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111桃園高中

回復 17# ChuCH 的帖子

求雙曲線\(-x^2+y^2=1\)及兩直線\(x=1\)、\(x=\sqrt{3}\)所圍封閉區域面積   
[解答]
\(\displaystyle \int_{1}^{\sqrt{3}}\sqrt{1+x^2}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec^3\theta d\theta=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec\theta d(\tan{\theta})\)
\(\displaystyle =\sec{\theta}\tan{\theta}-\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec{\theta}\tan^2{\theta}d\theta\)
\(\displaystyle =\sec{\theta}\tan{\theta}-\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\sec\theta(\sec^2\theta-1)d\theta\)
\(\displaystyle \Rightarrow\int\sec^3\theta d\theta =\frac{1}{2}(\sec\theta\tan\theta +\int sec\theta d\theta+C)\)
(\(\displaystyle \int sec\theta d\theta=\int \sec\theta \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}d\theta=\int \frac{1}{\sec\theta+\tan\theta} d(\sec\theta+\tan\theta)=\ln |\sec\theta+\tan\theta|+c\))
最後代入上下限\(\times 2\)

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2022-5-2 09:55

geogebra-export.png

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回復 19# thepiano 的帖子

感謝鋼琴老師

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回復 20# thepiano 的帖子

鋼琴老師客氣了!

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