fläche x,2x,-x+2

Lösung

fläche

x, 2 x, - x + 2

\frac{1}{3}

Dezimale

0.33333 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

x = 2 x : x = 0

x = - x + 2 : x = 1

2 x = - x + 2 : x = \frac{2}{3}

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

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

\int_{0}^{\frac{2}{3}} ∣ x - 2 x ∣ d x = \frac{2}{9}

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

= \frac{2}{9} + \frac{1}{9}

Vereinfache

= \frac{1}{3}

l a pl a ce t r an s f or m t e^{- 4 t} cos (5 t)

a re a - x^{4}, x^{2} - 2 x

\frac{d}{d x} (\frac{1 - 3 e ^{x}}{1 - 2 cos ( x )})

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

\int 3 e^{8 x} d x