integral from-1 to 2 of [4x^3-6x^2+1]

解答

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

0

求解步骤

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

\int_{- 1}^{2} 4 x^{3} d x = 15

\int_{- 1}^{2} 6 x^{2} d x = 18

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

= 15 - 18 + 3

化简

= 0

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