integral from 0 to 1 of (1-x^2)^{3/2}

解答

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

\frac{3 π}{16}

十进制

0.58904 \dots

求解步骤

\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})^{\frac{1}{2}} d x]_{0}^{1}

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

\int \frac{15}{x ^{2} + 1} d x

\frac{\partial}{\partial x} (3 x)

d er i v a t i v e e^{17 x}

\frac{d}{d x} (x e^{2 x^{2} - 4 x})

t an g e n t y = x^{3} - 2 x + 2, (4, 58)