fläche y= 2/x ,y=8x,y= 1/(8x)

Lösung

fläche

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8 x}

- \frac{15}{8} + 8 ln (2) + divergiert

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

\frac{2}{x} = 8 x : x = \frac{1}{2}, x = - \frac{1}{2}

\frac{2}{x} = \frac{1}{8 x} :

Keine Lösung für

x \in R

8 x = \frac{1}{8 x} : x = \frac{1}{8}, x = - \frac{1}{8}

= \int_{- \frac{1}{2}}^{- \frac{1}{8}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{1}{8}}^{\frac{1}{8}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{8}}^{\frac{1}{2}} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- \frac{1}{2}}^{- \frac{1}{8}} \frac{2}{x} - 8 x d x + \int_{- \frac{1}{8}}^{\frac{1}{8}} \frac{2}{x} - \frac{1}{8 x} d x + \int_{\frac{1}{8}}^{\frac{1}{2}} 8 x - \frac{2}{x} d x

\int_{- \frac{1}{2}}^{- \frac{1}{8}} \frac{2}{x} - 8 x d x = 4 ln (2) - \frac{15}{16}

\int_{- \frac{1}{8}}^{\frac{1}{8}} \frac{2}{x} - \frac{1}{8 x} d x =

divergiert

\int_{\frac{1}{8}}^{\frac{1}{2}} 8 x - \frac{2}{x} d x = - \frac{15}{16} + 4 ln (2)

= 4 ln (2) - \frac{15}{16} + divergiert - \frac{15}{16} + 4 ln (2)

Vereinfache

= - \frac{15}{8} + 8 ln (2) + divergiert

x \frac{d y}{d x} - y = x^{2} + y^{2}

\frac{4 d u}{d t} = u^{2}

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

y^{^{''}} - 4 y^{^{'}} + 8 y = 0

f (x) = arcsin (\frac{x}{4})