integral from 0 to 2 of (6x-6)(8x^2+3)

Lösung

\int_{0}^{2} (6 x - 6) (8 x^{2} + 3) d x

64

Schritte zur Lösung

\int_{0}^{2} (6 x - 6) (8 x^{2} + 3) d x

Wende die partielle Integration an

= [(6 x - 6) (\frac{8 x ^{3}}{3} + 3 x) - \int 6 (\frac{8 x ^{3}}{3} + 3 x) d x]_{0}^{2}

\int 6 (\frac{8 x ^{3}}{3} + 3 x) d x = 6 (\frac{2 x ^{4}}{3} + \frac{3 x ^{2}}{2})

= [(6 x - 6) (\frac{8 x ^{3}}{3} + 3 x) - 6 (\frac{2 x ^{4}}{3} + \frac{3 x ^{2}}{2})]_{0}^{2}

Multipliziere aus

(6 x - 6) (\frac{8 x ^{3}}{3} + 3 x) - 6 (\frac{2 x ^{4}}{3} + \frac{3 x ^{2}}{2}) : 12 x^{4} - 16 x^{3} + 9 x^{2} - 18 x

= [12 x^{4} - 16 x^{3} + 9 x^{2} - 18 x]_{0}^{2}

Berechne die Grenzen:

64

= 64

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