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Beliebt Rechnen >

fläche x=2y-y^2,x=y^2-4y

Lösung

fläche

x = 2 y - y^{2}, x = y^{2} - 4 y

Lösung

9

Schritte zur Lösung

Stelle

y

nach

x = 2 y - y^{2}

um:

y = 1 - - x + 1, y = - x + 1 + 1

Stelle

y

nach

x = y^{2} - 4 y

um:

y = 2 + x + 4, y = 2 - x + 4

f_{1} (x) = 1 - - x + 1

f_{2} (x) = - x + 1 + 1

Um die Schnittpunkte zu finden, löse

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

1 - - x + 1 = - x + 1 + 1 : x = 1

1 - - x + 1 = 2 + x + 4 :

Keine Lösung für

x \in R

1 - - x + 1 = 2 - x + 4 : x = 0

- x + 1 + 1 = 2 + x + 4 : x = - 3

- x + 1 + 1 = 2 - x + 4 :

Keine Lösung für

x \in R

2 + x + 4 = 2 - x + 4 : x = - 4

= \int_{- 4}^{- 3} ∣ f_{3} (x) - f_{4} (x) ∣ d x + \int_{- 3}^{0} ∣ f_{2} (x) - f_{4} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 4}^{- 3} 2 + x + 4 - (2 - x + 4) d x + \int_{- 3}^{0} - x + 1 + 1 - (2 - x + 4) d x + \int_{0}^{1} 1 - - x + 1 - (- x + 1 + 1) d x

\int_{- 4}^{- 3} 2 + x + 4 - (2 - x + 4) d x = \frac{4}{3}

\int_{- 3}^{0} - x + 1 + 1 - (2 - x + 4) d x = \frac{19}{3}

\int_{0}^{1} 1 - - x + 1 - (- x + 1 + 1) d x = \frac{4}{3}

= \frac{4}{3} + \frac{19}{3} + \frac{4}{3}

Vereinfache

= 9

Graph

Beliebte Beispiele

fläche x^2-x,x=4

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

derivative of 2cos(2x-1)

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

limit as x approaches 4 of sqrt(x)-x^2

x \to 4 lim (x - x^{2})

d/(dy)(4arctan(x/y))

\frac{d}{d y} (4 arctan (\frac{x}{y}))

(\partial)/(\partial x)(4sqrt(y^2+7x^2))

\frac{\partial}{\partial x} (4 y^{2} + 7 x^{2})