fläche 4x+y^2=32,x=y

Lösung

fläche

4 x + y^{2} = 32, x = y

72

Schritte zur Lösung

Stelle

y

nach

4 x + y^{2} = 32

um:

y = 2 8 - x, y = - 2 8 - x

Stelle

y

nach

x = y

um:

y = x

f_{1} (x) = 2 8 - x

f_{2} (x) = - 2 8 - x

Um die Schnittpunkte zu finden, löse

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

2 8 - x = - 2 8 - x : x = 8

2 8 - x = x : x = 4

- 2 8 - x = x : x = - 8

= \int_{- 8}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{4}^{8} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 8}^{4} - 2 8 - x - x d x + \int_{4}^{8} 2 8 - x - (- 2 8 - x) d x

\int_{- 8}^{4} - 2 8 - x - x d x = \frac{152}{3}

\int_{4}^{8} 2 8 - x - (- 2 8 - x) d x = \frac{64}{3}

= \frac{152}{3} + \frac{64}{3}

Vereinfache

= 72