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面积 x=4y-y^2,x=y^2-7y

解答

面积

x = 4 y - y^{2}, x = y^{2} - 7 y

解答

\frac{1331}{24}

+1

十进制

55.45833 \dots

求解步骤

对于

x = 4 y - y^{2}

将

y

移到一边:

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

对于

x = y^{2} - 7 y

将

y

移到一边:

y = \frac{7 + 49 + 4 x}{2}, y = \frac{7 - 49 + 4 x}{2}

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

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

要找到相交点，解

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

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

2 - - x + 4 = \frac{7 + 49 + 4 x}{2} : x \in R

无解

2 - - x + 4 = \frac{7 - 49 + 4 x}{2} : x = 0

- x + 4 + 2 = \frac{7 + 49 + 4 x}{2} : x = - \frac{33}{4}

- x + 4 + 2 = \frac{7 - 49 + 4 x}{2} : x \in R

无解

\frac{7 + 49 + 4 x}{2} = \frac{7 - 49 + 4 x}{2} : x = - \frac{49}{4}

= \int_{- \frac{49}{4}}^{- \frac{33}{4}} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{- \frac{33}{4}}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- \frac{49}{4}}^{- \frac{33}{4}} \frac{7 + 49 + 4 x}{2} - \frac{7 - 49 + 4 x}{2} d x + \int_{- \frac{33}{4}}^{0} - x + 4 + 2 - \frac{7 - 49 + 4 x}{2} d x + \int_{0}^{4} 2 - - x + 4 - (- x + 4 + 2) d x

\int_{- \frac{49}{4}}^{- \frac{33}{4}} \frac{7 + 49 + 4 x}{2} - \frac{7 - 49 + 4 x}{2} d x = \frac{32}{3}

\int_{- \frac{33}{4}}^{0} - x + 4 + 2 - \frac{7 - 49 + 4 x}{2} d x = \frac{273}{8}

\int_{0}^{4} 2 - - x + 4 - (- x + 4 + 2) d x = \frac{32}{3}

= \frac{32}{3} + \frac{273}{8} + \frac{32}{3}

化简

= \frac{1331}{24}

作图

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