2012年考研数二第10题解析

解析

$$\frac{n \times n}{n^{2}+n^{2}} \leqslant \frac{n}{1+n^{2}} + \frac{n}{2^{2}+n^{2}} + \cdot \cdot \cdot + \frac{n}{n^{2}+n^{2}} \leqslant \frac{n \times n}{1+n^{2}}$$

$$\frac{n \times n}{n^{2}+n^{2}} = \frac{1}{2};$$

$$\frac{n \times n}{1+n^{2}} = 1.$$

$$\frac{n}{i^{2} + n^{2}}, i = 1,2,3 … n.$$

$$\frac{n}{i^{2} + n^{2}} =$$

$$\frac{n}{\frac{i^{2}}{n^{2}} + \frac{n^{2}}{n^{2}}} \times \frac{1}{n^{2}} =$$

$$\frac{1}{1 + (\frac{i}{n})^{2}} \times \frac{1}{n}.$$

$$\frac{n}{1+n^{2}} + \frac{n}{2^{2}+n^{2}} + \cdot \cdot \cdot + \frac{n}{n^{2}+n^{2}} =$$

$$\sum_{i=1}^{n} \frac{1}{1 + (\frac{i}{n})^{2}} \times \frac{1}{n} \Rightarrow$$

$$\frac{1}{n} \sum_{i=1}^{n} \frac{1}{1 + (\frac{i}{n})^{2}}. ①$$

$$\int_{a}^{b} f(x)dx=$$

$$\lim_{\lambda \rightarrow 0} \sum_{i=1}^{n} f(\xi_{i}) \Delta x_{i}.$$

$$\frac{1}{n} \sum_{i=1}^{n} \frac{1}{1 + (\frac{i}{n})^{2}} =$$

$$\int_{0}^{1} (\frac{1}{1+x^{2}}) dx =$$

$$\arctan 1 – \arctan 0 =$$

$$\frac{\pi}{4} – 0 = \frac{\pi}{4}.$$

P.S:

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