fläche x=y,2x+y^2=8

Lösung

fläche

x = y, 2 x + y^{2} = 8

18

Schritte zur Lösung

Stelle

y

nach

x = y

um:

y = x

Stelle

y

nach

2 x + y^{2} = 8

um:

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

f_{1} (x) = x

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

Um die Schnittpunkte zu finden, löse

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

x = 8 - 2 x : x = 2

x = - 8 - 2 x : x = - 4

8 - 2 x = - 8 - 2 x : x = 4

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

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

\int_{- 4}^{2} x - (- 8 - 2 x) d x = \frac{38}{3}

\int_{2}^{4} - 8 - 2 x - 8 - 2 x d x = \frac{16}{3}

= \frac{38}{3} + \frac{16}{3}

Vereinfache

= 18

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

in v erse l a pl a ce \frac{1}{s ( s ^{2} + 4 s + 5 )}

\int \frac{1}{25 x ^{2} - 16} d x

in v erse l a pl a ce \frac{2 s + 3}{s ^{2} + 7}

d er i v a t i v e y = \frac{x}{x + 1}