integral from 0 to 2 of (x+1)(3x-4)

Lösung

\int_{0}^{2} (x + 1) (3 x - 4) d x

- 2

Schritte zur Lösung

\int_{0}^{2} (x + 1) (3 x - 4) d x

Wende die partielle Integration an

= [(3 x - 4) (\frac{x ^{2}}{2} + x) - \int 3 (\frac{x ^{2}}{2} + x) d x]_{0}^{2}

\int 3 (\frac{x ^{2}}{2} + x) d x = 3 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2})

= [(3 x - 4) (\frac{x ^{2}}{2} + x) - 3 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2})]_{0}^{2}

Berechne die Grenzen:

- 2

= - 2

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