面積 sqrt(x),x-2,0

解

面積

x, x - 2, 0

\frac{10}{3}

十進法表記

3.33333 \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 = x - 2 : x = 4

x = 0 : x = 0

x - 2 = 0 : x = 2

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

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

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

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

= \frac{4 2}{3} + \frac{16 - 4 2}{3} - 2

簡素化

= \frac{10}{3}