fläche 2x^2,8x^2,5-3x

Lösung

fläche

2 x^{2}, 8 x^{2}, 5 - 3 x

\frac{1097}{128}

Dezimale

8.5703125

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

2 x^{2} = 8 x^{2} : x = 0

2 x^{2} = 5 - 3 x : x = 1, x = - \frac{5}{2}

8 x^{2} = 5 - 3 x : x = \frac{5}{8}, x = - 1

= \int_{- \frac{5}{2}}^{- 1} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{\frac{5}{8}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{5}{8}}^{1} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{5}{2}}^{- 1} 2 x^{2} - (5 - 3 x) d x + \int_{- 1}^{0} 8 x^{2} - 2 x^{2} d x + \int_{0}^{\frac{5}{8}} 2 x^{2} - 8 x^{2} d x + \int_{\frac{5}{8}}^{1} 2 x^{2} - (5 - 3 x) d x

\int_{- \frac{5}{2}}^{- 1} 2 x^{2} - (5 - 3 x) d x = \frac{45}{8}

\int_{- 1}^{0} 8 x^{2} - 2 x^{2} d x = 2

\int_{0}^{\frac{5}{8}} 2 x^{2} - 8 x^{2} d x = \frac{125}{256}

\int_{\frac{5}{8}}^{1} 2 x^{2} - (5 - 3 x) d x = \frac{117}{256}

= \frac{45}{8} + 2 + \frac{125}{256} + \frac{117}{256}

Vereinfache

= \frac{1097}{128}

\frac{\partial}{\partial x} (4 x^{2} y + 7 x y^{3})

d er i v a t i v e f (x) = (3 x - 2) (4 x^{3} - x^{2} + 1)

x \to 2 lim (x^{2} - 6 x + 3)

\frac{d}{d x} (\frac{1}{1 + e ^{0.5 x}})

n = 1 \sum \infty n e^{- 6 n}