integral of xsqrt(b-x)

解答

\int x b - x d x

- 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}}) + C

求解步骤

\int x b - x d x

使用换元积分法

= \int - 2 (- u^{2} + b) u^{2} d u

提出常数:

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

= - 2 \cdot \int (- u^{2} + b) u^{2} d u

使用分布积分法

= - 2 (\frac{1}{3} u^{3} (- u^{2} + b) - \int - \frac{2 u ^{4}}{3} d u)

\int - \frac{2 u ^{4}}{3} d u = - \frac{2 u ^{5}}{15}

= - 2 (\frac{1}{3} u^{3} (- u^{2} + b) - (- \frac{2 u ^{5}}{15}))

u = b - x

代回

= - 2 (\frac{1}{3} (b - x)^{3} (- (b - x)^{2} + b) - (- \frac{2 ( b - x ) ^{5}}{15}))

化简

- 2 (\frac{1}{3} (b - x)^{3} (- (b - x)^{2} + b) - (- \frac{2 ( b - x ) ^{5}}{15})) : - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}})

= - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}})

解答补常数

= - 2 (\frac{1}{3} x (b - x)^{\frac{3}{2}} + \frac{2}{15} (b - x)^{\frac{5}{2}}) + C

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