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

Lösung

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

820

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{9} \frac{u ^{3}}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{9} u^{3} d u

Wende die Potenzregel an

= \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{9}

Berechne die Grenzen:

1640

= \frac{1}{2} \cdot 1640

Vereinfache

= 820

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