一、题目
已知：
$$
\begin{aligned}
I & = \\
& \lim \limits_{n \rightarrow \infty}\left(\frac{1+\sqrt{n^{2}-1^{2}}}{n^{2}}+\frac{2+\sqrt{n^{2}-2^{2}}}{n^{2}}+\cdots+\frac{n+\sqrt{n^{2}-n^{2}}}{n^{2}}\right)
\end{aligned}
$$
则：
$$
I \ = \ ?
$$
难度评级：
二、解析 
首先：
$$
\begin{aligned}
I \\
& = \lim \limits_{n \rightarrow \infty}\left(\frac{1+\sqrt{n^{2}-1^{2}}}{n^{2}}+\frac{2+\sqrt{n^{2}-2^{2}}}{n^{2}}+\cdots+\frac{n+\sqrt{n^{2}-n^{2}}}{n^{2}}\right) \\ \\
& = \lim \limits_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{i+\sqrt{n^{2}-i^{2}}}{n^{2}} \\ \\
& = \lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{i+\sqrt{n^{2}-i^{2}}}{n} \\ \\
& = \lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n}\left[\frac{i}{n}+\sqrt{1-\left(\frac{i}{n}\right)^{2}}\right] \\ \\
& \xlongequal{x = \frac{i}{n}} \int_{0}^{1} \left(x+\sqrt{1-x^{2}}\right) \mathrm{d} x \\ \\
& = \frac{1}{2}+\int_{0}^{1} \sqrt{1-x^{2}} \mathrm{~d} x \\ \\
& \xlongequal{x=\sin t} \frac{1}{2}+\int_{0}^{\frac{\pi}{2}} \cos ^{2} t \mathrm{~d} t \\ \\
& = \frac{1}{2}+\frac{1}{2} \cdot \frac{\pi}{2} \\ \\
& = \frac{1}{2}+\frac{\pi}{4}
\end{aligned}
$$