자연계열 2번

1. $\displaystyle f(x)=\frac{1}{1+x}$는 $\displaystyle [0,1]$에서 연속함수이다.Correction. 여기서 연속성을 사용하는 구간은 $\displaystyle [0,2]$이다. $\displaystyle [0,1]$은 $\displaystyle [0,2]$의 오기다. 그러나 원칙에 따라 원문 표기를 그대로 실었다.$\displaystyle x_{n}=\frac{1}{n}\sum_{k=1}^{2n}\frac{2n}{2n+2k-1}=\frac{1}{n}\sum_{k=1}^{2n}\frac{1}{1+\frac{2k-1}{2n}}=\frac{1}{n}\sum_{k=1}^{2n}f\left(\frac{2k-1}{2n}\right)$이다.$\displaystyle \lim_{n\to\infty}x_{n}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}f\left(\frac{2k-1}{2n}\right)=\int_{0}^{2}f(x)\,dx=\int_{0}^{2}\frac{1}{1+x}\,dx=\ln3.$

2. $\displaystyle \begin{aligned}\int_{x}^{\sqrt{3}x}f^{\prime}(t)\,dt&=\Biggl[f^{\prime}(t)t\Biggr]_{x}^{\sqrt{3}x}-\int_{x}^{\sqrt{3}x}f^{\prime\prime}(t)t\,dt\\&=\Biggl[f^{\prime}(t)t\Biggr]_{x}^{\sqrt{3}x}-\int_{x}^{\sqrt{3}x}\sin t\,dt=\Biggl[f^{\prime}(t)t\Biggr]_{x}^{\sqrt{3}x}+\Biggl[\cos t\Biggr]_{x}^{\sqrt{3}x}\\&=f^{\prime}(\sqrt{3}x)\sqrt{3}x-f^{\prime}(x)x+\cos(\sqrt{3}x)-\cos x\end{aligned}$$\displaystyle \begin{aligned}\lim_{x\to\frac{\pi}{\sqrt{3}}}\int_{x}^{\sqrt{3}x}f^{\prime}(t)\,dt&=\lim_{x\to\frac{\pi}{\sqrt{3}}}\{f^{\prime}(\sqrt{3}x)\sqrt{3}x-f^{\prime}(x)x+\cos(\sqrt{3}x)-\cos x\}\\&=\lim_{x\to\frac{\pi}{\sqrt{3}}}f^{\prime}(\sqrt{3}x)\lim_{x\to\frac{\pi}{\sqrt{3}}}\sqrt{3}x-\lim_{x\to\frac{\pi}{\sqrt{3}}}f^{\prime}(x)\lim_{x\to\frac{\pi}{\sqrt{3}}}x+\cos\pi-\cos\left(\frac{\pi}{\sqrt{3}}\right)\\&=\frac{2}{3}\pi-1-\cos\frac{\pi}{\sqrt{3}}=\frac{2}{3}\pi-0.76\end{aligned}$

3. 오른쪽 그림에서$\displaystyle \overline{\mathrm{AP}_{k}}^{2}=1^{2}+1^{2}-2\times1\times1\times\cos\left(\frac{\pi}{n+1}k\right)=2-2\cos\left(\frac{\pi}{n+1}k\right)$$\displaystyle \overline{\mathrm{P}_{k}\mathrm{B}}^{2}=4-\overline{\mathrm{AP}_{k}}^{2}=2+2\cos\left(\frac{\pi}{n+1}k\right)$이고, 따라서$\displaystyle \begin{aligned}S_{k}^{2}&=\left(\frac{1}{2}\times\overline{\mathrm{AP}_{k}}\times\overline{\mathrm{P}_{k}\mathrm{B}}\right)^{2}=\frac{1}{4}\left(2-2\cos\left(\frac{\pi}{n+1}k\right)\right)\left(2+2\cos\left(\frac{\pi}{n+1}k\right)\right)\\&=\sin^{2}\left(\frac{\pi}{n+1}k\right)\end{aligned}$

이므로$\displaystyle \begin{aligned}\lim_{n\to\infty}\frac{S_{1}^{2}+S_{2}^{2}+\cdots+S_{n}^{2}}{n+1}&=\lim_{n\to\infty}\sum_{k=1}^{n}\sin^{2}\left(\frac{\pi}{n+1}k\right)\frac{1}{n+1}=\lim_{n+1\to\infty}\sum_{k=1}^{n+1}\sin^{2}\left(\frac{\pi}{n+1}k\right)\frac{1}{n+1}\\&=\lim_{n+1\to\infty}\frac{1}{\pi}\sum_{k=1}^{n+1}\sin^{2}\left(0+\frac{\pi-0}{n+1}k\right)\frac{\pi-0}{n+1}=\frac{1}{\pi}\int_{0}^{\pi}\sin^{2}t\,dt\end{aligned}$이고, 부분적분법에 의해 $\displaystyle \int\sin^{2}t\,dt=\frac{1}{2}(t-\cos t\sin t)+C$이므로,구하는 값은 $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\sin^{2}t\,dt=\frac{1}{2\pi}\Biggl[t-\cos t\sin t\Biggr]_{0}^{\pi}=\frac{1}{2}$이다.