面积 6x, 2/3 x,40-x^2

解答

面积

6 x, \frac{2}{3} x, 40 - x^{2}

\frac{20192}{81}

十进制

249.28395 \dots

求解步骤

要找到相交点，解

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

6 x = \frac{2}{3} x : x = 0

6 x = 40 - x^{2} : x = - 10, x = 4

\frac{2}{3} x = 40 - x^{2} : x = - \frac{20}{3}, x = 6

= \int_{- 10}^{- \frac{20}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{20}{3}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{4} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{4}^{6} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 10}^{- \frac{20}{3}} 6 x - (40 - x^{2}) d x + \int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 6 x d x + \int_{0}^{4} 6 x - \frac{2}{3} x d x + \int_{4}^{6} \frac{2}{3} x - (40 - x^{2}) d x

\int_{- 10}^{- \frac{20}{3}} 6 x - (40 - x^{2}) d x = \frac{5300}{81}

\int_{- \frac{20}{3}}^{0} \frac{2}{3} x - 6 x d x = \frac{3200}{27}

\int_{0}^{4} 6 x - \frac{2}{3} x d x = \frac{128}{3}

\int_{4}^{6} \frac{2}{3} x - (40 - x^{2}) d x = \frac{68}{3}

= \frac{5300}{81} + \frac{3200}{27} + \frac{128}{3} + \frac{68}{3}

化简

= \frac{20192}{81}

\frac{d}{d x} (6 cos (x) sin (x))

\frac{\partial}{\partial x} (e^{5 y})

\int \frac{1}{x ^{2} \cdot 4 - x ^{2}} d x

x \to 0 - lim (\frac{5}{x} - \frac{5}{∣ x ∣})

x \to - 2 - lim (2 x + 2 + x)