integral from-1 to+1 of x^{99}P_{100}(x)

Lösung

\int_{- 1}^{+ 1} x^{99} P_{100} (x) d x

\frac{2 P _{100}}{101}

Schritte zur Lösung

\int_{- 1}^{1} x^{99} P_{100} x d x

Entferne die Konstante:

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

= P_{100} \cdot \int_{- 1}^{1} x x^{99} d x

Vereinfache

x x^{99} : x^{100}

= P_{100} \cdot \int_{- 1}^{1} x^{100} d x

Wende die Potenzregel an

= P_{100} [\frac{x ^{101}}{101}]_{- 1}^{1}

Berechne die Grenzen:

\frac{2}{101}

= P_{100} \frac{2}{101}

Vereinfache

= \frac{2 P _{100}}{101}

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