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

解答

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

\frac{32}{35}

十进制

0.91428 \dots

求解步骤

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

乘开

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

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

使用积分加法定则:

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

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

\int_{- 1}^{1} 1 d x = 2

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

\int_{- 1}^{1} 3 x^{4} d x = \frac{6}{5}

\int_{- 1}^{1} x^{6} d x = \frac{2}{7}

= 2 - 2 + \frac{6}{5} - \frac{2}{7}

化简

2 - 2 + \frac{6}{5} - \frac{2}{7} : \frac{32}{35}

= \frac{32}{35}

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