integral from-1 to 2 of (x-1)^2+1

解答

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

6

求解步骤

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

乘开

(x - 1)^{2} + 1 : x^{2} - 2 x + 2

= \int_{- 1}^{2} x^{2} - 2 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_{- 1}^{2} x^{2} d x - \int_{- 1}^{2} 2 x d x + \int_{- 1}^{2} 2 d x

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

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

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

= 3 - 3 + 6

化简

= 6

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