fläche 4x+y^2=12,y=x

Lösung

fläche

4 x + y^{2} = 12, y = x

\frac{64}{3}

Dezimale

21.33333 \dots

Schritte zur Lösung

Stelle

y

nach

4 x + y^{2} = 12

um:

y = 2 3 - x, y = - 2 3 - x

f_{1} (x) = 2 3 - x

f_{2} (x) = - 2 3 - x

Um die Schnittpunkte zu finden, löse

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

2 3 - x = - 2 3 - x : x = 3

2 3 - x = x : x = 2

- 2 3 - x = x : x = - 6

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

= \int_{- 6}^{2} - 2 3 - x - x d x + \int_{2}^{3} 2 3 - x - (- 2 3 - x) d x

\int_{- 6}^{2} - 2 3 - x - x d x = \frac{56}{3}

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

= \frac{56}{3} + \frac{8}{3}

Vereinfache

= \frac{64}{3}

x \to 0 lim (\frac{sin ( 2 x )}{3 x})

\int 4 π r^{2} d r

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

\int x 4 x^{2} - 5 d x

d er i v a t i v e f (x) = (3 x^{4} + 8)^{\frac{4}{3}}