面積 y=x,y=5x,y=-x+4

解

面積

y = x, y = 5 x, y = - x + 4

\frac{8}{3}

十進法表記

2.66666 \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 = 5 x : x = 0

x = - x + 4 : x = 2

5 x = - x + 4 : x = \frac{2}{3}

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

= \int_{0}^{\frac{2}{3}} ∣ x - 5 x ∣ d x + \int_{\frac{2}{3}}^{2} ∣ x - (- x + 4) ∣ d x

\int_{0}^{\frac{2}{3}} ∣ x - 5 x ∣ d x = \frac{8}{9}

\int_{\frac{2}{3}}^{2} ∣ x - (- x + 4) ∣ d x = \frac{16}{9}

= \frac{8}{9} + \frac{16}{9}

簡素化

= \frac{8}{3}