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面积 y^2=4x+4,y=4x-16

解答

面积

y^{2} = 4 x + 4, y = 4 x - 16

解答

\frac{243}{8}

+1

十进制

30.375

求解步骤

对于

y^{2} = 4 x + 4

将

y

移到一边:

y = 2 x + 1, y = - 2 x + 1

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

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

要找到相交点，解

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

2 x + 1 = - 2 x + 1 : x = - 1

2 x + 1 = 4 x - 16 : x = \frac{21}{4}

- 2 x + 1 = 4 x - 16 : x = 3

= \int_{- 1}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{\frac{21}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 1}^{3} 2 x + 1 - (- 2 x + 1) d x + \int_{3}^{\frac{21}{4}} 2 x + 1 - (4 x - 16) d x

\int_{- 1}^{3} 2 x + 1 - (- 2 x + 1) d x = \frac{64}{3}

\int_{3}^{\frac{21}{4}} 2 x + 1 - (4 x - 16) d x = \frac{217}{24}

= \frac{64}{3} + \frac{217}{24}

化简

= \frac{243}{8}

作图

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