fläche y=x,y=-x+3,y=0

Lösung

fläche

y = x, y = - x + 3, y = 0

\frac{9}{4}

Dezimale

2.25

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

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

x = 0 : x = 0

- x + 3 = 0 : x = 3

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

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

\int_{0}^{\frac{3}{2}} ∣ x - 0 ∣ d x = \frac{9}{8}

\int_{\frac{3}{2}}^{3} ∣ 0 - (- x + 3) ∣ d x = \frac{9}{8}

= \frac{9}{8} + \frac{9}{8}

Vereinfache

= \frac{9}{4}

\int (sin^{2} (2 x)) d x

\int \frac{1}{2} x + 3 d x

x \to 0 lim (\frac{10}{x ( x + 1 )})

\frac{d}{d x} (x + cot (x))

\int \frac{( ln ( z ) ) ^{5}}{z} d z