面积 9x,x^2-10,0

解答

面积

9 x, x^{2} - 10, 0

\frac{423}{2}

十进制

211.5

求解步骤

要找到相交点，解

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

9 x = x^{2} - 10 : x = 10, x = - 1

9 x = 0 : x = 0

x^{2} - 10 = 0 : x = 10, x = - 10

= \int_{- 10}^{- 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{10} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{10}^{10} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 10}^{- 1} x^{2} - 10 - 0 d x + \int_{- 1}^{0} ∣ 9 x - 0 ∣ d x + \int_{0}^{10} ∣ 9 x - 0 ∣ d x + \int_{10}^{10} x^{2} - 10 - 9 x d x

\int_{- 10}^{- 1} x^{2} - 10 - 0 d x = - \frac{- 1 + 10 10}{3} - 10 + 1010

\int_{- 1}^{0} ∣ 9 x - 0 ∣ d x = \frac{9}{2}

\int_{0}^{10} ∣ 9 x - 0 ∣ d x = 45

\int_{10}^{10} x^{2} - 10 - 9 x d x = - \frac{1000 - 10 10}{3} + 505 - 1010

= - \frac{- 1 + 10 10}{3} - 10 + 1010 + \frac{9}{2} + 45 - \frac{1000 - 10 10}{3} + 505 - 1010

化简

= \frac{423}{2}

\int e^{6 x} x^{3} d x

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