2018 Multi-University Training Contest 6-oval-and-rectangle(hdu 6362)-题解

一、题意

　　求椭圆内接矩形周长的期望。

二、推导过程

　　已知$c$，容易得出矩形弦长$d=4a\sqrt{1-\frac{c^2}{b^2}}$

　　接下来，矩形周长$p=4c+d=4c+4a\sqrt{1-\frac{c^2}{b^2}}$

　　那么，椭圆内接矩形周长的期望值$E=\frac{\int_{0}^{b}(4c+4a\sqrt{1-\frac{c^2}{b^2}})\, dc}{b}$

　　令$F(c)=\int_{0}^{b}(4c+4a\sqrt{1-\frac{c^2}{b^2}})\, dc$, 那么：

　　$F(c)=\int_{0}^{b}(4c+4a\sqrt{1-\frac{c^2}{b^2}})\, dc$

　　　　$=\int_{0}^{b}4c\,dc + \int_{0}^{b}4a\sqrt{1-\frac{c^2}{b^2}}\,dc$

　　　　$=2b^2 + \frac{4a}{b}\int_{0}^{b}\sqrt{b^2-c^2}\,dc$

　　　　$=2b^2 + \frac{4a}{b}(\frac{1}{2}(c\sqrt{b^2-c^2}+b^2*arcsin\frac{c}{b}))\,|_{0}^{b}$

　　　　$=2b^2 + \frac{2a}{b}(c\sqrt{b^2-c^2}+b^2*arcsin\frac{c}{b})\,|_{0}^{b}$

　　　　$=2b^2 + \frac{2a}{b}(b^2*\frac{\pi}{2})$

　　　　$=2b^2 + ab\pi$

　　所以，$E=\frac{F(c)}{b}=2b+a\pi$

转载于:https://www.cnblogs.com/fuzhihong0917/p/9445274.html

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