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受欢迎的微积分 >

integral from 0 to 1 of sqrt((1-x)/x)

解答

\int_{0}^{1} \frac{1 - x}{x} d x

解答

\frac{π}{2}

+1

十进制

1.57079 \dots

求解步骤

\int_{0}^{1} \frac{1 - x}{x} d x

使用换元积分法

= \int_{0}^{2} \frac{2 - u}{2 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{2} \frac{2 - u}{u} d u

使用分布积分法

= \frac{1}{2} [\frac{2 - u}{u} u - \int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u]_{0}^{2}

\int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u = - 2 arcsin (\frac{1}{2} u)

= \frac{1}{2} [\frac{2 - u}{u} u - (- 2 arcsin (\frac{1}{2} u))]_{0}^{2}

化简

= \frac{1}{2} [\frac{2 - u}{u} u + 2 arcsin (\frac{1}{2} u)]_{0}^{2}

带入上下限计算后得:

π

= \frac{1}{2} π

化简

= \frac{π}{2}

作图

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