integral from-3 to 1 of [(2x-2)(2x+2)]

Lösung

\int_{- 3}^{1} [(2 x - 2) (2 x + 2)] d x

\frac{64}{3}

Dezimale

21.33333 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [(2 x + 2) (x^{2} - 2 x) - \int 2 (x^{2} - 2 x) d x]_{- 3}^{1}

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

= [(2 x + 2) (x^{2} - 2 x) - 2 (\frac{x ^{3}}{3} - x^{2})]_{- 3}^{1}

Vereinfache

[(2 x + 2) (x^{2} - 2 x) - 2 (\frac{x ^{3}}{3} - x^{2})]_{- 3}^{1} : [\frac{1}{3} (4 x^{3} - 12 x)]_{- 3}^{1}

= [\frac{1}{3} (4 x^{3} - 12 x)]_{- 3}^{1}

Berechne die Grenzen:

\frac{64}{3}

= \frac{64}{3}

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