integral from 0 to 2 of x|x-1|

Lösung

\int_{0}^{2} x ∣ x - 1 ∣ d x

1

Schritte zur Lösung

\int_{0}^{2} x ∣ x - 1 ∣ d x

Entferne die Extremwerte

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

\int_{0}^{1} x (- x + 1) d x = \frac{1}{6}

\int_{1}^{2} x (x - 1) d x = \frac{5}{6}

= \frac{1}{6} + \frac{5}{6}

Vereinfache

\frac{1}{6} + \frac{5}{6} : 1

= 1

\int_{0}^{1} e^{x} 1 - x d x

\int_{0}^{2} e^{x^{2} - x} d x

\int_{0}^{y} \frac{1}{y} d y

\int_{0}^{3} x^{3} 81 - x^{4} d x

\int_{0}^{1} arccos (x) d x