面积 y= 3/x ,y=8x,y= 1/8 x

解答

面积

y = \frac{3}{x}, y = 8 x, y = \frac{1}{8} x

18 ln (2)

十进制

12.47664 \dots

求解步骤

要找到相交点，解

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

\frac{3}{x} = 8 x : x = \frac{3}{2 2}, x = - \frac{3}{2 2}

\frac{3}{x} = \frac{1}{8} x : x = 26, x = - 26

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

= \int_{- 26}^{- \frac{3}{2 2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{3}{2 2}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{3}{2 2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{3}{2 2}}^{26} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 26}^{- \frac{3}{2 2}} \frac{3}{x} - \frac{1}{8} x d x + \int_{- \frac{3}{2 2}}^{0} \frac{1}{8} x - 8 x d x + \int_{0}^{\frac{3}{2 2}} 8 x - \frac{1}{8} x d x + \int_{\frac{3}{2 2}}^{26} \frac{3}{x} - \frac{1}{8} x d x

\int_{- 26}^{- \frac{3}{2 2}} \frac{3}{x} - \frac{1}{8} x d x = 9 ln (2) - \frac{189}{128}

\int_{- \frac{3}{2 2}}^{0} \frac{1}{8} x - 8 x d x = \frac{189}{128}

\int_{0}^{\frac{3}{2 2}} 8 x - \frac{1}{8} x d x = \frac{189}{128}

\int_{\frac{3}{2 2}}^{26} \frac{3}{x} - \frac{1}{8} x d x = 9 ln (2) - \frac{189}{128}

= 9 ln (2) - \frac{189}{128} + \frac{189}{128} + \frac{189}{128} + 9 ln (2) - \frac{189}{128}

化简

= 18 ln (2)

\frac{d}{d x} (x^{3} + 8 x)

t a y l or \frac{1}{1 - x ^{2}}, 0

a re a y = 2 x, y = x, x = 0, x = 4

t an g e n t f (x) = 5 - 2 x^{2}, (1, 3)

\frac{d}{d x} (7 e^{- x} - 8 e^{- 7 x})