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受欢迎的微积分 >

integral of x^2sqrt(4-x)

解答

\int x^{2} 4 - x d x

解答

- \frac{2}{7} (4 - x)^{\frac{7}{2}} + \frac{16}{5} (4 - x)^{\frac{5}{2}} - \frac{32}{3} (4 - x)^{\frac{3}{2}} + C

求解步骤

\int x^{2} 4 - x d x

使用换元积分法

= \int - (- u + 4)^{2} u d u

提出常数:

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

= - \int (- u + 4)^{2} u d u

乘开

(- u + 4)^{2} u : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= - \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u^{\frac{5}{2}} d u - \int 8 u^{\frac{3}{2}} d u + \int 16 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

\int 16 u d u = \frac{32}{3} u^{\frac{3}{2}}

= - (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}})

u = 4 - x

代回

= - (\frac{2}{7} (4 - x)^{\frac{7}{2}} - \frac{16}{5} (4 - x)^{\frac{5}{2}} + \frac{32}{3} (4 - x)^{\frac{3}{2}})

化简

= - \frac{2}{7} (4 - x)^{\frac{7}{2}} + \frac{16}{5} (4 - x)^{\frac{5}{2}} - \frac{32}{3} (4 - x)^{\frac{3}{2}}

解答补常数

= - \frac{2}{7} (4 - x)^{\frac{7}{2}} + \frac{16}{5} (4 - x)^{\frac{5}{2}} - \frac{32}{3} (4 - x)^{\frac{3}{2}} + C

作图

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