integral from 0 to 1 of (sqrt(1-x^2))^3

解答

\int_{0}^{1} (1 - x^{2})^{3} d x

\frac{3 π}{16}

十进制

0.58904 \dots

求解步骤

\int_{0}^{1} (1 - x^{2})^{3} d x

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{2} = 2^{- 1}

= \int_{0}^{1} ((1 - x^{2})^{2^{- 1}})^{3} d x

化简

((1 - x^{2})^{2^{- 1}})^{3} : (1 - x^{2})^{\frac{3}{2}}

= \int_{0}^{1} (1 - x^{2})^{\frac{3}{2}} d x

使用分布积分法

= [x (1 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 1 - x^{2} d x]_{0}^{1}

\int - 3 x^{2} 1 - x^{2} d x = - \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

= [x (1 - x^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))))]_{0}^{1}

化简

= [x (1 - x^{2})^{\frac{3}{2}} + \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))]_{0}^{1}

带入上下限计算后得:

\frac{3 π}{16}

= \frac{3 π}{16}

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