面積 y= 1/(8x^2),y=x,y= x/8

解

面積

y = \frac{1}{8 x ^{2}}, y = x, y = \frac{x}{8}

\frac{3}{16}

十進法表記

0.1875

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}{8} : x = 1

x = \frac{x}{8} : x = 0

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

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

\int_{0}^{\frac{1}{2}} x - \frac{x}{8} d x = \frac{7}{64}

\int_{\frac{1}{2}}^{1} \frac{1}{8 x ^{2}} - \frac{x}{8} d x = \frac{5}{64}

= \frac{7}{64} + \frac{5}{64}

簡素化

= \frac{3}{16}