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受欢迎的微积分 >

integral from 0 to 3 of sqrt(x/(6-x))

解答

\int_{0}^{3} \frac{x}{6 - x} d x

解答

6 (\frac{π}{4} - \frac{1}{2})

+1

十进制

1.71238 \dots

求解步骤

\int_{0}^{3} \frac{x}{6 - x} d x

使用换元积分法

= \int_{6}^{3} - \frac{- u + 6}{u} d u

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

= - \int_{3}^{6} - \frac{- u + 6}{u} d u

提出常数:

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

= - (- \int_{3}^{6} \frac{- u + 6}{u} d u)

乘开

\frac{- u + 6}{u} : - 1 + \frac{6}{u}

= - (- \int_{3}^{6} - 1 + \frac{6}{u} d u)

使用换元积分法

= - (- \int_{2}^{1} - \frac{6 v - 1}{v ^{2}} d v)

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

= - (- (- \int_{1}^{2} - \frac{6 v - 1}{v ^{2}} d v))

提出常数:

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

= - (- (- (- 6 \cdot \int_{1}^{2} \frac{v - 1}{v ^{2}} d v)))

使用分布积分法

= - (- (- (- 6 [- \frac{v - 1}{v} - \int - \frac{1}{2 v v - 1} d v]_{1}^{2})))

\int - \frac{1}{2 v v - 1} d v = - arctan (v - 1)

= - (- (- (- 6 [- \frac{v - 1}{v} - (- arctan (v - 1))]_{1}^{2})))

化简

= 6 [- \frac{v - 1}{v} + arctan (v - 1)]_{1}^{2}

带入上下限计算后得:

\frac{π}{4} - \frac{1}{2}

= 6 (\frac{π}{4} - \frac{1}{2})

作图

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