integral from 0 to 1 of (1-x)^n

Lösung

\int_{0}^{1} (1 - x)^{n} d x

\frac{1}{n + 1}

Schritte zur Lösung

\int_{0}^{1} (1 - x)^{n} d x

Wende U-Substitution an

= \int_{1}^{0} - u^{n} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - u^{n} d u

Entferne die Konstante:

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

= - (- \int_{0}^{1} u^{n} d u)

Wende die Potenzregel an

= - (- [\frac{u ^{n + 1}}{n + 1}]_{0}^{1})

Vereinfache

= [\frac{u ^{n + 1}}{n + 1}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{n + 1}

= \frac{1}{n + 1}

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