fläche y=x,y=4x,y=-x+2

Lösung

fläche

y = x, y = 4 x, y = - x + 2

\frac{3}{5}

Dezimale

0.6

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

x = 4 x : x = 0

x = - x + 2 : x = 1

4 x = - x + 2 : x = \frac{2}{5}

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

= \int_{0}^{\frac{2}{5}} ∣ x - 4 x ∣ d x + \int_{\frac{2}{5}}^{1} ∣ x - (- x + 2) ∣ d x

\int_{0}^{\frac{2}{5}} ∣ x - 4 x ∣ d x = \frac{6}{25}

\int_{\frac{2}{5}}^{1} ∣ x - (- x + 2) ∣ d x = \frac{9}{25}

= \frac{6}{25} + \frac{9}{25}

Vereinfache

= \frac{3}{5}

\int 6 sin (3 x) d x

s im pl i f y \frac{x ^{2} - 3}{( x - 1 ) ^{2} + ( x - 1 ) ^{2}}

\int_{1}^{4} \frac{1}{9 + ( x - 1 ) ^{2}} d x

\frac{d}{d x} (24 x - 4 x)

\frac{d}{d x} (7 x - \frac{5}{x})