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受欢迎的微积分 >

integral from 0 to 1 of sqrt(2/(2-r^2))r

解答

\int_{0}^{1} \frac{2}{2 - r ^{2}} r d r

解答

2 - 2

+1

十进制

0.58578 \dots

求解步骤

\int_{0}^{1} \frac{2}{2 - r ^{2}} r d r

化简

\frac{2}{2 - r ^{2}} : \frac{2}{2 - r ^{2}}

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

提出常数:

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

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

化简

\frac{1}{2 - r ^{2}} r : \frac{r}{2 - r ^{2}}

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

使用换元积分法

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

提出常数:

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

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

使用幂法则

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

化简

2 (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{2})) : \frac{1}{2} [2 u]_{1}^{2}

= \frac{1}{2} [2 u]_{1}^{2}

带入上下限计算后得:

22 - 2

= \frac{1}{2} (22 - 2)

化简

= 2 - 2

作图

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