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

解答

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

\frac{9}{14}

十进制

0.64285 \dots

求解步骤

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

乘开

(1 - x^{3})^{2} : 1 - 2 x^{3} + x^{6}

= \int_{0}^{1} 1 - 2 x^{3} + x^{6} d x

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_{0}^{1} 1 d x - \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} x^{6} d x

\int_{0}^{1} 1 d x = 1

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

\int_{0}^{1} x^{6} d x = \frac{1}{7}

= 1 - \frac{1}{2} + \frac{1}{7}

化简

1 - \frac{1}{2} + \frac{1}{7} : \frac{9}{14}

= \frac{9}{14}

d er i v a t i v e g (x) = ln (2 x^{2} + 4)

d er i v a t i v e (6 t - 1) (4 t - 4)^{- 1}

\frac{d}{d x} (x^{4} - 2 a x^{2} - y + 3)

\int 2 x (x^{2} + 10)^{3} d x

\int e^{8 x} cos (5 x) d x