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人気のある微分積分 >

面積 x=sqrt(1/(12y)),y=x,x=16y

解

面積

x = \frac{1}{12 y}, y = x, x = 16 y

解

\frac{- 3 6 + 2 3 3 \cdot 2 ^{\frac{2}{3}}}{16}

+1

十進法表記

0.17260 \dots

解答ステップ

以下のために

y

を分離:

x = \frac{1}{12 y} : y = \frac{1}{12 x ^{2}}

以下のために

y

を分離:

x = 16 y : y = \frac{x}{16}

f_{1} (x) = \frac{1}{12 x ^{2}}

f_{2} (x) = x

交点を見つけるには以下を解く:

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

\frac{1}{12 x ^{2}} = x : x = \frac{1}{3 12}

\frac{1}{12 x ^{2}} = \frac{x}{16} : x = 3 \frac{4}{3}

x = \frac{x}{16} : x = 0

= \int_{0}^{\frac{1}{3 12}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 12}}^{3 \frac{4}{3}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 12}} x - \frac{x}{16} d x + \int_{\frac{1}{3 12}}^{3 \frac{4}{3}} \frac{1}{12 x ^{2}} - \frac{x}{16} d x

\int_{0}^{\frac{1}{3 12}} x - \frac{x}{16} d x = \frac{5 3 3}{64 3 2}

\int_{\frac{1}{3 12}}^{3 \frac{4}{3}} \frac{1}{12 x ^{2}} - \frac{x}{16} d x = \frac{- 8 \cdot 1 2 ^{\frac{2}{3}} 3 \frac{3}{4} + 32 \cdot 3 ^{\frac{2}{3}} 3 3 - 6 \cdot 1 2 ^{\frac{2}{3}} 3 \frac{2}{9} + 3}{96 \cdot 1 2 ^{\frac{2}{3}}}

= \frac{5 3 3}{64 3 2} + \frac{- 8 \cdot 1 2 ^{\frac{2}{3}} 3 \frac{3}{4} + 32 \cdot 3 ^{\frac{2}{3}} 3 3 - 6 \cdot 1 2 ^{\frac{2}{3}} 3 \frac{2}{9} + 3}{96 \cdot 1 2 ^{\frac{2}{3}}}

簡素化

= \frac{- 3 6 + 2 3 3 \cdot 2 ^{\frac{2}{3}}}{16}

グラフ

人気の例

(\partial)/(\partial x)(-(x/y))

\frac{\partial}{\partial x} (- (\frac{x}{y}))

derivative of x/(4-sqrt(16-x^2))

\frac{d}{d x} (\frac{x}{4 - 16 - x ^{2}})

tangent f(x)=5-2x^2,(-5,-45)

t an g e n t f (x) = 5 - 2 x^{2}, (- 5, - 45)

integral from 0 to 2 of (1+x)^{1/3}*2

\int_{0}^{2} (1 + x)^{\frac{1}{3}} \cdot 2 d x

xy'+(1+x*cot(x))y=0

x y^{'} + (1 + x \cdot cot (x)) y = 0