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

解答

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

\frac{8}{3}

十进制

2.66666 \dots

求解步骤

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

乘开

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

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

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

\int_{- 1}^{1} 2 x d x = 0

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

= \frac{2}{3} + 0 + 2

化简

\frac{2}{3} + 0 + 2 : \frac{8}{3}

= \frac{8}{3}

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