fläche 2x-x^2,8-4x,4x

Lösung

fläche

2 x - x^{2}, 8 - 4 x, 4 x

\frac{16}{3}

Dezimale

5.33333 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

2 x - x^{2} = 8 - 4 x : x = 2, x = 4

2 x - x^{2} = 4 x : x = - 2, x = 0

8 - 4 x = 4 x : x = 1

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

= \int_{- 2}^{0} 2 x - x^{2} - 4 x d x + \int_{0}^{1} 2 x - x^{2} - 4 x d x + \int_{1}^{2} 8 - 4 x - (2 x - x^{2}) d x + \int_{2}^{4} 2 x - x^{2} - (8 - 4 x) d x

\int_{- 2}^{0} 2 x - x^{2} - 4 x d x = \frac{4}{3}

\int_{0}^{1} 2 x - x^{2} - 4 x d x = \frac{4}{3}

\int_{1}^{2} 8 - 4 x - (2 x - x^{2}) d x = \frac{4}{3}

\int_{2}^{4} 2 x - x^{2} - (8 - 4 x) d x = \frac{4}{3}

= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} + \frac{4}{3}

Vereinfache

= \frac{16}{3}

\frac{\partial}{\partial t} (- 3 s^{2} t^{2})

\frac{d}{d x} ((10 x^{2} - 20 x + 20) e^{x})

\frac{d}{d x} (x (x^{2} + 8))

\int sin (x) + e^{x} d x

in v erse l a pl a ce \frac{( 3 s + 4 )}{( s + 2 ) ^{2}}