integral from-1 to 0 of (2x+1)^{19}

Lösung

\int_{- 1}^{0} (2 x + 1)^{19} d x

0

Schritte zur Lösung

\int_{- 1}^{0} (2 x + 1)^{19} d x

Wende U-Substitution an

= \int_{- 1}^{1} \frac{u ^{19}}{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}^{1} u^{19} d u

Wende die Potenzregel an

= \frac{1}{2} [\frac{u ^{20}}{20}]_{- 1}^{1}

Berechne die Grenzen:

0

= \frac{1}{2} \cdot 0

Vereinfache

= 0

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