integral from 0 to 1 of |x-y|

Lösung

\int_{0}^{1} ∣ x - y ∣ d y

- x + \frac{1}{2}

Schritte zur Lösung

\int_{0}^{1} ∣ x - y ∣ d y

Entferne die Extremwerte

= \int_{0}^{1} - x + y d y

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{1} x d y + \int_{0}^{1} y d y

\int_{0}^{1} x d y = x

\int_{0}^{1} y d y = \frac{1}{2}

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

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

\int_{3}^{7} \frac{1}{x ^{2} - 4} d x

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

\int_{0}^{2} 5 t d t

\int_{0}^{3} \frac{x}{1 + 4 x ^{2}} d x