integral from-1 to 2 of (3u-2)(u+1)

解答

\int_{- 1}^{2} (3 u - 2) (u + 1) d u

\frac{9}{2}

十进制

4.5

求解步骤

\int_{- 1}^{2} (3 u - 2) (u + 1) d u

乘开

(3 u - 2) (u + 1) : 3 u^{2} + u - 2

= \int_{- 1}^{2} 3 u^{2} + u - 2 d u

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} 3 u^{2} d u + \int_{- 1}^{2} u d u - \int_{- 1}^{2} 2 d u

\int_{- 1}^{2} 3 u^{2} d u = 9

\int_{- 1}^{2} u d u = \frac{3}{2}

\int_{- 1}^{2} 2 d u = 6

= 9 + \frac{3}{2} - 6

化简

9 + \frac{3}{2} - 6 : \frac{9}{2}

= \frac{9}{2}

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