fläche y=x,y= x/3 ,y=2

Lösung

fläche

y = x, y = \frac{x}{3}, y = 2

4

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

x = \frac{x}{3} : x = 0

x = 2 : x = 2

\frac{x}{3} = 2 : x = 6

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

= \int_{0}^{2} x - \frac{x}{3} d x + \int_{2}^{6} \frac{x}{3} - 2 d x

\int_{0}^{2} x - \frac{x}{3} d x = \frac{4}{3}

\int_{2}^{6} \frac{x}{3} - 2 d x = \frac{8}{3}

= \frac{4}{3} + \frac{8}{3}

Vereinfache

= 4

\int_{0}^{2} x^{2} 1 + x^{3} d x

\frac{d}{d x} (\frac{2}{1 + 4 x ^{2}})

\int \frac{x ^{2}}{36 + x ^{2}} d x

d er i v a t i v e x^{3} - 5

in v erse l a pl a ce \frac{3 s - 1}{s ^{2} - 1}