面積 f(x)=((25-x^2)^{1/2}), 4/3 x,5

解

面積

f (x) = ((25 - x^{2})^{\frac{1}{2}}), \frac{4}{3} x, 5

\frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})

十進法表記

1.33123 \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)

((25 - x^{2})^{\frac{1}{2}}) = \frac{4}{3} x : x = 3

((25 - x^{2})^{\frac{1}{2}}) = 5 : x = 0

\frac{4}{3} x = 5 : x = \frac{15}{4}

= \int_{0}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{\frac{15}{4}} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x + \int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x

\int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x = - \frac{25}{2} arcsin (\frac{3}{5}) + 9

\int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x = \frac{3}{8}

= - \frac{25}{2} arcsin (\frac{3}{5}) + 9 + \frac{3}{8}

簡素化

= \frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})