面积 y= 1/(20x^2),y=x,y= x/6

解答

面积

y = \frac{1}{20 x ^{2}}, y = x, y = \frac{x}{6}

\frac{9 \cdot 2 ^{\frac{2}{3}} 3 5 - 2 3 90 - 3 10 \cdot 3 ^{\frac{2}{3}}}{120}

十进制

0.09154 \dots

求解步骤

要找到相交点，解

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

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

\frac{1}{20 x ^{2}} = \frac{x}{6} : x = 3 \frac{3}{10}

x = \frac{x}{6} : x = 0

= \int_{0}^{\frac{1}{3 20}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 20}}^{3 \frac{3}{10}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

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

\int_{0}^{\frac{1}{3 20}} x - \frac{x}{6} d x = \frac{3 5}{24 3 2}

\int_{\frac{1}{3 20}}^{3 \frac{3}{10}} \frac{1}{20 x ^{2}} - \frac{x}{6} d x = \frac{- 4 3 450 + 12 \cdot 5 ^{\frac{2}{3}} 3 4 - 2 \cdot 1 5 ^{\frac{2}{3}} 3 2 + 1 0 ^{\frac{2}{3}}}{240 3 5}

= \frac{3 5}{24 3 2} + \frac{- 4 3 450 + 12 \cdot 5 ^{\frac{2}{3}} 3 4 - 2 \cdot 1 5 ^{\frac{2}{3}} 3 2 + 1 0 ^{\frac{2}{3}}}{240 3 5}

化简

= \frac{9 \cdot 2 ^{\frac{2}{3}} 3 5 - 2 3 90 - 3 10 \cdot 3 ^{\frac{2}{3}}}{120}

\frac{d}{d x} (a (e^{2 x} + 2 e^{2 x} x))

d er i v a t i v e f (x) = 3 x^{2} + 2 ln (8) (x)

y^{^{'}} = \frac{( x ^{5} )}{y}

\int_{0}^{3 π} 3 x^{2} sin (\frac{1}{6} x) d x

x \to 0 lim ((x)^{- 1 + x})