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面积 x=y^2-8y,x=3y-y^2

解答

面积

x = y^{2} - 8 y, x = 3 y - y^{2}

解答

\frac{1331}{24}

+1

十进制

55.45833 \dots

求解步骤

对于

x = y^{2} - 8 y

将

y

移到一边:

y = 4 + x + 16, y = 4 - x + 16

对于

x = 3 y - y^{2}

将

y

移到一边:

y = - \frac{- 3 + 9 - 4 x}{2}, y = - \frac{- 3 - 9 - 4 x}{2}

f_{1} (x) = 4 + x + 16

f_{2} (x) = 4 - x + 16

要找到相交点，解

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

4 + x + 16 = 4 - x + 16 : x = - 16

4 + x + 16 = - \frac{- 3 + 9 - 4 x}{2} : x \in R

无解

4 + x + 16 = - \frac{- 3 - 9 - 4 x}{2} : x = - \frac{55}{4}

4 - x + 16 = - \frac{- 3 + 9 - 4 x}{2} : x = 0

4 - x + 16 = - \frac{- 3 - 9 - 4 x}{2} : x \in R

无解

- \frac{- 3 + 9 - 4 x}{2} = - \frac{- 3 - 9 - 4 x}{2} : x = \frac{9}{4}

= \int_{- 16}^{- \frac{55}{4}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{55}{4}}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{\frac{9}{4}} ∣ f_{4} (x) - f_{3} (x) ∣ d x

= \int_{- 16}^{- \frac{55}{4}} 4 + x + 16 - (4 - x + 16) d x + \int_{- \frac{55}{4}}^{0} 4 - x + 16 - (- \frac{- 3 - 9 - 4 x}{2}) d x + \int_{0}^{\frac{9}{4}} - \frac{- 3 - 9 - 4 x}{2} - (- \frac{- 3 + 9 - 4 x}{2}) d x

\int_{- 16}^{- \frac{55}{4}} 4 + x + 16 - (4 - x + 16) d x = \frac{9}{2}

\int_{- \frac{55}{4}}^{0} 4 - x + 16 - (- \frac{- 3 - 9 - 4 x}{2}) d x = \frac{1115}{24}

\int_{0}^{\frac{9}{4}} - \frac{- 3 - 9 - 4 x}{2} - (- \frac{- 3 + 9 - 4 x}{2}) d x = \frac{9}{2}

= \frac{9}{2} + \frac{1115}{24} + \frac{9}{2}

化简

= \frac{1331}{24}

作图

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