integral from 0 to 2 of 1/8 (x+y)

Lösung

\int_{0}^{2} \frac{1}{8} (x + y) d y

\frac{x + 1}{4}

Schritte zur Lösung

\int_{0}^{2} \frac{1}{8} (x + y) d y

Entferne die Konstante:

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

= \frac{1}{8} \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}{8} (\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}{8} (2 x + 2)

Vereinfache

\frac{1}{8} (2 x + 2) : \frac{x + 1}{4}

= \frac{x + 1}{4}

f (x) = e^{2 x} - 1

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