fläche y=x,y=3x,y=-x+2

Lösung

fläche

y = x, y = 3 x, y = - x + 2

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

x = 3 x : x = 0

x = - x + 2 : x = 1

3 x = - x + 2 : x = \frac{1}{2}

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

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

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

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

= \frac{1}{4} + \frac{1}{4}

Vereinfache

= \frac{1}{2}

n = 0 \sum \infty n \cdot (\frac{3}{4})^{n}

\int \frac{( x + 1 )}{x ^{3} + x ^{2} + 6 x} d x

x^{^{'}} = 4 x - x^{3}

\frac{d}{d x} (\frac{x - 5}{x + 6})

\frac{d}{d x} (b x)