integral from-1 to 2 of x/(sqrt(x+2))

解答

\int_{- 1}^{2} \frac{x}{x + 2} d x

\frac{2}{3}

十进制

0.66666 \dots

求解步骤

\int_{- 1}^{2} \frac{x}{x + 2} d x

使用换元积分法

= \int_{1}^{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_{1}^{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_{1}^{2} u^{2} d u - \int_{1}^{2} 2 d u)

\int_{1}^{2} u^{2} d u = \frac{7}{3}

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

= 2 (\frac{7}{3} - 2)

化简

2 (\frac{7}{3} - 2) : \frac{2}{3}

= \frac{2}{3}

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