面积 1/4 x^2-1,y=8,y=2

解答

面积

\frac{1}{4} x^{2} - 1, y = 8, y = 2

72 - 83

十进制

58.14359 \dots

求解步骤

要找到相交点，解

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

8 = 2 :

假

8 = \frac{1}{4} x^{2} - 1 : x = 6, x = - 6

2 = \frac{1}{4} x^{2} - 1 : x = 23, x = - 23

= \int_{- 6}^{- 23} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 23}^{23} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{23}^{6} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 6}^{- 23} 8 - (\frac{1}{4} x^{2} - 1) d x + \int_{- 23}^{23} ∣ 2 - 8 ∣ d x + \int_{23}^{6} 8 - (\frac{1}{4} x^{2} - 1) d x

\int_{- 6}^{- 23} 8 - (\frac{1}{4} x^{2} - 1) d x = 36 - 163

\int_{- 23}^{23} ∣ 2 - 8 ∣ d x = 243

\int_{23}^{6} 8 - (\frac{1}{4} x^{2} - 1) d x = 36 - 163

= 36 - 163 + 243 + 36 - 163

化简

= 72 - 83

n = 4 \sum \infty 3 - 0. 5^{n}

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t a y l or sin (x), \frac{3 π}{2}

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