面积 y= 1/(18x^2),y=x,y= x/(20)

解答

面积

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{20}

\frac{59 3 18 - 4 \cdot 2 ^{\frac{2}{3}} 3 225 - 2 \cdot 1 0 ^{\frac{2}{3}} 3 9 + 3 2 \cdot 3 ^{\frac{2}{3}}}{720}

十进制

0.13793 \dots

求解步骤

要找到相交点，解

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

\frac{1}{18 x ^{2}} = x : x = \frac{1}{3 18}

\frac{1}{18 x ^{2}} = \frac{x}{20} : x = 3 \frac{10}{9}

x = \frac{x}{20} : x = 0

= \int_{0}^{\frac{1}{3 18}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 18}}^{3 \frac{10}{9}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 18}} x - \frac{x}{20} d x + \int_{\frac{1}{3 18}}^{3 \frac{10}{9}} \frac{1}{18 x ^{2}} - \frac{x}{20} d x

\int_{0}^{\frac{1}{3 18}} x - \frac{x}{20} d x = \frac{19}{40 \cdot 1 8 ^{\frac{2}{3}}}

\int_{\frac{1}{3 18}}^{3 \frac{10}{9}} \frac{1}{18 x ^{2}} - \frac{x}{20} d x = \frac{- 3 \frac{9}{10} + 3 18}{18} - \frac{6 \cdot 3 0 ^{\frac{2}{3}} - 3 6 \cdot 9 ^{\frac{2}{3}}}{1080 3 2 \cdot 2 ^{\frac{2}{3}}}

= \frac{19}{40 \cdot 1 8 ^{\frac{2}{3}}} + \frac{- 3 \frac{9}{10} + 3 18}{18} - \frac{6 \cdot 3 0 ^{\frac{2}{3}} - 3 6 \cdot 9 ^{\frac{2}{3}}}{1080 3 2 \cdot 2 ^{\frac{2}{3}}}

化简

= \frac{59 3 18 - 4 \cdot 2 ^{\frac{2}{3}} 3 225 - 2 \cdot 1 0 ^{\frac{2}{3}} 3 9 + 3 2 \cdot 3 ^{\frac{2}{3}}}{720}

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