4広島大学 2020年 数学 第4問解答・解説

$$\displaystyle \begin{aligned} (1)\quad &\int_0^{\pi/n}\sin nx\,dx=\left[-\frac{\cos nx}{n}\right]_0^{\pi/n}=\frac{1-\cos\pi}{n}=\frac{2}{n}.\\ (2)\quad &\int_0^\pi|\sin nx|\,dx=\frac{1}{n}\int_0^{n\pi}|\sin t|\,dt=\frac{1}{n}\sum\limits _{k=0}^{n-1}2=2. \\ &\int_{k\pi}^{(k+1)\pi}|\sin t|\,dt=\int_0^\pi\sin t\,dt=2\quad(0\le k\le n-1).\\ (3)\quad &y_{\max}=\min\left(\frac{1}{2},|\sin nx|\right),\qquad V=\pi\int_0^\pi y_{\max}^2\,dx\\ &=\frac{\pi}{n}\int_0^{n\pi}\min\left(\frac{1}{4},\sin^2t\right)\,dt=\pi\int_0^\pi\min\left(\frac{1}{4},\sin^2t\right)\,dt,\\ &\sin t=\frac{1}{2}\quad\text{すなわち}\quad t=\frac{\pi}{6},\frac{5\pi}{6}\quad(0\le t\le\pi),\\ &V=\pi\left(2\int_0^{\pi/6}\sin^2t\,dt+\frac{1}{4}\cdot\frac{2\pi}{3}\right)\\ &=\pi\left(2\left(\frac{\pi}{12}-\frac{\sqrt{3}}{8}\right)+\frac{\pi}{6}\right)=\frac{\pi^2}{3}-\frac{\sqrt{3}\pi}{4}.\\ (4)\quad &V=\pi\int_0^\pi(\sqrt{x}|\sin nx|)^2\,dx=\pi\int_0^\pi x\sin^2(nx)\,dx\\ &=\frac{\pi}{2}\left(\frac{\pi^2}{2}-\int_0^\pi x\cos 2nx\,dx\right),\\ &\int_0^\pi x\cos 2nx\,dx=\left[\frac{x\sin 2nx}{2n}+\frac{\cos 2nx}{4n^2}\right]_0^\pi=0,\\ &\text{したがって}\quad V=\frac{\pi^3}{4}. \end{aligned}$$