integral from-2 to 0 of 9x^2(2x^3+4)^2

Lösung

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

896

Schritte zur Lösung

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

Entferne die Konstante:

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

= 9 \cdot \int_{- 2}^{0} x^{2} (2 x^{3} + 4)^{2} d x

Wende U-Substitution an

= 9 \cdot \int_{- 12}^{4} \frac{u ^{2}}{6} d u

Entferne die Konstante:

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

= 9 \cdot \frac{1}{6} \cdot \int_{- 12}^{4} u^{2} d u

Wende die Potenzregel an

= 9 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{- 12}^{4}

Vereinfache

9 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{- 12}^{4} : \frac{3}{2} [\frac{u ^{3}}{3}]_{- 12}^{4}

= \frac{3}{2} [\frac{u ^{3}}{3}]_{- 12}^{4}

Berechne die Grenzen:

\frac{1792}{3}

= \frac{3}{2} \cdot \frac{1792}{3}

Vereinfache

= 896

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