面積 1/8 x, 1/(x^2),x

解

面積

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

\frac{3}{4}

十進法表記

0.75

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

\frac{1}{8} x = x : x = 0

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

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

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

\int_{0}^{1} \frac{1}{8} x - x d x = \frac{7}{16}

\int_{1}^{2} \frac{1}{x ^{2}} - \frac{1}{8} x d x = \frac{5}{16}

= \frac{7}{16} + \frac{5}{16}

簡素化

= \frac{3}{4}