面积 2/x ,8x, 1/8 x

解答

面积

\frac{2}{x}, 8 x, \frac{1}{8} x

12 ln (2)

十进制

8.31776 \dots

求解步骤

要找到相交点，解

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

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

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

8 x = \frac{1}{8} x : x = 0

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

= \int_{- 4}^{- \frac{1}{2}} \frac{2}{x} - \frac{1}{8} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{8} x - 8 x d x + \int_{0}^{\frac{1}{2}} 8 x - \frac{1}{8} x d x + \int_{\frac{1}{2}}^{4} \frac{2}{x} - \frac{1}{8} x d x

\int_{- 4}^{- \frac{1}{2}} \frac{2}{x} - \frac{1}{8} x d x = 6 ln (2) - \frac{63}{64}

\int_{- \frac{1}{2}}^{0} \frac{1}{8} x - 8 x d x = \frac{63}{64}

\int_{0}^{\frac{1}{2}} 8 x - \frac{1}{8} x d x = \frac{63}{64}

\int_{\frac{1}{2}}^{4} \frac{2}{x} - \frac{1}{8} x d x = 6 ln (2) - \frac{63}{64}

= 6 ln (2) - \frac{63}{64} + \frac{63}{64} + \frac{63}{64} + 6 ln (2) - \frac{63}{64}

化简

= 12 ln (2)

d er i v a t i v e - 5 x^{4} + 5 x^{3} - x^{2} + x

\int_{- 7}^{7} 1 + (x)^{2} d x

\int tan^{2} (3 t) d t

\frac{\partial}{\partial x} (x \cdot a + b)

t an g e n t y = \frac{x - 1}{x - 2} + 1, (3, 3)