面积 1/(x^2),x,8x

解答

面积

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

\frac{3}{2}

十进制

1.5

求解步骤

要找到相交点，解

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

\frac{1}{x ^{2}} = x : x = 1

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

x = 8 x : x = 0

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

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

\int_{0}^{\frac{1}{2}} ∣ x - 8 x ∣ d x = \frac{7}{8}

\int_{\frac{1}{2}}^{1} \frac{1}{x ^{2}} - x d x = \frac{5}{8}

= \frac{7}{8} + \frac{5}{8}

化简

= \frac{3}{2}

\int_{0}^{1} \frac{1}{( x - 2 ) ^{2}} d x

x \to 0 lim (\frac{e ^{x} cos ( x )}{4})

\frac{d}{d x} (\frac{e ^{x}}{4 + e ^{x}})

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2} - 36 x}{x ^{2} + 4 x})

in v erse l a pl a ce \frac{25}{x ( x ^{2} + 6 x + 25 )}