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

Lösung

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

\frac{1}{3} + \frac{4}{3} x

Schritte zur Lösung

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

Entferne die Konstante:

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

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

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

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

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

Vereinfache

\frac{2}{3} (\frac{1}{2} + 2 x) : \frac{1}{3} + \frac{4}{3} x

= \frac{1}{3} + \frac{4}{3} x

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

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