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

Lösung

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

- \frac{3}{2}

Dezimale

- 1.5

Schritte zur Lösung

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

Multipliziere aus

(y - 2) (2 y + 1) : 2 y^{2} - 3 y - 2

= \int_{0}^{3} 2 y^{2} - 3 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

= \int_{0}^{3} 2 y^{2} d y - \int_{0}^{3} 3 y d y - \int_{0}^{3} 2 d y

\int_{0}^{3} 2 y^{2} d y = 18

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

\int_{0}^{3} 2 d y = 6

= 18 - \frac{27}{2} - 6

Vereinfache

18 - \frac{27}{2} - 6 : - \frac{3}{2}

= - \frac{3}{2}

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