integral of (1-x)x^n

Lösung

\int (1 - x) x^{n} d x

\frac{x ^{n + 1}}{n + 1} - \frac{x ^{n + 2}}{n + 2} + C

Schritte zur Lösung

\int (1 - x) x^{n} d x

Multipliziere aus

(1 - x) x^{n} : x^{n} - x^{1 + n}

= \int x^{n} - x^{1 + n} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{n} d x - \int x^{1 + n} d x

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}

\int x^{1 + n} d x = \frac{x ^{n + 2}}{n + 2}

= \frac{x ^{n + 1}}{n + 1} - \frac{x ^{n + 2}}{n + 2}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{n + 1}}{n + 1} - \frac{x ^{n + 2}}{n + 2} + C

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