fläche 1/(12x^2),x, x/(10)

Lösung

fläche

\frac{1}{12 x ^{2}}, x, \frac{x}{10}

\frac{15 \cdot 2 ^{\frac{2}{3}} 3 3 - 2 3 150 - 3 6 \cdot 5 ^{\frac{2}{3}}}{120}

Dezimale

0.15334 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

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

\frac{1}{12 x ^{2}} = \frac{x}{10} : x = 3 \frac{5}{6}

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

= \int_{0}^{\frac{1}{3 12}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{3 12}}^{3 \frac{5}{6}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{\frac{1}{3 12}} x - \frac{x}{10} d x + \int_{\frac{1}{3 12}}^{3 \frac{5}{6}} \frac{1}{12 x ^{2}} - \frac{x}{10} d x

\int_{0}^{\frac{1}{3 12}} x - \frac{x}{10} d x = \frac{3 3 3}{40 3 2}

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

= \frac{3 3 3}{40 3 2} + \frac{- 4 3 450 + 20 \cdot 3 ^{\frac{2}{3}} 3 4 - 2 \cdot 1 5 ^{\frac{2}{3}} 3 2 + 6 ^{\frac{2}{3}}}{240 3 3}

Vereinfache

= \frac{15 \cdot 2 ^{\frac{2}{3}} 3 3 - 2 3 150 - 3 6 \cdot 5 ^{\frac{2}{3}}}{120}

16 y^{^{''''}} + 24 y^{^{''}} + 9 y = 0

\frac{\partial}{\partial x} (x^{3} - 3 y^{2})

x \to 4 lim (\frac{1}{x})

n = 0 \sum \infty 6 x^{n}

\frac{\partial}{\partial y} (x^{5} - 7 x^{3} y^{3} + y^{6})