integral from-1 to 1 of (-1+2t^2)^3

解答

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

- \frac{18}{35}

十进制

- 0.51428 \dots

求解步骤

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

乘开

(- 1 + 2 t^{2})^{3} : - 1 + 6 t^{2} - 12 t^{4} + 8 t^{6}

= \int_{- 1}^{1} - 1 + 6 t^{2} - 12 t^{4} + 8 t^{6} d t

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 t + \int_{- 1}^{1} 6 t^{2} d t - \int_{- 1}^{1} 12 t^{4} d t + \int_{- 1}^{1} 8 t^{6} d t

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

\int_{- 1}^{1} 6 t^{2} d t = 4

\int_{- 1}^{1} 12 t^{4} d t = \frac{24}{5}

\int_{- 1}^{1} 8 t^{6} d t = \frac{16}{7}

= - 2 + 4 - \frac{24}{5} + \frac{16}{7}

化简

- 2 + 4 - \frac{24}{5} + \frac{16}{7} : - \frac{18}{35}

= - \frac{18}{35}

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