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

Lösung

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

\frac{16}{35}

Dezimale

0.45714 \dots

Schritte zur Lösung

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

Multipliziere aus

(1 - x^{2})^{3} : 1 - 3 x^{2} + 3 x^{4} - x^{6}

= \int_{0}^{1} 1 - 3 x^{2} + 3 x^{4} - x^{6} 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_{0}^{1} 1 d x - \int_{0}^{1} 3 x^{2} d x + \int_{0}^{1} 3 x^{4} d x - \int_{0}^{1} x^{6} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 3 x^{4} d x = \frac{3}{5}

\int_{0}^{1} x^{6} d x = \frac{1}{7}

= 1 - 1 + \frac{3}{5} - \frac{1}{7}

Vereinfache

1 - 1 + \frac{3}{5} - \frac{1}{7} : \frac{16}{35}

= \frac{16}{35}

\int_{2}^{5} (6 x^{2} - 3 x + 5) d x

\int_{- 1}^{3} \frac{1}{3 x} d x

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

\int_{0}^{2} 2 t 4 t^{2} + 1 d t

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