面积 2/x ,8x, 1/(8x)

解答

面积

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

- \frac{15}{8} + 8 ln (2) + 发散

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 \in R

无解

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

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

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

\int_{- \frac{1}{2}}^{- \frac{1}{8}} \frac{2}{x} - 8 x d x = 4 ln (2) - \frac{15}{16}

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

发散

\int_{\frac{1}{8}}^{\frac{1}{2}} 8 x - \frac{2}{x} d x = - \frac{15}{16} + 4 ln (2)

= 4 ln (2) - \frac{15}{16} + 发散 - \frac{15}{16} + 4 ln (2)

化简

= - \frac{15}{8} + 8 ln (2) + 发散

\int (3 x - 3) e^{3 x^{2} - 6 x} d x

d er i v a t i v e y = 1 7^{x}

\frac{\partial}{\partial x} (sin (π (6 x - 5 y)))

d er i v a t i v e f (x) = \frac{3}{( x ^{2} - 5 x ) ^{2}}

d er i v a t i v e y = 1 cos (2) (x)