面積 1/(8x^2),x, x/(20)

解

面積

\frac{1}{8 x ^{2}}, x, \frac{x}{20}

\frac{19}{160} + \frac{- 8 3 50 + 41 3 2 \cdot 2 ^{\frac{2}{3}} - 4 3 2 \cdot 5 ^{\frac{2}{3}}}{320}

十進法表記

0.23684 \dots

解答ステップ

交点を見つけるには以下を解く:

f_{i} (x) = f_{j} (x)

\frac{1}{8 x ^{2}} = x : x = \frac{1}{2}

\frac{1}{8 x ^{2}} = \frac{x}{20} : x = 3 \frac{5}{2}

x = \frac{x}{20} : x = 0

= \int_{0}^{\frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{3 \frac{5}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{2}} x - \frac{x}{20} d x + \int_{\frac{1}{2}}^{3 \frac{5}{2}} \frac{1}{8 x ^{2}} - \frac{x}{20} d x

\int_{0}^{\frac{1}{2}} x - \frac{x}{20} d x = \frac{19}{160}

\int_{\frac{1}{2}}^{3 \frac{5}{2}} \frac{1}{8 x ^{2}} - \frac{x}{20} d x = \frac{- 20 \cdot 2 ^{\frac{2}{3}} 3 \frac{2}{5} + 41 \cdot 2 ^{\frac{2}{3}} - 4 \cdot 5 ^{\frac{2}{3}}}{160 \cdot 2 ^{\frac{2}{3}}}

= \frac{19}{160} + \frac{- 20 \cdot 2 ^{\frac{2}{3}} 3 \frac{2}{5} + 41 \cdot 2 ^{\frac{2}{3}} - 4 \cdot 5 ^{\frac{2}{3}}}{160 \cdot 2 ^{\frac{2}{3}}}

簡素化

= \frac{19}{160} + \frac{- 8 3 50 + 41 3 2 \cdot 2 ^{\frac{2}{3}} - 4 3 2 \cdot 5 ^{\frac{2}{3}}}{320}