面積 y=x,y=-x+3,y=0

解

面積

y = x, y = - x + 3, y = 0

\frac{9}{4}

十進法表記

2.25

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

x = 0 : x = 0

- x + 3 = 0 : x = 3

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

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

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

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

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

簡素化

= \frac{9}{4}