fläche y^2=2x+1,y=x-1

Lösung

fläche

y^{2} = 2 x + 1, y = x - 1

\frac{16}{3}

Dezimale

5.33333 \dots

Schritte zur Lösung

Stelle

y

nach

y^{2} = 2 x + 1

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 = - \frac{1}{2}

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

- 2 x + 1 = x - 1 : x = 0

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

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

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

\int_{0}^{4} 2 x + 1 - (x - 1) d x = \frac{14}{3}

= \frac{2}{3} + \frac{14}{3}

Vereinfache

= \frac{16}{3}

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