integral from 0 to (3q)/2 of 6y(1-y)

Lösung

\int_{0}^{\frac{3 q}{2}} 6 y (1 - y) d y

6 (\frac{9 q ^{2}}{8} - \frac{9 q ^{3}}{8})

Schritte zur Lösung

\int_{0}^{\frac{3 q}{2}} 6 y (1 - y) d y

Entferne die Konstante:

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

= 6 \cdot \int_{0}^{\frac{3 q}{2}} y (1 - y) d y

Multipliziere aus

y (1 - y) : y - y^{2}

= 6 \cdot \int_{0}^{\frac{3 q}{2}} y - y^{2} 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

= 6 (\int_{0}^{\frac{3 q}{2}} y d y - \int_{0}^{\frac{3 q}{2}} y^{2} d y)

\int_{0}^{\frac{3 q}{2}} y d y = \frac{9 q ^{2}}{8}

\int_{0}^{\frac{3 q}{2}} y^{2} d y = \frac{9 q ^{3}}{8}

= 6 (\frac{9 q ^{2}}{8} - \frac{9 q ^{3}}{8})

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