integral from 1 to 3 of (8x^3+6x-45)

解答

\int_{1}^{3} (8 x^{3} + 6 x - 45) d x

94

求解步骤

\int_{1}^{3} 8 x^{3} + 6 x - 45 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}^{3} 8 x^{3} d x + \int_{1}^{3} 6 x d x - \int_{1}^{3} 45 d x

\int_{1}^{3} 8 x^{3} d x = 160

\int_{1}^{3} 6 x d x = 24

\int_{1}^{3} 45 d x = 90

= 160 + 24 - 90

化简

= 94

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