fläche y= 1/(16x^2),y=x,y= x/(18)

Lösung

fläche

y = \frac{1}{16 x ^{2}}, y = x, y = \frac{x}{18}

\frac{3 2 ( 18 - 1 8 ^{\frac{2}{3}} )}{96}

Dezimale

0.14609 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

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

\frac{1}{16 x ^{2}} = \frac{x}{18} : x = \frac{3 9}{2}

x = \frac{x}{18} : x = 0

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

= \int_{0}^{\frac{1}{2 3 2}} x - \frac{x}{18} d x + \int_{\frac{1}{2 3 2}}^{\frac{3 9}{2}} \frac{1}{16 x ^{2}} - \frac{x}{18} d x

\int_{0}^{\frac{1}{2 3 2}} x - \frac{x}{18} d x = \frac{17}{144 \cdot 2 ^{\frac{2}{3}}}

\int_{\frac{1}{2 3 2}}^{\frac{3 9}{2}} \frac{1}{16 x ^{2}} - \frac{x}{18} d x = \frac{- 27 \cdot 2 ^{\frac{2}{3}} + 18 \cdot 2 ^{\frac{2}{3}} 3 18 + 3 9}{144 3 9 \cdot 2 ^{\frac{2}{3}}}

= \frac{17}{144 \cdot 2 ^{\frac{2}{3}}} + \frac{- 27 \cdot 2 ^{\frac{2}{3}} + 18 \cdot 2 ^{\frac{2}{3}} 3 18 + 3 9}{144 3 9 \cdot 2 ^{\frac{2}{3}}}

Vereinfache

= \frac{3 2 ( 18 - 1 8 ^{\frac{2}{3}} )}{96}

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