integral from 1 to 4 of (3x^2-2)^3

解答

\int_{1}^{4} (3 x^{2} - 2)^{3} d x

\frac{1850631}{35}

十进制

52875.17142 \dots

求解步骤

\int_{1}^{4} (3 x^{2} - 2)^{3} d x

乘开

(3 x^{2} - 2)^{3} : 27 x^{6} - 54 x^{4} + 36 x^{2} - 8

= \int_{1}^{4} 27 x^{6} - 54 x^{4} + 36 x^{2} - 8 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}^{4} 27 x^{6} d x - \int_{1}^{4} 54 x^{4} d x + \int_{1}^{4} 36 x^{2} d x - \int_{1}^{4} 8 d x

\int_{1}^{4} 27 x^{6} d x = \frac{442341}{7}

\int_{1}^{4} 54 x^{4} d x = \frac{55242}{5}

\int_{1}^{4} 36 x^{2} d x = 756

\int_{1}^{4} 8 d x = 24

= \frac{442341}{7} - \frac{55242}{5} + 756 - 24

化简

\frac{442341}{7} - \frac{55242}{5} + 756 - 24 : \frac{1850631}{35}

= \frac{1850631}{35}

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