面積 y= 1/(18x^2),y=x,y= x/(10)

解

面積

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{10}

\frac{- 2 3 450 \cdot 2 ^{\frac{2}{3}} + 10 3 2 \cdot 3 ^{\frac{2}{3}} + 20 3 18 - 3 2 \cdot 3 0 ^{\frac{2}{3}}}{360}

十進法表記

0.11702 \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}{18 x ^{2}} = x : x = \frac{1}{3 18}

\frac{1}{18 x ^{2}} = \frac{x}{10} : x = 3 \frac{5}{9}

x = \frac{x}{10} : x = 0

= \int_{0}^{\frac{1}{3 18}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 18}}^{3 \frac{5}{9}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 18}} x - \frac{x}{10} d x + \int_{\frac{1}{3 18}}^{3 \frac{5}{9}} \frac{1}{18 x ^{2}} - \frac{x}{10} d x

\int_{0}^{\frac{1}{3 18}} x - \frac{x}{10} d x = \frac{3 ^{\frac{2}{3}}}{20 \cdot 2 ^{\frac{2}{3}}}

\int_{\frac{1}{3 18}}^{3 \frac{5}{9}} \frac{1}{18 x ^{2}} - \frac{x}{10} d x = \frac{10 \cdot 2 ^{\frac{2}{3}} ( - 3 \frac{9}{5} + 3 18 ) - 3 ^{\frac{2}{3}} ( 1 0 ^{\frac{2}{3}} - 1 )}{180 \cdot 2 ^{\frac{2}{3}}}

= \frac{3 ^{\frac{2}{3}}}{20 \cdot 2 ^{\frac{2}{3}}} + \frac{10 \cdot 2 ^{\frac{2}{3}} ( - 3 \frac{9}{5} + 3 18 ) - 3 ^{\frac{2}{3}} ( 1 0 ^{\frac{2}{3}} - 1 )}{180 \cdot 2 ^{\frac{2}{3}}}

簡素化

= \frac{- 2 3 450 \cdot 2 ^{\frac{2}{3}} + 10 3 2 \cdot 3 ^{\frac{2}{3}} + 20 3 18 - 3 2 \cdot 3 0 ^{\frac{2}{3}}}{360}