面積 4x,x,x^3

解

面積

4 x, x, x^{3}

8

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)

4 x = x : x = 0

4 x = x^{3} : x = 0, x = - 2, x = 2

x = x^{3} : x = 0, x = - 1, x = 1

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

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

\int_{- 2}^{- 1} 4 x - x^{3} d x = \frac{9}{4}

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

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

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

= \frac{9}{4} + \frac{3}{2} + \frac{1}{4} + 4

簡素化

= 8