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受欢迎的微积分 >

integral from 2 to 3 of 9/(sqrt(3-x))

解答

\int_{2}^{3} \frac{9}{3 - x} d x

解答

18

求解步骤

\int_{2}^{3} \frac{9}{3 - x} d x

提出常数:

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

= 9 \cdot \int_{2}^{3} \frac{1}{3 - x} d x

使用换元积分法

= 9 \cdot \int_{1}^{0} - \frac{1}{u} d u

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

= 9 (- \int_{0}^{1} - \frac{1}{u} d u)

提出常数:

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

= 9 (- (- \int_{0}^{1} \frac{1}{u} d u))

使用幂法则

= 9 (- (- [2 u^{\frac{1}{2}}]_{0}^{1}))

化简

9 (- (- [2 u^{\frac{1}{2}}]_{0}^{1})) : 9 [2 u]_{0}^{1}

= 9 [2 u]_{0}^{1}

带入上下限计算后得:

2

= 9 \cdot 2

化简

= 18

作图

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