面積 sqrt(x), 1/2 x,9

解

面積

x, \frac{1}{2} x, 9

\frac{494}{3}

十進法表記

164.66666 \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)

x = \frac{1}{2} x : x = 4, x = 0

x = 9 : x = 81

\frac{1}{2} x = 9 : x = 18

= \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{18} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{18}^{81} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{4} x - \frac{1}{2} x d x + \int_{4}^{18} x - \frac{1}{2} x d x + \int_{18}^{81} x - 9 d x

\int_{0}^{4} x - \frac{1}{2} x d x = \frac{4}{3}

\int_{4}^{18} x - \frac{1}{2} x d x = - 362 + \frac{247}{3}

\int_{18}^{81} x - 9 d x = 81 + 362

= \frac{4}{3} - 362 + \frac{247}{3} + 81 + 362

簡素化

= \frac{494}{3}