面积 y=7x,y=x^2-8,y=0

解答

面积

y = 7 x, y = x^{2} - 8, y = 0

\frac{679}{6}

十进制

113.16666 \dots

求解步骤

要找到相交点，解

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

7 x = x^{2} - 8 : x = 8, x = - 1

7 x = 0 : x = 0

x^{2} - 8 = 0 : x = 22, x = - 22

= \int_{- 22}^{- 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{22} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{22}^{8} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 22}^{- 1} x^{2} - 8 - 0 d x + \int_{- 1}^{0} ∣ 7 x - 0 ∣ d x + \int_{0}^{22} ∣ 7 x - 0 ∣ d x + \int_{22}^{8} x^{2} - 8 - 7 x d x

\int_{- 22}^{- 1} x^{2} - 8 - 0 d x = - \frac{- 1 + 16 2}{3} - 8 + 162

\int_{- 1}^{0} ∣ 7 x - 0 ∣ d x = \frac{7}{2}

\int_{0}^{22} ∣ 7 x - 0 ∣ d x = 28

\int_{22}^{8} x^{2} - 8 - 7 x d x = - \frac{512 - 16 2}{3} + 260 - 162

= - \frac{- 1 + 16 2}{3} - 8 + 162 + \frac{7}{2} + 28 - \frac{512 - 16 2}{3} + 260 - 162

化简

= \frac{679}{6}

\frac{d}{d x} ((x^{- 2} + x)^{3} (4 x)^{- 2})

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