面積 y=sqrt(2x),y=3x-4,y=0

解

面積

y = 2 x, y = 3 x - 4, y = 0

2

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 = 3 x - 4 : x = 2

2 x = 0 : x = 0

3 x - 4 = 0 : x = \frac{4}{3}

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

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

\int_{0}^{\frac{4}{3}} 2 x - 0 d x = \frac{16 2}{9 3}

\int_{\frac{4}{3}}^{2} 3 x - 4 - 2 x d x = \frac{24 3 - 16 2}{9 3} - \frac{2}{3}

= \frac{16 2}{9 3} + \frac{24 3 - 16 2}{9 3} - \frac{2}{3}

簡素化

= 2