面积 2/x ,8x, 1/2 x

解答

面积

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

8 ln (2)

十进制

5.54517 \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}{2} x : x = 2, x = - 2

8 x = \frac{1}{2} x : x = 0

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

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

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

\int_{- \frac{1}{2}}^{0} \frac{1}{2} x - 8 x d x = \frac{15}{16}

\int_{0}^{\frac{1}{2}} 8 x - \frac{1}{2} x d x = \frac{15}{16}

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

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

化简

= 8 ln (2)

x \to 3 lim (2 - x)

\frac{d}{d x} (\frac{1 - cos ( x )}{1 + sin ( x )})

\frac{d}{d x} (- \frac{e ^{x - 1} ( x - 3 )}{( x - 1 ) ^{3}})

in v erse l a pl a ce e^{s} \frac{1}{s ^{2}}

x \to 1 lim (\frac{x ^{2} + x}{x + 1})