integral from 0 to 1 of (4x^3-21x^2)

解答

\int_{0}^{1} (4 x^{3} - 21 x^{2}) d x

- 6

求解步骤

\int_{0}^{1} 4 x^{3} - 21 x^{2} 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_{0}^{1} 4 x^{3} d x - \int_{0}^{1} 21 x^{2} d x

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

\int_{0}^{1} 21 x^{2} d x = 7

= 1 - 7

化简

= - 6

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