fläche x=(y^2)/4 ,x= y/2+2

Lösung

fläche

x = \frac{y ^{2}}{4}, x = \frac{y}{2} + 2

9

Schritte zur Lösung

Stelle

y

nach

x = \frac{y ^{2}}{4}

um:

y = 2 x, y = - 2 x

Stelle

y

nach

x = \frac{y}{2} + 2

um:

y = 2 x - 4

f_{1} (x) = 2 x

f_{2} (x) = - 2 x

Um die Schnittpunkte zu finden, löse

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

2 x = - 2 x : x = 0

2 x = 2 x - 4 : x = 4

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

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

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

\int_{0}^{1} 2 x - (- 2 x) d x = \frac{8}{3}

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

= \frac{8}{3} + \frac{19}{3}

Vereinfache

= 9

n = 0 \sum \infty n sin (nπ)

\int \frac{x}{x ( x + 4 )} d x

\frac{\partial}{\partial x} (e^{x} \cdot y)

d er i v a t i v e f (x) = x + \frac{7}{x}

\int x \cdot ln (1 + x) d x