fläche y^2=4x+4,y=4x-16

Lösung

fläche

y^{2} = 4 x + 4, y = 4 x - 16

\frac{243}{8}

Dezimale

30.375

Schritte zur Lösung

Stelle

y

nach

y^{2} = 4 x + 4

um:

y = 2 x + 1, y = - 2 x + 1

f_{1} (x) = 2 x + 1

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

Um die Schnittpunkte zu finden, löse

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

2 x + 1 = - 2 x + 1 : x = - 1

2 x + 1 = 4 x - 16 : x = \frac{21}{4}

- 2 x + 1 = 4 x - 16 : x = 3

= \int_{- 1}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{3}^{\frac{21}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 1}^{3} 2 x + 1 - (- 2 x + 1) d x + \int_{3}^{\frac{21}{4}} 2 x + 1 - (4 x - 16) d x

\int_{- 1}^{3} 2 x + 1 - (- 2 x + 1) d x = \frac{64}{3}

\int_{3}^{\frac{21}{4}} 2 x + 1 - (4 x - 16) d x = \frac{217}{24}

= \frac{64}{3} + \frac{217}{24}

Vereinfache

= \frac{243}{8}

x \to + 0 lim (e^{4 x + 2})

\frac{d}{d x} (tan^{n} (x))

\int e^{- x^{6}} (- 6 x^{5}) d x

\int \frac{2 x}{( x ^{2} + 4 ) ^{2}} d x

im pl i c i t \frac{d y}{d x}, - x + y - x^{3} = y^{3} + 4 x^{2}