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

解答

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

6 π

十进制

18.84955 \dots

求解步骤

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

使用分布积分法

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

化简

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

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

带入上下限计算后得:

6 π

= 6 π

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