面積 2x^2,8x^2,5-3x

解

面積

2 x^{2}, 8 x^{2}, 5 - 3 x

\frac{1097}{128}

十進法表記

8.5703125

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)

2 x^{2} = 8 x^{2} : x = 0

2 x^{2} = 5 - 3 x : x = 1, x = - \frac{5}{2}

8 x^{2} = 5 - 3 x : x = \frac{5}{8}, x = - 1

= \int_{- \frac{5}{2}}^{- 1} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{\frac{5}{8}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{5}{8}}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{5}{2}}^{- 1} 2 x^{2} - (5 - 3 x) d x + \int_{- 1}^{0} 8 x^{2} - 2 x^{2} d x + \int_{0}^{\frac{5}{8}} 2 x^{2} - 8 x^{2} d x + \int_{\frac{5}{8}}^{1} 2 x^{2} - (5 - 3 x) d x

\int_{- \frac{5}{2}}^{- 1} 2 x^{2} - (5 - 3 x) d x = \frac{45}{8}

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

\int_{0}^{\frac{5}{8}} 2 x^{2} - 8 x^{2} d x = \frac{125}{256}

\int_{\frac{5}{8}}^{1} 2 x^{2} - (5 - 3 x) d x = \frac{117}{256}

= \frac{45}{8} + 2 + \frac{125}{256} + \frac{117}{256}

簡素化

= \frac{1097}{128}