fläche x^3,x+6=y,(-x)/2 =y

Lösung

fläche

x^{3}, x + 6 = y, \frac{- x}{2} = y

22

Schritte zur Lösung

Stelle

y

nach

x + 6 = y

um:

y = x + 6

Stelle

y

nach

\frac{- x}{2} = y

um:

y = - \frac{x}{2}

f_{1} (x) = x + 6

f_{2} (x) = - \frac{x}{2}

Um die Schnittpunkte zu finden, löse

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

x + 6 = - \frac{x}{2} : x = - 4

x + 6 = x^{3} : x = 2

- \frac{x}{2} = x^{3} : x = 0

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

= \int_{- 4}^{0} x + 6 - (- \frac{x}{2}) d x + \int_{0}^{2} x + 6 - x^{3} d x

\int_{- 4}^{0} x + 6 - (- \frac{x}{2}) d x = 12

\int_{0}^{2} x + 6 - x^{3} d x = 10

= 12 + 10

Vereinfache

= 22

\frac{\partial}{\partial y} (ln (x y))

\int (\frac{1}{x x ^{2} - 1}) d x

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{- x}

\frac{d}{d θ} (cos (θ) - 2 sin (θ))

x \to 0 - lim (a x^{2} + \frac{b}{2})