面积 y= 1/(18x^2),y=x,y= x/(10)

解答

面积

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{10}

\frac{- 2 3 450 \cdot 2 ^{\frac{2}{3}} + 10 3 2 \cdot 3 ^{\frac{2}{3}} + 20 3 18 - 3 2 \cdot 3 0 ^{\frac{2}{3}}}{360}

十进制

0.11702 \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}{10} : x = 3 \frac{5}{9}

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

= \int_{0}^{\frac{1}{3 18}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 18}}^{3 \frac{5}{9}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 18}} x - \frac{x}{10} d x + \int_{\frac{1}{3 18}}^{3 \frac{5}{9}} \frac{1}{18 x ^{2}} - \frac{x}{10} d x

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

\int_{\frac{1}{3 18}}^{3 \frac{5}{9}} \frac{1}{18 x ^{2}} - \frac{x}{10} d x = \frac{10 \cdot 2 ^{\frac{2}{3}} ( - 3 \frac{9}{5} + 3 18 ) - 3 ^{\frac{2}{3}} ( 1 0 ^{\frac{2}{3}} - 1 )}{180 \cdot 2 ^{\frac{2}{3}}}

= \frac{3 ^{\frac{2}{3}}}{20 \cdot 2 ^{\frac{2}{3}}} + \frac{10 \cdot 2 ^{\frac{2}{3}} ( - 3 \frac{9}{5} + 3 18 ) - 3 ^{\frac{2}{3}} ( 1 0 ^{\frac{2}{3}} - 1 )}{180 \cdot 2 ^{\frac{2}{3}}}

化简

= \frac{- 2 3 450 \cdot 2 ^{\frac{2}{3}} + 10 3 2 \cdot 3 ^{\frac{2}{3}} + 20 3 18 - 3 2 \cdot 3 0 ^{\frac{2}{3}}}{360}

\int \frac{x - 4}{x ^{2}} d x

d er i v a t i v e f (x) = 6 e^{x} (cos (x) - sin (x))

d er i v a t i v e 3 t^{7} + 4 t^{5} - t

\frac{d}{d x} (ln (\frac{4 x ^{2} - 9}{2 x - 3}))

x \to 1 + lim (x^{2} - x)