面積 y= 1/(14x^2),y=x,y= x/(10)

解

面積

y = \frac{1}{14 x ^{2}}, y = x, y = \frac{x}{10}

\frac{30 3 14 - 2 \cdot 2 ^{\frac{2}{3}} 3 350 - 1 0 ^{\frac{2}{3}} 3 14}{280}

十進法表記

0.13836 \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}{14 x ^{2}} = x : x = \frac{1}{3 14}

\frac{1}{14 x ^{2}} = \frac{x}{10} : x = 3 \frac{5}{7}

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

= \int_{0}^{\frac{1}{3 14}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 14}}^{3 \frac{5}{7}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 14}} x - \frac{x}{10} d x + \int_{\frac{1}{3 14}}^{3 \frac{5}{7}} \frac{1}{14 x ^{2}} - \frac{x}{10} d x

\int_{0}^{\frac{1}{3 14}} x - \frac{x}{10} d x = \frac{9}{20 \cdot 1 4 ^{\frac{2}{3}}}

\int_{\frac{1}{3 14}}^{3 \frac{5}{7}} \frac{1}{14 x ^{2}} - \frac{x}{10} d x = \frac{- 2 3 1225 \cdot 2 ^{\frac{2}{3}} + 10 \cdot 1 4 ^{\frac{2}{3}} 3 2 - 7 0 ^{\frac{2}{3}} + 7 ^{\frac{2}{3}}}{140 3 7 \cdot 2 ^{\frac{2}{3}}}

= \frac{9}{20 \cdot 1 4 ^{\frac{2}{3}}} + \frac{- 2 3 1225 \cdot 2 ^{\frac{2}{3}} + 10 \cdot 1 4 ^{\frac{2}{3}} 3 2 - 7 0 ^{\frac{2}{3}} + 7 ^{\frac{2}{3}}}{140 3 7 \cdot 2 ^{\frac{2}{3}}}

簡素化

= \frac{30 3 14 - 2 \cdot 2 ^{\frac{2}{3}} 3 350 - 1 0 ^{\frac{2}{3}} 3 14}{280}