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受欢迎的微积分 >

integral of r^3sqrt(4-r^2)

解答

\int r^{3} 4 - r^{2} d r

解答

\frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}} + C

求解步骤

\int r^{3} 4 - r^{2} d r

使用换元积分法

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

提出常数:

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

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

乘开

(- u^{2} + 4) u^{2} : - u^{4} + 4 u^{2}

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

\int u^{4} d u = \frac{u ^{5}}{5}

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

= - (- \frac{u ^{5}}{5} + \frac{4 u ^{3}}{3})

u = 4 - r^{2}

代回

= - (- \frac{( 4 - r ^{2} ) ^{5}}{5} + \frac{4 ( 4 - r ^{2} ) ^{3}}{3})

化简

- (- \frac{( 4 - r ^{2} ) ^{5}}{5} + \frac{4 ( 4 - r ^{2} ) ^{3}}{3}) : \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}}

= \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}}

解答补常数

= \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}} + C

作图

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