integral from 0 to 2 of sqrt((4-x^2)^3)

解答

\int_{0}^{2} (4 - x^{2})^{3} d x

3 π

十进制

9.42477 \dots

求解步骤

\int_{0}^{2} (4 - x^{2})^{3} d x

(4 - x^{2})^{3} = (4 - x^{2})^{\frac{3}{2}},

假定

(4 - x^{2}) \geq 0

= \int_{0}^{2} (4 - x^{2})^{\frac{3}{2}} d x

使用分布积分法

= [x (4 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 4 - x^{2} d x]_{0}^{2}

\int - 3 x^{2} 4 - x^{2} d x = - 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= [x (4 - x^{2})^{\frac{3}{2}} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))))]_{0}^{2}

化简

= [x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))]_{0}^{2}

带入上下限计算后得:

3 π

= 3 π

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