integral from 0 to 3 of kx(3-x)

Lösung

\int_{0}^{3} k x (3 - x) d x

\frac{9}{2} k

Schritte zur Lösung

\int_{0}^{3} k x (3 - x) d x

Entferne die Konstante:

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

= k \cdot \int_{0}^{3} x (3 - x) d x

Multipliziere aus

x (3 - x) : 3 x - x^{2}

= k \cdot \int_{0}^{3} 3 x - x^{2} d x

Wende die Summenregel an:

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

= k (\int_{0}^{3} 3 x d x - \int_{0}^{3} x^{2} d x)

\int_{0}^{3} 3 x d x = \frac{27}{2}

\int_{0}^{3} x^{2} d x = 9

= k (\frac{27}{2} - 9)

Füge

\frac{27}{2} - 9

zusammen:

\frac{9}{2}

= \frac{9}{2} k

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