integral from 0 to 2 of 1/3 (x+y)

Lösung

\int_{0}^{2} \frac{1}{3} (x + y) d y

\frac{1}{3} (2 x + 2)

Schritte zur Lösung

\int_{0}^{2} \frac{1}{3} (x + y) d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{0}^{2} 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

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

\int_{0}^{2} x d y = 2 x

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

= \frac{1}{3} (2 x + 2)

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