面积 1/(16x^2),x, x/(10)

解答

面积

\frac{1}{16 x ^{2}}, x, \frac{x}{10}

\frac{6 \cdot 5 ^{\frac{2}{3}} 3 10 - 3 2 \cdot 1 0 ^{\frac{2}{3}} - 4 \cdot 5 ^{\frac{2}{3}}}{160}

十进制

0.12658 \dots

求解步骤

要找到相交点，解

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

\frac{1}{16 x ^{2}} = x : x = \frac{1}{2 3 2}

\frac{1}{16 x ^{2}} = \frac{x}{10} : x = \frac{3 5}{2}

x = \frac{x}{10} : x = 0

= \int_{0}^{\frac{1}{2 3 2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2 3 2}}^{\frac{3 5}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{2 3 2}} x - \frac{x}{10} d x + \int_{\frac{1}{2 3 2}}^{\frac{3 5}{2}} \frac{1}{16 x ^{2}} - \frac{x}{10} d x

\int_{0}^{\frac{1}{2 3 2}} x - \frac{x}{10} d x = \frac{9}{80 \cdot 2 ^{\frac{2}{3}}}

\int_{\frac{1}{2 3 2}}^{\frac{3 5}{2}} \frac{1}{16 x ^{2}} - \frac{x}{10} d x = \frac{- 1 + 3 10}{8 3 5} - \frac{1 0 ^{\frac{2}{3}} - 1}{80 \cdot 2 ^{\frac{2}{3}}}

= \frac{9}{80 \cdot 2 ^{\frac{2}{3}}} + \frac{- 1 + 3 10}{8 3 5} - \frac{1 0 ^{\frac{2}{3}} - 1}{80 \cdot 2 ^{\frac{2}{3}}}

化简

= \frac{6 \cdot 5 ^{\frac{2}{3}} 3 10 - 3 2 \cdot 1 0 ^{\frac{2}{3}} - 4 \cdot 5 ^{\frac{2}{3}}}{160}

\int (3 x - 1) (x + 5) d x

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

\frac{d}{d x} (arccos (\frac{4}{x}))

\int 7 x d x

y^{^{''}} - 2 y^{^{'}} + 1 = 0