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受欢迎的微积分 >

面积 x=y^2-4y,x=2y-y^2

解答

面积

x = y^{2} - 4 y, x = 2 y - y^{2}

解答

9

求解步骤

对于

x = y^{2} - 4 y

将

y

移到一边:

y = 2 + x + 4, y = 2 - x + 4

对于

x = 2 y - y^{2}

将

y

移到一边:

y = 1 - - x + 1, y = - x + 1 + 1

f_{1} (x) = 2 + x + 4

f_{2} (x) = 2 - x + 4

要找到相交点，解

f_{i} (x) = f_{j} (x)

2 + x + 4 = 2 - x + 4 : x = - 4

2 + x + 4 = 1 - - x + 1 : x \in R

无解

2 + x + 4 = - x + 1 + 1 : x = - 3

2 - x + 4 = 1 - - x + 1 : x = 0

2 - x + 4 = - x + 1 + 1 : x \in R

无解

1 - - x + 1 = - x + 1 + 1 : x = 1

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

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

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

\int_{- 3}^{0} 2 - x + 4 - (- x + 1 + 1) d x = \frac{19}{3}

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

= \frac{4}{3} + \frac{19}{3} + \frac{4}{3}

化简

= 9

作图

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