frac{n+2}{n}\right) + \dots + \mathrm{f}\left(\frac{2n}{n}\right) \right\}$ represents the total area of these $n$ rectangles.

3. Explain the limiting process to link it to the definite integral.

As $n \to \infty$, the width of each rectangle $\frac{1}{n} \to 0$ and the number of rectangles approaches infinity. The total area of the rectangles approaches the exact area under the continuous curve $y = \mathrm{f}(x)$ between $x = 1$ and $x = 2$.

By definition, this exact area is precisely given by the definite integral $\int_{1}^{2} \mathrm{f}(x) \,\mathrm{d}x$.

1. Express the term in the summation in the form $\mathrm{f}\left(\frac{n+r}{n}\right)$ to identify the function $\mathrm{f}$.

$\frac{n}{n+r} = \frac{1}{\frac{n+r}{n}}$
By comparing this with the terms $\mathrm{f}\left(\frac{n+r}{n}\right)$ in the sum from part (i), we can identify that $\mathrm{f}(x) = \frac{1}{x}$.

2. Set up the definite integral based on the result established in part (i).

$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{n}{n+r} = \int_{1}^{2} \frac{1}{x} \,\mathrm{d}x$

3. Evaluate the definite integral exactly.

$\int_{1}^{2} \frac{1}{x} \,\mathrm{d}x = \left[ \ln x \right]_{1}^{2}$
$= \ln 2 - \ln 1$
$= \ln 2$