integral from 0 to 2 of x/(sqrt(x+2))

解答

\int_{0}^{2} \frac{x}{x + 2} d x

2 (\frac{8 - 2 2}{3} - 4 + 22)

十进制

1.10456 \dots

求解步骤

\int_{0}^{2} \frac{x}{x + 2} d x

使用换元积分法

= \int_{2}^{2} 2 (u^{2} - 2) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{2} u^{2} - 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

= 2 (\int_{2}^{2} u^{2} d u - \int_{2}^{2} 2 d u)

\int_{2}^{2} u^{2} d u = \frac{8 - 2 2}{3}

\int_{2}^{2} 2 d u = 4 - 22

= 2 (\frac{8 - 2 2}{3} - (4 - 22))

化简

= 2 (\frac{8 - 2 2}{3} - 4 + 22)

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