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面积 x=-2y^2+3y-5,x+3y=-13

解答

面积

x = - 2 y^{2} + 3 y - 5, x + 3 y = - 13

解答

\frac{125}{3}

+1

十进制

41.66666 \dots

求解步骤

对于

x = - 2 y^{2} + 3 y - 5

将

y

移到一边:

y = - \frac{- 3 + - 8 x - 31}{4}, y = - \frac{- 3 - - 8 x - 31}{4}

对于

x + 3 y = - 13

将

y

移到一边:

y = \frac{- 13 - x}{3}

f_{1} (x) = - \frac{- 3 + - 8 x - 31}{4}

f_{2} (x) = - \frac{- 3 - - 8 x - 31}{4}

要找到相交点，解

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

- \frac{- 3 + - 8 x - 31}{4} = - \frac{- 3 - - 8 x - 31}{4} : x = - \frac{31}{8}

- \frac{- 3 + - 8 x - 31}{4} = \frac{- 13 - x}{3} : x = - 10

- \frac{- 3 - - 8 x - 31}{4} = \frac{- 13 - x}{3} : x = - 25

= \int_{- 25}^{- 10} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 10}^{- \frac{31}{8}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 25}^{- 10} - \frac{- 3 - - 8 x - 31}{4} - \frac{- 13 - x}{3} d x + \int_{- 10}^{- \frac{31}{8}} - \frac{- 3 + - 8 x - 31}{4} - (- \frac{- 3 - - 8 x - 31}{4}) d x

\int_{- 25}^{- 10} - \frac{- 3 - - 8 x - 31}{4} - \frac{- 13 - x}{3} d x = \frac{219}{8}

\int_{- 10}^{- \frac{31}{8}} - \frac{- 3 + - 8 x - 31}{4} - (- \frac{- 3 - - 8 x - 31}{4}) d x = \frac{343}{24}

= \frac{219}{8} + \frac{343}{24}

化简

= \frac{125}{3}

作图

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