integral from 0 to 5 of 2x^3-3x^2+4x+1

解答

\int_{0}^{5} 2 x^{3} - 3 x^{2} + 4 x + 1 d x

\frac{485}{2}

十进制

242.5

求解步骤

\int_{0}^{5} 2 x^{3} - 3 x^{2} + 4 x + 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_{0}^{5} 2 x^{3} d x - \int_{0}^{5} 3 x^{2} d x + \int_{0}^{5} 4 x d x + \int_{0}^{5} 1 d x

\int_{0}^{5} 2 x^{3} d x = \frac{625}{2}

\int_{0}^{5} 3 x^{2} d x = 125

\int_{0}^{5} 4 x d x = 50

\int_{0}^{5} 1 d x = 5

= \frac{625}{2} - 125 + 50 + 5

化简

\frac{625}{2} - 125 + 50 + 5 : \frac{485}{2}

= \frac{485}{2}

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